physics_solutions_manual_

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SolutionsManual

1AGlencoeProgramStudentEditionForensicsLaboratoryManual,TeacherTeacherWraparoundEditionEditionSupplementalProblemsTeacherChapterResourcesAdditionalChallengeProblemsMiniLabWorksheetsPre-AP/CriticalThinkingProblemsPhysicsLabWorksheetsPhysicsTestPrep:StudyingfortheStudyGuideEnd-of-CourseExam,StudentEditionSectionQuizzesPhysicsTestPrep:StudyingfortheReinforcementEnd-of-CourseExam,TeacherEditionEnrichmentConnectingMathtoPhysicsTransparencyMastersSolutionsManualTransparencyWorksheetsChapterAssessmentTechnologyAnswerKeyMakerTeacherClassroomResourcesExamView®ProTeachingTransparenciesInteractiveChalkboardLaboratoryManual,StudentEditionMcGraw-HillLearningNetworkLaboratoryManual,TeacherEditionStudentWorks™CD-ROMProbewareLaboratoryManual,StudentTeacherWorks™CD-ROMEditionphysicspp.comWebsiteProbewareLaboratoryManual,TeacherEditionForensicsLaboratoryManual,StudentEditionCopyright©byTheMcGraw-HillCompanies,Inc.Allrightsreserved.Permissionisgrantedtoreproducethematerialcontainedhereinontheconditionthatsuchmaterialberepro-ducedonlyforclassroomuse;beprovidedtostudents,teachers,andfamilieswithoutcharge;andbeusedsolelyinconjunctionwiththePhysics:PrinciplesandProblemsprogram.Anyotherreproduction,foruseorsale,isprohibitedwithoutpriorwrittenpermissionofthepublisher.Sendallinquiriesto:Glencoe/McGraw-Hill8787OrionPlaceColumbus,Ohio43240ISBN0-07-865893-4PrintedintheUnitedStatesofAmerica123456789045090807060504

2ContentsTotheTeacher......................ivChapter1Chapter17APhysicsToolkit....................1ReflectionandMirrors..............357Chapter2Chapter18RepresentingMotion................15RefractionandLenses..............377Chapter3Chapter19AcceleratedMotion.................29InterferenceandDiffraction.........399Chapter4Chapter20ForcesinOneDimension............61StaticElectricity...................413Chapter5Chapter21ForcesinTwoDimensions...........87ElectricFields.....................427Chapter6Chapter22MotioninTwoDimensions.........115CurrentElectricity.................445Chapter7Chapter23Gravitation.......................141SeriesandParallelCircuits..........463Chapter8Chapter24RotationalMotion.................169MagneticFields...................485Chapter9Chapter25MomentumandItsConservation....193ElectromagneticInduction..........501Chapter10Chapter26Energy,Work,andSimpleMachines..225Electromagnetism..................517Chapter11Chapter27EnergyandItsConservation.........247QuantumTheory..................531Chapter12Chapter28ThermalEnergy...................271TheAtom........................545Chapter13Chapter29StatesofMatter...................287Solid-StateElectronics..............559Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter14Chapter30VibrationsandWaves..............311NuclearPhysics...................573Chapter15AppendixBSound...........................329AdditionalProblems...............591Chapter16FundamentalsofLight.............345Physics:PrinciplesandProblemsContentsiii

3TotheTeacherTheSolutionsManualisacomprehensiveguidetothequestionsandproblemsintheStudentEditionofPhysics:PrinciplesandProblems.ThisincludesthePracticeProblems,SectionReviews,ChapterAssessments,andChallengeProblemsforeachchapter,aswellastheAdditionalProblemsthatappearinAppendixBoftheStudentEdition.TheSolutionsManualrestateseveryquestionandproblemsothatyoudonothavetolookbackatthetextwhenreviewingproblemswithstudents.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ivTotheTeacherPhysics:PrinciplesandProblems

4CHAPTER1APhysicsToolkit6.Convert5021centimeterstokilometers.PracticeProblems1m1km5021cm1.1MathematicsandPhysics100cm1000mpages3–105.021102kmpage57.Howmanysecondsareinaleapyear?Foreachproblem,givetherewrittenequationyouwoulduseandtheanswer.24h60min60s366days1day1h1min1.Alightbulbwitharesistanceof50.0ohms31,622,400sisusedinacircuitwitha9.0-voltbattery.Whatisthecurrentthroughthebulb?8.Convertthespeed5.30m/stokm/h.V9.0voltI0.18ampere5.30m60s60min1kmR50.0ohms1s1min1h1000m2.Anobjectwithuniformaccelerationa,19.08km/hstartingfromrest,willreachaspeedofvintimetaccordingtotheformulavat.page8WhatistheaccelerationofabicyclistwhoSolvethefollowingproblems.acceleratesfromrestto7m/sin4s?9.a.6.201cm7.4cm0.68cmv7m/s2a1.75m/s12.0cmt4s6.201cm3.Howlongwillittakeascooteraccelerating7.4cmat0.400m/s2togofromresttoaspeedof0.68cm4.00m/s?12.0cmv4.00m/s26.281cmt10.0sa0.400m/s226.3cmafterroundingb.1.6km1.62m1200cm4.Thepressureonasurfaceisequaltotheforcedividedbythearea:PF/A.1.6km1600mA53-kgwomanexertsaforce(weight)of1.62m1.62m520Newtons.Ifthepressureexertedon1200cm12mtheflooris32,500N/m2,whatisthearea1613.62mofthesolesofhershoes?1600mor1.6kmafterroundingF520N2A0.016mP32,500N/m210.a.10.8g8.264g10.8g8.264gpage7Usedimensionalanalysistocheckyourequation2.536gCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.beforemultiplying.2.5gafterrounding5.Howmanymegahertzis750kilohertz?1000Hz1MHz750kHz1kHz1,000,000Hz0.75MHzPhysics:PrinciplesandProblemsSolutionsManual1

5Chapter1continuedb.4.75m0.4168ma.Substitutethevaluesintotheequationyouwilluse.Aretheunitscorrect?4.75m0.4168mFBqv4.3332m(4.5kg/As2)(1.601019As)4.33mafterrounding(2.4105m/s)11.a.139cm2.3cmForcewillbemeasuredinkgm/s2,320cm2or3.2102cm2whichiscorrect.b.Thevaluesarewritteninscientificb.3.2145km4.23kmnotation,m10n.Calculatethe10npart13.6km2oftheequationtoestimatethesizeoftheanswer.12.a.13.78g11.3mL10191051014;theanswerwill1.22g/mLbeabout201014,or21013.b.18.21g4.4cm3c.Calculateyouranswer.Checkitagainst4.1g/cm3yourestimatefrompartb.1.71013kgm/s2SectionReviewd.Justifythenumberofsignificantdigitsinyouranswer.1.1MathematicsandPhysicsTheleast-precisevalueis4.5T,withpages3–102significantdigits,sotheanswerispage10roundedto2significantdigits.13.MathWhyareconceptsinphysicsdescribedwithformulas?16.MagnetismRewriteFBqvtofindvintermsofF,q,andB.Theformulasareconciseandcanbeusedtopredictnewdata.FvCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Bq14.MagnetismTheforceofamagneticfield17.CriticalThinkingAnacceptedvalueforonacharged,movingparticleisgivenbytheaccelerationduetogravityis9.801m/s2.FBqv,whereFistheforceinkgm/s2,qisInanexperimentwithpendulums,youthechargeinAs,andvisthespeedinm/s.calculatethatthevalueis9.4m/s2.ShouldBisthestrengthofthemagneticfield,theacceptedvaluebetossedouttoaccom-measuredinteslas,T.Whatis1teslamodateyournewfinding?Explain.describedinbaseunits?2hasbeenNo.Thevalue9.801m/sFBqv,soBFestablishedbymanyotherexperiments,qvandtodiscardthefindingyouwouldkgm/s2kghavetoexplainwhytheywerewrong.T(As)(m/s)As2Thereareprobablysomefactorsaffectingyourcalculation,suchas1T1kg/As2frictionandhowpreciselyyoucanmeasurethedifferentvariables.15.MagnetismAprotonwithcharge1.601019Asismovingat2.4105m/sthroughamagneticfieldof4.5T.Youwanttofindtheforceontheproton.2SolutionsManualPhysics:PrinciplesandProblems

6Chapter1continuedd.HowpreciseisthemeasureoftheSectionReviewheightofonebox?Of12boxes?1.2Measurementnearesttenthofacm;nearesttenthpages11–14ofacmpage1423.CriticalThinkingYourfriendstatesina18.AccuracySomewoodenrulersdonotstartreportthattheaveragetimerequiredtowith0attheedge,buthaveitsetinafewcirclea1.5-mitrackwas65.414s.Thiswasmillimeters.Howcouldthisimprovethemeasuredbytiming7lapsusingaclockaccuracyoftheruler?withaprecisionof0.1s.HowmuchAstheedgeoftherulergetswornawayconfidencedoyouhaveintheresultsofovertime,thefirstmillimeterortwoofthereport?Explain.thescalewouldalsobewornawayifAresultcanneverbemoreprecisethanthescalestartedattheedge.theleastprecisemeasurement.The19.ToolsYoufindamicrometer(atoolusedcalculatedaveragelaptimeexceedstomeasureobjectstothenearest0.01mm)theprecisionpossiblewiththeclock.thathasbeenbadlybent.Howwoulditcomparetoanew,high-qualitymeterstickintermsofitsprecision?Itsaccuracy?PracticeProblemsItwouldbemoreprecisebutless1.3GraphingDataaccurate.pages15–19page1820.ParallaxDoesparallaxaffecttheprecision24.Themassvaluesofspecifiedvolumesofofameasurementthatyoumake?Explain.puregoldnuggetsaregiveninTable1-4.No,itdoesn’tchangethefinenessofthedivisionsonitsscale.Table1-4MassofPureGoldNuggets21.ErrorYourfriendtellsyouthathisheightis182cm.Inyourownwords,explaintheVolume(cm3)Mass(g)rangeofheightsimpliedbythisstatement.1.019.4Hisheightwouldbebetween181.5and2.038.6182.5cm.Precisionofameasurement3.058.1isone-halfthesmallestdivisionontheinstrument.Theheight182cmwould4.077.4range0.5cm.5.096.5a.Plotmassversusvolumefromthe22.PrecisionAboxhasalengthof18.1cmandvaluesgiveninthetableanddrawawidthof19.2cm,anditis20.3cmtall.thecurvethatbestfitsallpoints.a.Whatisitsvolume?7.05103cm3100b.Howpreciseisthemeasureoflength?80Ofvolume?60Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.nearesttenthofacm;nearestMass(g)4010cm320c.Howtallisastackof12oftheseboxes?012345243.6cmVolume(cm3)Physics:PrinciplesandProblemsSolutionsManual3

7Chapter1continuedb.Describetheresultingcurve.27.PredictUsetherelationillustratedinastraightlineFigure1-16todeterminethemassrequiredtostretchthespring15cm.c.Accordingtothegraph,whattypeofrelationshipexistsbetweenthemassof16gthepuregoldnuggetsandtheirvolume?28.PredictUsetherelationinFigure1-18toTherelationshipislinear.predictthecurrentwhentheresistanceisd.Whatisthevalueoftheslopeofthis16ohms.graph?Includetheproperunits.7.5Ay96.5g19.4gslopex5.0cm31.0cm329.CriticalThinkingInyourownwords,3explainthemeaningofashallowerline,or19.3g/cmasmallerslopethantheoneinFigure1-16,e.Writetheequationshowingmassasainthegraphofstretchversustotalmassforfunctionofvolumeforgold.adifferentspring.m(19.3g/cm3)VThespringwhoselinehasasmallerf.Writeawordinterpretationfortheslopeslopeisstiffer,andthereforerequiresoftheline.moremasstostretchitonecentimeter.Themassforeachcubiccentimeterofpuregoldis19.3g.ChapterAssessmentConceptMappingSectionReviewpage241.3GraphingData30.Completethefollowingconceptmapusingpages15–19thefollowingterms:hypothesis,graph,mathematicalmodel,dependentvariable,page19measurement.25.MakeaGraphGraphthefollowingdata.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Timeistheindependentvariable.hypothesisTime(s)05101520253035Speed(m/s)1210864222experiment128measurementSpeed(m/s)4dependentindependent010203040variablevariableTime(s)26.InterpretaGraphWhatwouldbethegraphmeaningofanonzeroy-intercepttoagraphoftotalmassversusvolume?Thereisanonzerototalmasswhenthemathematicalvolumeofthematerialiszero.Thismodelcouldhappenifthemassvalueincludesthematerial’scontainer.4SolutionsManualPhysics:PrinciplesandProblems

8Chapter1continuedMasteringConcepts1b.m1000page2431.Describeascientificmethod.(1.1)millimeterIdentifyaproblem;gatherinformationc.1000maboutitbyobservingandexperimenting;kilometeranalyzetheinformationtoarriveatananswer.37.Toconvert1.8htominutes,bywhatconversionfactorshouldyoumultiply?(1.1)32.Whyismathematicsimportanttoscience?60min,becausetheunitswillcancel(1.1)1hMathematicsallowsyoutobequantita-correctly.tive,tosay“howfast,”notjust“fast.”38.Solveeachproblem.Givethecorrectnumber33.WhatistheSIsystem?(1.1)ofsignificantdigitsintheanswers.(1.1)TheInternationalSystemofUnits,orSI,a.4.667104g3.02105gisabase10systemofmeasurement3.49105gthatisthestandardinscience.Thebaseb.(1.70102J)(5.922104cm3)unitsarethemeter,kilogram,second,2.87105J/cm3kelvin,mole,ampere,andcandela.39.Whatdeterminestheprecisionofa34.Howarebaseunitsandderivedunitsmeasurement?(1.2)related?(1.1)theprecisionofameasuringdevice,Thederivedunitsarecombinationsofwhichislimitedbythefinestdivisionthebaseunits.onitsscale35.Supposeyourlabpartnerrecordeda40.Howdoesthelastdigitdifferfromthemeasurementas100g.(1.1)otherdigitsinameasurement?(1.2)a.WhyisitdifficulttotellthenumberofThefinaldigitisestimated.significantdigitsinthismeasurement?Zerosarenecessarytoindicatethe41.Acar’sodometermeasuresthedistancemagnitudeofthevalue,butthereisfromhometoschoolas3.9km.Usingnowayofknowingwhetherornotthestringonamap,youfindthedistancetobeinstrumentusedtomeasuretheval-4.2km.Whichanswerdoyouthinkismoreuesactuallymeasuredthezeros.Theaccurate?Whatdoesaccuratemean?(1.2)zerosmayserveonlytolocatethe1.Themostaccuratemeasureistheb.Howcanthenumberofsignificantmeasureclosesttotheactualdistance.digitsinsuchanumberbemadeclear?TheodometerisprobablymoreWritethenumberinscientificaccurateasitactuallycoveredthenotation,includingonlythedistance.Themapisamodelmadesignificantdigits.frommeasurements,soyourmeasure-mentsfromthemaparemoreremoved36.Givethenameforeachofthefollowingfromtherealdistance.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.multiplesofthemeter.(1.1)142.Howdoyoufindtheslopeofalineara.m100graph?(1.3)centimeterTheslopeofalineargraphistheratiooftheverticalchangetothehorizontalchange,orriseoverrun.Physics:PrinciplesandProblemsSolutionsManual5

9Chapter1continued43.Foradriver,thetimebetweenseeinga46.GiventheequationFmv2/R,whatstoplightandsteppingonthebrakesisrelationshipexistsbetweeneachofthecalledreactiontime.Thedistancetraveledfollowing?(1.3)duringthistimeisthereactiondistance.a.FandRReactiondistanceforagivendriverandinverserelationshipvehicledependslinearlyonspeed.(1.3)b.Fandma.Wouldthegraphofreactiondistanceversusspeedhaveapositiveoralinearrelationshipnegativeslope?c.FandvPositive.Asspeedincreases,quadraticrelationshipreactiondistanceincreases.b.AdriverwhoisdistractedhasalongerApplyingConceptsreactiontimethanadriverwhoisnot.pages25–26Wouldthegraphofreactiondistance47.Figure1-21givestheheightabovetheversusspeedforadistracteddriverhavegroundofaballthatisthrownupwardalargerorsmallerslopethanforafromtheroofofabuilding,forthefirstnormaldriver?Explain.1.5sofitstrajectory.Whatistheball’sLarger.Thedriverwhowasdistractedheightatt0?Predicttheball’sheightatwouldhavealongerreactiontimet2sandatt5s.andthusagreaterreactiondistanceHeightofBallv.Timeatagivenspeed.44.Duringalaboratoryexperiment,the25temperatureofthegasinaballoonisvaried20andthevolumeoftheballoonismeasured.Whichquantityistheindependent15variable?WhichquantityisthedependentHeight(m)10variable?(1.3)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Temperatureistheindependentvari-5able;volumeisthedependentvariable.0123445.WhattypeofrelationshipisshowninTime(s)Figure1-20?Givethegeneralequationfor■Figure1-21thistypeofrelation.(1.3)Whent0andt2,theball’sheightywillbeabout20m.Whent5,theballwillhavelandedonthegroundsotheheightwillbe0m.48.Isascientificmethodonesetofclearlydefinedsteps?Supportyouranswer.Thereisnodefiniteorderofspecificsteps.However,whateverapproachxisused,italwaysincludescloseobservation,controlledexperimentation,■Figure1-20summarizing,checking,andrechecking.quadratic;yax2bxc6SolutionsManualPhysics:PrinciplesandProblems

10Chapter1continued49.ExplainthedifferencebetweenascientificStates.Findasmanyveryshortandverytheoryandascientificlaw.longexamplesasyoucan.Ascientificlawisaruleofnature,wheresampleanswer:ascientifictheoryisanexplanationofhalf-lifeofpolonium-194,0.7s;thescientificlawbasedonobservation.timebetweenheartbeats,0.8s;timetoAtheoryexplainswhysomethinghap-walkbetweenphysicsclassandmathpens;alawdescribeswhathappens.class,2.4min;lengthofschoolyear,180days;timebetweenelectionsfor50.DensityThedensityofasubstanceisitstheU.S.HouseofRepresentatives,massperunitvolume.2years;timebetweenU.S.presidentiala.Giveapossiblemetricunitfordensity.elections,4years;ageoftheUnitedpossibleanswersincludeg/cm3orStates,(about)230yearskg/m354.SpeedofLightTwostudentsmeasureb.Istheunitfordensityabaseunitorathespeedoflight.Oneobtainsderivedunit?(3.0010.001)108m/s;theotherobtainsderivedunit(2.9990.006)108m/s.a.Whichismoreprecise?51.Whatmetricunitwouldyouusetomeasure(3.0010.001)108m/seachofthefollowing?b.Whichismoreaccurate?a.thewidthofyourhand(2.9990.006)108m/scmb.thethicknessofabookcover55.Youmeasurethedimensionsofadeskasmm132cm,83cm,and76cm.Thesumofc.theheightofyourclassroomthesemeasuresis291cm,whiletheproductis8.3105cm3.Explainhowthesignificantmdigitsweredeterminedineachcase.d.thedistancefromyourhometoyourInadditionandsubtraction,youaskclassroomwhatplacetheleastprecisemeasureiskmknownto:inthiscase,tothenearestcm.Sotheanswerisroundedtothe52.SizeMakeachartofsizesofobjects.nearestcm.InmultiplicationandLengthsshouldrangefromlessthan1mmdivision,youlookatthenumberoftoseveralkilometers.Samplesmightsignificantdigitsintheleastpreciseincludethesizeofacell,thedistancelightanswer:inthiscase,2.Sotheansweristravelsin1s,andtheheightofaroom.roundedto2significantdigits.sampleanswer:radiusoftheatom,51011m;virus,56.MoneySupposeyoureceive$5.00atthe107m;thicknessofpaper,0.1mm;beginningofaweekandspend$1.00eachwidthofpaperbackbook,10.7cm;dayforlunch.Youprepareagraphoftheheightofadoor,1.8m;widthoftown,amountyouhaveleftattheendofeachday7.8km;radiusofEarth,6106m;foroneweek.WouldtheslopeofthisgraphdistancetotheMoon,4108mbepositive,zero,ornegative?Why?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.negative,becausethechangein53.TimeMakeachartoftimeintervals.verticaldistanceisnegativeforaSampleintervalsmightincludethetimepositivechangeinhorizontaldistancebetweenheartbeats,thetimebetweenpresidentialelections,theaveragelifetimeofahuman,andtheageoftheUnitedPhysics:PrinciplesandProblemsSolutionsManual7

11Chapter1continued57.Dataareplottedonagraph,andthevaluec.Whatisthenumberofsignificantdigitsonthey-axisisthesameforeachvalueofforthetotalmass?theindependentvariable.Whatisthefourslope?Why?Howdoesydependonx?d.WhyisthenumberofsignificantdigitsZero.Thechangeinverticaldistanceisdifferentforthetotalmassandthezero.ydoesnotdependonx.individualmasses?Whenaddingmeasurements,the58.DrivingThegraphofbrakingdistanceprecisionmatters:bothmassesareversuscarspeedispartofaparabola.Thus,theequationiswrittendav2bvc.knowntothenearesthundredthofagram,sothetotalshouldbegiventoThedistance,d,hasunitsinmeters,andthenearesthundredthofagram.velocity,v,hasunitsinmeters/second.SignificantdigitssometimesareHowcouldyoufindtheunitsofa,b,andc?gainedwhenadding.Whatwouldtheybe?Theunitsineachtermoftheequation61.HistoryAristotlesaidthatthespeedofamustbeinmetersbecausedistance,d,fallingobjectvariesinverselywiththedensityismeasuredinmeters.ofthemediumthroughwhichitfalls.av2a(m/s)2,soaisins2/m;a.AccordingtoAristotle,wouldarockfallbvb(m/s),sobisins1.fasterinwater(density1000kg/m3),orinair(density1kg/m3)?59.HowlongistheleafinFigure1-22?Lowerdensitymeansfasterspeed,Includetheuncertaintyinyoursotherockfallsfasterinair.measurement.b.Howfastwouldarockfallinavacuum?Basedonthis,whywouldAristotlesaythattherecouldbenosuchthingasavacuum?BecauseavacuumwouldhaveaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.zerodensity,therockshouldfallinfinitelyfast.Nothingcanfallthatfast.62.Explainthedifferencebetweenahypothesis■Figure1-22andascientifictheory.8.3cm0.05cm,or83mm0.5mmAscientifictheoryhasbeentestedandsupportedmanytimesbeforeit60.Themassesoftwometalblocksarebecomesaccepted.Ahypothesisisanmeasured.BlockAhasamassof8.45gideaabouthowthingsmightwork—itandblockBhasamassof45.87g.hasmuchlesssupport.a.Howmanysignificantdigitsare63.Giveanexampleofascientificlaw.expressedinthesemeasurements?Newton’slawsofmotion,lawofconser-A:three;B:fourvationofenergy,lawofconservationofb.WhatisthetotalmassofblockApluscharge,lawofreflectionblockB?54.32g8SolutionsManualPhysics:PrinciplesandProblems

12Chapter1continued64.WhatreasonmighttheancientGreeks68.Addorsubtractasindicated.havehadnottoquestionthehypothesisa.5.80109s3.20108sthatheavierobjectsfallfasterthanlighter9s6.1210objects?Hint:Didyoueverquestionwhichb.4.87106m1.93106mfallsfaster?2.94106mAirresistanceaffectsmanylightobjects.Withoutcontrolledexperi-c.3.14105kg9.36105kgments,theireverydayobservationstold1.250104kgthemthatheavierobjectsdidfallfaster.d.8.12107g6.20106g7.50107g65.MarsExplainwhatobservationsledtochangesinscientists’ideasaboutthesurface69.RankthefollowingmassmeasurementsofMars.fromleasttogreatest:11.6mg,1021g,Astelescopesimprovedandlater0.000006kg,0.31mg.probesweresentintospace,scientists0.31mg,1021µg,0.000006kg,11.6mggainedmoreinformationaboutthesurface.Whentheinformationdid70.Statethenumberofsignificantdigitsinnotsupportoldhypotheses,theeachofthefollowingmeasurements.hypotheseschanged.a.0.00003m66.AgraduatedcylinderismarkedeverymL.1Howpreciseameasurementcanyoumakeb.64.01fmwiththisinstrument?40.5mLc.80.001mMasteringProblems5d.0.720gpages26–281.1MathematicsandPhysics367.Converteachofthefollowinge.2.40106kgmeasurementstometers.3a.42.3cmf.6108kg0.423m1b.6.2pmg.4.071016m6.21012m3c.21km2.1104m71.Addorsubtractasindicated.a.16.2m5.008m13.48md.0.023mm2.3105m34.7mb.5.006m12.0077m8.0084me.214m2.14104m25.022mc.78.05cm232.046cm2f.57nmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.46.00cm25.7108md.15.07kg12.0kg3.1kgPhysics:PrinciplesandProblemsSolutionsManual9

13Chapter1continued72.Multiplyordivideasindicated.78.Howpreciseameasurementcouldyoua.(6.21018m)(4.71010m)makewiththescaleshowninFigure1-23?2.9109m2b.(5.6107m)/(2.81012s)2.0105m/sc.(8.1104km)(1.6103km)1.3106km2d.(6.5105kg)/(3.4103m3)1.9102kg/m373.GravityTheforceduetogravityisFmgwhereg9.80m/s2.■Figure1-23a.Findtheforceduetogravityona0.5g41.63-kgobject.408kgm/s279.GivethemeasureshownonthemeterinFigure1-24aspreciselyasyoucan.Includeb.Theforceduetogravityonanobjectis632kgm/s2.Whatisitsmass?theuncertaintyinyouranswer.64.5kg231474.DimensionalAnalysisPressureismea-suredinpascals,where1Pa1kg/ms2.5A0WillthefollowingexpressiongiveaACLASSApressureinthecorrectunits?(0.55kg)(2.1m/s)9.8m/s23Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.No;itisinkg/s1.2Measurement■Figure1-2475.Awatertankhasamassof3.64kgwhenit3.60.1Aisemptyandamassof51.8kgwhenitisfilledtoacertainlevel.Whatisthemassof80.Estimatetheheightofthenearestdoorthewaterinthetank?frameincentimeters.Thenmeasureit.48.2kgHowaccuratewasyourestimate?Howprecisewasyourestimate?Howprecise76.Thelengthofaroomis16.40m,itswidthwasyourmeasurement?Whyarethetwois4.5m,anditsheightis3.26m.Whatprecisionsdifferent?volumedoestheroomenclose?Astandardresidentialdoorframeheight2.4102m3is80inches,whichisabout200cm.Theprecisiondependsonthemeasurement77.Thesidesofaquadrangularplotoflandinstrumentused.are132.68m,48.3m,132.736m,and48.37m.Whatistheperimeteroftheplot?362.1m10SolutionsManualPhysics:PrinciplesandProblems

14Chapter1continued81.BaseUnitsGivesixexamplesofquantitiesMassofyoumightmeasureinaphysicslab.IncludeThreeSubstancestheunitsyouwoulduse.800Sample:distance,cm;volume,mL;700mass,g;current,A;time,s;600temperature,°CC50082.TemperatureThetemperaturedropsfrom400Mass(g)B24°Cto10°Cin12hours.300200a.Findtheaveragetemperaturechangeper100hour.A1.2°C/h01020304050b.Predictthetemperaturein2morehoursVolume(cm3)ifthetrendcontinues.■Figure1-258°C84.Duringaclassdemonstration,aphysicsc.Couldyouaccuratelypredicttheinstructorplacedamassonahorizontaltemperaturein24hours?tablethatwasnearlyfrictionless.TheNo.Temperatureisunlikelytocon-instructorthenappliedvarioushorizontaltinuefallingsharplyandsteadilyforcestothemassandmeasuredthethatlong.distanceittraveledin5secondsforeachforceapplied.TheresultsoftheexperimentareshowninTable1-5.1.3GraphingData83.Figure1-25showsthemassesofthreeTable1-5substancesforvolumesbetween0and60cm3.DistanceTraveledwithDifferentForcesa.Whatisthemassof30cm3ofeachForce(N)Distance(cm)substance?5.024(a)80g,(b)260g,(c)400g10.049b.Ifyouhad100gofeachsubstance,whatwouldbetheirvolumes?15.075(a)36cm3,(b)11cm3,(c)7cm320.099c.Inoneortwosentences,describethe25.0120meaningoftheslopesofthelinesin30.0145thisgraph.Thesloperepresentstheincreaseda.Plotthevaluesgiveninthetableandmassofeachadditionalcubicdrawthecurvethatbestfitsallpoints.centimeterofthesubstance.160d.Whatisthey-interceptofeachline?120Whatdoesitmean?They-interceptis(0,0).Itmeansthat803,thereisnoneoftheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.whenV0cmDistance(cm)40substancepresent(m0g).0.010.020.030.0Force(N)Physics:PrinciplesandProblemsSolutionsManual11

15Chapter1continuedb.Describetheresultingcurve.b.Describetheresultingcurve.astraightlineahyperbolac.Usethegraphtowriteanequationc.Accordingtothegraph,whatistherelatingthedistancetotheforce.relationshipbetweenmassandthed4.9Faccelerationproducedbyaconstantforce?d.Whatistheconstantintheequation?Finditsunits.Accelerationvariesinverselywithmass.Theconstantis4.9andhasunitscm/N.d.Writetheequationrelatingaccelerationtomassgivenbythedatainthegraph.e.Predictthedistancetraveledwhena22.0-Nforceisexertedontheobject12amfor5s.e.Findtheunitsoftheconstantinthe108cmor110cmusing2significantequation.digits2kgm/s85.Thephysicsinstructorfromthepreviousf.Predicttheaccelerationofan8.0-kgproblemchangedtheprocedure.Themassmass.wasvariedwhiletheforcewaskeptcon-1.5m/s2stant.Timeanddistanceweremeasured,andtheaccelerationofeachmasswascal-86.Duringanexperiment,astudentmeasuredculated.Theresultsoftheexperimentarethemassof10.0cm3ofalcohol.ThestudentshowninTable1-6.thenmeasuredthemassof20.0cm3ofalcohol.Inthisway,thedatainTable1-7Table1-6werecollected.AccelerationofDifferentMassesTable1-7Mass(kg)Acceleration(m/s2)TheMassValuesof1.012.0SpecificVolumesofAlcoholCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.05.9Volume(cm3)Mass(g)3.04.110.07.94.03.020.015.85.02.530.023.76.02.040.031.650.039.6a.Plotthevaluesgiveninthetableanddrawthecurvethatbestfitsallpoints.a.Plotthevaluesgiveninthetableand)12.0drawthecurvethatbestfitsallthepoints.240.09.030.06.020.0Mass(g)3.0Acceleration(m/s10.00.02.04.06.00.0Mass(kg)10.020.030.040.050.060.0Volume(cm3)12SolutionsManualPhysics:PrinciplesandProblems

16Chapter1continuedb.Describetheresultingcurve.90.Youaregiventhefollowingmeasurementsastraightlineofarectangularbar:length2.347m,thickness3.452cm,height2.31mm,c.Usethegraphtowriteanequationmass1659g.Determinethevolume,inrelatingthevolumetothemassofthecubicmeters,anddensity,ing/cm3,ofthealcohol.beam.Expressyourresultsinproperform.m0.79Vvolume1.87104m3,or187cm3;d.Findtheunitsoftheslopeofthegraph.3density8.87g/cmWhatisthenamegiventothisquantity?g/cm3;density91.Adropofwatercontains1.71021mole-e.Whatisthemassof32.5cm3ofcules.Ifthewaterevaporatedattherateofalcohol?onemillionmoleculespersecond,howmanyyearswouldittakeforthedropto25.7gcompletelyevaporate?MixedReview1.71021molecules1.71015spage281,000,000molecules1s87.Arrangethefollowingnumbersfrommost15s)1h1day1yprecisetoleastprecise(1.7103600s24h365days0.0034m45.6m1234m5.4107y0.0034m,45.6m,1234m92.A17.6-gramsampleofmetalisplacedina88.Figure1-26showsthetoroidal(doughnut-3ofgraduatedcylindercontaining10.0cmshaped)interiorofthenow-dismantled3,water.Ifthewaterlevelrisesto12.20cmTokamakFusionTestReactor.Explainwhyawhatisthedensityofthemetal?widthof80mwouldbeanunreasonablemvalueforthewidthofthetoroid.WhatdensityVwouldbeareasonablevalue?17.6g12.20cm310.0cm38.00g/cm3ThinkingCriticallypage2893.ApplyConceptsIthasbeensaidthatfoolscanaskmorequestionsthanthewisecananswer.Inscience,itisfrequentlythecasethatonewisepersonisneededtoasktherightquestionratherthantoanswerit.■Figure1-26Explain.80metersisequivalenttoabout260feet,The“right”questionisonethatpointswhichwouldbeverylarge.10meterstofruitfulresearchandtootherwouldbeamorereasonablevalue.questionsthatcanbeanswered.89.Youarecrackingacodeandhave94.ApplyConceptsFindtheapproximatemassCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.discoveredthefollowingconversionofwaterinkilogramsneededtofillacon-factors:1.23longs23.0mediums,andtainerthatis1.40mlongand0.600mwide74.5mediums645shorts.Howmanytoadepthof34.0cm.Reportyourresulttoshortsareequaltoonelong?onesignificantdigit.(Useareferencesource23.0med645shorttofindthedensityofwater.)1long162shorts1.23long74.5medPhysics:PrinciplesandProblemsSolutionsManual13

17Chapter1continuedV99.Explainhowimprovedprecisioninw(140cm)(60.0cm)(34.0cm)285,600cm3.Becausethedensityofmeasuringtimewouldhaveledtowateris1.00g/cm3,themassofwatermoreaccuratepredictionsabouthowinkilogramsis286kg.anobjectfalls.Answerswillvary.Forexample,students95.AnalyzeandConcludeAcontainerofgasmightsuggestthatimprovedprecisionwithapressureof101kPahasavolumecanleadtobetterobservations.of324cm3andamassof4.00g.Ifthepressureisincreasedto404kPa,whatisthedensityofthegas?PressureandvolumeChallengeProblemareinverselyproportional.page17PressureandvolumeareinverselyAnobjectissuspendedfromspring1,andtheproportional.Sincethepressureis1spring’selongation(thedistanceitstretches)isX1.4timesgreater,thevolumewillbe4Thenthesameobjectisremovedfromthefirstoftheoriginalvolume.springandsuspendedfromasecondspring.The324cm33elongationofspring2isX81.0cm2.X2isgreaterthanX1.41.Onthesameaxes,sketchthegraphsofthe4.00g0.0494g/cm3massversuselongationforbothsprings.81.0cm396.DesignanExperimentHowhighcanyouthrowaball?Whatvariablesmightaffecttheanswertothisquestion?X2massofball,footing,practice,andXconditioningX197.CalculateIftheSunsuddenlyceasedtoshine,howlongwouldittakeEarthtoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.becomedark?(YouwillhavetolookuptheFspeedoflightinavacuumandthedistancefromtheSuntoEarth.)Howlongwouldit2.Istheoriginincludedinthegraph?WhyortakethesurfaceofJupitertobecomedark?whynot?dE1.4961011mYes;theorigincorrespondstotEv3.00108m/s0elongationwhentheforceis0.499s8.31min3.Whichslopeissteeper?dJ7.781011mtJv3.00108m/sTheslopeforX2issteeper.2593s43.2min4.Atagivenmass,X21.6X1.IfXWritinginPhysics25.3cm,whatisX1?Xpage2821.6X198.Researchanddescribeatopicinthehistory5.3cm1.6X1ofphysics.Explainhowideasaboutthe3.3cmX1topicchangedovertime.Besuretoincludethecontributionsofscientistsandtoevaluatetheimpactoftheircontributionsonscientificthoughtandtheworldoutsidethelaboratory.Answerswillvary.14SolutionsManualPhysics:PrinciplesandProblems

18CHAPTER2RepresentingMotion4.CriticalThinkingUsetheparticlemodelSectionReviewtodrawmotiondiagramsfortworunners2.1PicturingMotioninarace,whenthefirstrunnercrossesthepages31–33finishlineastheotherrunneristhree-fourthsofthewaytothefinishline.page331.MotionDiagramofaRunnerUsetheRunner2t0t1t2t3t4particlemodeltodrawamotiondiagramforabikeriderridingataconstantpace.Runner1t0t1t2t3t4StartFinish2.MotionDiagramofaBirdUsetheparti-clemodeltodrawasimplifiedmotiondia-SectionReviewgramcorrespondingtothemotiondiagraminFigure2-4foraflyingbird.Whatpoint2.2WhereandWhen?onthebirddidyouchoosetorepresentit?pages34–37page375.DisplacementTheparticlemodelforacartravelingonaninterstatehighwayisshownbelow.Thestartingpointisshown.HereThereMakeacopyoftheparticlemodel,anddrawavectortorepresentthedisplacement■Figure2-4ofthecarfromthestartingtimetotheend3.MotionDiagramofaCarUsetheparticleofthethirdtimeinterval.modeltodrawasimplifiedmotiondiagramHereTherecorrespondingtothemotiondiagraminFigure2-5foracarcomingtoastopata6.DisplacementTheparticlemodelforaboystopsign.Whatpointonthecardidyouwalkingtoschoolisshownbelow.usetorepresentit?HomeSchoolMakeacopyoftheparticlemodel,anddrawvectorstorepresentthedisplacementbetweeneachpairofdots.HomeSchool■Figure2-5Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual15

19Chapter2continued7.PositionTwostudentscomparedthePracticeProblemspositionvectorstheyeachhaddrawnonamotiondiagramtoshowthepositionofa2.3Position-TimeGraphsmovingobjectatthesametime.Theyfoundpages38–42thattheirvectorsdidnotpointinthesamepage39direction.Explain.Forproblems9–11,refertoFigure2-13.Apositionvectorgoesfromtheorigintotheobject.Whentheoriginsaredif-150.0ferent,thepositionvectorsaredifferent.Ontheotherhand,adisplacementvec-100.0torhasnothingtodowiththeorigin.50.08.CriticalThinkingAcartravelsstraightPosition(m)0.0alongthestreetfromthegrocerystoreto1.03.05.07.0thepostoffice.Torepresentitsmotionyou50.0Time(s)useacoordinatesystemwithitsoriginatthegrocerystoreandthedirectionthecaris■Figure2-13movinginasthepositivedirection.Your9.Describethemotionofthecarshownbyfriendusesacoordinatesystemwithitsori-thegraph.ginatthepostofficeandtheoppositeThecarbeginsatapositionof125.0mdirectionasthepositivedirection.Wouldandmovestowardtheorigin,arrivingatthetwoofyouagreeonthecar’sposition?theorigin5.0safteritbeginsmoving.Displacement?Distance?ThetimeintervalThecarcontinuesbeyondtheorigin.thetriptook?Explain.Thetwostudentsshouldagreeonthe10.Drawamotiondiagramthatcorrespondstodisplacement,distance,andtimeintervalthegraph.forthetrip,becausethesethreequanti-tiesareindependentofwheretheorigint00.0st55.0sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofthecoordinatesystemisplaced.Thetwostudentswouldnotagreeonthe125.0m0.0mcar’sposition,becausethepositionisdmeasuredfromtheoriginofthecoordi-natesystemtothelocationofthecar.11.Answerthefollowingquestionsaboutthecar’smotion.Assumethatthepositived-directioniseastandthenegatived-directioniswest.a.Whenwasthecar25.0meastoftheorigin?at4.0sb.Wherewasthecarat1.0s?100.0m12.Describe,inwords,themotionofthetwopedestriansshownbythelinesinFigure2-14.Assumethatthepositivedirec-tioniseastonBroadStreetandtheoriginistheintersectionofBroadandHighStreets.16SolutionsManualPhysics:PrinciplesandProblems

20Chapter2continuedBroadSt.page41Forproblems14–17,refertothefigureinEastExampleProblem2.200.0A150.0100.0Position(m)HighSt.50.0BPosition(m)0.02.04.06.08.010.012.0WestTime(s)Time(s)■Figure2-14■ExampleProblem2FigurePedestrianAstartswestofHighStreet14.Whateventoccurredatt0.0s?andwalkseast(thepositivedirection).RunnerApassedtheorigin.PedestrianBbeginseastofHighStreetandwalkswest(thenegativedirection).15.Whichrunnerwasaheadatt48.0s?SometimeafterBcrossesHighStreet,runnerBAandBpasseachother.Sometimeaftertheypass,PedestrianAcrosses16.WhenrunnerAwasat0.0m,wherewasHighStreet.runnerB?at50.0m13.Odinawalkeddownthehallatschoolfromthecafeteriatothebandroom,a17.HowfarapartwererunnersAandBatdistanceof100.0m.Aclassofphysicst20.0s?studentsrecordedandgraphedherapproximately30mpositionevery2.0s,notingthatshemoved2.6mevery2.0s.WhenwasOdinainthe18.Juanitagoesforawalk.Sometimelater,herfollowingpositions?friendHeatherstartstowalkafterher.Theira.25.0mfromthecafeteriamotionsarerepresentedbytheposition-19stimegraphsinFigure2-16.b.25.0mfromthebandroom6.058s5.0ac.CreateagraphshowingOdina’sit4.0anmotion.Jur3.0eathe100.0H2.080.0Position(km)1.060.040.00.00.51.01.52.0Distancefromcafeteria(m)20.0Time(h)0.0010.030.050.070.0■Figure2-16Time(s)a.HowlonghadJuanitabeenwalkingCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.whenHeatherstartedherwalk?6.0minPhysics:PrinciplesandProblemsSolutionsManual17

21Chapter2continuedb.WillHeathercatchuptoJuanita?Howt00.0st77.0scanyoutell?No.ThelinesrepresentingJuanita’sandHeather’smotionsgetfarther0m140mapartastimeincreases.Thelineswillnotintersect.Forproblems21–23,refertoFigure2-18.21.TimeUsetheposition-timegraphoftheSectionReviewhockeypucktodeterminewhenitwas10.0mbeyondtheorigin.2.3Position-TimeGraphs0.5spages38–42page4222.DistanceUsetheposition-timegraphof19.Position-TimeGraphFromtheparticlethehockeypucktodeterminehowfaritmodelinFigure2-17ofababycrawlingmovedbetween0.0sand5.0s.acrossakitchenfloor,plotaposition-time100mgraphtorepresenthismotion.Thetimeintervalbetweensuccessivedotsis1s.23.TimeIntervalUsetheposition-timegraphforthehockeypucktodeterminehowmuchtimeittookforthepucktogofrom40m020406080100120140160beyondtheoriginto80mbeyondtheorigin.Position(cm)2.0s■Figure2-1716024.CriticalThinkingLookattheparticle140modelandposition-timegraphshownin120100Figure2-19.Dotheydescribethesame8060motion?Howdoyouknow?DonotPosition(m)40confusethepositioncoordinatesysteminCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.200theparticlemodelwiththehorizontalaxis12345678intheposition-timegraph.Thetimeinter-Time(s)valsintheparticlemodelare2s.20.MotionDiagramCreateaparticlemodelfromtheposition-timegraphofahockey010puckglidingacrossafrozenpondinPosition(m)Figure2-18.12140120100880604Position(m)Position(m)402000.01.02.03.04.05.06.07.012345Time(s)Time(s)■Figure2-19■Figure2-1818SolutionsManualPhysics:PrinciplesandProblems

22Chapter2continuedNo,theydon’tdescribethesame26.Describe,inwords,themotionofthecruisemotion.Althoughbothobjectsaretrav-shipinthepreviousproblem.elinginthepositivedirection,oneisTheshipismovingtothenorthatamovingmorequicklythantheother.speedof0.33m/s.Studentscanciteanumberofdifferentspecificexamplesfromthegraphand27.ThegraphinFigure2-23representstheparticlemodeltobackthisup.motionofabicycle.Determinethebicycle’saveragespeedandaveragevelocity,anddescribeitsmotioninwords.PracticeProblems202.4HowFast?15pages43–4710page4525.ThegraphinFigure2-22describesthe5Position(km)motionofacruiseshipduringitsvoyagethroughcalmwaters.Thepositive051015202530d-directionisdefinedtobesouth.Time(min)■Figure2-23Time(s)Becausethebicycleismovinginthe12340positivedirection,theaveragespeedandaveragevelocityarethesame.Usingthepoints(0.0min,0.0km)and1(15.0min,10.0km),Position(m)d2vtdd■Figure2-2221t2t1a.Whatistheship’saveragespeed?10.0km0.0kmUsingthepoints(0.0s,0.0m)and15.0min0.0min(3.0s,1.0m)0.67km/mindvv0.67km/mininthepositivetdirectiondd21Thebicycleismovinginthepositivet2t1directionataspeedof0.67km/min.1.0m0.0m3.0s0.0s0.33m/s0.33m/sb.Whatisitsaveragevelocity?Theaveragevelocityistheslopeoftheline,includingthesign,soitisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.33m/sor0.33m/snorth.Physics:PrinciplesandProblemsSolutionsManual19

23Chapter2continued28.WhenMarilyntakesherpetdogforawalk,SlopeA2thedogwalksataveryconsistentpaceof3SlopeB20.55m/s.Drawamotiondiagramandposi-Slopetion-timegraphtorepresentMarilyn’sdogC1Slopewalkingthe19.8-mdistancefrominfrontD1ofherhousetothenearestfirehydrant.30.AverageVelocityRankthegraphsaccord-MotionDiagramingtoaveragevelocity,fromgreatestaveraget00st636svelocitytoleastaveragevelocity.Specificallyindicateanyties.B,D,C,A0.0m19.8mHouseHydrantSlopeA23SlopePosition-TimeGraphB220.0SlopeC115.0SlopeD110.031.InitialPositionRankthegraphsaccordinghouse(m)5.0totheobject’sinitialposition,frommostPositionfrompositivepositiontomostnegativeposition.061218243036Specificallyindicateanyties.WouldyourTime(s)rankingbedifferentifyouhadbeenaskedtodotherankingaccordingtoinitialdistancefromtheorigin?SectionReviewA,C,B,D.Yes,therankingfromgreatesttoleastdistancewouldbeA,C,D,B.2.4HowFast?pages43–4732.AverageSpeedandAverageVelocitypage47ExplainhowaveragespeedandaverageCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Forproblems29–31,refertoFigure2-25.velocityarerelatedtoeachother.29.AverageSpeedRanktheposition-timeAveragespeedistheabsolutevalueofgraphsaccordingtotheaveragespeedofthetheaveragevelocity.Speedisonlyaobject,fromgreatestaveragespeedtoleastmagnitude,whilevelocityisamagni-averagespeed.Specificallyindicateanyties.tudeandadirection.BDAPosition(m)CTime(s)■Figure2-25Forspeedusetheabsolutevalue,there-foreA,B,CD20SolutionsManualPhysics:PrinciplesandProblems

24Chapter2continued33.CriticalThinkingInsolvingaphysics37.Thefollowingquantitiesdescribelocationproblem,whyisitimportanttocreateoritschange:position,distance,anddis-pictorialandphysicalmodelsbeforetryingplacement.Brieflydescribethedifferencestosolveanequation?amongthem.(2.2)Answerswillvary,butherearesomeofPositionanddisplacementaredifferenttheimportantpoints.Drawingthemod-fromdistancebecausepositionanddis-elsbeforewritingdowntheequationplacementbothcontaininformationhelpsyoutogettheproblemsituationaboutthedirectioninwhichanobjectorganizedinyourhead.It’sdifficulttohasmoved,whiledistancedoesnot.writedowntheproperequationifyouDistanceanddisplacementaredifferentdon’thaveaclearpictureofhowthingsfrompositionbecausetheydescribearesituatedand/ormoving.Also,youhowanobject’slocationhaschangedchoosethecoordinatesysteminthisduringatimeinterval,wherepositionstep,andthisisessentialinmakingtellsexactlywhereanobjectislocatedsureyouusethepropersignsontheataprecisetime.quantitiesyouwillsubstituteintotheequationlater.38.Howcanyouuseaclocktofindatimeinterval?(2.2)ReadtheclockatthebeginningandChapterAssessmentendoftheintervalandsubtractthebeginningtimefromtheendingtime.ConceptMappingpage5239.In-lineSkatingHowcanyouusethe34.Completetheconceptmapbelowusingtheposition-timegraphsfortwoin-lineskatersfollowingterms:words,equivalentrepresenta-todetermineifandwhenonein-lineskatertions,position-timegraph.willpasstheotherone?(2.3)Drawthetwographsonthesamesetofaxes.Oneinlineskaterwillpassthewordsotherifthelinesrepresentingeachoftheirmotionsintersect.Thepositioncoordinateofthepointwherethelinesequivalentdataposition-timemotionintersectisthepositionwheretherepresentationstablegraphdiagrampassingoccurs.40.WalkingVersusRunningAwalkerandaMasteringConceptsrunnerleaveyourfrontdooratthesamepage52time.Theymoveinthesamedirectionat35.Whatisthepurposeofdrawingamotiondifferentconstantvelocities.Describethediagram?(2.1)position-timegraphsofeach.(2.4)AmotiondiagramgivesyouapictureBotharestraightlinesthatstartattheofmotionthathelpsyouvisualizesameposition,buttheslopeofthedisplacementandvelocity.runner’slineissteeper.36.Underwhatcircumstancesisitlegitimateto41.Whatdoestheslopeofaposition-timeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.treatanobjectasapointparticle?(2.1)graphmeasure?(2.4)Anobjectcanbetreatedasapointvelocityparticleifinternalmotionsarenotimportantandiftheobjectissmallincomparisontothedistanceitmoves.Physics:PrinciplesandProblemsSolutionsManual21

25Chapter2continued42.Ifyouknowthepositionsofamoving46.Figure2-26isagraphoftwopeoplerunning.objectattwopointsalongitspath,andyoualsoknowthetimeittookfortheobjecttoBregetfromonepointtotheother,canyounnuRdeterminetheparticle’sinstantaneousRunnerAvelocity?Itsaveragevelocity?Explain.(2.4)ItispossibletocalculatetheaveragePosition(m)velocityfromtheinformationgiven,butitisnotpossibletofindtheinstanta-neousvelocity.Time(s)■Figure2-26ApplyingConceptsa.DescribethepositionofrunnerApage52relativetorunnerBatthey-intercept.43.TestthefollowingcombinationsandRunnerAhasaheadstartbyexplainwhyeachdoesnothavetheproper-fourunits.tiesneededtodescribetheconceptofveloc-b.Whichrunnerisfaster?ity:dt,dt,dt,t/d.RunnerBisfaster,asshownbydtincreaseswheneithertermthesteeperslope.increases.Thesignofdtdependsc.WhatoccursatpointPandbeyond?upontherelativesizesofdandt.dtincreaseswheneitherincreas-RunnerBpassesrunnerAatpointes.t/ddecreaseswithincreasingP.displacementandincreaseswithincreasingtimeinterval,whichisback-47.Theposition-timegraphinFigure2-27wardsfromvelocity.showsthemotionoffourcowswalkingfromthepasturebacktothebarn.Rankthe44.FootballWhencanafootballbeconsid-cowsaccordingtotheiraveragevelocity,eredapointparticle?fromslowesttofastest.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.AfootballcanbetreatedasapointparticleifitsrotationsarenotimportanteandifitissmallincomparisontotheislBessieEdistanceitmoves—fordistancesofdalin1yardormore.MooylloD45.WhencanafootballplayerbetreatedasaPosition(m)pointparticle?AfootballplayercanbetreatedasapointparticleifhisorherinternalTime(s)motionsarenotimportantandifheor■Figure2-27sheissmallincomparisontothedis-Moolinda,Dolly,Bessie,Elsietanceheorshemoves—fordistancesofseveralyardsormore.22SolutionsManualPhysics:PrinciplesandProblems

26Chapter2continued48.Figure2-28isaposition-timegraphforarabbitrunningawayfromadog.321Position(m)0123Time(s)■Figure2-28a.Describehowthisgraphwouldbedifferentiftherabbitrantwiceasfast.Theonlydifferenceisthattheslopeofthegraphwouldbetwiceassteep.b.Describehowthisgraphwouldbedif-ferentiftherabbitranintheoppositedirection.Themagnitudeoftheslopewouldbethesame,butitwouldbenegative.MasteringProblems2.4HowFast?page53Level149.Abiketravelsataconstantspeedof4.0m/sfor5.0s.Howfardoesitgo?dvt(4.0m/s)(5s)20m50.AstronomyLightfromtheSunreachesEarthin8.3min.Thespeedoflightis3.00108m/s.HowfarisEarthfromtheSun?dvt8m/s)(8.3min)60s(3.00101min1.51011mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual23

27Chapter2continuedLevel2pages53–5451.Acarismovingdownastreetat55km/h.ALevel1childsuddenlyrunsintothestreet.Ifit54.CyclingAcyclistmaintainsaconstanttakesthedriver0.75storeactandapplyvelocityof5.0m/s.Attimet0.0s,thethebrakes,howmanymeterswillthecarcyclistis250mfrompointA.havemovedbeforeitbeginstoslowdown?a.Plotaposition-timegraphofthedvtcyclist’slocationfrompointAat10.0-s1000m1h(55km/h)(0.75s)intervalsfor60.0s.1km3600s11m55050052.Norajogsseveraltimesaweekandalwayskeepstrackofhowmuchtimesheruns450eachtimeshegoesout.Onedaysheforgets400totakeherstopwatchwithherandwonders350ifthere’sawayshecanstillhavesomeideaPosition(m)300ofhertime.Asshepassesaparticularbank,sheremembersthatitis4.3kmfromher250house.Sheknowsfromherprevious200trainingthatshehasaconsistentpaceof0.010.020.030.040.050.060.04.0m/s.HowlonghasNorabeenjoggingTime(s)whenshereachesthebank?b.Whatisthecyclist’spositionfrompointdvtAat60.0s?1000m550m(4.3km)d1kmtc.Whatisthedisplacementfromthev4.0m/sstartingpositionat60.0s?1075s550m250m3.0102m1min(1075s)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.60s55.Figure2-29isaparticlemodelfora18minchickencasuallywalkingacrosstheroad.Timeintervalsareevery0.1s.Drawthecorrespondingposition-timegraphandLevel3equationtodescribethechicken’smotion.53.DrivingYouandafriendeachdrive50.0km.Youtravelat90.0km/h;yourThissideTheothersidefriendtravelsat95.0km/h.HowlongwillyourfriendhavetowaitforyouattheendTimeintervalsare0.1s.ofthetrip?■Figure2-29dvtd50.0kmdt1v90.0km/hTheotherside0.556hd50.0kmt2v95.0km/h0.526h60mint1t2(0.556h0.526h)1hThissidet1.8min01.9sMixedReview56.Figure2-30showsposition-timegraphsfor24SolutionsManualPhysics:PrinciplesandProblems

28Chapter2continuedJosziandHeikepaddlingcanoesinalocal250river.CarB2001817016CarAJoszi15014120Heike1210010Position(km)508Position(km)601.01.63.041.42.02Time(h)00.51.01.52.02.5Time(h)b.Bothcarspassedagasstation120km■Figure2-30fromtheschool.Whendideachcarpassthegasstation?Calculatethetimesa.Atwhattime(s)areJosziandHeikeinandshowthemonyourgraph.thesameplace?1.0hd120kmt1.6hA75km/hAb.HowmuchtimedoesJoszispendontheriverbeforehepassesHeike?d120kmt1.4hB85km/h45minBc.Whereontheriverdoesitappearthat58.Drawaposition-timegraphfortwocarstheremightbeaswiftcurrent?travelingtothebeach,whichis50kmfromfrom6.0to9.0kmfromtheoriginschool.Atnoon,CarAleavesastorethatis10kmclosertothebeachthantheschoolisandmovesat40km/h.CarBstartsfromLevel2schoolat12:30P.M.andmovesat100km/h.57.DrivingBothcarAandcarBleaveschoolWhendoeseachcargettothebeach?whenastopwatchreadszero.CarAtravelsataconstant75km/h,andcarBtravelsata50constant85km/h.40a.Drawaposition-timegraphshowingCarA30themotionofbothcars.Howfarare20CarBthetwocarsfromschoolwhenthestop-watchreads2.0h?Calculatethedis-Position(m)10tancesandshowthemonyourgraph.0Noon12101220123012401250100PMdAvAt(75km/h)(2.0h)Time150kmBothcarsarriveatthebeachat1:00P.M.dBvBt(85km/h)(2.0h)Level3Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.170km59.Twocarstravelalongastraightroad.Whenastopwatchreadst0.00h,carAisatdA48.0kmmovingataconstant36.0km/h.Later,whenthewatchreadst0.50h,carBisatdB0.00kmmovingPhysics:PrinciplesandProblemsSolutionsManual25

29Chapter2continuedat48.0km/h.Answerthefollowingques-60.Figure2-31showstheposition-timegraphtions,first,graphicallybycreatingaposi-depictingJim’smovementupanddownthetion-timegraph,andsecond,algebraicallyaisleatastore.Theoriginisatoneendofbywritingequationsforthepositionstheaisle.dAanddBasafunctionofthestopwatch14.0time,t.a.WhatwillthewatchreadwhencarB12.0passescarA?10.0300.08.06.0200.0Position(m)4.0150.0CarA2.0150.0Position(km)CarB0.0010.020.030.040.050.060.0100.0Time(s)50.0■Figure2-31a.WriteastorydescribingJim’smovements1.000.001.002.003.004.005.006.007.00atthestorethatwouldcorrespondtotheTime(h)motionrepresentedbythegraph.Answerswillvary.Carspasswhenthedistancesareequal,db.WhendoesJimhaveapositionof6.0m?AdBdfrom8.0to18.0s,53.0to56.0s,andA48.0km(36.0km/h)tat43.0sanddB0(48.0km/h)(t0.50h)c.Howmuchtimepassesbetweenwhenso48.0km(36.0km/h)tJimenterstheaisleandwhenhegetstoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(48.0km/h)(t0.50h)apositionof12.0m?WhatisJim’saver-(48.0km)(36.0km/h)tagevelocitybetween37.0sand46.0s?(48.0km/h)t24kmt33.0s72km(12.0km/h)tUsingthepoints(37.0s,12.0m)andt6.0h(46.0s,3.00m)b.AtwhatpositionwillcarBpasscarA?dfdi3.00m12.0mdvA48.0km(36.0km/h)(6.0h)tfti46.0s37.0s2.6102km1.00m/sc.Whenthecarspass,howlongwillithavebeensincecarAwasatthereferencepoint?dvtd48.0kmsot1.33hv36.0km/hCarAhasstarted1.33hbeforetheclockstarted.t6.0h1.33h7.3h26SolutionsManualPhysics:PrinciplesandProblems

30Chapter2continuedThinkingCriticallyExplanation:Assumeyouwanttotravel90kmin1h.page54Ifyoucoverthefirsthalfofthedistance61.ApplyCalculatorsMembersofaphysicsat48km/h,thenyou’vegone45kminclassstood25mapartandusedstopwatch-destomeasurethetimewhichacartraveling0.9375h(becauset).Thismeansvonthehighwaypassedeachperson.Theiryouhaveused93.75%ofyourtimefordataareshowninTable2-3.thefirsthalfofthedistanceleaving6.25%ofthetimetogotheremainingTable2-345km.Positionv.Time45kmvTime(s)Position(m)0.0625h0.00.0720km/h1.325.063.DesignanExperimentEverytimeapar-2.750.0ticularredmotorcycleisdrivenpastyour3.675.0friend’shome,hisfatherbecomesangrybecausehethinksthemotorcycleisgoing5.1100.0toofastfortheposted25mph(40km/h)5.9125.0speedlimit.Describeasimpleexperiment7.0150.0youcoulddotodeterminewhetherornotthemotorcycleisspeedingthenexttimeit8.6175.0isdrivenpastyourfriend’shouse.10.3200.0Thereareactuallyseveralgoodpossibil-Useagraphingcalculatortofitalinetoaitiesforanswersonthisone.Twothatposition-timegraphofthedataandtoplotshouldbeamongthemostpopulararethisline.Besuretosetthedisplayrangeofbrieflydescribedhere.1)Getseveralthegraphsothatallthedatafitonit.Findpeopletogetherandgiveeveryoneatheslopeoftheline.Whatwasthespeedofwatch.Synchronizethewatchesandthecar?standalongthestreetseparatedbyaconsistentdistance,maybe10morso.200.0Whenthemotorcyclepasses,haveeach150.0personrecordthetime(atleasttoan100.0accuracyofseconds)thatthemotorcy-clecrossedinfrontofthem.Plotaposi-50.0Position(m)tiontimegraph,andcomputetheslope0.02.04.06.08.010.012.0ofthebest-fitline.Iftheslopeisgreaterthan25mph,themotorcycleisspeed-Time(s)ing.2)Getsomeonewithadriver’sTheslopeofthelineandthespeedoflicensetodriveacaralongthestreetatthecarare19.7m/s.25mphinthesamedirectionasyouexpectthemotorcycletogo.Ifthe62.ApplyConceptsYouplanacartripformotorcyclegetsclosertothecar(ifthewhichyouwanttoaverage90km/h.Youdistancebetweenthemdecreases),theCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.coverthefirsthalfofthedistanceatanmotorcycleisspeeding.Ifthedistanceaveragespeedofonly48km/h.Whatmustbetweenthemstaysthesame,theyouraveragespeedbeinthesecondhalfofmotorcycleisdrivingatthespeedlimit.thetriptomeetyourgoal?Isthisreason-Ifthedistanceincreases,themotorcycleable?Notethatthevelocitiesarebasedonisdrivinglessthanthespeedlimit.halfthedistance,nothalfthetime.720km/h;No64.InterpretGraphsIsitpossibleforanPhysics:PrinciplesandProblemsSolutionsManual27

31Chapter2continuedobject’sposition-timegraphtobeahori-CumulativeReviewzontalline?Averticalline?Ifyouanswerpage54yestoeithersituation,describetheassociat-67.Converteachofthefollowingtimemea-edmotioninwords.surementstoitsequivalentinseconds.Itispossibletohaveahorizontalline(Chapter1)asaposition-timegraph;thiswoulda.58nsindicatethattheobject’spositionisnot5.8108schanging,orinotherwords,thatitisnotmoving.Itisnotpossibletohaveab.0.046Gsposition-timegraphthatisavertical4.6107sline,becausethiswouldmeanthec.9270msobjectismovingataninfinitespeed.9.27sWritinginPhysicsd.12.3kspage541.23104s65.Physicistshavedeterminedthatthespeedoflightis3.00108m/s.Howdidtheyarrive68.Statethenumberofsignificantdigitsinthefollowingmeasurements.(Chapter1)atthisnumber?Readaboutsomeoftheseriesofexperimentsthatweredonetoa.3218kgdeterminelight’sspeed.Describehowthe4experimentaltechniquesimprovedtomakeb.60.080kgtheresultsoftheexperimentsmoreaccurate.5Answerswillvary.Galileoattemptedtoc.801kgdeterminethespeedoflightbutwasunsuccessful.DanishastronomerOlaus3Roemersuccessfullymeasuredthed.0.000534kgspeedoflightin1676byobservingthe3eclipsesofthemoonsofJupiter.HisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.estimatewas140,000miles/s(225,30869.Usingacalculator,Chrisobtainedthefol-km/s).Manyotherssincehavetriedtolowingresults.Rewritetheanswertoeachmeasureitmoreaccuratelyusingrotat-operationusingthecorrectnumberofsig-ingtoothedwheels,rotatingmirrorsandnificantdigits.(Chapter1)theKerrcellshutter.a.5.32mm2.1mm7.4200000mm7.4mm66.Somespeciesofanimalshavegoodendurance,whileothershavetheabilitytob.13.597m3.65m49.62905m2moveveryquickly,butforonlyashort49.6m2amountoftime.Usereferencesourcestoc.83.2kg12.804kg70.3960000kgfindtwoexamplesofeachqualityand70.4kgdescribehowitishelpfultothatanimal.Answerswillvary.Examplesofanimalswithhighendurancetooutlastpreda-torsorpreyincludemules,bears,andcoyotes.Animalswiththespeedtoquicklyescapepredatorsorcapturepreyincludecheetahs,antelopesanddeer.28SolutionsManualPhysics:PrinciplesandProblems

32CHAPTER3AcceleratedMotion3.Refertothev-tgraphofthetoytrainPracticeProblemsinFigure3-6toanswerthefollowing3.1Accelerationquestions.pages57–6412.0page6110.01.Adogrunsintoaroomandseesacatatthe8.0otherendoftheroom.Thedoginstantly6.0stopsrunningbutslidesalongthewood4.0flooruntilhestops,byslowingdownwithaVelocity(m/s)2.0constantacceleration.Sketchamotiondia-0.0gramforthissituation,andusethevelocity10.020.030.040.0vectorstofindtheaccelerationvector.Time(s)■Figure3-6123Timea.Whenisthetrain’sspeedconstant?intervalv1v2v3Velocity5.0to15.0sb.Duringwhichtimeintervalisthetrain’sPositionStartStopaccelerationpositive?v2v10.0to5.0sac.Whenisthetrain’saccelerationmostnegative?2.Figure3-5isav-tgraphforStevenashe15.0to20.0swalksalongthemidwayatthestatefair.Sketchthecorrespondingmotiondiagram,4.RefertoFigure3-6tofindtheaveragecompletewithvelocityvectors.accelerationofthetrainduringthefollow-ingtimeintervals.a.0.0sto5.0sva2v1t2t110.0m/s0.0m/s5.0s0.0sVelocity(m/s)2.0m/s2011234567890b.15.0sto20.0sTime(s)v■Figure3-5a2v1t2t14.0m/s10.0m/sTime(s)0123456819020.0s15.0sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.71.2m/s2c.0.0sto40.0sva2v1t2t1Physics:PrinciplesandProblemsSolutionsManual29

33Chapter3continued0.0m/s0.0m/sa.Whatistheaverageaccelerationofthe40.0s0.0sbuswhilebraking?0.0m/s2vat5.Plotav-tgraphrepresentingthefollowing0.0m/s25m/s28.3m/smotion.Anelevatorstartsatrestfromthe3.0sgroundfloorofathree-storyshoppingb.Ifthebustooktwiceaslongtostop,mall.Itacceleratesupwardfor2.0satahowwouldtheaccelerationcomparerateof0.5m/s2,continuesupatacon-withwhatyoufoundinparta?stantvelocityof1.0m/sfor12.0s,and2)halfasgreat(4.2m/sthenexperiencesaconstantdownwardaccelerationof0.25m/s2for4.0sasit10.Rohithhasbeenjoggingtothebusstopforreachesthethirdfloor.2.0minat3.5m/swhenhelooksathiswatchandseesthathehasplentyoftime1.0beforethebusarrives.Overthenext10.0s,heslowshispacetoaleisurely0.75m/s.Whatwashisaverageaccelerationduringthis10.0s?Velocity(m/s)vat0.05.010.015.020.00.75m/s3.5m/sTime(s)10.0s0.28m/s2page6411.Iftherateofcontinentaldriftwereto6.Aracecar’svelocityincreasesfrom4.0m/sabruptlyslowfrom1.0cm/yrto0.5cm/yrto36m/sovera4.0-stimeinterval.Whatisoverthetimeintervalofayear,whatwoulditsaverageacceleration?betheaverageacceleration?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.av36m/s4.0m/s8.0m/s2v0.5cm/yr1.0cm/yrt4.0sat1.0yr0.5cm/yr27.Theracecarinthepreviousproblemslowsfrom36m/sto15m/sover3.0s.Whatisitsaverageacceleration?SectionReviewv15m/s36m/s2a7.0m/st3.0s3.1Accelerationpages57–648.Acariscoastingbackwardsdownhillatapage64speedof3.0m/swhenthedrivergetsthe12.Velocity-TimeGraphWhatinformationenginestarted.After2.5s,thecarismovingcanyouobtainfromavelocity-timegraph?uphillat4.5m/s.Ifuphillischosenasthepositivedirection,whatisthecar’saverageThevelocityatanytime,thetimeatacceleration?whichtheobjecthadaparticularvelocity,thesignofthevelocity,andthev4.5m/s(3.0m/s)2a3.0m/sdisplacement.t2.5s13.Position-TimeandVelocity-TimeGraphs9.Abusismovingat25m/swhenthedriverTwojoggersrunataconstantvelocityofstepsonthebrakesandbringsthebustoa7.5m/stowardtheeast.Attimet0,onestopin3.0s.30SolutionsManualPhysics:PrinciplesandProblems

34Chapter3continuedis15meastoftheoriginandtheotheris16.AverageVelocityandAverageAcceleration15mwest.Acanoeistpaddlesupstreamat2m/sanda.Whatwouldbethedifference(s)inthethenturnsaroundandfloatsdownstreamatposition-timegraphsoftheirmotion?4m/s.Theturnaroundtimeis8s.Bothlineswouldhavethesamea.Whatistheaveragevelocityofthecanoe?slope,buttheywouldrisefromtheChooseacoordinatesystemwithd-axisatdifferentpoints,15m,thepositivedirectionupstream.and15m.vvivfb.Whatwouldbethedifference(s)intheir2velocity-timegraphs?2m/s(4m/s)Theirvelocity-timegraphswouldbe2identical.1m/sb.Whatistheaverageaccelerationofthe14.VelocityExplainhowyouwoulduseacanoe?velocity-timegraphtofindthetimeatvwhichanobjecthadaspecifiedvelocity.atDraworimagineahorizontallineatthevvfispecifiedvelocity.Findthepointwheretthegraphintersectsthisline.Dropa(4m/s)(2m/s)linestraightdowntothet-axis.This8swouldbetherequiredtime.0.8m/s215.Velocity-TimeGraphSketchavelocity-time17.CriticalThinkingApoliceofficerclockedgraphforacarthatgoeseastat25m/sforadrivergoing32km/hoverthespeedlimit100s,thenwestat25m/sforanother100s.justasthedriverpassedaslowercar.Both25driverswereissuedspeedingtickets.Thejudgeagreedwiththeofficerthatbothwereguilty.ThejudgementwasissuedbasedonTime(s)theassumptionthatthecarsmusthave100200beengoingthesamespeedbecausetheyVelocity(m/s)wereobservednexttoeachother.Arethe25judgeandthepoliceofficercorrect?Explainwithasketch,amotiondiagram,andaposition-timegraph.No,theyhadthesameposition,notvelocity.Tohavethesamevelocity,theywouldhavehadtohavethesamerela-tivepositionforalengthoftime.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual31

35Chapter3continuedSketchPosition-TimeGraphMotiondiagramPassing4positiont0t1t2t3t43Fastercar2Position10Slowercart0t1t2t3t4t1t2t3t4TimePracticeProblems3.2MotionwithConstantAccelerationpages65–71page6518.Agolfballrollsupahilltowardaminiature-golfhole.Assumethatthedirectiontowardtheholeispositive.a.Ifthegolfballstartswithaspeedof2.0m/sandslowsataconstantrateof0.50m/s2,whatisitsvelocityafter2.0s?vfviat2.0m/s(0.50m/s2)(2.0s)1.0m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Whatisthegolfball’svelocityiftheconstantaccelerationcontinuesfor6.0s?vfviat2.0m/s(0.50m/s2)(6.0s)1.0m/sc.Describethemotionofthegolfballinwordsandwithamotiondiagram.Theball’svelocitysimplydecreasedinthefirstcase.Inthesecondcase,theballslowedtoastopandthenbeganrollingbackdownthehill.Timeinterval1234VelocitydPositionVelocityTimeinterval6519.Abusthatistravelingat30.0km/hspeedsupataconstantrateof3.5m/s2.Whatvelocitydoesitreach6.8slater?vfviat2)(6.8s)1km3600s30.0km/h(3.5m/s1000m1h120km/h32SolutionsManualPhysics:PrinciplesandProblems

36Chapter3continued20.Ifacaracceleratesfromrestataconstantb.2.0s5.5m/s2,howlongwillittakeforthecartoAt2.0s,v78m/sreachavelocityof28m/s?c.2.5svfviatAt2.5s,v80m/svvsotfia23.Usedimensionalanalysistoconvertan28m/s0.0m/sairplane’sspeedof75m/stokm/h.5.5m/s23600s1km2km/h(75m/s)2.7105.1s1h1000m21.Acarslowsfrom22m/sto3.0m/sata24.Aposition-timegraphforaponyrunningconstantrateof2.1m/s2.HowmanyinafieldisshowninFigure3-12.Drawsecondsarerequiredbeforethecaristhecorrespondingvelocity-timegraphusingtravelingat3.0m/s?thesametimescale.vfviatyvvsotfia3.0m/s22m/s2.1m/s2xTime(s)9.0sDisplacement(m)■Figure3-12page67yPosition-timegraph22.UseFigure3-11todeterminethevelocityofanairplanethatisspeedingupateachofSpeedsthefollowingtimes.up82Time(s)80Velocity(m/s)xDisplacement(m)Stopsand78SpeedsSlowsturnsaroundupdown7674Velocity(m/s)25.Acarisdrivenataconstantvelocityof7225m/sfor10.0min.Thecarrunsoutofgas70andthedriverwalksinthesamedirection0.01.02.03.0at1.5m/sfor20.0mintothenearestgasTime(s)station.Thedrivertakes2.0mintofilla■Figure3-11gasolinecan,thenwalksbacktothecaratGraphBrepresentsconstantspeed.So1.2m/sandeventuallydriveshomeatgraphAshouldbeusedforthefollow-25m/sinthedirectionoppositethatofingcalculations.theoriginaltrip.a.1.0sa.Drawav-tgraphusingsecondsasyourCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.timeunit.Calculatethedistancethedri-At1.0s,v74m/sverwalkedtothegasstationtofindthetimeittookhimtowalkbacktothecar.Physics:PrinciplesandProblemsSolutionsManual33

37Chapter3continuedv25fviat20vfvi0.0m/s1.75m/st28.8s15a0.20m/s1027.Aracecartravelsonaracetrackat44m/s5andslowsataconstantratetoavelocity0Time(s)of22m/sover11s.Howfardoesitmove5500duringthistime?Velocity(m/s)10001500200025003000350040001015v(vfvi)v222025dvt(vvfi)tdistancethedriverwalkedtothe2gasstation:(22m/s44m/s)(11s)dvt2(1.5m/s)(20.0min)(60s/min)1.2102m1800m28.Acaracceleratesataconstantratefrom1.8km15m/sto25m/swhileittravelsadistancetimetowalkbacktothecar:of125m.Howlongdoesittaketoachieved1800mthisspeed?t1500s25minv1.2m/sv(vfvi)b.Drawaposition-timegraphforthev22situationusingtheareasunderthedvtvelocity-timegraph.(vvfi)t20,0002Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2dt15,000(vfvi)(2)(125m)10,00025m/s15m/s25sPosition(m)500029.Abikeriderpedalswithconstantaccelera-tiontoreachavelocityof7.5m/soveratimeof4.5s.Duringtheperiodofaccelera-010002000300040005000tion,thebike’sdisplacementis19m.WhatTime(s)wastheinitialvelocityofthebike?page69v(vfvi)26.Askateboarderismovingataconstantv22velocityof1.75m/swhenshestartsupan(vvdvtfi)tinclinethatcauseshertoslowdownwitha2constantaccelerationof0.20m/s2.How2dmuchtimepassesfromwhenshebeginstosovitvfslowdownuntilshebeginstomovebackdowntheincline?(2)(19m)4.5s7.5m/s0.94m/s34SolutionsManualPhysics:PrinciplesandProblems

38Chapter3continuedpage7130.Amanrunsatavelocityof4.5m/sfor15.0min.Whengoingupanincreasinglysteephill,heslowsdownataconstantrateof0.05m/s2for90.0sandcomestoastop.Howfardidherun?1dv(v1t122fv2i)t21(4.5m/s)(15.0min)(60s/min)(0.0m/s4.5m/s)(90.0s)24.3103m31.Sekaziislearningtorideabikewithouttrainingwheels.Hisfatherpusheshimwithaconstantaccelerationof0.50m/s2for6.0s,andthenSekazicontinuesat3.0m/sforanother6.0sbeforefalling.WhatisSekazi’sdisplacement?Solvethisproblembyconstructingavelocity-timegraphforSekazi’smotionandcomput-ingtheareaunderneaththegraphedline.3.012Velocity(m/s)6.012.0Time(s)Part1:Constantacceleration:1d(3.0m/s)(6.0s)129.0mPart2:Constantvelocity:d2(3.0m/s)(12.0s6.0s)18mThusdd1d29.0m18m27m32.Youstartyourbicyclerideatthetopofahill.Youcoastdownthehillatacon-stantaccelerationof2.00m/s2.Whenyougettothebottomofthehill,youaremovingat18.0m/s,andyoupedaltomaintainthatspeed.Ifyoucontinueatthisspeedfor1.00min,howfarwillyouhavegonefromthetimeyouleftthehilltop?Part1:Constantacceleration:v2v22a(dfifdi)anddi0.00mv2v2sodfif2asincevi0.00m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2vdff2a(18.0m/s)2(2)(2.00m/s2)81.0mPhysics:PrinciplesandProblemsSolutionsManual35

39Chapter3continuedPart2:Constantvelocity:d3m2vt(18.0m/s)(60.0s)1.0810Thusdd1d281.0m1.08103m1.16103m33.Suneeistrainingforanupcoming5.0-kmrace.Shestartsouthertrainingrunbymovingataconstantpaceof4.3m/sfor19min.Thensheacceleratesataconstantrateuntilshecrossesthefinishline,19.4slater.Whatisheraccelerationduringthelastportionofthetrainingrun?Part1:Constantvelocity:dvt(4.3m/s)(19min)(60s/min)4902mPart2:Constantacceleration:12datfdivit22(dfdivit)(2)(5.0103m4902m(4.3m/s)(19.4s))at2(19.4s)20.077m/s2SectionReview3.2MotionwithConstantAccelerationCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pages65–71page7134.AccelerationAwomandrivingataspeedof23m/sseesadeerontheroadaheadandappliesthebrakeswhensheis210mfromthedeer.Ifthedeerdoesnotmoveandthecarstopsrightbeforeithitsthedeer,whatistheaccelerationprovidedbythecar’sbrakes?v2v22a(dfifdi)v2v2afi2(ddfi)0.0m/s(23m/s)2(2)(210m)1.3m/s235.DisplacementIfyouweregiveninitialandfinalvelocitiesandtheconstantaccelerationofanobject,andyouwereaskedtofindthedisplacement,whatequationwouldyouuse?v2v22adfif36SolutionsManualPhysics:PrinciplesandProblems

40Chapter3continued36.DistanceAnin-lineskaterfirstacceleratesfrom0.0m/sto5.0m/sin4.5s,thencontinuesatthisconstantspeedforanother4.5s.Whatisthetotaldistancetrav-eledbythein-lineskater?Acceleratingvvdif(tfvtf2f)0.0m/s5.0m/s(4.5s)211.25mConstantspeeddfvftf(5.0m/s)(4.5s)22.5mtotaldistance11.25m22.5m34m37.FinalVelocityAplanetravelsadistanceof5.0102mwhilebeingaccelerateduniformlyfromrestattherateof5.0m/s2.Whatfinalvelocitydoesitattain?v2v22a(dfffdi)anddi0,sov2v22adfffvf(0.0m/s)22(5.0m/s2)(5.0102m)71m/s38.FinalVelocityAnairplaneaccelerateduniformlyfromrestattherateof5.0m/s2for14s.Whatfinalvelocitydiditattain?vfviatf0(5.0m/s2)(14s)7.0101m/s39.DistanceAnairplanestartsfromrestandacceleratesataconstant3.00m/s2for30.0sbeforeleavingtheground.a.Howfardiditmove?12datfvitf2f(0.0m/s)(30.0s)212)(30.0s)2(3.00m/s21.35103mb.Howfastwastheairplanegoingwhenittookoff?vfviatfCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.0m/s(3.00m/s2)(30.0s)90.0m/sPhysics:PrinciplesandProblemsSolutionsManual37

41Chapter3continued40.GraphsAsprinterwalksuptothestartingblocksataconstantspeedandpositionsherselfforthestartoftherace.Shewaitsuntilshehearsthestartingpistolgooff,andthenacceleratesrapidlyuntilsheattainsaconstantvelocity.Shemaintainsthisvelocityuntilshecrossesthefinishline,andthensheslowsdowntoawalk,takingmoretimetoslowdownthanshedidtospeedupatthebeginningoftherace.Sketchavelocity-timeandaposition-timegraphtorepresenthermotion.Drawthemoneabovetheotheronthesametimescale.Indicateonyourp-tgraphwherethestartingblocksandfinishlineare.PositionTimeVelocityTimeStartingFinishblocksline41.CriticalThinkingDescribehowyoucouldcalculatetheaccelerationofanauto-mobile.Specifythemeasuringinstrumentsandtheproceduresthatyouwoulduse.Onepersonreadsastopwatchandcallsouttimeintervals.Anotherpersonreadsthespeedometerateachtimeandrecordsit.Plotspeedversustimeandfindtheslope.PracticeProblems3.3FreeFallpages72–75Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page7442.Aconstructionworkeraccidentallydropsabrickfromahighscaffold.a.Whatisthevelocityofthebrickafter4.0s?Sayupwardisthepositivedirection.v2fviat,ag9.80m/sv2)(4.0s)f0.0m/s(9.80m/s39m/swhentheupwarddirectionispositiveb.Howfardoesthebrickfallduringthistime?12dvatit2012)(4.0s)2(9.80m/s278mThebrickfalls78m.43.Supposeforthepreviousproblemyouchooseyourcoordinatesystemsothattheoppositedirectionispositive.38SolutionsManualPhysics:PrinciplesandProblems

42Chapter3continueda.Whatisthebrick’svelocityafter4.0s?Nowthepositivedirectionisdownward.v2fviat,ag9.80m/sv2)(4.0s)f0.0m/s(9.80m/s39m/swhenthedownwarddirectionispositiveb.Howfardoesthebrickfallduringthistime?dv1at2,ag9.80m/s2it2(0.0m/s)(4.0s)1(9.80m/s2)(4.0s)2278mThebrickstillfalls78m.44.Astudentdropsaballfromawindow3.5mabovethesidewalk.Howfastisitmovingwhenithitsthesidewalk?v2v22ad,agandvfii0sovf2gd(2)(9.80m/s2)(3.5m)8.3m/s45.Atennisballisthrownstraightupwithaninitialspeedof22.5m/s.Itiscaughtatthesamedistanceabovetheground.a.Howhighdoestheballrise?ag,andatthemaximumheight,vf0v2v22adbecomesfiv22gdiv22i(22.5m/s)d25.8m2g(2)(9.80m/s2)b.Howlongdoestheballremainintheair?Hint:Thetimeittakestheballtoriseequalsthetimeittakestofall.Calculatetimetoriseusingvfviat,withagandvf0vi22.5m/st2.30sg9.80m/s2Thetimetofallequalsthetimetorise,sothetimetoremainintheairisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tair2trise(2)(2.30s)4.60s46.YoudecidetoflipacointodeterminewhethertodoyourphysicsorEnglishhomeworkfirst.Thecoinisflippedstraightup.a.Ifthecoinreachesahighpointof0.25mabovewhereyoureleasedit,whatwasitsinitialspeed?Physics:PrinciplesandProblemsSolutionsManual39

43Chapter3continuedv2v22adfivivf22gdwhereagandvf0attheheightofthetoss,sovi(0.0m/s)2(2)(9.80m/s2)(0.25m)2.2m/sb.Ifyoucatchitatthesameheightasyoureleasedit,howmuchtimediditspendintheair?vfviatwhereagvi2.2m/sandvf2.2m/svvfitg2.2m/s2.2m/s29.80m/s0.45sSectionReview3.3FreeFallpages72–75page7547.MaximumHeightandFlightTimeAccelerationduetogravityonMarsisaboutone-thirdthatonEarth.SupposeyouthrowaballupwardwiththesamevelocityonMarsasonEarth.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Howwouldtheball’smaximumheightcomparetothatonEarth?Atmaximumheight,vf0,v2isod,orthreetimeshigher.f2gb.Howwoulditsflighttimecompare?12,orTimeisfoundfromdgtf2f2dtf.Distanceismultipliedby3andgisdividedby3,fgsotheflighttimewouldbethreetimesaslong.48.VelocityandAccelerationSupposeyouthrowaballstraightupintotheair.Describethechangesinthevelocityoftheball.Describethechangesintheaccelerationoftheball.40SolutionsManualPhysics:PrinciplesandProblems

44Chapter3continuedVelocityisreducedataconstantrateastheballtravelsupward.Atitshighestpoint,velocityiszero.Astheballbeginstodrop,thevelocitybeginstoincreaseinthenegativedirectionuntilitreachestheheightfromwhichitwasinitiallyreleased.Atthatpoint,theballhasthesamespeedithaduponrelease.Theaccelerationisconstantthroughouttheball’sflight.49.FinalVelocityYoursisterdropsyourhousekeysdowntoyoufromthesecondfloorwindow.Ifyoucatchthem4.3mfromwhereyoursisterdroppedthem,whatisthevelocityofthekeyswhenyoucatchthem?Upwardispositivev2v22adwhereagivvi22gd(0.0m/s)2(2)(9.80m/s2)(4.3m)9.2m/s50.InitialVelocityAstudenttryingoutforthefootballteamkicksthefootballstraightupintheair.Theballhitshimonthewaybackdown.Ifittook3.0sfromthetimewhenthestudentpuntedtheballuntilhegetshitbytheball,whatwasthefootball’sinitialvelocity?Chooseacoordinatesystemwithupasthepositivedirectionandtheoriginatthepunter.Choosetheinitialtimeatthepuntandthefinaltimeatthetopofthefootball’sflight.vfviatfwhereagvivfgtf0.0m/s(9.80m/s2)(1.5s)15m/s51.MaximumHeightWhenthestudentinthepreviousproblemkickedthefoot-ball,approximatelyhowhighdidthefootballtravel?v2v22a(d)whereagfiv2v2fid2g(0.0m/s)2(15m/s)2(2)(9.80m/s2)11m52.CriticalThinkingWhenaballisthrownverticallyupward,itcontinuesupwardCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.untilitreachesacertainposition,andthenitfallsdownward.Atthathighestpoint,itsvelocityisinstantaneouslyzero.Istheballacceleratingatthehighestpoint?Deviseanexperimenttoproveordisproveyouranswer.Theballisaccelerating;itsvelocityischanging.Takeastrobephototomeasureitsposition.Fromphotos,calculatetheball’svelocity.Physics:PrinciplesandProblemsSolutionsManual41

45Chapter3continuedChapterAssessment3025ConceptMapping20page801553.Completethefollowingconceptmapusing10thefollowingsymbolsorterms:d,velocity,Velocity(m/s)5m/s2,v,m,acceleration.05101520253035Time(s)Quantitiesofmotion■Figure3-16Thecarstartsfromrestandincreasespositionvelocityaccelerationitsspeed.Asthecar’sspeedincreases,thedrivershiftsgears.dva57.Whatdoestheslopeofthetangenttothecurveonavelocity-timegraphmeasure?mm/sm/s2(3.1)instantaneousaccelerationMasteringConceptspage8058.Canacartravelingonaninterstatehighway54.Howarevelocityandaccelerationrelated?haveanegativevelocityandapositiveaccel-(3.1)erationatthesametime?Explain.CantheAccelerationisthechangeinvelocitycar’svelocitychangesignswhileitistravelingdividedbythetimeintervalinwhichitwithconstantacceleration?Explain.(3.1)occurs:itistherateofchangeofYes,acar’svelocityispositiveornega-velocity.tivewithrespecttoitsdirectionofmotionfromsomepointofreference.55.Giveanexampleofeachofthefollowing.OnedirectionofmotionisdefinedasCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(3.1)positive,andvelocitiesinthatdirectiona.anobjectthatisslowingdown,buthasareconsideredpositive.Theoppositeapositiveaccelerationdirectionofmotionisconsiderednega-tive;allvelocitiesinthatdirectionareifforwardisthepositivedirection,anegative.Anobjectundergoingpositivecarmovingbackwardatdecreasingaccelerationiseitherincreasingitsspeedvelocityinthepositivedirectionorb.anobjectthatisspeedingup,buthasareducingitsvelocityinthenegativenegativeaccelerationdirection.Acar’svelocitycanchangeinthesamecoordinatesystem,acarsignswhenexperiencingconstantmovingbackwardatincreasingacceleration.Forexample,itcanbespeedtravelingright,whiletheaccelerationistotheleft.Thecarslowsdown,stops,56.Figure3-16showsthevelocity-timegraphandthenstartsacceleratingtotheleft.foranautomobileonatesttrack.Describehowthevelocitychangeswithtime.(3.1)59.Canthevelocityofanobjectchangewhenitsaccelerationisconstant?Ifso,giveanexample.Ifnot,explain.(3.1)Yes,thevelocityofanobjectcanchangewhenitsaccelerationiscon-stant.Example:droppingabook.The42SolutionsManualPhysics:PrinciplesandProblems

46Chapter3continuedlongeritdrops,thefasteritgoes,butStudentanswerswillvary.Someexam-theaccelerationisconstantatg.plesareasteelball,arock,andaper-sonfallingthroughsmalldistances.60.Ifanobject’svelocity-timegraphisastraightlineparalleltothet-axis,whatcanyoucon-ApplyingConceptscludeabouttheobject’sacceleration?(3.1)pages80–81Whenthevelocity-timegraphisaline66.Doesacarthatisslowingdownalwayshaveparalleltothet-axis,theaccelerationisanegativeacceleration?Explain.zero.No,ifthepositiveaxispointsinthedirectionoppositethevelocity,the61.Whatquantityisrepresentedbytheareaaccelerationwillbepositive.underavelocity-timegraph?(3.2)thechangeindisplacement67.CroquetAcroquetball,afterbeinghitbyamallet,slowsdownandstops.Dothe62.Writeasummaryoftheequationsforvelocityandaccelerationoftheballhaveposition,velocity,andtimeforanobjectthesamesigns?experiencingmotionwithuniformNo,theyhaveoppositesigns.acceleration.(3.2)(vv68.Ifanobjecthaszeroacceleration,doesitfi)tfameanitsvelocityiszero?Giveanexample.vNo,a0whenvelocityisconstant.fviatfvvvvfi69.Ifanobjecthaszerovelocityatsome22instant,isitsaccelerationzero?Giveandvtexample.vvfiNo,aballrollinguphillhaszerovelocityt2attheinstantitchangesdirection,butassumingti0,thenitsaccelerationisnonzero.ttf70.Ifyouweregivenatableofvelocitiesofanvfviobjectatvarioustimes,howwouldyoufinddtf2outwhethertheaccelerationwasconstant?Drawavelocity-timegraphandsee63.Explainwhyanaluminumballandasteelwhetherthecurveisastraightlineorballofsimilarsizeandshape,droppedvcalculateaccelerationsusingafromthesameheight,reachthegroundattthesametime.(3.3)andcomparetheanswerstoseeiftheyarethesame.Allobjectsacceleratetowardthegroundatthesamerate.71.ThethreenotchesinthegraphinFigure3-16occurwherethedriverchanged64.Givesomeexamplesoffallingobjectsforgears.Describethechangesinvelocityandwhichairresistancecannotbeignored.accelerationofthecarwhileinfirstgear.Is(3.3)theaccelerationjustbeforeagearchangeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Studentanswerswillvary.Somelargerorsmallerthantheaccelerationjustexamplesaresheetsofpaper,afterthechange?Explainyouranswer.parachutes,leaves,andfeathers.65.Givesomeexamplesoffallingobjectsforwhichairresistancecanbeignored.(3.3)Physics:PrinciplesandProblemsSolutionsManual43

47Chapter3continuedVelocityincreasesrapidlyatfirst,thenmoreslowly.Accelerationisgreatestatthebeginningbutisreducedasvelocityincreases.Eventually,itisnec-essaryforthedrivertoshiftintosecondgear.Theaccelerationissmallerjustbeforethegearchangebecausetheslopeislessatthatpointonthegraph.Oncethedrivershiftsandthegearsengage,accelerationandtheslopeofthecurveincrease.72.UsethegraphinFigure3-16anddeterminethetimeintervalduringwhichtheaccelerationislargestandthetimeintervalduringwhichtheaccelerationissmallest.Theaccelerationislargestduringanintervalstartingatt0andlasting1abouts.Itissmallestbeyond33s.273.Explainhowyouwouldwalktoproduceeachoftheposition-timegraphsinFigure3-17.CFDEABGHDisplacementDisplacementTimeTime■Figure3-17Walkinthepositivedirectionataconstantspeed.Walkinthepositivedirectionatanincreasingspeedforashorttime;keepwalkingatamoderatespeedfortwicethatamountoftime;slowdownoverashorttimeandstop;remainstopped;andturnaroundandrepeattheprocedureuntiltheoriginalpositionisreached.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.74.Drawavelocity-timegraphforeachofthegraphsinFigure3-18.DisplacementDisplacementDisplacementTimeTimeTime■Figure3-18TimeVelocityVelocityVelocityTimeTime44SolutionsManualPhysics:PrinciplesandProblems

48Chapter3continued75.Anobjectshotstraightuprisesfor7.0sb.IftheballonJupiterwerethrownwithbeforeitreachesitsmaximumheight.Asec-aninitialvelocitythatisthreetimesondobjectfallingfromresttakes7.0stogreater,howwouldthisaffectyourreachtheground.Comparetheanswertoparta?displacementsofthetwoobjectsduringthisWithvf0,thevaluedfisdirectlytimeinterval.proportionaltothesquareofinitialBothobjectstraveledthesamedis-(3v22i)velocity,v.tance.Theobjectthatisshotstraighti.Thatis,dfvf2gupwardrisestothesameheightfromOnEarth,aninitialvelocitythreewhichtheotherobjectfell.timesgreaterresultsinaballrisingninetimeshigher.OnJupiter,how-76.TheMoonThevalueofgontheMoonisever,theheightofninetimeshigherone-sixthofitsvalueonEarth.wouldbereducedtoonlythreea.Wouldaballthatisdroppedbyantimeshigherbecauseofdf’sinverseastronauthitthesurfaceoftheMoonrelationshiptoagthatisthreewithagreater,equal,orlesserspeedtimesgreater.thanthatofaballdroppedfromthesameheighttoEarth?78.RockAisdroppedfromacliffandrockBisthrownupwardfromthesameposition.TheballwillhittheMoonwithalesserspeedbecausetheaccelera-a.Whentheyreachthegroundatthetionduetogravityislessonthebottomofthecliff,whichrockhasaMoon.greatervelocity?b.Wouldittaketheballmore,less,orRockBhitsthegroundwithaequaltimetofall?greatervelocity.Theballwilltakemoretimetofall.b.Whichhasagreateracceleration?Theyhavethesameacceleration,77.JupiterTheplanetJupiterhasaboutthreetheaccelerationduetogravity.timesthegravitationalaccelerationofEarth.c.Whicharrivesfirst?SupposeaballisthrownverticallyupwardrockAwiththesameinitialvelocityonEarthandonJupiter.NeglecttheeffectsofJupiter’sMasteringProblemsatmosphericresistanceandassumethatgravityistheonlyforceontheball.3.1Accelerationpages81–82a.HowdoesthemaximumheightreachedbytheballonJupitercomparetotheLevel1maximumheightreachedonEarth?79.Acarisdrivenfor2.0hat40.0km/h,thenforanother2.0hat60.0km/hinthesameTherelationshipbetweendandgis(v2v2)direction.fianinverseone:d.f2ga.Whatisthecar’saveragevelocity?Ifgincreasesbythreetimes,orTotaldistance:(v2v2)80.0km120.0km200.0kmfi1d,d.f2(3g)fchangesby3totaltimeis4.0hours,so,Therefore,aballonJupiterwouldd200.0km1km/hv5.010Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1t4.0hrisetoaheightofthatonEarth.3b.Whatisthecar’saveragevelocityifitisdriven1.0102kmateachofthetwospeeds?Physics:PrinciplesandProblemsSolutionsManual45

49Chapter3continuedTotaldistanceis2.03102km;1.0102km1.0102kmtotaltime40.0km/h60.0km/h4.2hd2.0102kmsovt4.2h48km/h80.Findtheuniformaccelerationthatcausesacar’svelocitytochangefrom32m/sto96m/sinan8.0-speriod.vatvv21t96m/s32m/s28.0m/s8.0s81.Acarwithavelocityof22m/sisaccelerateduniformlyattherateof1.6m/s2for6.8s.Whatisitsfinalvelocity?vfviatf22m/s(1.6m/s2)(6.8s)33m/s82.RefertoFigure3-19tofindtheaccelerationofthemovingobjectateachofthefollowingtimes.30.0Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.20.010.0Velocity(m/s)0.05.010.015.020.025.030.0Time(s)■Figure3-19a.duringthefirst5.0softravelvat30.0m/s0.0m/s5.0s6.0m/s2b.between5.0sand10.0svat30.0m/s30.0m/s5.0s0.0m/s246SolutionsManualPhysics:PrinciplesandProblems

50Chapter3continuedc.between10.0sand15.0svat20.0m/s30.0m/s5.0s2.0m/s2d.between20.0sand25.0svat0.0m/s20.0m/s5.0s4.0m/s2Level283.Plotavelocity-timegraphusingtheinformationTable3-4inTable3-4,andanswerthefollowingquestions.Velocityv.Time16.0Time(s)Velocity(m/s)12.00.004.001.008.008.002.0012.04.00Time(s)3.0014.00.00Velocity(m/s)2.004.006.008.0010.012.04.0016.04.005.0016.08.006.0014.07.0012.0a.Duringwhattimeintervalistheobject8.008.00speedingup?Slowingdown?9.004.00speedingupfrom0.0sto4.0s;slowing10.00.00downfrom5.0sto10.0s11.04.00b.Atwhattimedoestheobjectreversedirection?12.08.00at10.0sc.Howdoestheaverageaccelerationoftheobjectintheintervalbetween0.0sand2.0sdifferfromtheaverageaccelerationintheintervalbetween7.0sand12.0s?vatbetween0.0sand2.0s:12.0m/s4.0m/s2a4.0m/s2.0s0.0sbetween7.0sand12.0s:a8.0m/s12.0m/s4.0m/s2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12.0s7.0sPhysics:PrinciplesandProblemsSolutionsManual47

51Chapter3continued84.Determinethefinalvelocityofaproton3.2MotionwithConstantAccelerationthathasaninitialvelocityof2.35105m/spage82andthenisaccelerateduniformlyinanLevel1electricfieldattherateof1.101012m/s287.RefertoFigure3-19tofindthedistancefor1.50107s.traveledduringthefollowingtimeintervals.vfviatfa.t0.0sandt5.0s2.35105m/s1AreaIbh(1.101012m/s2)(1.50107s)27.0104m/s1(5.0s)(30.0m/s)2Level375m85.SportsCarsMarcoislookingforausedb.t5.0sandt10.0ssportscar.HewantstobuytheonewiththeAreaIIbhgreatestacceleration.CarAcangofrom0(10.0s5.0s)(30.0m/s)m/sto17.9m/sin4.0s;carBcanacceleratefrom0m/sto22.4m/sin3.5s;andcarC150mcangofrom0to26.8m/sin6.0s.Rankthec.t10.0sandt15.0sthreecarsfromgreatestaccelerationtoleast,1specificallyindicatinganyties.AreaIIIAreaIVbhbh2CarA:(15.0s10.0s)(20.0m/s)av17.9m/s0m/s4.5m/s21t4.0s0.0s2(15.0s10.0s)(10.0m/s)CarB:125mav22.4m/s0m/s6.4m/s2d.t0.0sandt25.0st3.5s0.0sAreaIAreaIICarC:(AreaIIIAreaIV)AreaVIVv26.8m/s0m/s2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a4.5m/s75m150m125mt6.0s0.0s1CarBhasthegreatestaccelerationofbhbh26.4m/s2.Usingsignificantdigits,carAandcarCtiedat4.5m/s2.75m150m125m(20.0s15.0s)(20.0m/s)86.SupersonicJetAsupersonicjetflyingat1145m/sexperiencesuniformaccelerationat(25.0s20.0s)2therateof23.1m/s2for20.0s.5.0102ma.Whatisitsfinalvelocity?vfviatfLevel2145m/s(23.1m/s2)(20.0s)88.Adragsterstartingfromrestacceleratesat49m/s2.Howfastisitgoingwhenithas607m/straveled325m?b.Thespeedofsoundinairis331m/s.v2v22a(dWhatistheplane’sspeedintermsoffifdi)thespeedofsound?v2fvi2a(dfdi)607m/sN331m/s1.83timesthespeedofsound48SolutionsManualPhysics:PrinciplesandProblems

52Chapter3continued(0.0m/s)2(2)(49m/s2)(325m0.0m)180m/s89.Acarmovesat12m/sandcoastsupahillwithauniformaccelerationof1.6m/s2.a.Whatisitsdisplacementafter6.0s?12datfvitf2f(12m/s)(6.0s)1(1.6m/s2)(6.0s)2243mb.Whatisitsdisplacementafter9.0s?12datfvitf2f(12m/s)(9.0s)12)(9.0s)2(1.6m/s243mThecarisonthewaybackdownthehill.Theodometerwillshowthatthecartraveled45mupthehill2mbackdown47m.90.RaceCarAracecarcanbeslowedwithaconstantaccelerationof11m/s2.a.Ifthecarisgoing55m/s,howmanymeterswillittravelbeforeitstops?v2v22adfifv2v2fidf2a(0.0m/s)2(55m/s)2(2)(11m/s2)1.4102mb.Howmanymeterswillittaketostopacargoingtwiceasfast?v2v22adfifv2v2fidf2a(0.0m/s)2(110m/s)2550m,(2)(11m/s2)whichisabout4timeslongerthanwhengoinghalfthespeed.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.91.Acaristraveling20.0m/swhenthedriverseesachildstandingontheroad.Shetakes0.80storeact,thenstepsonthebrakesandslowsat7.0m/s2.Howfardoesthecargobeforeitstops?reactiondisplacementdr(20.0m/s)(0.80s)16mPhysics:PrinciplesandProblemsSolutionsManual49

53Chapter3continuedv2v2thereforefidf2a1t0anda2policetvspeeder0brakingdisplacement2vspeeder(0.0m/s)2(20.0m/s)2tdapoliceb(2)(7.0m/s2)29m(2)(30.0m/s)7.0m/s2totaldisplacementis8.6sdrdb16m29m45mAftert8.6s,thepolicecar’svelocityLevel3was92.AirplaneDeterminethedisplacementofavfviatplanethatexperiencesuniformaccelerationfrom66m/sto88m/sin12s.0.0m/s(7.0m/s2)(8.6s)(vfvi)t6.0101m/sdfvt2(88m/s66m/s)(12s)95.RoadBarrierThedriverofacargoing90.0km/hsuddenlyseesthelightsofabar-2rier40.0mahead.Ittakesthedriver0.75s9.2102mtoapplythebrakes,andtheaverageacceler-ationduringbrakingis10.0m/s2.93.Howfardoesaplaneflyin15swhileitsvelocityischangingfrom145m/stoa.Determinewhetherthecarhitsthe75m/satauniformrateofacceleration?barrier.Thecarwilltravel(vvdvtfi)tdvt(25.0m/s)(0.75s)218.8m(Roundoffattheend.)(75m/s145m/s)(15m/s)beforethedriverappliesthebrakes.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2Convertkm/htom/s.1.6103m(90.0km/h)(1000m/km)v94.PoliceCarAspeedingcaristravelingatai3600s/hconstantspeedof30.0m/swhenitpassesa25.0m/sstoppedpolicecar.Thepolicecaracceler-v2v22a(datesat7.0m/s2.Howfastwillitbegoingfifdi)whenitcatchesupwiththespeedingcar?v2v2dfidf2adispeedervspeedert(0.0m/s)2(25.0m/s)21218.8mdpolicevipolicetapolicet(2)(10.0m/s2)2125.0101m,yesithitsthebarriervaspeedertvipolicet2policetsincevipolice0then12vaspeedert2policet01a2v2policetspeedert10t2apolicetvspeeder50SolutionsManualPhysics:PrinciplesandProblems

54Chapter3continuedb.Whatisthemaximumspeedatwhichvfviatfwhereagthecarcouldbemovingandnothitthevbarrier40.0mahead?Assumethattheigtfaccelerationdoesn’tchange.0.0m/s(9.80m/s2)(8.0s)d78m/s(downward)totaldconstantddeceleratingb.Whatisthestone’sdisplacementduring40.0mthistime?dcvt(0.75s)vChoosethecoordinatesystemto02v2v2havetheoriginwherethestoneisdd2a2(10.0m/s2)atrestandpositivetobeupward.v212whereag20.0m/s2dfvit2atfv212vgt40m(0.75s)v2it2f20.0m/sv2(15m/s)v800m2/s200.0m1(9.80m/s2)(8.0s)22Usingthequadraticequation:2m3.110v22m/s(Thesenseoftheprob-lemexcludesthenegativevalue.)99.Abagisdroppedfromahoveringhelicopter.Thebaghasfallenfor2.0s.Whatisthebag’s3.3FreeFallvelocity?Howfarhasthebagfallen?page82Velocity:Level1v96.Astudentdropsapennyfromthetopofafviatfwhereagtoweranddecidesthatshewillestablishavigtfcoordinatesysteminwhichthedirectionofthepenny’smotionispositive.Whatisthe0.0m/s(9.80m/s2)(2.0s)signoftheaccelerationofthepenny?2.0101m/sThedirectionofthevelocityispositive,Displacement:andvelocityisincreasing.Therefore,1dat2whereagtheaccelerationisalsopositive.fvitf2f1297.Supposeanastronautdropsafeatherfromvitf2gt1.2mabovethesurfaceoftheMoon.Ifthe0.0m12)(2.0s)2accelerationduetogravityontheMoonis(9.80m/s21.62m/s2downward,howlongdoesittake2.0101mthefeathertohittheMoon’ssurface?d2(0m/s)t2fvitfatffatfLevel22df(2)(1.2m)100.Youthrowaballdownwardfromawindowtfa(1.62m/s2)1.2sataspeedof2.0m/s.Howfastwillitbemovingwhenithitsthesidewalk2.5m98.Astonethatstartsatrestisinfreefallbelow?for8.0s.ChooseacoordinatesystemwiththeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Calculatethestone’svelocityafter8.0s.positivedirectiondownwardandtheoriginatthepointwheretheballleavesyourhand.Physics:PrinciplesandProblemsSolutionsManual51

55Chapter3continuedv2v22adfifwhereagvfvi22gdf(2.0m/s)2(2)(9.80m/s2)(2.5m)7.3m/s101.Ifyouthrowtheballinthepreviousproblemupinsteadofdown,howfastwillitbemovingwhenithitsthesidewalk?Choosethesamecoordinatesystem.v2v22adfifwhereagvfvi22gdf(2.0m/s)2(2)(9.80m/s2)(2.5m)7.3m/s(dfisthedisplacement,notthetotaldistancetraveled.)Level3102.BeanbagYouthrowabeanbagintheairandcatchit2.2slater.a.Howhighdiditgo?Chooseacoordinatesystemwiththeupwarddirectionpositiveandtheoriginatthepointwherethebeanbagleftyourhand.Assumethatyoucatchthebeanbagatthesameplacewhereyouthrewit.Therefore,thetimetoreachthemaximumheightishalfofthetimeintheair.Choosetitobethetimewhenthebeanbagleftyourhandandtftobethetimeatthemaximumheight.Eachformulathatyouknowincludesvi,soyouwillhavetoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.calculatethatfirst.vfviatfwhereagvivfgtf0.0m/s(9.80m/s2)(1.1s)11m/sNowyoucanuseanequationthatincludesthedisplacement.d1at2whereagfdivitf2f12dgtivitf2f0.0m(11m/s)(1.1s)12)(1.1s)2(9.80m/s26.2mb.Whatwasitsinitialvelocity?vi11m/s52SolutionsManualPhysics:PrinciplesandProblems

56Chapter3continuedvvMixedReviewdiftfvtf2fpages82–840.0m/s4.0m/sLevel1(4.0s)2103.Aspaceshipfarfromanystarorplanet8.0mexperiencesuniformaccelerationfrom65.0m/sto162.0m/sin10.0s.Howfar106.Aweatherballoonisfloatingataconstantdoesitmove?heightaboveEarthwhenitreleasesapackChoosingacoordinatesystemwithofinstruments.theoriginatthepointwherethespeeda.Ifthepackhitsthegroundwithais65.0m/sandgivenvi65.0m/s,velocityof73.5m/s,howfardidthevf162.0m/s,andtf10.0sandpackfall?needingdf,weusetheformulawithv2v22adfiftheaveragevelocity.v2v21fidfdi(vivf)tfdf2a21(73.5m/s)2(0.00m/s)2df02(65.0m/s162.0m/s)(10.0s)2)(2)(9.80m/s1.14103m276m104.Figure3-20isastrobephotoofab.Howlongdidittakeforthepacktofall?horizontallymovingball.Whatinformationvfviatfwhereagaboutthephotowouldyouneedandwhatvvmeasurementswouldyoumaketoestimatefitfgtheacceleration?73.5m/s0.00m/s9.80m/s27.50sLevel2■Figure3-20107.BaseballAbaseballpitcherthrowsafast-Youneedtoknowthetimebetweenballataspeedof44m/s.Theaccelerationflashesandthedistancebetweentheoccursasthepitcherholdstheballinhisfirsttwoimagesandthedistancehandandmovesitthroughanalmostbetweenthelasttwo.Fromthese,youstraight-linedistanceof3.5m.Calculategettwovelocities.Betweenthesetwotheacceleration,assumingthatitisvelocities,atimeintervaloftsecondsconstantanduniform.Comparethisaccel-occurred.Dividethedifferenceerationtotheaccelerationduetogravity.betweenthetwovelocitiesbyt.v2v22adfif105.BicycleAbicycleacceleratesfrom0.0m/sv2v2fito4.0m/sin4.0s.Whatdistancedoesita2dftravel?(44m/s)202m/s2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.810(2)(3.5m)2.8102m/s229,or29timesg9.80m/s2Physics:PrinciplesandProblemsSolutionsManual53

57Chapter3continued108.Thetotaldistanceasteelballrollsdownv2v22adfifaninclineatvarioustimesisgiveninTable3-5.orv22adffv232Table3-5f(3.510m/s)a2d(2)(0.020m)Distancev.TimefTime(s)Distance(m)3.1108m/s20.00.0b.Overwhattimeintervaldoestheaccelerationtakeplace?1.02.0(v2.08.0fvi)td23.018.02df(2)(0.020m)4.032.0tvv3.5103m/s0.0m/sfi5.050.011106sa.Drawaposition-timegraphofthe11microsecondsmotionoftheball.Whensettingup110.SledsRocket-poweredsledsareusedtotheaxes,usefivedivisionsforeachtesttheresponsesofhumanstoaccelera-10moftravelonthed-axis.Usefivetion.Startingfromrest,onesledcanreachdivisionsfor1softimeonthet-axis.aspeedof444m/sin1.80sandcanbe50.0broughttoastopagainin2.15s.a.Calculatetheaccelerationofthesled40.0whenstarting,andcompareittothe30.0magnitudeoftheaccelerationduetogravity,9.80m/s2.20.0Position(m)vvv2110.0attCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Topof0.01.02.03.04.05.0444m/s0.00m/sincline1.80sTime(s)247m/s2b.Calculatethedistancetheballhasrolledattheendof2.2s.247m/s225timesg9.80m/s2After2.2secondstheballhasrolledapproximately10m.b.Findtheaccelerationofthesledasitisbrakingandcompareittothemagni-109.Engineersaredevelopingnewtypesoftudeoftheaccelerationduetogravity.gunsthatmightsomedaybeusedtovvv21launchsatellitesasiftheywerebullets.attOnesuchguncangiveasmallobjecta0.00m/s444m/svelocityof3.5km/swhilemovingit2.15sthroughadistanceofonly2.0cm.207m/s2a.Whataccelerationdoesthegungive207m/s2thisobject?21timesg9.80m/s254SolutionsManualPhysics:PrinciplesandProblems

58Chapter3continued111.Thevelocityofacarchangesoveran8.0-s1dbhbhtimeperiod,asshowninTable3-6.21Table3-62(5.0s)(20.0m/s0.0m/s)Velocityv.Time(8.0s5.0s)(20.0m/s)Time(s)Velocity(m/s)110m0.00.0e.Findtheslopeofthelinebetween1.04.0t0.0sandt4.0s.Whatdoesthis2.08.0sloperepresent?3.012.04.016.0v16.0m/s0.00m/sa5.020.0t4.0s0.0s6.020.04.0m/s2,acceleration7.020.0f.Findtheslopeofthelinebetween8.020.0t5.0sandt7.0s.Whatdoesthisa.Plotthevelocity-timegraphoftheslopeindicate?motion.v20.0m/s20.0m/sat7.0s5.0s20.02,constantvelocity0.0m/s16.012.0Level38.0112.Atruckisstoppedatastoplight.WhentheVelocity(m/s)lightturnsgreen,thetruckacceleratesat4.02.5m/s2.Atthesameinstant,acarpassesthetruckgoing15m/s.Whereandwhen0.01.02.03.04.05.06.07.08.0doesthetruckcatchupwiththecar?Time(s)Car:b.Determinethedisplacementofthecardfdivtfduringthefirst2.0s.dcardivcartfvcartfFindtheareaunderthev-tcurve.0(15m/s)t1fdbh2Truck:1(2.0s)(8.0m/s0.0m/s)d1at22fdivitf2f8.0m12dac.Whatdisplacementdoesthecarhavetruck2trucktfduringthefirst4.0s?0012)t22(2.5m/sfFindtheareaunderthev-tcurve.Whenthetruckcatchesup,the1dbh2displacementsareequal.112Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2(4.0s)(16.0m/s0.0m/s)vcartf2atrucktf32m12v0a2trucktfcartfd.Whatisthedisplacementofthecarduringtheentire8.0s?10taf2trucktfvcarFindtheareaunderthev-tcurve.Physics:PrinciplesandProblemsSolutionsManual55

59Chapter3continuedtherefore10.01taf0and2trucktfvcar05.02vcar0.0tfatruck5.0FistDisplacement(cm)(2)(15m/s)0.02.5m/s5.012s10.0df(15m/s)tfVelocity(m/s)15.05.010.015.020.025.030.0(15m/s)(12s)Time(ms)180m■Figure3-22113.SafetyBarriersHighwaysafetyengineersa.Usethevelocity-timegraphtodescribebuildsoftbarriers,suchastheoneshownthemotionofGeorge’sfistduringtheinFigure3-21,sothatcarshittingthemfirst10ms.willslowdownatasaferate.ApersonThefistmovesdownwardatwearingasafetybeltcanwithstandanabout13m/sforabout4ms.accelerationof3.0102m/s2.HowthickItthensuddenlycomestoahaltshouldbarriersbetosafelystopacarthat(accelerates).hitsabarrierat110km/h?b.Estimatetheslopeofthevelocity-timegraphtodeterminetheaccelerationofhisfistwhenitsuddenlystops.v0m/s(13m/s)at7.5ms4.0ms3.7103m/s2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.Expresstheaccelerationasamultipleofthegravitationalacceleration,■Figure3-21g9.80m/s2.(110km/h)(1000m/km)3.7103m/s22vi3600s/h31m/s23.8109.80m/svf2vi22adfTheaccelerationisabout380g.withv22add.Determinetheareaunderthevelocity-f0m/s,vif,ortimecurvetofindthedisplacementofv22thefistinthefirst6ms.Comparethisi(31m/s)df2a(2)(3.0102m/s2)withtheposition-timegraph.1.6mthickTheareacanbeapproximatedbyarectangle:114.KarateTheposition-timeandvelocity-(13m/s)(0.006s)8cmtimegraphsofGeorge’sfistbreakingaThisisinagreementwiththewoodenboardduringkaratepracticeareposition-timegraphwherethehandshowninFigure3-22.movesfrom8cmto0cm,foranetdisplacementof8cm.56SolutionsManualPhysics:PrinciplesandProblems

60Chapter3continued115.CargoAhelicopterisrisingat5.0m/schangethedistancethecartmoves,whenabagofitscargoisdropped.Thebecausetheaccelerationisalwaysbagfallsfor2.0s.thesame:g.a.Whatisthebag’svelocity?117.AnalyzeandConcludeWhichhasthevfviatfwhereaggreateracceleration:acarthatincreasesitsvspeedfrom50km/hto60km/h,orabikeigtfthatgoesfrom0km/hto10km/hinthe5.0m/s(9.80m/s2)(2.0s)sametime?Explain.15m/svvfib.Howfarhasthebagfallen?atd1at2whereagfvitf2f60km/h50km/hForcar,at12vgtitf2f10km/ht(5.0m/s)(2.0s)10km/h0km/h12)(2.0s)2Forbike,at(9.80m/s210km/h1.0101mtThebaghasfallen1.0101mThechangeinvelocityisthesame.c.Howfarbelowthehelicopteristhebag?118.AnalyzeandConcludeAnexpresstrain,travelingat36.0m/s,isaccidentallyside-Thehelicopterhasrisentrackedontoalocaltraintrack.Thed2)(2.0s)expressengineerspotsalocaltrainexactlyfvitf(5.0m/s1.00102maheadonthesametrackand1.0101mtravelinginthesamedirection.ThelocalThebagis1.0101mbelowtheengineerisunawareofthesituation.Theoriginand2.0101mbelowtheexpressengineerjamsonthebrakesandhelicopter.slowstheexpresstrainataconstantrateof3.00m/s2.IfthespeedofthelocalThinkingCriticallytrainis11.0m/s,willtheexpresstrainpage84beabletostopintime,orwilltherebe116.ApplyCBLsDesignalabtomeasuretheacollision?Tosolvethisproblem,takedistanceanacceleratedobjectmovesoverthepositionoftheexpresstrainwhentime.Useequaltimeintervalssothatyoutheengineerfirstsightsthelocaltraincanplotvelocityovertimeaswellasdis-asapointoforigin.Next,keepingintance.Apulleyattheedgeofatablewithamindthatthelocaltrainhasexactlyamassattachedisagoodwaytoachieve1.00102mlead,calculatehowfareachuniformacceleration.Suggestedmaterialstrainisfromtheoriginattheendoftheincludeamotiondetector,CBL,labcart,12.0sitwouldtaketheexpresstrainstring,pulley,C-clamp,andmasses.tostop(accelerateat3.00m/s2fromGeneratedistance-timeandvelocity-time36m/sto0m/s).Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.graphsusingdifferentmassesonthea.Onthebasisofyourcalculations,pulley.Howdoesthechangeinmassaffectwouldyouconcludethatacollisionyourgraphs?willoccur?Students’labswillvary.StudentsExpress:shouldfindthatachangeinthemassd1at2fvitf2fovertheedgeofthetablewillnot(36.0m/s)(12.0s)Physics:PrinciplesandProblemsSolutionsManual57

61Chapter3continued1(3.00m/s2)(12.0s)22432m216m216mLocal:d2fdivitfatf100m(11.0m/s)(12.0s)0232mOnthisbasis,nocollisionwilloccur.b.Thecalculationsthatyoumadedonotallowforthepossibilitythatacolli-sionmighttakeplacebeforetheendofthe12srequiredfortheexpresstraintocometoahalt.Tocheckthis,takethepositionoftheexpresstrainwhentheengineerfirstsightsthelocaltrainasthepointoforiginandcalcu-latethepositionofeachtrainattheendofeachsecondafterthesighting.Makeatableshowingthedistanceofeachtrainfromtheoriginattheendofeachsecond.Plotthesepositionsonthesamegraphanddrawtwolines.Useyourgraphtocheckyouranswertoparta.t(s)d(Local)d(Express)d(m)(m)2501111352122662003133954144120150Local51551451006166162Position(m)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Express71771795081881929199203t1021021024681012Time(s)1122121512232216Theycollidebetween6and7s.WritinginPhysicspage84119.ResearchanddescribeGalileo’scontributionstophysics.Studentanswerswillvary.AnswersshouldincludeGalileo’sexperimentsdemonstratinghowobjectsaccelerateastheyfall.AnswersmightincludehisuseofatelescopetodiscoverthemoonsofJupiterandtheringsofSaturn,andhisrelianceonexperimentalresultsratherthanauthority.58SolutionsManualPhysics:PrinciplesandProblems

62Chapter3continued120.ResearchthemaximumaccelerationaPosition-TimeGraphhumanbodycanwithstandwithoutblack-400ingout.Discusshowthisimpactsthedesignofthreecommonentertainmentor300transportationdevices.200Answerswillvary.Becausehumanscanexperiencenegativeeffects,like100blackouts,thedesignersofrollerPosition(m)coastersneedtostructurethedown-0wardslopesinsuchawaythatthecoasterdoesnotreachaccelerations100thatcauseblackouts.Likewise,engi-284610neersworkingonbullettrains,eleva-Time(s)tors,orairplanesneedtodesigntheMotionDiagramsysteminsuchawaythatallowstheobjecttorapidlyacceleratetohighspeeds,withoutcausingthepassen-gerstoblackout.ChallengeProblempage75CumulativeReviewYounoticeawaterballoonfallpastyourclass-page84roomwindow.Youestimatethatittookthe121.Solvethefollowingproblems.Expressyourballoonabouttsecondstofallthelengthoftheanswersinscientificnotation.(Chapter1)windowandthatthewindowisaboutymetersa.6.2104m5.7103mhigh.Supposetheballoonstartedfromrest,approximatelyhowhighabovethetopofthe6.3103mwindowwasitreleased?Youranswershouldbeb.8.7108km3.4107mintermsoft,y,g,andnumericalconstants.8.4108kmDownispositive.Workthisproblemintwoc.(9.21105cm)(1.83108cm)stages.Stage1isfallingthedistanceDto1.69104cm2thetopofthewindow.Stage2isfallingthed.(2.63106m)/(4.08106s)distanceyfromthetopofthewindowtothebottomofthewindow.6.451013m/sStage1:theoriginisatthetopofthefall.122.Theequationbelowdescribesthemotionv2v22a(df1i1f1di1)ofanobject.Createthecorrespondingposition-timegraphandmotiondiagram.02g(D0)Thenwriteaphysicsproblemthatcouldvf12gDbesolvedusingthatequation.Becreative.d(35.0m/s)t5.0m(Chapter2)Stage2:theoriginisatthetopofthewindow.Graphandmotiondiagramindicate12datf2di1vi1tf22f2constantvelocitymotionwithavelocityof35.0m/sandinitialposition12y0vgtf1t2of5.0m.AnswerswillvaryfortheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.create-a-problempart.1202gD(t)gt2ygt2gDt21ygt2D2gt2Physics:PrinciplesandProblemsSolutionsManual59

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64CHAPTER4ForcesinOneDimension3.AcablepullsacrateataconstantspeedPracticeProblemsacrossahorizontalsurface.Thesurfacepro-4.1ForceandMotionvidesaforcethatresiststhecrate’smotion.pages87–95Systempage89Foreachofthefollowingsituations,specifythesystemanddrawamotiondiagramandafree-bodydia-gram.Labelallforceswiththeiragents,andindicatethedirectionoftheaccelerationandofthenetforce.xDrawvectorsofappropriatelengths.vvvv1.Aflowerpotfallsfreelyfromawindowsill.(Ignoreanyforcesduetoairresistance.)a0SystemFfrictionFpulloncrateoncrateFnet0y4.Aropeliftsabucketataconstantspeed.(Ignoreairresistance.)vaFnetyFEarth’smassonflowerpotvFropeonbucket2.AskydiverfallsdownwardthroughtheairSystematconstantvelocity.(Theairexertsanvupwardforceontheperson.)a0Fnet0vyvFairresistancevondiverFEarth’smassonbucketva0Fnet05.Aropelowersabucketataconstantspeed.vSystemFEarth’smass(Ignoreairresistance.)ondiveryvFropeonbucketvSystemva0Fnet0vCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FEarth’smassonbucketPhysics:PrinciplesandProblemsSolutionsManual61

65Chapter4continuedpage9310.InertiaCanyoufeeltheinertiaofapencil?6.Twohorizontalforces,225Nand165N,Ofabook?Ifyoucan,describehow.areexertedonacanoe.IftheseforcesareYes,youcanfeeltheinertiaofeitherappliedinthesamedirection,findthenetobjectbyusingyourhandtogiveeitherhorizontalforceonthecanoe.objectanacceleration;thatis,trytoF2Nchangetheobjectsvelocity.net225N165N3.9010inthedirectionofthetwoforces11.Free-BodyDiagramDrawafree-bodydia-gramofabagofsugarbeingliftedbyyour7.Ifthesametwoforcesasintheprevioushandataconstantspeed.Specificallyidentifyproblemareexertedonthecanoeinoppo-thesystem.Labelallforceswiththeiragentssitedirections,whatisthenethorizontalandmakethearrowsthecorrectlengths.forceonthecanoe?Besuretoindicatethedirectionofthenetforce.F1Nnet225N165N6.010Fhandonbaginthedirectionofthelargerforce8.ThreeconfusedsleighdogsaretryingtopullasledacrosstheAlaskansnow.Alutiapullseastwithaforceof35N,Sewardalsoa0pullseastbutwithaforceof42N,andbigSugarKodiakpullswestwithaforceof53N.Whatisthenetforceonthesled?Identifyeastaspositiveandthesledasthesystem.SystemFEarth’smassonbagFnetFAlutiaonsledFSewardonsled12.DirectionofVelocityIfyoupushabookFintheforwarddirection,doesthismeanitsKodiakonsledvelocityhastobeforward?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.35N42N53NNo,itcouldbemovingbackwardand24Nyouwouldbereducingthatvelocity.Fnet24Neast13.Free-BodyDiagramDrawafree-bodydia-gramofawaterbucketbeingliftedbyaropeSectionReviewatadecreasingspeed.Specificallyidentify4.1ForceandMotionthesystem.Labelallforceswiththeiragentsandmakethearrowsthecorrectlengths.pages87–95page959.ForceIdentifyeachofthefollowingaseithera,b,orc:weight,mass,inertia,thepushofahand,thrust,resistance,airresis-Fropeonbuckettance,springforce,andacceleration.a.acontactforceb.afieldforceac.notaforceweight(b),mass(c),inertia(c),pushofahand(a),thrust(a),resistance(a),Systemairresistance(a),springforce(a),accel-FEarth’smassonbucketeration(c)62SolutionsManualPhysics:PrinciplesandProblems

66Chapter4continued14.CriticalThinkingAforceof1Nisthe18.InFigure4-8,theblockhasamassofonlyforceexertedonablock,andtheaccel-1.2kgandthespherehasamassof3.0kg.erationoftheblockismeasured.WhentheWhatarethereadingsonthetwoscales?sameforceistheonlyforceexertedona(Neglectthemassesofthescales.)secondblock,theaccelerationisthreetimesaslarge.Whatcanyouconcludeaboutthemassesofthetwoblocks?BecausemF/aandtheforcesarethesame,themassofthesecondblockisone-thirdthemassofthefirstblock.PracticeProblems4.2UsingNewton’sLawspages96–101■Figure4-8page97Bottomscale:Identifythesphereasthe15.Youplaceawatermelononaspringscaleatsystemandupaspositive.thesupermarket.Ifthemassofthewatermel-FnetFscaleonsphereonis4.0kg,whatisthereadingonthescale?FThescalereadstheweightofthewater-Earth’smassonspherema0melon:FscaleonsphereFEarth’smassonsphereF2)39Ngmg(4.0kg)(9.80m/smsphereg16.Kamariaislearninghowtoice-skate.She2)(3.0kg)(9.80m/swantshermothertopullheralongsothatshehasanaccelerationof0.80m/s2.If29NKamaria’smassis27.2kg,withwhatforceTopscale:Identifytheblockasthesys-doeshermotherneedtopullher?(Neglecttemandupaspositive.anyresistancebetweentheiceandFnetFtopscaleonblockKamaria’sskates.)F2)22NFbottomscaleonblocknetma(27.2kg)(0.80m/sFEarth’smassonblock17.TaruandReikosimultaneouslygrabama00.75-kgpieceofropeandbegintuggingonitinoppositedirections.IfTarupullswithFtopscaleonblockFbottomscaleonblockaforceof16.0Nandtheropeacceleratesawayfromherat1.25m/s2,withwhatFEarth’smassonblockforceisReikopulling?FbottomscaleonblockIdentifyReiko’sdirectionaspositivemblockgandtheropeasthesystem.29N(1.2kg)FnetFRiekoonropeFTaruonropema(9.80m/s2)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FRiekoonropemaFTaruonrope41N(0.75kg)(1.25m/s2)16.0N17NPhysics:PrinciplesandProblemsSolutionsManual63

67Chapter4continuedpage100c.Itspeedsupat2.00m/s2whilemoving19.OnEarth,ascaleshowsthatyouweighdownward.585N.Acceleratingdownward,a.Whatisyourmass?soa2.00m/s2Thescalereads585N.SincethereisFnoacceleration,yourweightequalsscaleFnetFgthedownwardforceofgravity:mamgFgmgm(ag)Fg585N(75.0kg)(2.00m/s2som59.7kgg9.80m/s29.80m/s2)b.WhatwouldthescalereadontheMoon(g1.60m/s2)?585Nd.Itmovesdownwardatconstantspeed.Onthemoon,gchanges:Constantspeed,soFgmgMoona0andFnet0(59.7kg)(1.60m/s2)FscaleFgmg95.5N(75.0kg)(9.80m/s2)20.UsetheresultsfromExampleProblem2to735Nanswerquestionsaboutascaleinaneleva-e.Itslowstoastopataconstantmagni-toronEarth.Whatforcewouldbeexertedtudeofacceleration.bythescaleonapersoninthefollowingConstantaccelerationa,thoughsituations?thesignofadependsonthedirec-a.Theelevatormovesatconstantspeed.tionofthemotionthatisending.Constantspeed,soa0andFFscaleFnetFgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.net0.mamgFscaleFg(75.0kg)(a)mg(75.0kg)(9.80m/s2)(75.0kg)(9.80m/s2)735N(75.0kg)(a)735Nb.Itslowsat2.00m/s2whilemovingupward.Slowingwhilemovingupward,soa2.00m/s2FscaleFnetFgmamgm(ag)(75.0kg)(2.00m/s29.80m/s2)585N64SolutionsManualPhysics:PrinciplesandProblems

68Chapter4continuedSectionReviewConstantVelocityFscale4.2UsingNewton’sLawspages96–101page10121.LunarGravityComparetheforceholdinga10.0-kgrockonEarthandontheMoon.FgTheaccelerationduetogravityontheapparentweightrealweightMoonis1.62m/s2.ToholdtherockonEarth:SlowingWhileRising/FSpeedingUpWhileDescendingnetFEarthonrockFholdonrock0FscaleFholdonrockFEarthonrockmgEarth(10.0kg)(9.80m/s2)98.0NFgToholdtherockontheMoon:apparentweightrealweightFnetFMoononrockFholdonrock0FSpeedingUpWhileRising/holdonrockFMoononrockmgMoonSlowingWhileDescending(10.0kg)(1.62m/s2)Fscale16.2N22.RealandApparentWeightYoutakearideinafastelevatortothetopofatallbuildingandridebackdownwhilestandingonabathroomscale.DuringwhichpartsoftheFgridewillyourapparentandrealweightsbeapparentweightrealweightthesame?Duringwhichpartswillyourapparentweightbelessthanyourreal23.AccelerationTecle,withamassof65.0kg,weight?Morethanyourrealweight?Sketchisstandingbytheboardsatthesideofanfree-bodydiagramstosupportyouranswers.ice-skatingrink.HepushesofftheboardsApparentweightandrealweightarethewithaforceof9.0N.Whatishisresultingsamewhenyouaretravelingeitheruporacceleration?downataconstantvelocity.ApparentIdentifyTecleasthesystemandtheweightislessthanrealweightwhenthedirectionawayfromtheboardsaselevatorisslowingwhilerisingorspeed-positive.Theicecanbetreatedasaingupwhiledescending.Apparentresistance-freesurface.weightisgreaterwhenspeedingupwhilerisingorslowingwhilegoingFnetFboardsonTeclemadown.FaboardsonTeclem9.0NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.65.0kg0.14m/s2awayfromtheboardsPhysics:PrinciplesandProblemsSolutionsManual65

69Chapter4continued24.MotionofanElevatorYouareridinginupwardforceduetoairresistanceontheanelevatorholdingaspringscalewithaparachute.Theupwardacceleration1-kgmasssuspendedfromit.Youlookatcausesthedriver’sdownwardvelocitytothescaleandseethatitreads9.3N.What,decrease.Newton’ssecondlawsaysthatifanything,canyouconcludeabouttheanetforceinacertaindirectionwillelevator’smotionatthattime?resultinanaccelerationinthatdirectionIftheelevatorisstationaryormoving(Fnetma).ataconstantvelocity,thescaleshould27.CriticalThinkingYouhaveajobatameatread9.80N.Becausethescalereadsawarehouseloadinginventoryontotrucksforlighterweight,theelevatormustbeshipmenttogrocerystores.Eachtruckhasaacceleratingdownward.Tofindtheweightlimitof10,000Nofcargo.Youpushexactacceleration:identifyupasposi-eachcrateofmeatalongalow-resistancetiveandthe1-kgmassasthesystem.rollerbelttoascaleandweighitbeforeFnetFscaleon1kgmovingitontothetruck.However,rightFafteryouweigha1000-Ncrate,thescaleEarth’smasson1kgmabreaks.DescribeawayinwhichyoucouldFascaleon1kgFEarth’smasson1kgapplyNewton’slawstofigureoutthemapproximatemassesoftheremainingcrates.9.3N9.80NAnswersmayvary.Onepossibleanswer1kgisthefollowing:Youcanneglectresis-0.5m/s2tanceifyoudoallyourmaneuveringon2downwardtherollerbelt.Becauseyouknowthe0.5m/sweightofthe1000Ncrate,youcanuseitasyourstandard.Pullonthe1000N25.MassMarcosisplayingtug-of-warwithhiscratewithaparticularforcefor1s,catusingastuffedtoy.Atoneinstantduringestimateitsvelocity,andcalculatethethegame,Marcospullsonthetoywithaaccelerationthatyourforcegavetoit.forceof22N,thecatpullsintheoppositeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Next,pullonacrateofunknownmassdirectionwithaforceof19.5N,andthetoyexperiencesanaccelerationof6.25m/s2.withasclosetothesameforceasyoucanfor1s.Estimatethecrate’svelocityWhatisthemassofthetoy?andcalculatetheaccelerationyourIdentifythetoyasthesystemandtheforcegavetoit.Theforceyoupulleddirectiontowardhiscatasthepositivewithoneachcratewillbethenetforcedirection.ineachcase.FnetFMarcosontoyFcatontoymaFnet1000-NcrateFnetunknowncrateFMarcosontoyFcatontoy(1000N)(am1000-Ncrate)(munk)(aunk)a(1000N)(a22N19.5Nm1000-Ncrate)unka6.25m/s2unk0.40kg26.AccelerationAskydiverfallsataconstantspeedinthespread-eagleposition.Afterheopenshisparachute,istheskydiveracceler-ating?Ifso,inwhichdirection?ExplainyouranswerusingNewton’slaws.Yes,forawhilethediverisacceleratingupwardbecausethereisanadditional66SolutionsManualPhysics:PrinciplesandProblems

70Chapter4continued31.AsuitcasesitsonastationaryairportluggagePracticeProblemscart,asinFigure4-13.Drawafree-body4.3InteractionForcesdiagramforeachobjectandspecificallyindi-pages102–107cateanyinteractionpairsbetweenthetwo.page10428.Youliftarelativelylightbowlingballwithyourhand,acceleratingitupward.Whataretheforcesontheball?Whatforcesdoestheballexert?Whatobjectsaretheseforcesexertedon?Theforcesontheballaretheforceofyourhandandthegravitationalforceof■Figure4-13Earth’smass.TheballexertsaforceonyourhandandagravitationalforceonSuitcaseCartEarth.AlltheseforcesareexertedonFsurfaceoncartyourhand,ontheball,oronEarth.FcartonsuitcaseFEarth’smass29.Abrickfallsfromaconstructionscaffold.oncartIdentifyanyforcesactingonthebrick.AlsoFsuitcaseoncartFEarth’smassidentifyanyforcesthatthebrickexertsandonsuitcasetheobjectsonwhichtheseforcesareexerted.(Airresistancemaybeignored.)page106Theonlyforceactingonthebrickisthe32.YouarehelpingtorepairaroofbyloadinggravitationalattractionofEarth’smass.equipmentintoabucketthatworkershoistThebrickexertsanequalandoppositetotherooftop.IftheropeisguaranteednotforceonEarth.tobreakaslongasthetensiondoesnotexceed450Nandyoufillthebucketuntilit30.Youtossaballupintheair.Drawafree-hasamassof42kg,whatisthegreatestbodydiagramfortheballwhileitisstillaccelerationthattheworkerscangivethemovingupward.Identifyanyforcesactingbucketastheypullittotheroof?ontheball.AlsoidentifyanyforcesthattheIdentifythebucketasthesystemandballexertsandtheobjectsonwhichtheseupaspositive.forcesareexerted.FnetFropeonbucketFEarth’smassonbucketFEarth’smassonballmaFTheonlyforceactingontheballistheropeonbucketFEarth’smassonbucketamforceofEarth’smassontheball,whenignoringairresistance.TheballexertsFropeonbucketmganequalandoppositeforceonEarth.m450N(42kg)(9.80m/s2)42kgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.91m/s2Physics:PrinciplesandProblemsSolutionsManual67

71Chapter4continued33.DiegoandMikaaretryingtofixatireonForthebottomropewiththepositiveDiego’scar,buttheyarehavingtroubleget-directionupward:tingthetireloose.Whentheypulltogether,FnetFbottomropeonbottomblockMikawithaforceof23NandDiegowithaforceof31N,theyjustbarelygetthetiretoFEarth’smassonbottomblockbudge.Whatisthemagnitudeofthestrengthma0oftheforcebetweenthetireandthewheel?FbottomropeonbottomblockIdentifythetireasthesystemandthedirectionofpullingaspositive.FEarth’smassonbottomblockFmgnetFwheelontireFMikaontireF(5.0kg)(9.80m/s2)Diegoontirema049NFForthetoprope,withthepositivedirec-wheelontireFMikaontireFDiegoontiretionupward:23N31NF54NnetFtopropeontopblockFbottomropeontopblockFSectionReviewEarth’smassontopblock4.3InteractionForcesma0Fpages102–107topropeontopblockpage107FEarth’smassontopblock34.ForceHoldabookmotionlessinyourhandintheair.IdentifyeachforceanditsFbottomropeontopblockinteractionpaironthebook.mgFbottomropeontopblockTheforcesonthebookaredownwardCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.forceofgravityduetothemassof(5.0kg)(9.80m/s2)49NEarthandtheupwardforceofthehand.98NTheforceofthebookonEarthandtheforceofthebookonthehandarethe37.TensionIfthebottomblockinproblemotherhalvesoftheinteractionpairs.36hasamassof3.0kgandthetensioninthetopropeis63.0N,calculatetheten-35.ForceLowerthebookfromproblem34atsioninthebottomropeandthemassofincreasingspeed.Doanyoftheforcesortheirthetopblock.interaction-pairpartnerschange?Explain.ForthebottomropewiththepositiveYes,theforceofthehandonthebookdirectionupward:becomessmallersothereisadown-Fwardacceleration.TheforceofthebooknetFbottomropeonbottomblockalsobecomessmaller;youcanfeelFEarth’smassonbottomblockthat.Theinteractionpairpartnersma0remainthesame.Fbottomropeonbottomblock36.TensionAblockhangsfromtheceilingbyFamasslessrope.AsecondblockisattachedEarth’smassonbottomblocktothefirstblockandhangsbelowiton(3.0kg)(9.80m/s2)anotherpieceofmasslessrope.Ifeachofthetwoblockshasamassof5.0kg,whatis29Nthetensionineachrope?68SolutionsManualPhysics:PrinciplesandProblems

72Chapter4continuedForthetopmasswiththepositiveChapterAssessmentdirectionupward:FConceptMappingnetFtopropeontopblockpage112Fbottomropeontopblock40.CompletethefollowingconceptmapusingFEarth’smassontopblockthefollowingtermandsymbols:normal,Fma0T,Fg.FEarth’smassontopblockmgforceFtopropeontopblockFbottomropeontopblocktensionnormalgravitymFtopropeontopblockFbottomropeontopblockgFTFNFg63.0N29N9.80m/s23.5kgMasteringConceptspage11238.NormalForcePolomahandsa13-kgbox41.Aphysicsbookismotionlessonthetopofto61-kgStephanie,whostandsonaplat-atable.Ifyougiveitahardpushwithyourform.Whatisthenormalforceexertedbyhand,itslidesacrossthetableandslowlytheplatformonStephanie?comestoastop.UseNewton’slawstoIdentifyStephanieasthesystemandanswerthefollowingquestions.(4.1)positivetobeupward.a.WhydoesthebookremainmotionlessFbeforetheforceofyourhandisapplied?netFplatformonStephanieAnobjectatresttendstostayatFboxonStephanierestifnooutsideforceactsonit.FEarth’smassonStephanieb.WhydoesthebookbegintomovewhenFyourhandpusheshardenoughonit?platformonStephanieTheforcefromyourhandisgreaterFboxonStephaniethananyopposingforcesuchasfric-Ftion.Withanetforceonit,thebookEarth’smassonStephanieslidesinthedirectionofthenetforce.mboxgmStephaniegc.Underwhatconditionswouldthebook(13kg)(9.80m/s2)(61kg)(9.80m/s2)remaininmotionataconstantspeed?7.3102NThebookwouldremaininmotionifthenetforceactingonitiszero.39.CriticalThinkingAcurtainpreventstwotug-of-warteamsfromseeingeachother.42.CyclingWhydoyouhavetopushharderOneteamtiesitsendoftheropetoatree.onthepedalsofasingle-speedbicycletoIftheotherteampullswitha500-Nforce,startitmovingthantokeepitmovingataCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.whatisthetension?Explain.constantvelocity?(4.1)Thetensionwouldbe500N.TheropeisAlargeforceisrequiredtoacceleratetheinequilibrium,sothenetforceonitismassofthebicycleandrider.Oncethezero.Theteamandthetreeexertequaldesiredconstantvelocityisreached,aforcesinoppositedirections.muchsmallerforceissufficienttoover-cometheever-presentfrictionalforces.Physics:PrinciplesandProblemsSolutionsManual69

73Chapter4continued43.Supposethattheaccelerationofanobjectis47.Arockisdroppedfromabridgeintoaval-zero.Doesthismeanthattherearenoley.Earthpullsontherockandacceleratesforcesactingonit?Giveanexamplesup-itdownward.AccordingtoNewton’sthirdportingyouranswer.(4.2)law,therockmustalsobepullingonEarth,No,itonlymeanstheforcesactingonityetEarthdoesnotseemtoaccelerate.arebalancedandthenetforceiszero.Explain.(4.3)Forexample,abookonatableisnotTherockdoespullonEarth,butEarth’smovingbuttheforceofgravitypullsenormousmasswouldundergoonlyadownonitandthenormalforceoftheminuteaccelerationasaresultofsuchtablepushesuponitandtheseforcesasmallforce.Thisaccelerationwouldarebalanced.goundetected.44.BasketballWhenabasketballplayerdrib-48.Ramonpushesonabedthathasbeenblesaball,itfallstothefloorandbouncespushedagainstawall,asinFigure4-17.up.Isaforcerequiredtomakeitbounce?Drawafree-bodydiagramforthebedandWhy?Ifaforceisneeded,whatistheagentidentifyalltheforcesactingonit.Makeainvolved?(4.2)separatelistofalltheforcesthatthebedYes,itsvelocitychangeddirection;appliestootherobjects.(4.3)thus,itwasacceleratedandaforceisrequiredtoacceleratethebasketball.Theagentisthefloor.45.Beforeaskydiveropensherparachute,shemaybefallingatavelocityhigherthantheterminalvelocitythatshewillhaveaftertheparachuteopens.(4.2)a.Describewhathappenstohervelocity■Figure4-17assheopenstheparachute.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BecausetheforceofairresistanceFflooronbedsuddenlybecomeslarger,thevelocityofthediverdropssuddenly.b.Describetheskydiver’svelocityfromwhenherparachutehasbeenopenforatimeuntilsheisabouttoland.FwallonbedFRamononbedTheforceofairresistanceandthegravitationalforceareequal.Theirsumiszero,sothereisnolongeranyFgacceleration.Theskydivercontinuesdownwardataconstantvelocity.Forcesthatbedappliestootherobjects:46.Ifyourtextbookisinequilibrium,whatcanFyousayabouttheforcesactingonit?(4.2)bedonRamon,FbedonEarth,Fbedonfloor,Ifthebookisinequilibrium,thenetFbedonwallforceiszero.Theforcesactingonthebookarebalanced.70SolutionsManualPhysics:PrinciplesandProblems

74Chapter4continued49.Figure4-18showsablockinfourdifferent52.BaseballAsluggerswingshisbatandhitssituations.Rankthemaccordingtothemag-abaseballpitchedtohim.Drawfree-bodynitudeofthenormalforcebetweenthediagramsforthebaseballandthebatatblockandthesurface,greatesttoleast.themomentofcontact.Specificallyindi-Specificallyindicateanyties.(4.3)cateanyinteractionpairsbetweenthetwodiagrams.(4.3)FballonbatFbatteronbatFbatonballApplyingConceptspages112–11353.WhiplashIfyouareinacarthatisstruckfrombehind,youcanreceiveaseriousneck■Figure4-18injurycalledwhiplash.a.UsingNewton’slaws,explainwhatfromlefttoright:second>fourth>happenstocausesuchaninjury.third>firstThecarissuddenlyaccelerated50.Explainwhythetensioninamasslessropeforward.Theseatacceleratesyourisconstantthroughoutit.(4.3)body,butyourneckhastoacceler-ateyourhead.ThiscanhurtyourIfyoudrawafree-bodydiagramforanyneckmuscles.pointontherope,therewillbetwoten-sionforcesactinginoppositedirec-b.Howdoesaheadrestreducewhiplash?tions.FnetFupFdownma0Theheadrestpushesonyourhead,(becauseitismassless).Therefore,acceleratingitinthesamedirectionFupFdown.AccordingtoNewton’sasthecar.thirdlaw,theforcethattheadjoiningpieceofropeexertsonthispointis54.SpaceShouldastronautschoosepencilsequalandoppositetotheforcethatthiswithhardorsoftleadformakingnotesinpointexertsonit,sotheforcemustbespace?Explain.constantthroughout.Asoftleadpencilwouldworkbetterbecauseitwouldrequirelessforceto51.AbirdsitsontopofastatueofEinstein.makeamarkonthepaper.Themagni-Drawfree-bodydiagramsforthebirdandtudeoftheinteractionforcepaircouldthestatue.Specificallyindicateanyinterac-pushtheastronautawayfromthepaper.tionpairsbetweenthetwodiagrams.(4.3)FEarthonstatueFstatueonbirdCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Fg,birdFbirdonstatueFg,statuePhysics:PrinciplesandProblemsSolutionsManual71

75Chapter4continued55.Whenyoulookatthelabeloftheproduct58.Youtossaballstraightupintotheair.inFigure4-19togetanideaofhowmucha.Drawafree-bodydiagramfortheballattheboxcontains,doesittellyouitsmass,threepointsduringitsmotion:ontheweight,orboth?Wouldyouneedtomakewayup,attheverytop,andonthewayanychangestothislabeltomakeitcorrectdown.SpecificallyidentifytheforcesforconsumptionontheMoon?actingontheballandtheiragents.OnthewayupFEarth’smassonballAtthetopFEarth’smassonball■Figure4-19OnthewaydownTheouncestellyoutheweightinEnglishunits.Thegramstellyouthemassinmetricunits.ThelabelwouldneedtoFEarth’smassonballread“2oz”tobecorrectontheMoon.Thegramswouldremainunchanged.b.Whatisthevelocityoftheballattheverytopofthemotion?56.Fromthetopofatallbuilding,youdrop0m/stwotable-tennisballs,onefilledwithairc.Whatistheaccelerationoftheballatandtheotherwithwater.Bothexperiencethissamepoint?airresistanceastheyfall.Whichballreach-Becausetheonlyforceactingonitesterminalvelocityfirst?Dobothhittheisthegravitationalattractionofgroundatthesametime?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Earth,a9.80m/s2.Thelighter,air-filledtabletennisballreachesterminalvelocityfirst.ItsmassMasteringProblemsislessforthesameshapeandsize,so4.1ForceandMotionthefrictionforceofupwardairresis-tancebecomesequaltothedownwardpage113forceofmgsooner.BecausetheforceLevel1ofgravityonthewater-filledtable-tennis59.Whatisthenetforceactingona1.0-kgballball(moremass)islarger,itsterminalinfree-fall?velocityislarger,anditstrikestheFnetFgmggroundfirst.(1.0kg)(9.80m/s2)57.Itcanbesaidthat1kgequals2.2lb.What9.8Ndoesthisstatementmean?Whatwouldbetheproperwayofmakingthecomparison?ItmeansthatonEarth’ssurface,theweightof1kgisequivalentto2.2lb.Youshouldcomparemassestomassesandweightstoweights.Thus9.8Nequals2.2lb.72SolutionsManualPhysics:PrinciplesandProblems

76Chapter4continued60.SkatingJoyceandEfuaareskating.Joyce65.ThreeobjectsaredroppedsimultaneouslypushesEfua,whosemassis40.0-kg,withafromthetopofatallbuilding:ashotput,forceof5.0N.WhatisEfua’sresultinganair-filledballoon,andabasketball.acceleration?a.RanktheobjectsintheorderinwhichFmatheywillreachterminalvelocity,fromFfirsttolast.amballoon,basketball,shotput5.0Nb.Ranktheobjectsaccordingtotheorder40.0kginwhichtheywillreachtheground,0.12m/s2fromfirsttolast.shotput,basketball,balloon61.Acarofmass2300kgslowsdownataratec.Whatistherelationshipbetweenyourof3.0m/s2whenapproachingastopsign.answerstopartsaandb?WhatisthemagnitudeofthenetforceTheyareinversesofeachother.causingittoslowdown?Fma66.Whatistheweightinpoundsofa100.0-N(2300kg)(3.0m/s2)woodenshippingcase?1kg2.2lb6.9103N(100.0N)22lb9.80N1kg62.BreakingtheWishboneAfter67.Youplacea7.50-kgtelevisiononaspringThanksgiving,KevinandGamalusethescale.Ifthescalereads78.4N,whatistheturkey’swishbonetomakeawish.IfKevinaccelerationduetogravityatthatlocation?pullsonitwithaforce0.17NlargerthantheFgmgforceGamalpullswithintheoppositedirec-Ftion,andthewishbonehasamassof13g,ggmwhatisthewishbone’sinitialacceleration?78.4NFa7.50kgm0.17N10.5m/s20.013kgLevel213m/s268.DragRacingA873-kg(1930-lb)dragster,startingfromrest,attainsaspeedof26.3m/s(58.9mph)in0.59s.4.2UsingNewton’sLawsa.Findtheaverageaccelerationofthepages113–114dragsterduringthistimeinterval.Level163.Whatisyourweightinnewtons?avtF2)(m)gmg(9.80m/s(26.3m/s0.0m/s)Answerswillvary.0.59s64.MotorcycleYournewmotorcycleweighs45m/s2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2450N.Whatisitsmassinkilograms?b.WhatisthemagnitudeoftheaveragenetFgmgforceonthedragsterduringthistime?Fg2450NFmamg9.80m/s2(873kg)(45m/s2)2.50102N3.9104NPhysics:PrinciplesandProblemsSolutionsManual73

77Chapter4continuedc.Assumethatthedriverhasamassofe.Itslowstoastopwhilemovingdown-68kg.Whathorizontalforcedoesthewardwithaconstantacceleration.seatexertonthedriver?DependsonthemagnitudeoftheFma(68kg)(45m/s2)acceleration.3NF3.110scaleFgFslowingmgma69.AssumethatascaleisinanelevatoronEarth.Whatforcewouldthescaleexertonm(ga)a53-kgpersonstandingonitduringthe(53kg)(9.80m/s2a)followingsituations?a.Theelevatormovesupataconstant70.Agrocerysackcanwithstandamaximumspeed.of230Nbeforeitrips.Willabagholding15kgofgroceriesthatisliftedfromtheFscaleFgcheckoutcounteratanaccelerationofmg7.0m/s2hold?2)UseNewton’ssecondlawF(53kg)(9.80m/snetma.5.2102NIfFgroceries230,thenthebagrips.b.Itslowsat2.0m/s2whilemovingFgroceriesmgroceriesagroceriesupward.mSlowswhilemovinguporspeedsgroceriesgupwhilemovingdown,mgroceries(agroceriesg)FscaleFgFslowing(15kg)(7.0m/s29.80m/s2)mgma250Nm(ga)Thebagdoesnothold.(53kg)(9.80m/s22.0m/s2)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.71.A0.50-kgguineapigisliftedupfromthe4.1102Nground.Whatisthesmallestforceneeded2whilemovingtoliftit?Describeitsresultingmotion.c.Itspeedsupat2.0m/sFdownward.liftFgSlowswhilemovinguporspeedsmgupwhilemovingdown,(0.50kg)(9.80m/s2)FscaleFgFspeeding4.9NmgmaItwouldmoveataconstantspeed.m(ga)Level3(53kg)(9.80m/s22.0m/s2)72.AstronomyOnthesurfaceofMercury,the4.1102Ngravitationalaccelerationis0.38timesitsvalueonEarth.d.Itmovesdownwardataconstantspeed.a.Whatwoulda6.0-kgmassweighonFscaleFgMercury?mgFgmg(0.38)(53kg)(9.80m/s2)(6.0kg)(9.80m/s2)(0.38)5.2102N22N74SolutionsManualPhysics:PrinciplesandProblems

78Chapter4continuedb.Ifthegravitationalaccelerationonthe4.3InteractionForcessurfaceofPlutois0.08timesthatofpage114Mercury,whatwoulda7.0-kgmassLevel1weighonPluto?75.A6.0-kgblockrestsontopofa7.0-kgFgmg(0.38)(0.08)block,whichrestsonahorizontaltable.2)(0.38)(0.08)a.Whatistheforce(magnitudeanddirec-(7.0kg)(9.80m/stion)exertedbythe7.0-kgblockonthe2.1N6.0-kgblock?F73.A65-kgdiverjumpsoffofa10.0-mtower.netNmga.Findthediver’svelocitywhenhehitsFNF7-kgblockon6-kgblockthewater.mgv2v22gdfi(6.0kg)(9.80m/s2)vi0m/s59N;thedirectionisupward.sovf2gdb.Whatistheforce(magnitudeanddirec-tion)exertedbythe6.0-kgblockonthe2(9.80m/s2)(10.0m)7.0-kgblock?14.0m/sequalandoppositetothatinparta;b.Thedivercomestoastop2.0mbelowtherefore,59Ndownwardthesurface.Findthenetforceexertedbythewater.76.RainAraindrop,withmass2.45mg,fallsv2v22adtotheground.Asitisfalling,whatmagni-fitudeofforcedoesitexertonEarth?v2viFf0,soa2draindroponEarthFgmgandFmamv2(0.00245kg)(9.80m/s2)i2d2N2.4010(65kg)(14.0m/s)22(2.0m)77.A90.0-kgmananda55-kgmanhavea3.2103Ntug-of-war.The90.0-kgmanpullsontheropesuchthatthe55-kgmanacceleratesat0.025m/s2.Whatforcedoestheropeexert74.CarRacingAracecarhasamassof710kg.Itstartsfromrestandtravels40.0min3.0s.onthe90.0-kgman?Thecarisuniformlyacceleratedduringthesameinmagnitudeastheforcetheentiretime.Whatnetforceisexertedonit?ropeexertsonthe55-kgman:12Fma(55kg)(0.025m/s2)1.4Ndv0t2atSincev00,2daandFma,sot2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2mdFt2(2)(710kg)(40.0m)(3.0s)26.3103NPhysics:PrinciplesandProblemsSolutionsManual75

79Chapter4continuedLevel2FnetFmiddleblockontopblock78.Malelionsandhumansprinterscanbothaccelerateatabout10.0m/s2.IfatypicalFEarth’smassontopblocklionweighs170kgandatypicalsprinterma0weighs75kg,whatisthedifferenceintheFforceexertedonthegroundduringaracemiddleblockontopblockbetweenthesetwospecies?FEarth’smassontopblockUseNewton’ssecondlaw,Fnetma.mtopblockgThedifferencebetweenF(4.6kg)(9.80m/s2)lionandFhumanisF45NlionFhumanmThenormalforceisbetweenthebottomlionalionmhumanahumanandmiddleblock;themiddleblockis(170kg)(10.0m/s2)thesystem;upwardispositive.(75kg)(10.0m/s2)FnetFbottomblockonmiddleblock9.5102NFtopblockonmiddleblockF79.A4500-kghelicopteracceleratesupwardatEarth’smassonmiddleblock2.0m/s2.Whatliftforceisexertedbythema0aironthepropellers?FtopblockonmiddleblockmaFnetFapplFgFapplmgFmiddleblockontopblocksoFapplmamgm(ag)Fbottomblockonmiddleblock(4500kg)((2.0m/s2)Fmiddleblockontopblock(9.8m/s2))FEarth’smassonmiddleblockCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5.3104NFmiddleblockontopblockLevel3mmiddleblockg80.Threeblocksarestackedontopofone45N(1.2kg)(9.80m/s2)another,asinFigure4-20.Thetopblockhasamassof4.6kg,themiddleonehasa57Nmassof1.2kg,andthebottomonehasaThenormalforceisbetweenthebottommassof3.7kg.Identifyandcalculateanyblockandthesurface;thebottomblocknormalforcesbetweentheobjects.isthesystem;upwardispositive.FnetFsurfaceonbottomblockFmiddleblockonbottomblockFEarth’smassonbottomblockma0■Figure4-20Thenormalforceisbetweenthetopandmiddleblocks;thetopblockisthesystem;upwardispositive.76SolutionsManualPhysics:PrinciplesandProblems

80Chapter4continuedFFsurfaceonbottomblockFthrustthrustma,soamFmiddleblockonbottomblockFgmgFEarth’smassonbottomblockFgmFgmiddleblockonbottomblockmbottomblockgvfviatandvi0,so57N(3.7kg)(9.80m/s2)vfta93NvfFthrustMixedReviewmpages114–115vmfLevel1Fthrust81.Thedragsterinproblem68completeda12d402.3-m(0.2500-mi)runin4.936s.Ifthefdivit2atcarhadaconstantacceleration,whatwasitsdivi0,soaccelerationandfinalvelocity?12d12f2atdfdivit2at1Fthrustvfm2d2mFivi0,sothrust2df1v2maft22Fthrust(2)(402.3m)22Fg(4.936s)vf1g33.02m/s22Fthrust1dfdi(vfvi)t22.75106N2(79.2m/s)9.80m/s21d6.35106Nivi0,so22d139mvfft83.Thedragsterinproblem68crossedthefinish(2)(402.3m)linegoing126.6m/s.Doestheassumption4.936sofconstantaccelerationholdtrue?What163.0m/sotherpieceofevidencecouldyouusetodetermineiftheaccelerationwasconstant?Level282.JetA2.75106-Ncatapultjetplaneis126.6m/sisslowerthanfoundinreadyfortakeoff.Ifthejet’senginessupplyproblem81,sotheaccelerationcannotaconstantthrustof6.35106N,howmuchbeconstant.Further,theaccelerationrunwaywillitneedtoreachitsminimuminthefirst0.59swas45m/s2,nottakeoffspeedof285km/h?33.02m/s2.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1hvf(285km/h)(1000m/km)3600s79.2m/sPhysics:PrinciplesandProblemsSolutionsManual77

81Chapter4continued84.Supposea65-kgboyanda45-kggirlusea86.BaseballAsabaseballisbeingcaught,itsmasslessropeinatug-of-waronanicy,speedgoesfrom30.0m/sto0.0m/sinresistance-freesurfaceasinFigure4-21.about0.0050s.ThemassofthebaseballisIftheaccelerationofthegirltowardthe0.145kg.boyis3.0m/s2,findthemagnitudeofthea.Whatisthebaseball’sacceleration?accelerationoftheboytowardthegirl.vafvitfti0.0m/s30.0m/s0.0050s0.0s6.0103m/s2?3.0m/s2b.Whatarethemagnitudeanddirectionoftheforceactingonit?Fma(0.145kg)(6.0103m/s2)■Figure4-218.7102NF1,2F1,2,som1a1m2a2(oppositedirectionofthevelocityofmanda2a2theball)1m1c.Whatarethemagnitudeanddirection(45kg)(3.0m/s2)oftheforceactingontheplayerwho(65kg)caughtit?2.1m/s2Samemagnitude,oppositedirection(indirectionofvelocityofball)85.SpaceStationPratishweighs588Nandisweightlessinaspacestation.IfshepushesLevel3offthewallwithaverticalaccelerationof87.AirHockeyAnair-hockeytableworksby3.00m/s2,determinetheforceexertedbypumpingairthroughthousandsoftinyCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.thewallduringherpushoff.holesinatabletosupportlightpucks.Thisallowsthepuckstomovearoundoncush-UseNewton’ssecondlawtoobtainionsofairwithverylittleresistance.OneofPratish’smass,mPratish.UseNewton’sthesepuckshasamassof0.25kgandisthirdlawFAFBmAaAmBaB.pushedalongbya12.0-Nforcefor9.0s.Fa.Whatisthepuck’sacceleration?gmPratishgFmaFFwallonPratishFPratishonwallammPratishaPratish12.0NF0.25kggaPratishg48m/s2(588N)(3.00m/s2)b.Whatisthepuck’sfinalvelocity?9.80m/s2vfviat1.80102Nvi0,sovfat(48m/s2)(9.0s)4.3102m/s78SolutionsManualPhysics:PrinciplesandProblems

82Chapter4continued88.Astudentstandsonabathroomscaleinan89.WeatherBalloonTheinstrumentsattachedelevatoratrestonthe64thfloorofabuild-toaweatherballooninFigure4-22haveaing.Thescalereads836N.massof5.0kg.Theballoonisreleasedanda.Astheelevatormovesup,thescaleread-exertsanupwardforceof98Nontheingincreasesto936N.Findtheacceler-instruments.ationoftheelevator.FnetFgFelevatorFelevatorFnetFgmaFgm,sog98NFnetFgaFgg5.0kgg(FnetFg)Fg■Figure4-22(9.80m/s2)(963N836N)a.Whatistheaccelerationoftheballoon836Nandinstruments?1.17m/s2FnetFapplFgb.Astheelevatorapproachesthe74thFapplmgfloor,thescalereadingdropsto782N.Whatistheaccelerationoftheelevator?98N(5.0kg)(9.80m/s2)FnetFgFelevator49N(up)FelevatorFnetFgmaFnetamFgm,so49Ng5.0kgFnetFg9.8m/s2aFggb.Aftertheballoonhasacceleratedfor10.0s,theinstrumentsarereleased.g(FnetFg)WhatisthevelocityoftheinstrumentsFgatthemomentoftheirrelease?(9.80m/s2)(782N836N)vat836N(9.8m/s2)(10.0s)0.633m/s298m/s(up)c.Usingyourresultsfrompartsaandb,c.Whatnetforceactsontheinstrumentsexplainwhichchangeinvelocity,start-aftertheirrelease?ingorstopping,takesthelongertime.justtheinstrumentweight,49NStopping,becausethemagnitudeofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(down)theaccelerationislessandvtaPhysics:PrinciplesandProblemsSolutionsManual79

83Chapter4continuedd.Whendoesthedirectionoftheinstru-91.Twoblocks,masses4.3kgand5.4kg,arements’velocityfirstbecomedownward?pushedacrossafrictionlesssurfacebyaThevelocitybecomesnegativeafterhorizontalforceof22.5N,asshowninitpassesthroughzero.Thus,useFigure4-23.vfvigt,wherevf0,orvtig(98m/s)(9.80m/s2)1.0101safterrelease90.Whenahorizontalforceof4.5Nactsonablockonaresistance-freesurface,itpro-■Figure4-23ducesanaccelerationof2.5m/s2.Supposeasecond4.0-kgblockisdroppedontothea.Whatistheaccelerationoftheblocks?first.Whatisthemagnitudeoftheaccelera-Identifythetwoblockstogetherastionofthecombinationifthesameforcethesystem,andrightaspositive.continuestoact?AssumethatthesecondFnetma,andmm1m2blockdoesnotslideonthefirstblock.FFmafirstblockainitialm1m2Fm22.5Nfirstblockainitial4.3kg5.4kgFmbothblocksafinal2.3m/s2totheright(mfirstblockmsecondblock)afinalb.Whatistheforceofthe4.3-kgblockonthe5.4-kgblock?Fso,aCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.finalmIdentifythe5.4-kgblockasthefirstblockmsecondblocksystemandrightaspositive.FFFnetF4.3-kgblockon5.4-kgblockmasecondblockinitialma4.5N2)(5.4kg)(2.3m/s4.5N4.0kg2.5m/s212Ntotheright0.78m/s2c.Whatistheforceofthe5.4-kgblockonthe4.3-kgblock?AccordingtoNewton’sthirdlaw,thisshouldbeequalandoppositetotheforcefoundinpartb,sotheforceis12Ntotheleft.80SolutionsManualPhysics:PrinciplesandProblems

84Chapter4continued92.Twoblocks,oneofmass5.0kgandtheotherofmass3.0kg,aretiedtogetherwithamasslessropeasinFigure4-24.Thisropeisstrungoveramassless,resis-tance-freepulley.Theblocksarereleasedfromrest.Findthefollowing.a.thetensionintheropeb.theaccelerationoftheblocksHint:youwillneedtosolvetwosimultaneousequations.3.0kg5.0kg■Figure4-24Equation1comesfromafree-bodydiagramforthe5.0-kgblock.Downispositive.FnetFEarth’smasson5.0-kgblockFropeon5.0-kgblockm5.0-kgblocka(1)Equation2comesfromafree-bodydiagramforthe3.0-kgblock.Upispositive.FnetFropeon3.0-kgblockFEarth’smasson3.0-kgblockm3.0-kgblocka(2)Theforcesoftheropeoneachblockwillhavethesamemagnitude,becausethetensionisconstantthroughouttherope.CallthisforceT.FEarth’smasson5.0-kgblockTm5.0-kgblocka(1)TFEarth’smasson3.0-kgblockm3.0-kgblocka(2)Solveequation2forTandplugintoequation1:m5.0-kgblockaFEarth’smasson5.0-kgblockFEarth’smasson3.0-kgblockm3.0-kgblockaFEarth’smasson5.0-kgblockFEarth’smasson3.0-kgblockam5.0-kgblockm3.0-kgblock(m5.0-kgblockm3.0-kgblock)gm3.0-kgblockm5.0-kgblock(5.0kg3.0kg)(9.80m/s2)3.0kg5.0kgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.4m/s2Physics:PrinciplesandProblemsSolutionsManual81

85Chapter4continuedSolveequation2forT:FormA,mAaFTmAgTFEarth’smasson3.0-kgblockFormB,mBaFTmBgm3.0-kgblockaFTmBgmBamB(ga)m3.0-kgblockgm3.0-kgblockaSubstitutingintotheequationform3.0-kgblock(ga)mAgives(3.0kg)(9.80m/s22.4m/s2)mAamBgmBamAg37Nor(mAmB)a(mBmA)gmThinkingCriticallyThereforeaBmAgmAmBpages115–1163.0kg2.0kg93.FormulateModelsA2.0-kgmass,mA,2.0kg3.0kganda3.0-kgmass,mB,areconnectedtoalightweightcordthatpassesoverafriction-(9.80m/s2)lesspulley.Thepulleyonlychangesthe2upward2.0m/sdirectionoftheforceexertedbytherope.Thehangingmassesarefreetomove.94.UseModelsSupposethatthemassesinChoosecoordinatesystemsforthetwoproblem93arenow1.00kgand4.00kg.masseswiththepositivedirectionbeingFindtheaccelerationofthelargermass.upformAanddownformB.ma.Createapictorialmodel.aBmAgmAmB4.00kg1.00kg(9.80m/s2)1.00kg4.00kgx5.88m/s2downwardmA95.InferTheforceexertedona0.145-kgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.baseballbyabatchangesfrom0.0NtomB1.0104Nin0.0010s,thendropsbacktozerointhesameamountoftime.Thebase-b.Createaphysicalmodelwithmotionballwasgoingtowardthebatat25m/s.andfree-bodydiagrams.a.Drawagraphofforceversustime.Whatistheaverageforceexertedontheballbythebat?FTx1.0104amAaFTmAg0.5104mBForce(N)mBg0.00.51.01.52.0Time(ms)1c.WhatistheaccelerationofthesmallerFave2Fpeakmass?14N)maF(1.010netwheremisthetotalmass2beingaccelerated.5.0103N82SolutionsManualPhysics:PrinciplesandProblems

86Chapter4continuedb.Whatistheaccelerationoftheball?FT2m2aFT1Fnet5.0103N2)6.0Na(4.0kg)(3.0m/sm0.145kg18N3.4104m/s2c.Whatisthefinalvelocityoftheball,97.CritiqueUsingtheExampleProblemsinassumingthatitreversesdirection?thischapterasmodels,writeasolutiontovthefollowingproblem.Ablockofmassfviat3.46kgissuspendedfromtwovertical25m/s(3.4104m/s2)ropesattachedtotheceiling.Whatisthe(0.0020s)tensionineachrope?43m/s96.ObserveandInferThreeblocksthatareTTconnectedbymasslessstringsarepulledalongafrictionlesssurfacebyahorizontal3.46kgforce,asshowninFigure4-25.FT1FT26.0kgFg4.0kg2.0kgm1AnalyzeandSketchtheProblemm3F36.0Nm21Drawfree-bodydiagramsfortheblockandchooseupwardtobepositive■Figure4-25SolvefortheUnknowna.Whatistheaccelerationofeachblock?Known:Sincetheyallmovetogether,themblock3.46kgaccelerationisthesameforall3blocks.Unknown:FmaFrope1onblockFrope2onblock?(m2SolvefortheUnknown1m2m3)aUseNewton’ssecondlawtofindtheFatensionintheropesm1m2m3Fnet2Frope1onblock36N2.0kg4.0kg6.0kgFEarth’smassonblock3.0m/s2ma0b.WhatarethetensionforcesineachofFFEarth’smassonblockthestrings?rope1onblock2Hint:Drawaseparatefree-bodydiagrammgFforeachblock.rope1onblock2Fnetma(3.46kg)(9.80m/s2)2FFT2m3aCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.17.0NFT2FT1m2aFT1m1a(2.0kg)(3.0m/s2)6.0NPhysics:PrinciplesandProblemsSolutionsManual83

87Chapter4continued3EvaluatetheAnswerIdentifythemassasthesystemand•Aretheunitscorrect?Nisthecorrectupwardaspositive.unitforatension,sinceitisaforce.FnetFscaleonmassFgma0•Doesthesignmakesense?Theposi-FtivesignindicatesthatthetensionisscaleonmassFgpullingupwards.Fscaleonmassmg•Isthemagnituderealistic?WewouldPluto:Fscaleonmassexpectthemagnitudetobeonthesameorderastheblock’sweight.(5.00kg)(0.30m/s2)1.5N98.ThinkCriticallyBecauseofyourphysicsknowledge,youareservingasascientificMercury:Fscaleonmassconsultantforanewscience-fictionTV2)(5.00kg)(3.7m/sseriesaboutspaceexploration.Inepisode3,theheroine,MistyMoonglow,hasbeen19Naskedtobethefirstpersontorideinanew99.ApplyConceptsDevelopaCBLlab,usinginterplanetarytransportforuseinoursolaramotiondetector,thatgraphsthedistancesystem.Shewantstobesurethatthetrans-afree-fallingobjectmovesoverequalinter-portactuallytakeshertotheplanetsheisvalsoftime.Alsographvelocityversustime.supposedtobegoingto,sosheneedstoCompareandcontrastyourgraphs.Usingtakeatestingdevicealongwithhertoyourvelocitygraph,determinetheaccelera-measuretheforceofgravitywhenshetion.Doesitequalg?arrives.Thescriptwritersdon’twanthertojustdropanobject,becauseitwillbehardStudentlabswillvarywithequipmenttodepictdifferentaccelerationsoffallingavailableanddesigns.p-tgraphsandobjectsonTV.Theythinkthey’dlikesome-v-tgraphsshouldreflectuniformaccel-thinginvolvingascale.Itisyourjobtoeration.TheaccelerationshouldbedesignaquickexperimentMistycancon-closetog.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ductinvolvingascaletodeterminewhichplanetinoursolarsystemshehasarrivedWritinginPhysicson.Describetheexperimentandincludepage116whattheresultswouldbeforPluto100.ResearchNewton’scontributionstophysics(g0.30m/s2),whichiswheresheissup-andwriteaone-pagesummary.Doyouthinkposedtogo,andMercury(g3.70m/s2),histhreelawsofmotionwerehisgreatestwhichiswheresheactuallyendsup.accomplishments?Explainwhyorwhynot.Answerswillvary.HereisonepossibleAnswerswillvary.Newton’scontribu-answer:Sheshouldtakeaknownmass,tionsshouldincludehisworkonlightsay5.00-kg,withherandplaceitontheandcolor,telescopes,astronomy,lawsscale.Sincethegravitationalforceofmotion,gravity,andperhapsdependsuponthelocalaccelerationcalculus.Oneargumentinfavorofhisduetogravity,thescalewillreadadif-threelawsofmotionbeinghisgreatestferentnumberofnewtons,dependingaccomplishmentsisthatmechanicsisonwhichplanetsheison.Thefollowingbasedonthefoundationoftheselaws.analysisshowshowtofigureoutwhatHisadvancesintheunderstandingofthescalewouldreadonagivenplanet:theconceptofgravitymaybesuggest-edashisgreatestaccomplishmentinsteadofhisthreelawsofmotion.84SolutionsManualPhysics:PrinciplesandProblems

88Chapter4continued101.Review,analyze,andcritiqueNewton’sTaro’shousetoyours,whenshouldyoufirstlaw.Canweprovethislaw?Explain.expecttheskiertopassyourhouse?Besuretoconsidertheroleofresistance.(Chapter2)Answerswillvary.Newton’sfirstlawofddvt,ortmotioninvolvesanobjectwhosenetvforcesarezero.Iftheobjectisatrest,itd5.2km5.2103mremainsatrest;ifitisinmotion,itwill1000m1hcontinuetomoveinthesamedirectionv(8.0km/h)1km3600sataconstantvelocity.Onlyaforceact-ingonanobjectatrestcancauseitto2.2m/smove.Likewise,onlyaforceactingon5.2103mtanobjectinmotioncancauseitto2.2m/schangeitsdirectionorspeed.Thetwo2.4103scases(objectatrest,objectinmotion)39mincouldbeviewedastwodifferentframesofreference.Thislawcanbedemon-Theskiershouldpassyourhouseatstrated,butitcannotbeproven.8:250:399:04A.M.102.Physicistsclassifyallforcesintofourfunda-104.Figure4-26isaposition-timegraphofmentalcategories:gravitational,electro-themotionoftwocarsonaroad.magnetic,strongnuclear,andweaknuclear.(Chapter3)Investigatethesefourforcesanddescribethesituationsinwhichtheyarefound.PositionofTwoCars12Answerswillvary.ThestrongnuclearBforcehasaveryshortrangeandiswhat6holdsprotonsandneutronstogetherinAthenucleusofanatom.TheweakDistance(m)02468nuclearforceismuchlessstrongthanthestrongnuclearforceandisinvolvedTime(s)inradioactivedecay.Theelectromagnet-■Figure4-26icforceisinvolvedinholdingatomsa.Atwhattime(s)doesonecarpasstheandmoleculestogetherandisbasedonother?theattractionofoppositecharges.3s,8sGravityisalong-rangeforcebetweentwoormoremasses.b.Whichcarismovingfasterat7.0s?carACumulativeReviewc.Atwhattime(s)dothecarshavethepage116samevelocity?103.Cross-CountrySkiingYourfriendis5strainingforacross-countryskiingrace,andd.OverwhattimeintervaliscarBspeed-youandsomeotherfriendshaveagreedtoingupallthetime?providehimwithfoodandwateralonghistrainingroute.Itisabitterlycoldday,sononenoneofyouwantstowaitoutsidelongere.OverwhattimeintervaliscarBslow-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.thanyouhaveto.Taro,whosehouseistheingdownallthetime?stopbeforeyours,callsyouat8:25A.M.to3sto10stellyouthattheskierjustpassedhishouseandisplanningtomoveatanaveragespeedof8.0km/h.Ifitis5.2kmfromPhysics:PrinciplesandProblemsSolutionsManual85

89Chapter4continued105.RefertoFigure4-26tofindthe2.Ittakestheglider1.3stopassthroughainstantaneousspeedforthefollowing:secondgate.Whatisthedistancebetween(Chapter3)thetwogates?a.carBat2.0s12dfdivit2at0m/sb.carBat9.0sLetdipositionoffirstgate0.0m0m/sd0.0m(0.25m/s)(1.3s)c.carAat2.0s1(0.80m/s2)(1.3s)221m/s1.0mChallengeProblem3.The0.40-Nforceisappliedbymeansofastringattachedtotheglider.Theotherendpage100ofthestringpassesoveraresistance-freeAnair-trackgliderpassesthroughaphotoelectricpulleyandisattachedtoahangingmass,gateataninitialspeedof0.25m/s.Asitpassesm.Howbigism?throughthegate,aconstantforceof0.40NisFgmmassgappliedtothegliderinthesamedirectionasitsmotion.Thegliderhasamassof0.50kg.Fgm1.Whatistheaccelerationoftheglider?massgFma0.40NF9.80m/s2am4.1102kg0.40N0.50kg4.Deriveanexpressionforthetension,T,in2thestringasafunctionofthemass,M,of0.80m/stheglider,themass,m,ofthehangingCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mass,andg.TmgMa86SolutionsManualPhysics:PrinciplesandProblems

90CHAPTER5ForcesinTwoDimensionsPracticeProblems5.1Vectorspages119–125page1211.Acarisdriven125.0kmduewest,then65.0kmduesouth.Whatisthemagnitudeofitsdisplacement?Solvethisproblembothgraphicallyandmathematically,andcheckyouranswersagainsteachother.2A2B2R125.0kmRA2B265.0km(65.0km)2(125.0km)2141km141km2.Twoshopperswalkfromthedoorofthemalltotheircar,whichis250.0mdownalaneofcars,andthenturn90°totherightandwalkanadditional60.0m.Whatisthemagnitudeofthedisplacementoftheshoppers’carfromthemalldoor?Solvethisproblembothgraphicallyandmathematically,andcheckyouranswersagainsteachother.2A2B2RMall250.0mdoorRA2B260.0m257mCar(250.0m)2(60.0m)2257m3.Ahikerwalks4.5kminonedirection,thenmakesa45°turntotherightandwalksanother6.4km.Whatisthemagnitudeofherdisplacement?R2A2B22ABcosRA2B22ABcos(4.5km)2(6.4km)22(4.5km)(6.4km)(cos135°)1.0101km4.Anantiscrawlingonthesidewalk.Atonemoment,itismovingsouthadistanceof5.0mm.Itthenturnssouthwestandcrawls4.0mm.Whatisthemagnitudeoftheant’sdisplacement?R2A2B22ABcosCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.RA2B22ABcos(5.0mm)2(4.0mm)22(5.0mm)(4.0mm)(cos135°)8.3mmPhysics:PrinciplesandProblemsSolutionsManual87

91Chapter5continuedpage125Solveproblems5–10algebraically.Youmayalsochoosetosolvesomeofthemgraphicallytocheckyouranswers.5.Sudhirwalks0.40kminadirection60.0°westofnorth,thengoes0.50kmduewest.Whatishisdisplacement?Identifynorthandwestasthepositivedirections.d1Wd1sin(0.40km)(sin60.0°)0.35kmd1Nd1cos(0.40km)(cos60.0°)0.20kmd2W0.50kmd2N0.00kmRWd1Wd2W0.35km0.50km0.85kmRNd1Nd2N0.20km0.00km0.20kmRRW2RN2(0.85km)2(0.20km)20.87kmR1WtanRN10.85kmtan0.20km77°R0.87kmat77°westofnorth6.AfuaandChrissyaregoingtosleepovernightintheirtreehouseandareCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.usingsomeropestopullupaboxcontainingtheirpillowsandblankets,whichhaveatotalmassof3.20kg.Thegirlsstandondifferentbranches,asshowninFigure5-6,andpullattheanglesandwiththeforcesindicated.Findthex-andy-componentsofthenetforceonthebox.Hint:Drawafree-bodydiagramsothatyoudonotleaveoutaforce.Identifyupandrightaspositive.FAonbox,xFAonboxcosA(20.4N)(cos120°)■Figure5-610.2NFAonbox,yFAonboxsinA(20.4N)(sin120°)17.7NFConbox,xFConboxcosA(17.7N)(cos55°)88SolutionsManualPhysics:PrinciplesandProblems

92Chapter5continued10.2NFcombinedFrope1onswingcosFConbox,yFConboxsinAFrope2onswingcos(17.7N)(sin55°)2Frope2onswingcos14.5N(2)(2.28N)(cos13.0°)Fg,x0.0N4.44NupwardFg,ymg9.Couldavectoreverbeshorterthanoneof(3.20kg)(9.80m/s2)itscomponents?Equalinlengthtooneof31.4Nitscomponents?Explain.FItcouldneverbeshorterthanoneofitsnetonbox,xFAonbox,xcomponents,butifitliesalongeitherFConbox,xFg,xthex-ory-axis,thenoneofitscompo-nentsequalsitslength.10.2N10.2N0.0N0.0N10.Inacoordinatesysteminwhichthex-axisisFnetonbox,yFAonbox,yeast,forwhatrangeofanglesisthex-compo-nentpositive?Forwhatrangeisitnegative?FConbox,yFg,yThex-componentispositiveforangles17.7N14.5N31.4Nlessthan90°andforanglesgreater0.8Nthan270°.It’snegativeforanglesgreaterthan90°butlessthan270°.Thenetforceis0.8Nintheupwarddirection.7.Youfirstwalk8.0kmnorthfromhome,thenSectionReviewwalkeastuntilyourdisplacementfromhome5.1Vectorsis10.0km.Howfareastdidyouwalk?pages119–125Theresultantis10.0km.Usingthepage125PythagoreanTheorem,thedistanceeast11.Distancev.DisplacementIsthedistanceisthatyouwalkequaltothemagnitudeofR2A2B2,soyourdisplacement?GiveanexamplethatBR2A2supportsyourconclusion.22Notnecessarily.Forexample,youcould(10.0km)(8.0km)walkaroundtheblock(onekmper6.0kmside).Yourdisplacementwouldbezero,butthedistancethatyouwalkwouldbe8.Achild’sswingisheldupbytworopestied4kilometers.toatreebranchthathangs13.0°fromthevertical.Ifthetensionineachropeis2.28N,12.VectorDifferenceSubtractvectorKfromwhatisthecombinedforce(magnitudeandvectorL,showninFigure5-7.direction)ofthetworopesontheswing?Theforcewillbestraightup.BecauseCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.theanglesareequal,thehorizontalforceswillbeequalandoppositeandcancelout.ThemagnitudeofthisverticalforceisPhysics:PrinciplesandProblemsSolutionsManual89

93Chapter5continued16.CriticalThinkingAboxismovedthrough5.0onedisplacementandthenthroughasec-onddisplacement.Themagnitudesofthe4.0M37.0°Ktwodisplacementsareunequal.Couldthedisplacementshavedirectionssuchthattheresultantdisplacementiszero?Supposetheboxwasmovedthroughthreedisplace-6.0Lmentsofunequalmagnitude.Couldthe■Figure5-7resultantdisplacementbezero?Support6.0(4.0)10.0totherightyourconclusionwithadiagram.No,butiftherearethree13.ComponentsFindthecomponentsofdisplacements,thesumcanvectorM,showninFigure5-7.bezeroifthethreevectorsMxmcosformatrianglewhenthey(5.0)(cos37.0°)areplacedtip-to-tail.Also,thesumofthreedisplace-4.0totherightmentscanbezerowithoutMymsinformingatriangleifthesumoftwo(5.0)(sin37.0°)displacementsinonedirectionequals3.0upwardthethirdintheoppositedirection.14.VectorSumFindthesumofthethreevectorsshowninFigure5-7.PracticeProblemsRxKxLxMx5.2Friction4.06.04.0pages126–1306.0page128RyKyLyMy17.Agirlexertsa36-NhorizontalforceassheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pullsa52-Nsledacrossacementsidewalk0.00.03.0atconstantspeed.Whatisthecoefficientof3.0kineticfrictionbetweenthesidewalkandRRx2Ry2themetalsledrunners?Ignoreairresistance.F6.023.02Nmg52N6.7Sincethespeedisconstant,thefrictionforceequalstheforceexertedbytheR1ygirl,36N.tanRxFfkFN13tanF6fsokF27°NR6.7at27°36N52N15.CommutativeOperationsTheorderin0.69whichvectorsareaddeddoesnotmatter.Mathematicianssaythatvectoradditionis18.Youneedtomovea105-kgsofatoadiffer-commutative.Whichordinaryarithmeticentlocationintheroom.Ittakesaforceofoperationsarecommutative?Whicharenot?102Ntostartitmoving.Whatisthecoeffi-cientofstaticfrictionbetweenthesofaandAdditionandmultiplicationarecommu-thecarpet?tative.Subtractionanddivisionarenot.90SolutionsManualPhysics:PrinciplesandProblems

94Chapter5continuedF5.8NfsFN0.58Ff1.0101NsFNFf,afterk,afterFNFfmg(0.06)(1.0101N)102N0.6N(105kg)(9.80m/s2)0.0991page13022.A1.4-kgblockslidesacrossaroughsurface19.Mr.Amesisdraggingaboxfullofbookssuchthatitslowsdownwithanaccelerationfromhisofficetohiscar.Theboxandof1.25m/s2.Whatisthecoefficientofbookstogetherhaveacombinedweightofkineticfrictionbetweentheblockandthe134N.Ifthecoefficientofstaticfrictionsurface?betweenthepavementandtheboxis0.55,howhardmustMr.AmespushtheboxinFnetkFNordertostartitmoving?makmgFAmesonboxFfrictionakgsFN1.25m/s2smg29.80m/s(0.55)(134N)0.12874N23.Youhelpyourmommovea41-kgbookcasetoadifferentplaceinthelivingroom.If20.Supposethatthesledinproblem17isyoupushwithaforceof65Nandtherestingonpackedsnow.Thecoefficientofbookcaseacceleratesat0.12m/s2,whatiskineticfrictionisnowonly0.12.Ifapersonthecoefficientofkineticfrictionbetweenweighing650Nsitsonthesled,whatforcethebookcaseandthecarpet?isneededtopullthesledacrossthesnowatconstantspeed?FnetFkFNFkmgmaAtconstantspeed,appliedforceequalsFmafrictionforce,sokmgFfkFN65N(41kg)(0.12m/s2)(0.12)(52N650N)(41kg)(9.80m/s2)84N0.1521.Supposethataparticularmachineina24.Ashuffleboarddiskisacceleratedtoaspeedfactoryhastwosteelpiecesthatmustrubof5.8m/sandreleased.Ifthecoefficientofagainsteachotherataconstantspeed.kineticfrictionbetweenthediskandtheBeforeeitherpieceofsteelhasbeentreatedconcretecourtis0.31,howfardoesthedisktoreducefriction,theforcenecessarytogetgobeforeitcomestoastop?Thecourtsarethemtoperformproperlyis5.8N.Afterthe15.8mlong.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pieceshavebeentreatedwithoil,whatwillIdentifythedirectionofthedisk’smotionbetherequiredforce?aspositive.FindtheaccelerationoftheFf,beforek,beforeFNdiskduetotheforceoffriction.Ff,beforeFnetkFNkmgmasoFNk,beforeakgPhysics:PrinciplesandProblemsSolutionsManual91

95Chapter5continuedThenusetheequationv2v2(0.0m/s)(23m/s)2fi2a(d(2)(0.41)(9.80m/s2)fdi)tofindthedistance.Letdi0andsolvefordf.66m,sohehitsthebranchbeforev2v2hecanstop.dfif2av2v2fiSectionReview(2)(kg)(0.0m/s)2(5.8m/s)25.2Friction(2)(0.31)(9.80m/s2)pages126–1305.5mpage13027.FrictionInthissection,youlearnedabout25.Considertheforcepushingtheboxinstaticandkineticfriction.HowaretheseExampleProblem4.Howlongwouldittwotypesoffrictionsimilar?Whatarethetakeforthevelocityoftheboxtodoubletodifferencesbetweenstaticandkineticfriction?2.0m/s?TheyaresimilarinthattheybothactinTheinitialvelocityis1.0m/s,thefinaladirectionoppositetothemotion(orvelocityis2.0m/s,andtheaccelerationintendedmotion)andtheybothresultis2.0m/s2,sofromtwosurfacesrubbingagainsteachvvother.Botharedependentonthenormalafi;letttti0andsolvefortf.forcebetweenthesetwosurfaces.Staticfifrictionapplieswhenthereisnorelativevvtfimotionbetweenthetwosurfaces.Kineticfafrictionisthetypeoffrictionwhenthere2.0m/s1.0m/sisrelativemotion.Thecoefficientof2.0m/s2staticfrictionbetweentwosurfacesis0.50sgreaterthanthecoefficientofkineticfric-tionbetweenthosesametwosurfaces.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.26.KeMinisdrivingalongonarainynightat23m/swhenheseesatreebranchlying28.FrictionAtaweddingreception,youacrosstheroadandslamsonthebrakesnoticeasmallboywholookslikehismasswhenthebranchis60.0minfrontofhim.isabout25kg,runningpartwayacrosstheIfthecoefficientofkineticfrictionbetweendancefloor,thenslidingonhiskneesuntilthecar’slockedtiresandtheroadis0.41,hestops.Ifthekineticcoefficientoffrictionwillthecarstopbeforehittingthebranch?betweentheboy’spantsandthefloorisThecarhasamassof2400kg.0.15,whatisthefrictionalforceactingonChoosepositivedirectionasdirectionhimasheslides?ofcar’smovement.FfrictionkFNFnetkFNkmgmakmgakg(0.15)(25kg)(9.80m/s2)Thenusetheequationv2v237Nfi2a(dfdi)tofindthedistance.29.VelocityDerekisplayingcardswithhisLetdi0andsolvefordf.friends,anditishisturntodeal.Acardhas22amassof2.3g,anditslides0.35malongvvdfif2athetablebeforeitstops.Ifthecoefficient22ofkineticfrictionbetweenthecardandthevvfitableis0.24,whatwastheinitialspeedof(2)(g)kthecardasitleftDerek’shand?92SolutionsManualPhysics:PrinciplesandProblems

96Chapter5continuedIdentifythedirectionofthecard’skFNFontablemamovementaspositiveFmaontableFnetkFNkmgmakmgakg25N(13kg)(0.26m/s)v(13kg)(9.80m/s2)fdi0so0.17vi2adfAllyoucanconcludeaboutthecoeffi-2(kg)dfcientofstaticfrictionisthatitis2)(0.between2(0.24)(9.80m/s35m)Fontable1.3m/ssmg30.ForceThecoefficientofstaticfriction20N(13kg)(9.80m/s2)betweena40.0-kgpicnictableandthegroundbelowitis0.43m.Whatisthe0.16greatesthorizontalforcethatcouldbeFontableexertedonthetablewhileitremainsandsmgstationary?25NF(13kg)(9.80m/s2)fsFN0.20smg(0.43)(40.0kg)(9.80m/s2)1.7102NPracticeProblems31.AccelerationRyanismovingtoanew5.3ForceandMotioninapartmentandputsadresserinthebackofTwoDimensionshispickuptruck.Whenthetruckacceleratespages131–135forward,whatforceacceleratesthedresser?Underwhatcircumstancescouldthedresserpage133slide?Inwhichdirection?33.Anantclimbsatasteadyspeedupthesideofitsanthill,whichisinclined30.0°fromFrictionbetweenthedresserandthethevertical.Sketchafree-bodydiagramfortruckacceleratesthedresserforward.theant.Thedresserwillslidebackwardiftheforceacceleratingitisgreaterthanxsmg.FyfFN32.CriticalThinkingYoupusha13-kgtableinthecafeteriawithahorizontalforceofFgy20N,butitdoesnotmove.Youthenpushitwithahorizontalforceof25N,anditFgxFgacceleratesat0.26m/s2.What,ifanything,canyouconcludeaboutthecoefficientsof60.0°Fgstaticandkineticfriction?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Fromtheslidingportionofyourexperi-mentyoucandeterminethatthecoeffi-cientofkineticfrictionbetweenthetableandthefloorisFfFontableF2Physics:PrinciplesandProblemsSolutionsManual93

97Chapter5continued34.ScottandBeccaaremovingafoldingtable37.Asuitcaseisonaninclinedplane.Atwhatoutofthesunlight.Acupoflemonade,angle,relativetothevertical,willthecom-withamassof0.44kg,isonthetable.Scottponentofthesuitcase’sweightparalleltoliftshisendofthetablebeforeBeccadoes,theplanebeequaltohalftheperpendicularandasaresult,thetablemakesanangleofcomponentofitsweight?15.0°withthehorizontal.Findthecompo-Fg,parallelFgsin,whentheangleisnentsofthecup’sweightthatareparallelwithrespecttothehorizontalandperpendiculartotheplaneofthetable.FFg,perpendicularFgcos,whentheg,parallelFgsinangleiswithrespecttothehorizontal(0.44kg)(9.80m/s2)(sin15.0°)Fg,perpendicular2Fg,parallel1.1NFg,perpendicularF2g,perpendicularFgcosFg,parallel(0.44kg)(9.80m/s2)FgcosF(cos15.0°)gsin4.2N1tan35.Kohana,whohasamassof50.0kg,isat11tanthedentist’sofficehavingherteethcleaned,2asshowninFigure5-14.Ifthecomponent26.6°relativetothehorizontal,orofherweightperpendiculartotheplaneof63.4°relativetotheverticaltheseatofthechairis449N,atwhatanglepage135isthechairtilted?38.ConsiderthecrateontheinclineinExampleProblem5.a.Calculatethemagnitudeoftheacceleration.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FamFgsin■Figure5-14mFmgsing,perpendicularFgcosmgcosmFcos1g,perpendiculargsinmg(9.80m/s2)(sin30.0°)1449Ncos2)2(50.0kg)(9.80m/s4.90m/s23.6°b.After4.00s,howfastwillthecratebemoving?36.Fernando,whohasamassof43.0kg,slidesvvafi;letvdownthebanisterathisgrandparents’house.ttiti0.fiIfthebanistermakesanangleof35.0°withSolveforvthehorizontal,whatisthenormalforcef.betweenFernandoandthebanister?vfatfFNmgcos(4.90m/s2)(4.00s)(43.0kg)(9.80m/s2)(cos35.0°)19.6m/s345N94SolutionsManualPhysics:PrinciplesandProblems

98Chapter5continued39.IftheskierinExampleProblem6wereonSectionReviewa31°downhillslope,whatwouldbethemagnitudeoftheacceleration?5.3ForceandMotioninSinceag(sincos),TwoDimensionsa(9.80m/s2)(sin31°(0.15)(cos31°))pages131–1353.8m/s2page13542.ForcesOnewaytogetacarunstuckisto40.Stacie,whohasamassof45kg,startsdowntieoneendofastrongropetothecarandaslidethatisinclinedatanangleof45°theotherendtoatree,thenpushtheropewiththehorizontal.Ifthecoefficientofatitsmidpointatrightanglestotherope.kineticfrictionbetweenStacie’sshortsandDrawafree-bodydiagramandexplainwhytheslideis0.25,whatisheracceleration?evenasmallforceontheropecanexertaFlargeforceonthecar.Stacie’sweightparallelwithslideFfmaThevectorsshowninthefreebodyFStacie’sweightparallelwithslideFfadiagramindicatethatevenasmallmforceperpendiculartotheropecanmgsinkFNincreasethetensionintheropemenoughtoovercomethefrictionforce.mgsinkmgcosSinceF2Tsin(whereistheanglembetweentherope’soriginalpositionandg(sinitsdisplacedposition),kcos)F(9.80m/s2)[sin45°(0.25)(cos45°)]T2sin5.2m/s2Forsmallervaluesof,thetension,T,willincreasegreatly.41.Aftertheskieronthe37°hillinExampleProblem6hadbeenmovingfor5.0s,thefrictionofthesnowsuddenlyincreasedandFemadethenetforceontheskierzero.WhatFcaristhenewcoefficientoffriction?Ffrictionag(sinkcos)agsingkcosIfa0,0gsingkcoskcossinsinkcossin37°kcos37°0.75Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual95

99Chapter5continued43.MassAlargescoreboardissuspendedfromtheceilingofasportsarenaby10strongcables.Sixofthecablesmakeanangleof8.0°withtheverticalwhiletheotherfourmakeanangleof10.0°.Ifthetensionineachcableis1300.0N,whatisthescoreboard’smass?Fnet,ymay0Fnet,yFcablesonboardFg6Fcablecos64Fcablecos4mg06Fcablecos64Fcablecos4mg6(1300.0N)(cos8.0°)4(1300.0N)(cos10.0°)9.80m/s21.31103kg44.AccelerationA63-kgwaterskierispulledupa14.0°inclinebyaropeparalleltotheinclinewithatensionof512N.Thecoefficientofkineticfrictionis0.27.Whatarethemagnitudeanddirectionoftheskier’sacceleration?FNmgcosFropeonskierFgFfmaFropeonskiermgsinkmgcosmaFropeonskiermgsinkmgcosam512N(63kg)(9.80m/s2)(sin14.0°)(0.27)(63kg)(9.80m/s2)(cos14.0°)63kg3.2m/s2,uptheinclineCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.45.EquilibriumYouarehangingapaintingusingtwolengthsofwire.Thewireswillbreakiftheforceistoogreat.ShouldyouhangthepaintingasshowninFigures5-15aor5-15b?Explain.FgFigure5-15b;F■Figure5-15aT2sin,soFTgetssmallerasgetslarger,andislargerin5-15b.■Figure5-15b46.CriticalThinkingCanthecoefficientoffrictioneverhaveavaluesuchthataskierwouldbeabletoslideuphillataconstantvelocity?Explainwhyorwhynot.Assumetherearenootherforcesactingontheskier.No,becauseboththefrictionalforceopposingthemotionoftheskierandthecomponentofEarth’sgravityparalleltotheslopepointdownhill,notuphill.96SolutionsManualPhysics:PrinciplesandProblems

100Chapter5continued52.ExplainthemethodthatyouwouldusetoChapterAssessmentsubtracttwovectorsgraphically.(5.1)ConceptMappingReversethedirectionofthesecondpage140vectorandthenaddthem.47.Completetheconceptmapbelowbylabel-ingthecircleswithsine,cosine,ortangentto53.Explainthedifferencebetweenthesetwoindicatewhethereachfunctionispositivesymbols:AandA.(5.1)ornegativeineachquadrant.Aisthesymbolforthevectorquantity.Aisthesignedmagnitude(length)ofthevector.Quadrant54.ThePythagoreantheoremusuallyiswrittenc2a2b2.IfthisrelationshipisusedIIIIIIIVinvectoraddition,whatdoa,b,andcrepresent?(5.1)aandbrepresentthelengthsoftwovectorsthatareattherightanglestosinecosinecosinesinecosinesinetangenttangenttangentcosinetangentoneanother.crepresentsthelengthtangentofthesumofthetwovectors.MasteringConcepts55.Whenusingacoordinatesystem,howisthepage140angleordirectionofavectordetermined48.Describehowyouwouldaddtwovectorswithrespecttotheaxesofthecoordinategraphically.(5.1)system?(5.1)Makescaledrawingsofarrowsrepresent-Theangleismeasuredcounterclock-ingthevectorquantities.Placethewisefromthex-axis.arrowsforthequantitiestobeaddedtip-to-tail.Drawanarrowfromthetailofthe56.Whatisthemeaningofacoefficientoffirsttothetipofthelast.Measurethefrictionthatisgreaterthan1.0?Howlengthofthatarrowandfinditsdirection.wouldyoumeasureit?(5.2)Thefrictionalforceisgreaterthanthe49.Whichofthefollowingactionsispermissi-normalforce.Youcanpulltheobjectblewhenyougraphicallyaddonevectortoalongthesurface,measuringtheforceanother:movingthevector,rotatingtheneededtomoveitatconstantspeed.vector,orchangingthevector’slength?(5.1)Alsomeasuretheweightoftheobject.allowed:movingthevectorwithoutchanginglengthordirection57.CarsUsingthemodeloffrictiondescribedinthistextbook,wouldthefrictionbetween50.Inyourownwords,writeacleardefinitionatireandtheroadbeincreasedbyawideoftheresultantoftwoormorevectors.Doratherthananarrowtire?Explain.(5.2)notexplainhowtofindit;explainwhatitItwouldmakenodifference.Frictionrepresents.(5.1)doesnotdependuponsurfacearea.Theresultantisthevectorsumoftwoormorevectors.Itrepresentsthequan-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.58.Describeacoordinatesystemthatwouldtitythatresultsfromaddingthevectors.besuitablefordealingwithaprobleminwhichaballisthrownupintotheair.(5.3)51.HowistheresultantdisplacementaffectedOneaxisisvertical,withthepositivewhentwodisplacementvectorsareaddeddirectioneitherupordown.inadifferentorder?(5.1)Itisnotaffected.Physics:PrinciplesandProblemsSolutionsManual97

101Chapter5continued59.IfacoordinatesystemissetupsuchthatOnecomponentisparalleltothethepositivex-axispointsinadirection30°inclinedsurfaceandtheotherisabovethehorizontal,whatshouldbetheperpendiculartoit.anglebetweenthex-axisandthey-axis?Whatshouldbethedirectionofthepositive65.Forabookonaslopingtable,describey-axis?(5.3)whathappenstothecomponentoftheThetwoaxesmustbeatrightangles.weightforceparalleltothetableandtheThepositivey-axispoints30°awayforceoffrictiononthebookasyoufromtheverticalsothatitisatrightincreasetheanglethatthetablemakesanglestothex-axis.withthehorizontal.(5.3)a.Whichcomponentsofforce(s)increase60.Explainhowyouwouldsetupacoordinatewhentheangleincreases?systemformotiononahill.(5.3)AsyouincreasetheanglethetableFormotiononahill,thevertical(y)axismakeswiththehorizontal,thecom-isusuallysetupperpendicular,ornor-ponentofthebook’sweightforcemal,tothesurfaceofthehill.alongthetableincreases.b.Whichcomponentsofforce(s)decrease?61.Ifyourtextbookisinequilibrium,whatcanWhentheangleincreases,thecom-yousayabouttheforcesactingonit?(5.3)ponentoftheweightforcenormaltoThenetforceactingonthebookiszero.thetabledecreasesandthefrictionforcedecreases.62.Cananobjectthatisinequilibriumbemoving?Explain.(5.3)ApplyingConceptsYes,Newton’sfirstlawpermitsmotionpages140–141aslongastheobject’svelocityiscon-66.Avectorthatis1cmlongrepresentsadis-stant.Itcannotaccelerate.placementof5km.Howmanykilometers63.Whatisthesumofthreevectorsthat,whenarerepresentedbya3-cmvectordrawntoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.placedtiptotail,formatriangle?Ifthesethesamescale?vectorsrepresentforcesonanobject,what5km(3cm)15kmdoesthisimplyabouttheobject?(5.3)1cmThevectorsumofforcesforminga67.MowingtheLawnIfyouarepushingaclosedtriangleiszero.Ifthesearethelawnmoweracrossthegrass,asshowninonlyforcesactingontheobject,thenetFigure5-16,canyouincreasethehorizon-forceontheobjectiszeroandthetalcomponentoftheforcethatyouexertobjectisinequilibrium.onthemowerwithoutincreasingthemag-nitudeoftheforce?Explain.64.Youareaskedtoanalyzethemotionofabookplacedonaslopingtable.(5.3)FFa.Describethebestcoordinatesystemforyanalyzingthemotion.Setupthey-axisperpendicularFtothesurfaceofthetableandthexx-axispointinguphillandparalleltothesurface.b.Howarethecomponentsoftheweightofthebookrelatedtotheangleofthetable?■Figure5-1698SolutionsManualPhysics:PrinciplesandProblems

102Chapter5continuedYes,lowerthehandletomaketheangle73.UnderwhatconditionscanthePythagoreanbetweenthehandleandthehorizontaltheorem,ratherthanthelawofcosines,besmaller.usedtofindthemagnitudeofaresultantvector?68.Avectordrawn15mmlongrepresentsaThePythagoreantheoremcanbeusedvelocityof30m/s.Howlongshouldyouonlyifthetwovectorstobeaddedaredrawavectortorepresentavelocityofatrightanglestooneanother.20m/s?15mm(20m/s)10mm74.Aprobleminvolvesacarmovingupahill,30m/ssoacoordinatesystemischosenwiththe69.Whatisthelargestpossibledisplacementpositivex-axisparalleltothesurfaceoftheresultingfromtwodisplacementswithmag-hill.Theproblemalsoinvolvesastonethatnitudes3mand4m?Whatisthesmallestisdroppedontothecar.Sketchtheproblempossibleresultant?Drawsketchestoandshowthecomponentsofthevelocitydemonstrateyouranswers.vectorofthestone.Thelargestis7m;thesmallestis1m.Onecomponentisinthenegativex-direction,theotherinthenegative3m4my-direction,assumingthatthepositivedirectionpointsupward,perpendicular7mtothehill.4m3mvx1mvy70.Howdoestheresultantdisplacementv+xchangeastheanglebetweentwovectorsincreasesfrom0°to180°?+yTheresultantincreases.71.AandBaretwosidesofarighttriangle,wheretanA/B.a.Whichsideofthetriangleislongeriftanisgreaterthan1.0?75.PullingaCartAccordingtolegend,ahorseAislonger.learnedNewton’slaws.Whenthehorsewasb.Whichsideislongeriftanislesstoldtopullacart,itrefused,sayingthatthan1.0?ifitpulledthecartforward,accordingtoBislonger.Newton’sthirdlaw,therewouldbeanequalc.Whatdoesitmeaniftanisequalforcebackwards;thus,therewouldbebal-to1.0?ancedforces,and,accordingtoNewton’sAandBareequalinlength.secondlaw,thecartwouldnotaccelerate.Howwouldyoureasonwiththishorse?72.TravelingbyCarAcarhasavelocityofTheequalandoppositeforcesreferredCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.50km/hinadirection60°northofeast.AtoinNewton’sthirdlawareactingoncoordinatesystemwiththepositivex-axisdifferentobjects.Thehorsewouldpullpointingeastandapositivey-axispointingonthecart,andthecartwouldpullnorthischosen.Whichcomponentoftheonthehorse.Thecartwouldhaveanvelocityvectorislarger,xory?unbalancednetforceonit(neglectingThenorthwardcomponent(y)islonger.friction)andwouldthusaccelerate.Physics:PrinciplesandProblemsSolutionsManual99

103Chapter5continued76.TennisWhenstretchingatennisnetMasteringProblemsbetweentwoposts,itisrelativelyeasyto5.1Vectorspulloneendofthenethardenoughtopages141–142removemostoftheslack,butyouneedawinchtotakethelastbitofslackoutofLevel1thenettomakethetopalmostcompletely79.CarsAcarmoves65kmdueeast,horizontal.Whyisthistrue?then45kmduewest.Whatisitstotaldisplacement?Whenstretchingthenetbetweenthetwoposts,thereisnoperpendicular65km(45km)2.0101kmcomponentupwardtobalancethed2.0101km,eastweightofthenet.Alltheforceexertedonthenetishorizontal.Stretchingthe80.Findthehorizontalandverticalcompo-nettoremovethelastbitofslacknentsofthefollowingvectors,asshowninrequiresgreatforceinordertoreduceFigure5-17.theflexibilityofthenetandtoincreasetheinternalforcesthatholdittogether.B(3.0)A(3.0)E(5.0)77.Theweightofabookonaninclinedplanecanberesolvedintotwovectorcompo-nents,onealongtheplane,andtheotherF(5.0)C(6.0)D(4.0)perpendiculartoit.a.Atwhatanglearethecomponents■Figure5-17equal?a.E45°ExEcosb.Atwhatangleistheparallelcomponent(5.0)(cos45°)equaltozero?3.50°EyEsinCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.Atwhatangleistheparallelcomponent(5.0)(sin45°)equaltotheweight?90°3.5b.F78.TVTowersThetransmittingtowerofaTVFxFcosstationishelduprightbyguywiresthat(5.0)(cos225°)extendfromthetopofthetowertothe3.5ground.Theforcealongtheguywirescanberesolvedintotwoperpendicularcompo-FyFsinnents.Whichoneislarger?(5.0)(sin225°)Thecomponentperpendiculartothe3.5groundislargeriftheanglebetweenc.AtheguywireandhorizontalisgreaterAthan45°.xAcos(3.0)(cos180°)3.0AyAsin(3.0)(sin180°)0.0100SolutionsManualPhysics:PrinciplesandProblems

104Chapter5continued81.Graphicallyfindthesumofthefollowing83.Youwalk30msouthand30meast.Findpairsofvectors,whoselengthsanddirec-themagnitudeanddirectionoftheresul-tionsareshowninFigure5-17.tantdisplacementbothgraphicallyanda.DandAalgebraically.R2A2B2DR(30m)2(30m)2A40mR(1.0)30mtan130mb.CandD45°CDR40m,45°eastofsouthR(10.0)c.CandA4530mSR42CA30mER(3.0)Thedifferenceintheanswersisduetod.EandFsignificantdigitsbeingconsideredinthecalculation.EF84.HikingAhiker’stripconsistsofthreeseg-R0.0ments.PathAis8.0kmlongheading60.0°northofeast.PathBis7.0kmlonginaLevel2directiondueeast.PathCis4.0kmlong82.Graphicallyaddthefollowingsetsofheading315°counterclockwisefromeast.vectors,asshowninFigure5-17.a.Graphicallyaddthehiker’sdisplace-a.A,C,andDmentsintheorderA,B,C.ACBCDARR(7.0)b.A,B,andEb.Graphicallyaddthehiker’sdisplace-mentsintheorderC,B,A.ERR(6.5)ACBABc.B,D,andFc.WhatcanyouconcludeabouttheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.resultingdisplacements?DYoucanaddvectorsinanyorder.BTheresultisalwaysthesame.FRPhysics:PrinciplesandProblemsSolutionsManual101

105Chapter5continued85.WhatisthenetforceactingontheringinBxBcosBFigure5-18?(128N)(cos30.0°)y111N500.0N400.0NByBsinB(128N)(sin30.0°)64N40.0°50.0°RxxAxBx64N111N■Figure5-1847NR2A2B2RyAyByRA2B20N64N(500.0N)2(400.0N)264N640.3NRRx2Ry2tanA(47N)2(64N)2B79N1AtanBR1ytanR1500.0xtan400.0164tan51.34°fromB47Thenetforceis640.3Nat51.34°54°86.WhatisthenetforceactingontheringinLevel3Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Figure5-19?87.AShipatSeaAshipatseaisdueintoaport500.0kmduesouthintwodays.yHowever,aseverestormcomesinandblowsit100.0kmdueeastfromitsoriginal128Nposition.Howfaristheshipfromitsdesti-nation?Inwhatdirectionmustittravelto128N30.0°reachitsdestination?64Nx2A2B2RR(100.0km)2(500.0km)2■Figure5-19509.9kmA128N64NR64Ntan1yRAxxAcosA1500.0(64N)(cos180°)tan100.064N78.69°AyAsinAR509.9km,78.69°southofwest(64N)(sin180°)0N102SolutionsManualPhysics:PrinciplesandProblems

106Chapter5continued88.SpaceExplorationAdescentvehicleland-5.2FrictioningonMarshasaverticalvelocitytowardpage142thesurfaceofMarsof5.5m/s.AtthesameLevel1time,ithasahorizontalvelocityof3.5m/s.90.Ifyouuseahorizontalforceof30.0Ntoa.Atwhatspeeddoesthevehiclemoveslidea12.0-kgwoodencrateacrossaflooralongitsdescentpath?ataconstantvelocity,whatisthecoefficientR2A2B2ofkineticfrictionbetweenthecrateandthefloor?R(5.5m/s)2(3.5m/s)2FfkFNkmgFhorizontalvR6.5m/sFhorizontalb.Atwhatanglewiththeverticalisthiskmgpath?30.0N1Ry(12.0kg)(9.80m/s2)tanRx0.25515.5tan3.591.A225-kgcrateispushedhorizontallywitha58°fromhorizontal,whichis32°forceof710N.Ifthecoefficientoffrictionis0.20,calculatetheaccelerationofthecrate.fromverticalmaFnetFapplFf89.NavigationAlfredoleavescampand,usingwhereFacompass,walks4kmE,then6kmS,fkFNkmg3kmE,5kmN,10kmW,8kmN,and,Thereforefinally,3kmS.Attheendofthreedays,heFkmgapplislost.Bydrawingadiagram,computehowamfarAlfredoisfromcampandwhichdirec-710N(0.20)(225kg)(9.80m/s2)tionheshouldtaketogetbacktocamp.225kgTakenorthandeasttobepositive1.2m/s2directions.North:6km5km8km3km4km.East:4kmLevel23km10km3km.Thehikeris92.Aforceof40.0Nacceleratesa5.0-kgblock4kmnorthand3kmwestofcamp.at6.0m/s2alongahorizontalsurface.Toreturntocamp,thehikermustgo3kmeastand4kmsouth.a.Howlargeisthefrictionalforce?R2A2B2maFnetFapplFfsoFR(3km)2(4km)2fFapplma40.0N(5.0kg)(6.0m/s2)5kmR1.0101N1ytanRb.Whatisthecoefficientoffriction?xF14kmfkFNkmgtan3kmFsofkmgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.53°1.0101NR5km,53°southofeast(5.0kg)(9.80m/s2)0.20Physics:PrinciplesandProblemsSolutionsManual103

107Chapter5continued93.MovingAppliancesYourfamilyjusthada5.3ForceandMotioninTwoDimensionsnewrefrigeratordelivered.Thedeliverymanpages142–143hasleftandyourealizethattherefrigeratorLevel1isnotquiteintherightposition,soyou95.Anobjectinequilibriumhasthreeforcesplantomoveitseveralcentimeters.Iftheexertedonit.A33.0-Nforceactsat90.0°refrigeratorhasamassof180kg,thecoeffi-fromthex-axisanda44.0-Nforceactsatcientofkineticfrictionbetweenthebottom60.0°fromthex-axis.Whatarethemagni-oftherefrigeratorandtheflooris0.13,andtudeanddirectionofthethirdforce?thestaticcoefficientoffrictionbetweenFirst,findthemagnitudeofthesumofthesesamesurfacesis0.21,howharddothesetwoforces.Theequilibrantwillyouhavetopushhorizontallytogetthehavethesamemagnitudebutoppositerefrigeratortostartmoving?direction.FonfridgeFfrictionF133.0N,90.0°sFNF244.0N,60.0°smgF3?(0.21)(180kg)(9.80m/s2)F1xF1cos1370N(33.0N)(cos90.0°)Level30.0N94.StoppingataRedLightYouaredrivingF1yF1sin1a2500.0-kgcarataconstantspeedof14.0m/salongawet,butstraight,level(33.0N)(sin90.0°)road.Asyouapproachanintersection,33.0Nthetrafficlightturnsred.YouslamontheF2xF2cos2brakes.Thecar’swheelslock,thetiresbeginskidding,andthecarslidestoahaltina(44.0N)(cos60.0°)distanceof25.0m.Whatisthecoefficient22.0NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofkineticfrictionbetweenyourtiresandF2yF2sin2thewetroad?F(44.0N)(sin60.0°)fkFNma38.1Nm(v2v2)fikmg2dwherevf0F3xF1xF2x(Theminussignindicatestheforceis0.0N22.0Nactingoppositetothedirectionof22.0Nmotion.)F3yF1yF2yv2i33.0N38.1Nk2dg71.1N(14.0m/s)22)F3F3x2F3y22(25.0m)(9.80m/s0.400(22.0N)2(71.1N)274.4N104SolutionsManualPhysics:PrinciplesandProblems

108Chapter5continuedForequilibrium,thesumofthecompo-F5xF5cos5(50.0N)(cos60.0°)nentsmustequalzero,so25.0NFtan13y180.0°F5yF5sin5(50.0N)(sin60.0°)F3x43.3N171.1Ntan180.0°F22.0N6xF1xF2xF3xF4xF5x253°0.0N40.0N0.0NF(40.0N)25.0N374.4N,253°25.0NLevel2F96.Fiveforcesactonanobject:(1)60.0Nat6yF1yF2yF3yF4yF5y90.0°,(2)40.0Nat0.0°,(3)80.0Nat60.0N0.0N(80.0N)270.0°,(4)40.0Nat180.0°,and0.0N43.3N(5)50.0Nat60.0°.Whatarethemagni-23.3Ntudeanddirectionofasixthforcethatwouldproduceequilibrium?F6F6x2F6y2Solutionsbycomponents22(25.0N)(23.3N)F160.0N,90.0°34.2NFF240.0N,0.0°16y180.0°6tanFF6x380.0N,270.0°123.3NF440.0N,180.0°tan180.0°25.0NF550.0N,60.0°223°F6?F634.2N,223°F1xF1cos197.AdvertisingJoewishestohangasignweighing7.50102NsothatcableA,(60.0N)(cos90.0°)0.0Nattachedtothestore,makesa30.0°angle,F1yF1sin1(60.0N)(sin90.0°)asshowninFigure5-20.CableBishori-60.0Nzontalandattachedtoanadjoiningbuild-ing.WhatisthetensionincableB?F2xF2cos2(40.0N)(cos0.0°)40.0NF30.0°A2yF2sin2(40.0N)(sin0.0°)0.0NBF3xF3cos3(80.0N)(cos270.0°)0.0NF3yF3sin3(80.0N)(sin270.0°)80.0N■Figure5-20Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.F4xF4cos4(40.0N)(cos180.0°)40.0NF4yF4sin4(40.0N)(sin180.0°)0.0NPhysics:PrinciplesandProblemsSolutionsManual105

109Chapter5continuedSolutionbycomponents.ThesumofFparallelFgsinthecomponentsmustequalzero,so(215N)(sin35.0°)FAyFg0123NsoFAyFgLevel37.50102N100.EmergencyRoomYouareshadowingaFAyFAsin60.0°nurseintheemergencyroomofalocalhospital.AnorderlywheelsinapatientFAysoFwhohasbeeninaveryseriousaccidentAsin60.0°andhashadseverebleeding.Thenurse7.50102Nquicklyexplainstoyouthatinacaselikesin60.0°this,thepatient’sbedwillbetiltedwith866NtheheaddownwardtomakesurethebrainAlso,Fgetsenoughblood.Shetellsyouthat,forBFA0,somostpatients,thelargestanglethatthebedFBFAcanbetiltedwithoutthepatientbeginningFtoslideoffis32.0°fromthehorizontal.Acos60.0°(866N)(cos60.0°)a.Onwhatfactororfactorsdoesthisangleoftiltingdepend?433N,rightThecoefficientofstaticfriction98.Astreetlampweighs150N.Itissupportedbetweenthepatientandthebed’sbytwowiresthatformanangleof120.0°sheets.witheachother.Thetensionsinthewiresb.Findthecoefficientofstaticfrictionareequal.betweenatypicalpatientandthebed’sa.Whatisthetensionineachwiresup-sheets.portingthestreetlamp?FgparalleltobedmgsinCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Fg2TsinFfFgsFNsoT2sinsmgcos150N(2)(sin30.0°)mgsinso1.5102Nsmgcossinb.Iftheanglebetweenthewiressupport-cosingthestreetlampisreducedto90.0°,tanwhatisthetensionineachwire?Ftan32.0°gT2sin0.625150N(2)(sin45°)1.1102N99.A215-Nboxisplacedonaninclinedplanethatmakesa35.0°anglewiththehorizon-tal.Findthecomponentoftheweightforceparalleltotheplane’ssurface.106SolutionsManualPhysics:PrinciplesandProblems

110Chapter5continued101.Twoblocksareconnectedbyastringoverafric-tionless,masslesspulleysuchthatoneisrestingonaninclinedplaneandtheotherishangingoverthetopedgeoftheplane,asshowninFigure5-21.Thehangingblockhasamassof16.0kg,andtheoneontheplanehasamassof8.0kg.Thecoefficientofkineticfrictionbetweentheblockandtheinclinedplaneis0.23.Theblocksarereleasedfromrest.37.0°■Figure5-21a.Whatistheaccelerationoftheblocks?FmbothaFghangingFplaneFfplanemhanginggFgplanesinkFgplanecossoambothmhanginggmplanegsinkmplanegcosmbothg(mhangingmplanesinkmplanecos)mhangingmplane(9.80ms2)(16.0kg(8.0kg)(sin37.0°)(0.23)(8.0kg)(cos37.0°))(16.0kg80kg)4.0m/s2b.Whatisthetensioninthestringconnectingtheblocks?FTFgFamgmam(ga)(16.0kg)(9.80m/s24.0m/s2)93N102.InFigure5-22,ablockofmassMispushedwithsuchaforce,F,thatthesmallerblockofmmassmdoesnotslidedownthefrontofit.ThereisnofrictionbetweenthelargerblockFMandthesurfacebelowit,butthecoefficientofstaticfrictionbetweenthetwoblocksiss.FindanexpressionforFintermsofM,m,s,andg.Smallerblock:Ff,MonmsFN,Monmmg■Figure5-22mgFmaN,MonmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.sgasPhysics:PrinciplesandProblemsSolutionsManual107

111Chapter5continuedLargerblock:b.Whatforcewillbeneededtostartthesledmoving?F–FN,monMMaFfsFNmgMgFsssFgg(0.30)(4.90102N)F(mM)s1.5102NMixedReviewc.Whatforceisneededtokeepthesledmovingataconstantvelocity?pages143–144FLevel1fsFN103.ThescaleinFigure5-23isbeingpulledsFgonbythreeropes.Whatnetforcedoesthescaleread?(0.10)(4.90102N)49N,kineticfrictiond.Oncemoving,whattotalforcemustbeappliedtothesledtoaccelerateitat3.0m/s2?maFnetFapplFfsoFapplmaFf27.0°27.0°(50.0kg)(3.0m/s2)49N2.0102NLevel275.0N75.0N105.MythologySisyphuswasacharacterinGreekmythologywhowasdoomedin150.0NHadestopushabouldertothetopofaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.steepmountain.Whenhereachedthetop,■Figure5-7theboulderwouldslidebackdowntheFindthey-componentofthetwosidemountainandhewouldhavetostartallropesandthenaddthemtothemiddleoveragain.AssumethatSisyphusslidestherope.boulderupthemountainwithoutbeingFyFcosabletorollit,eventhoughinmostver-sionsofthemyth,herolledit.(75.0N)(cos27.0°)a.Ifthecoefficientofkineticfriction66.8Nbetweentheboulderandthemountain-Fy,totalFy,leftFy,middleFy,rightsideis0.40,themassoftheboulderis66.8N150.0N66.8N20.0kg,andtheslopeofthemountainisaconstant30.0°,whatistheforce283.6NthatSisyphusmustexertonthebouldertomoveitupthemountainatacon-104.SleddingAsledwithamassof50.0kgisstantvelocity?pulledalongflat,snow-coveredground.Thestaticfrictioncoefficientis0.30,andFSonrockFgtoslopeFfthekineticfrictioncoefficientis0.10.FSonrockmgsina.Whatdoesthesledweigh?F2)kmgcosma0gmg(50.0kg)(9.80m/s4.90102NFSonrockmgsinkmgcos108SolutionsManualPhysics:PrinciplesandProblems

112Chapter5continuedmg(sinkcos)(20.0kg)(9.80m/s2)(sin30.0°(0.40)(cos30.0°))166Nb.IfSisyphuspushestheboulderatavelocityof0.25m/sandittakeshim8.0htoreachthetopofthemountain,whatisthemythicalmountain’sverticalheight?hdsinvtsin(0.25m/s)(8.0h)(3600s/h)(sin30.0°)3.6103m3.6kmLevel3106.LandscapingAtreeisbeingtransportedonaflatbedtrailerbyalandscaper,asshowninFigure5-24.Ifthebaseofthetreeslidesonthetreewillthetrailer,falloverandbedamaged.Ifthecoefficientofstaticfrictionbetweenthetreeandthetraileris0.50,whatistheminimumstoppingdistanceofthetruck,travelingat55km/h,ifitistoaccelerateuniformlyandnothavethetreeslideforwardandfallonthetrailer?■Figure5-24FtruckFfsFNsmgmaasmgmsg(0.50)(9.80m/s2)4.9m/s2v2v22adwithvfif0,v2sodi2a1000m1h2(55km/h)1km3600s(2)(4.9ms2)24mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual109

113Chapter5continuedThinkingCriticallypage144107.UseModelsUsingtheExampleProblemsinthischapterasmodels,writeanexampleproblemtosolvethefollowingproblem.Includethefollowingsections:AnalyzeandSketchtheProblem,SolvefortheUnknown(withacompletestrategy),andEvaluatetheAnswer.Adriverofa975-kgcartraveling25m/sputsonthebrakes.Whatistheshortestdistanceitwilltakeforthecartostop?Assumethattheroadisconcrete,theforceoffrictionoftheroadonthetiresisconstant,andthetiresdonotslip.didf25m/sFN+xvFfFnetaFgAnalyzeandSketchtheProblem•Chooseacoordinatesystemwithapositiveaxisinthedirectionofmotion.•Drawamotiondiagram.•Labelvanda.•Drawthefree-bodydiagram.Known:Unknown:di0df?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.vi25m/svf0m975kgs0.80SolvefortheUnknownSolveNewton’ssecondlawfora.FnetmaFfmaSubstituteFfFnetFNmaSubstituteFfFNmgmaSubstituteFNmgagUsetheexpressionforaccelerationtosolvefordistance.v2v22a(dfifdi)v2v2dfifdi2av2v2dfiSubstituteagi(2)(g)110SolutionsManualPhysics:PrinciplesandProblems

114Chapter5continued(0.0m/s)2(25m/s)2Substituted0.0mi0.0m,vf0.0m/s,(2)(0.65)(9.80m/s)2v2i25m/s,0.65,g9.80m/s49m108.AnalyzeandConcludeMargaretMary,Doug,andKakoareatalocalamusementparkandseeanattractioncalledtheGiantSlide,whichissimplyaverylongandhighinclinedplane.Visitorsattheamusementparkclimbalongflightofstepstothetopofthe27°inclinedplaneandaregivencanvassacks.Theysitonthesacksandslidedownthe70-m-longplane.Atthetimewhenthethreefriendswalkpasttheslide,a135-kgmananda20-kgboyareeachatthetoppreparingtoslidedown.“Iwonderhowmuchlesstimeitwilltakethemantoslidedownthanitwilltaketheboy,”saysMargaretMary.“Ithinktheboywilltakelesstime,”saysDoug.“You’rebothwrong,”saysKako.“Theywillreachthebottomatthesametime.”a.Performtheappropriateanalysistodeterminewhoiscorrect.FnetFgFfFgsinkFNmgsinkmgcosmaag(sinkcos),sotheaccelerationisindependentofthemass.Theywilltie,soKakoiscorrect.b.Ifthemanandtheboydonottakethesameamountoftimetoreachtheoftheslide,calculatehowmanysecondsofdifferencetherewillbebetweenthetwotimes.Theywillreachthebottomatthesametime.WritinginPhysicspage144109.Investigatesomeofthetechniquesusedinindustrytoreducethefrictionbetweenvariouspartsofmachines.Describetwoorthreeofthesetechniquesandexplainthephysicsofhowtheywork.Answerswillvaryandmayincludelubricantsandreductionofthenormalforcetoreducetheforceoffriction.110.OlympicsInrecentyears,manyOlympicathletes,suchassprinters,swimmers,skiers,andspeedskaters,haveusedmodifiedequipmenttoreducetheeffectsoffrictionandairorwaterdrag.Researchapieceofequipmentusedbyoneofthesetypesofathletesandthewayithaschangedovertheyears.Explainhowphysicshasimpactedthesechanges.Answerswillvary.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual111

115Chapter5continuedCumulativeReviewChallengeProblempage144page132111.AddorsubtractasindicatedandstatetheFindtheequilibrantforthefollowingforces.answerwiththecorrectnumberofsignifi-Fcantdigits.(Chapter1)161.0Nat17.0°northofeastF238.0Nat64.0°northofeasta.85.26g4.7gF354.0Nat8.0°westofnorth90.0gF493.0Nat53.0°westofnorthb.1.07km0.608kmF565.0Nat21.0°southofwestF1.68km6102.0Nat15.0°westofsouthFc.186.4kg57.83kg726.0NsouthF877.0Nat22.0°eastofsouth128.6kgF951.0Nat33.0°eastofsouthd.60.08s12.2sF1082.0Nat5.0°southofeast47.9sy112.Yourideyourbikefor1.5hatanaverage43velocityof10km/h,thenfor30minat215km/h.Whatisyouraveragevelocity?1(Chapter3)xAveragevelocityisthetotaldisplace-10mentdividedbythetotaltime.579ddvfittfi8v1t1v2t2dit1t2ti6diti0,soCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.F1x(61.0N)(cos17.0°)58.3Nvv1t1v2t2t1t2F1y(61.0N)(sin17.0°)17.8N(10kmh)(1.5h)(15kmh)(0.5h)F1.5h0.5h2x(38.0N)(cos64.0°)16.7N10km/hF2y(38.0N)(sin64.0°)34.2N113.A45-NforceisexertedintheupwardF3x(54.0N)(sin8.0°)7.52Ndirectionona2.0-kgbriefcase.WhatistheFaccelerationofthebriefcase?(Chapter4)3y(54.0N)(cos8.0°)53.5NFnetFappliedFgFappliedmgF4x(93.0N)(sin53.0°)74.3NmaF4y(93.0N)(cos53.0°)56.0NFappliedmgsoamF5x(65.0N)(cos21.0°)60.7N45N(2.0kg)(9.80ms2)F5y(65.0N)(sin21.0°)23.3N2.0kg13m/s2F6x(102N)(sin15.0°)26.4NF6y(102N)(cos15.0°)98.5N112SolutionsManualPhysics:PrinciplesandProblems

116Chapter5continued10F7x0.0NFFyiyi1F7y26.0N107.65NF8x(77.0N)(sin22.0°)28.8NF2(F2R(Fx)y)F8y(77.0N)(cos22.0°)71.4N(44.38N)2(107.65N)2F9x(51.0N)(sin33.0°)27.8N116NF9y(51.0N)(cos33.0°)42.8NF1yRtanFxF10x(82.0N)(cos5.0°)81.7N1107.65NFtan10y(82.0N)(sin5.0°)7.15N44.38N67.6°10FxFixFequilibrant116Nat112.4°i144.38N116Nat22.4°WofNCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual113

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118CHAPTER6MotioninTwoDimensions2.LucyandherfriendareworkingatanPracticeProblemsassemblyplantmakingwoodentoy6.1ProjectileMotiongiraffes.Attheendoftheline,thegiraffespages147–152gohorizontallyofftheedgeoftheconveyorbeltandfallintoaboxbelow.Iftheboxispage1500.6mbelowtheleveloftheconveyorbelt1.Astoneisthrownhorizontallyataspeedand0.4mawayfromit,whatmustbetheof5.0m/sfromthetopofacliffthatishorizontalvelocityofgiraffesastheyleave78.4mhigh.theconveyorbelt?a.Howlongdoesittakethestonetoreachthebottomofthecliff?2yxvxtvxg12Sincevgty0,yvyt2xsovx122ybecomesygt2g2y0.4mortg(2)(0.6m)(2)(78.4m)9.80m/s29.80m/s21m/s4.00s3.Youarevisitingafriendfromelementaryb.Howfarfromthebaseofthecliffdoesschoolwhonowlivesinasmalltown.Onethestonehittheground?localamusementistheice-creamparlor,xvxtwhereStan,theshort-ordercook,slideshis(5.0m/s)(4.00s)completedice-creamsundaesdownthecounterataconstantspeedof2.0m/sto2.0101mtheservers.(Thecounteriskeptverywellc.Whatarethehorizontalandverticalpolishedforthispurpose.)Iftheserverscomponentsofthestone’svelocityjustcatchthesundaes7.0cmfromtheedgeofbeforeithitstheground?thecounter,howfardotheyfallfromthevedgeofthecountertothepointatwhichx5.0m/s.Thisisthesameastheinitialhorizontalspeedbecausethetheserverscatchthem?accelerationofgravityinfluencesxvxt;onlytheverticalmotion.Forthexverticalcomponent,usevvtvigtxwithvvyandvi,theinitialvertical12componentofvelocity,zero.ygt2Att4.00s1x2gCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2vxvygt(9.80m/s2)(4.0s)1(9.80m/s2)0.070m222.0m/s39.2m/s0.0060mor0.60cmPhysics:PrinciplesandProblemsSolutionsManual115

119Chapter6continuedpage1524.AplayerkicksafootballfromgroundTrajectory25levelwithaninitialvelocityof27.0m/s,60.0°30.0°abovethehorizontal,asshowninFigure6-4.Findeachofthe(m)yfollowing.Assumethatair30.0°resistanceisnegligible.0x(m)60a.theball’shangtime■Figure6-4vyvisinWhenitlands,yv120.yt2gtTherefore,2v2yttg2vytg2vinisg(2)(27.0m/s)(sin30.0°)9.80m/s22.76sb.theball’smaximumheightMaximumheightoccursathalfthe“hangtime,”or1.38s.Thus,12yvgtyt2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12visint2gt(27.0m/s)(sin30.0°)(1.38s)1(9.80m/s2)(1.38s)229.30mc.theball’srangeDistance:vxvicosxvxt(vicos)(t)(27.0m/s)(cos30.0°)(2.76s)64.5m5.Theplayerinproblem4thenkickstheballwiththesamespeed,butat60.0°fromthehorizontal.Whatistheball’shangtime,range,andmaximumheight?FollowingthemethodofPracticeProblem4,Hangtime:2vintisg(2)(27.0m/s)(sin60.0°)29.80m/s4.77s116SolutionsManualPhysics:PrinciplesandProblems

120Chapter6continuedDistance:xvicost(27.0m/s)(cos60.0°)(4.77s)64.4mMaximumheight:1att(4.77s)2.38s212yvgtisint212)(2.38s)2(27.0m/s)(sin60.0°)(2.38s)(9.80m/s227.9m6.Arockisthrownfroma50.0-m-highcliffwithaninitialvelocityof7.0m/satanangleof53.0°abovethehorizontal.Findthevelocityvectorforwhenithitsthegroundbelow.vxvicosvyvisingt2yvisinggvisin2ygvvx2vy2(vicos)2(visin2yg)22((7.0m/s)cos53.0°)2(7.0m/s)(sin53.0°)(2)(50.0m)(9.80m/s2)37m/sv1ytanvxtan1visini2ygvicosi1(7.0m/s)(sin53.0°)(2)(50.0m)(9.80m/s2)tan(7.0m/s)(cos53.0°)83°fromhorizontalSectionReview6.1ProjectileMotionCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pages147–152page1527.ProjectileMotionTwobaseballsarepitchedhorizontallyfromthesameheight,butatdifferentspeeds.Thefasterballcrosseshomeplatewithinthestrikezone,buttheslowerballisbelowthebatter’sknees.Whydoesthefasterballnotfallasfarastheslowerone?Physics:PrinciplesandProblemsSolutionsManual117

121Chapter6continuedThefasterballisintheairashortertime,andthusgainsasmallerverticalvelocity.8.Free-BodyDiagramAnicecubeslideswithoutfrictionacrossatableataconstantvelocity.Itslidesoffthetableandlandsonthefloor.Drawfree-bodyandmotiondiagramsoftheicecubeattwopointsonthetableandattwopointsintheair.Free-BodyDiagramsMotionDiagramsOnthetableIntheairOnthetableIntheairFFa0NNaFgFgFgFg9.ProjectileMotionAsoftballistossedintotheairatanangleof50.0°withtheverticalataninitialvelocityof11.0m/s.Whatisitsmaximumheight?v2v22a(dfiyfdi);ag,di0Atmaximumheightvf0,sov2iydf2g(vos)2ic2g((11.0m/s)(cos50.0°))2(2)(9.80m/s2)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.55m10.ProjectileMotionAtennisballisthrownoutawindow28mabovethegroundataninitialvelocityof15.0m/sand20.0°belowthehorizontal.Howfardoestheballmovehorizontallybeforeithitstheground?xv0xt,butneedtofindtFirst,determinevyf:v2v22gyyfyivyfvyi22gy(visin)22gy((15.0m/s)(sin20.0°))2(2)(9.80m/s2)(28m)24.0m/sNowusevyfvyigttofindt.vvyfyitgvvyfising118SolutionsManualPhysics:PrinciplesandProblems

122Chapter6continued2.40m/s(15.0m/s)(sin20.0°)9.80m/s21.92sxvxit(vicos)(t)(15.0m/s)(cos20.0°)(1.92s)27.1m11.CriticalThinkingSupposethatanobjectisthrownwiththesameinitialvelocityanddirectiononEarthandontheMoon,wheregisone-sixththatonEarth.Howwillthefollowingquantitieschange?a.vxwillnotchangeb.theobject’stimeofflight2vywillbelarger;tgc.ymaxwillbelargerd.RwillbelargerPracticeProblems6.2CircularMotionpages153–156page15612.Arunnermovingataspeedof8.8m/sroundsabendwitharadiusof25m.Whatisthecentripetalaccelerationoftherunner,andwhatagentexertsforceontherunner?v2(8.8m/s)22,thefrictionalforceofthetrackactingona3.1m/scr25mtherunner’sshoesexertstheforceontherunner.13.Acarracingonaflattracktravelsat22m/saroundacurvewitha56-mradius.Findthecar’scentripetalacceleration.Whatminimumcoefficientofstaticfrictionbetweenthetiresandroadisnecessaryforthecartoroundthecurvewithoutslipping?v2(22m/s)22a8.6m/scr56mRecallFfFN.ThefrictionforcemustsupplythecentripetalforcesoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Ffmac.ThenormalforceisFNmg.ThecoefficientoffrictionmustbeatleastFfmacac8.6m/s20.88FNmgg9.80m/s2Physics:PrinciplesandProblemsSolutionsManual119

123Chapter6continued14.Anairplanetravelingat201m/smakesab.Whatisthedirectionofthenetforceturn.Whatisthesmallestradiusofthethatisactingonyou?circularpath(inkm)thatthepilotcanThenetforceactingonyourbodyismakeandkeepthecentripetalaccelerationtotherightunder5.0m/s2?c.Whatexertsthisforce?v2v2(201m/s)2Theforceisexertedbythecar’sa,sor8.1kmcrac5.0m/s2seat.15.A45-kgmerry-go-roundworkerstandson18.CentripetalForceIfa40.0-gstoneistheride’splatform6.3mfromthecenter.whirledhorizontallyontheendofaIfherspeedasshegoesaroundthecircle0.60-mstringataspeedof2.2m/s,whatisis4.1m/s,whatistheforceoffrictionthetensioninthestring?necessarytokeepherfromfallingofftheFTmacplatform?mv2mv2(45kg)(4.1m/s)2rF120NfFcr6.3m(0.0400kg)(22m/s)20.60mSectionReview0.32N6.2CircularMotion19.CentripetalAccelerationAnewspaperpages153–156articlestatesthatwhenturningacorner,page156adrivermustbecarefultobalancethe16.UniformCircularMotionWhatisthecentripetalandcentrifugalforcestokeepdirectionoftheforcethatactsonthefromskidding.Writealettertotheeditorclothesinthespincycleofawashingthatcritiquesthisarticle.machine?Whatexertstheforce?ThelettershouldstatethatthereisanTheforceistowardthecenterofthetub.accelerationbecausethedirectionofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thewallsofthetubexerttheforceonthevelocityischanging;therefore,theclothes.Ofcourse,thewholepointistheremustbeanetforceinthedirec-thatsomeofthewaterintheclothestionofthecenterofthecircle.Theroadgoesoutthroughholesinthewallofthesuppliesthatforceandthefrictiontubratherthanmovingtowardthecenter.betweentheroadandthetiresallowstheforcetobeexertedonthetires.The17.Free-BodyDiagramYouaresittinginthecar’sseatexertstheforceonthedriverbackseatofacargoingaroundacurvetothethataccelerateshimorhertowardtheright.Sketchmotionandfree-bodydiagramscenterofthecircle.Thenotealsotoanswerthefollowingquestions.shouldmakeitclearthatcentrifugalforceisnotarealforce.20.CentripetalForceAbowlingballhasaamassof7.3kg.IfyoumoveitaroundaFcirclewitharadiusof0.75mataspeednetvof2.5m/s,whatforcewouldyouhavetoexertonit?a.WhatisthedirectionofyourFacceleration?netmacYourbodyisacceleratedtothemv2right.r120SolutionsManualPhysics:PrinciplesandProblems

124Chapter6continued(7.3kg)(2.5m/s)2so,vw/gvb/gvb/w0.75m0.5m/s2.5m/s61N2.0m/s;againsttheboat21.CriticalThinkingBecauseofEarth’sdaily25.Anairplanefliesduenorthat150km/hrelativerotation,youalwaysmovewithuniformtotheair.Thereisawindblowingat75km/hcircularmotion.Whatistheagentthattotheeastrelativetotheground.Whatisthesuppliestheforcethatacceleratesyou?plane’sspeedrelativetotheground?Howdoesthismotionaffectyourapparentweight?vvp2vw2Earth’sgravitysuppliestheforcethatacceleratesyouincircularmotion.Your(150km/h)2(75km/h)2uniformcircularmotiondecreasesyour1.7102km/happarentweight.SectionReviewPracticeProblems6.3RelativeVelocity6.3RelativeVelocitypages157–159pages157–159page159page15926.RelativeVelocityAfishingboatwitha22.Youareridinginabusmovingslowlymaximumspeedof3m/srelativetothethroughheavytrafficat2.0m/s.Youhurrywaterisinariverthatisflowingat2m/s.tothefrontofthebusat4.0m/srelativetoWhatisthemaximumspeedtheboatcanthebus.Whatisyourspeedrelativetotheobtainrelativetotheshore?Theminimumstreet?speed?Givethedirectionoftheboat,rela-vy/gvb/gvy/btivetotheriver’scurrent,forthemaximumspeedandtheminimumspeedrelativeto2.0m/s4.0m/stheshore.6.0m/srelativetostreetThemaximumspeedrelativetotheshoreiswhentheboatmovesatmaxi-23.Rafiispullingatoywagonthroughthemumspeedinthesamedirectionasneighborhoodataspeedof0.75m/s.theriver’sflow:Acaterpillarinthewagoniscrawlingtowardtherearofthewagonatarateofvb/svb/wvw/s2.0cm/s.Whatisthecaterpillar’svelocity3m/s2m/srelativetotheground?5m/svc/gvw/gvc/wTheminimumspeedrelativetothe0.75m/s0.02m/sshoreiswhentheboatmovesintheoppositedirectionoftheriver’sflow0.73m/swiththesamespeedastheriver:24.Aboatisroweddirectlyupriverataspeedvb/svb/wvw/sof2.5m/srelativetothewater.ViewersonCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3m/s(2m/s)theshoreseethattheboatismovingatonly0.5m/srelativetotheshore.Whatis1m/sthespeedoftheriver?Isitmovingwithoragainsttheboat?vb/gvb/wvw/g;Physics:PrinciplesandProblemsSolutionsManual121

125Chapter6continued27.RelativeVelocityofaBoatApowerboatheadsduenorthwestat13m/srelativetothewateracrossariverthatflowsduenorthat5.0m/s.Whatisthevelocity(bothmagnitudeanddirection)ofthemotorboatrelativetotheshore?vRvRN2vRW2(vbNvrN)2(vbWvrW)2(vbsinvr)2(vbcos)2((13m/s)(sin45°)5.0m/s)2((13m/s)(cos45°))217m/sv1RWtanvRNv1bcostanvbsinvr1(13m/s)(cos45°)tan(3m/s)(sin45°)5.0m/s33°vR17m/s,33°westofnorth28.RelativeVelocityAnairplanefliesduesouthat175km/hrelativetotheair.Thereisawindblowingat85km/htotheeastrelativetotheground.Whataretheplane’sspeedanddirectionrelativetotheground?vR(175km/h)2(85km/h)2190km/h1175km/htan64°85km/hvR190km/h,64°southofeastCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.29.APlane’sRelativeVelocityAnairplanefliesduenorthat235km/hrelativetotheair.Thereisawindblowingat65km/htothenortheastrelativetotheground.Whataretheplane’sspeedanddirectionrelativetotheground?vRvRE2vRN2(vpEvaE)2(vpNvwN)2(vwcos)2(vpvwsin)2((65km/h)(cos45°))2(235km/h(65km/h)(sin45°))2280km/hv1RNtanvREv1pvasintanvacos1235km/h(65km/h)(sin45°)tan(65km/h)(cos45°)72°northofeast280km/h,72°northofeast122SolutionsManualPhysics:PrinciplesandProblems

126Chapter6continued30.RelativeVelocityAnairplanehasaspeedof285km/hrelativetotheair.Thereisawindblowingat95km/hat30.0°northofeastrelativetoEarth.Inwhichdirectionshouldtheplaneheadtolandatanairportduenorthofitspresentlocation?Whatistheplane’sspeedrelativetotheground?Totravelnorth,theeastcomponentsmustbeequalandopposite.vpWcospv,sopRv1pWpcosvpRv1wEcoswcosvpR1(95km/h)(cos30.0°)cos285km/h73°northofwestvpRvpNvwNvpsinpvwsinw(285km/h)(sin107°)(95km/h)(sin30.0°)320km/h31.CriticalThinkingYouarepilotingaboatacrossafast-movingriver.Youwanttoreachapierdirectlyoppositeyourstartingpoint.Describehowyouwouldnavigatetheboatintermsofthecomponentsofyourvelocityrelativetothewater.Youshouldchoosethecomponentofyourvelocityalongthedirectionoftherivertobeequalandoppositetothevelocityoftheriver.ChapterAssessmentConceptMappingpage16432.Usethefollowingtermstocompletetheconceptmapbelow:constantspeed,horizontalpartofprojectilemotion,constantacceleration,relative-velocitymotion,uniformcircularmotion.CategoriesofMotionconstantconstantconstantaccelerationvelocityspeedCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.verticalparthorizontalrelative-velocityuniformcircularofprojectilepartofprojectilemotionmotionmotionmotionPhysics:PrinciplesandProblemsSolutionsManual123

127Chapter6continuedMasteringConceptspage16433.ConsiderthetrajectoryofthecannonballshowninFigure6-11.(6.1)BACDE■Figure6-11Upispositive,downisnegative.a.Whereisthemagnitudeofthevertical-velocitycomponentlargest?ThegreatestverticalvelocityoccursatpointA.b.Whereisthemagnitudeofthehorizontal-velocitycomponentlargest?Neglectingairresistance,thehorizontalvelocityatallpointsisthesame.Horizontalvelocityisconstantandindependentofverticalvelocity.c.Whereisthevertical-velocitysmallest?TheleastverticalvelocityoccursatpointE.d.Whereisthemagnitudeoftheaccelerationsmallest?Themagnitudeoftheaccelerationisthesameeverywhere.34.Astudentisplayingwitharadio-controlledracecaronthebalconyofasixth-floorapartment.Anaccidentalturnsendsthecarthroughtherailingandovertheedgeofthebalcony.DoesthetimeittakesthecartofalldependupontheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.speedithadwhenitleftthebalcony?(6.1)No,thehorizontalcomponentofmotiondoesnotaffecttheverticalcomponent.35.Anairplanepilotflyingatconstantvelocityandaltitudedropsaheavycrate.Ignoringairresistance,wherewilltheplaneberelativetothecratewhenthecratehitstheground?Drawthepathofthecrateasseenbyanobserverontheground.(6.1)Theplanewillbedirectlyoverthecratewhenthecratehitstheground.Bothhavethesamehorizontalvelocity.Thecratewilllooklikeitismovinghorizontallywhilefallingverticallytoanobserverontheground.36.Canyougoaroundacurvewiththefollowingaccelerations?Explain.a.zeroaccelerationNo,goingaroundacurvecausesachangeindirectionofvelocity.Thus,theaccelerationcannotbezero.b.constantacceleration(6.2)No,themagnitudeoftheaccelerationmaybeconstant,butthedirectionoftheaccelerationchanges.124SolutionsManualPhysics:PrinciplesandProblems

128Chapter6continued37.Toobtainuniformcircularmotion,how41.BaseballAbatterhitsapop-upstraightmustthenetforcedependonthespeedupoverhomeplateataninitialvelocityofthemovingobject?(6.2)of20m/s.TheballiscaughtbythecatcherCircularmotionresultswhentheatthesameheightthatitwashit.Atwhatdirectionoftheforceisconstantlyvelocitydoestheballlandinthecatcher’sperpendiculartotheinstantaneousmitt?Neglectairresistance.velocityoftheobject.20m/s,wherethenegativesignindicatesdown38.Ifyouwhirlayo-yoaboutyourheadinahorizontalcircle,inwhatdirectionmusta42.FastballInbaseball,afastballtakesaboutforceactontheyo-yo?Whatexertsthe1storeachtheplate.Assumingthatsucha2force?(6.2)pitchisthrownhorizontally,comparetheTheforceisalongthestringtowardthe1distancetheballfallsinthefirstswithcenterofthecirclethattheyo-yofol-4lows.Thestringexertstheforce.1thedistanceitfallsintheseconds.4Becauseoftheaccelerationduetograv-39.Whyisitthatacartravelingintheoppositeity,thebaseballfallsagreaterdistancedirectionasthecarinwhichyouareriding1onthefreewayoftenlookslikeitismovingduringthesecondsthanduringthe41fasterthanthespeedlimit?(6.3)firsts.4Themagnitudeoftherelativevelocity43.Youthrowarockhorizontally.Inasecondofthatcartoyourcarcanbefoundbyhorizontalthrow,youthrowtherockharderaddingthemagnitudesofthetwocars’andgiveitevenmorespeed.velocitiestogether.Sinceeachcarprobablyismovingatclosetothea.Howwillthetimeittakestherocktospeedlimit,theresultingrelativeveloci-hitthegroundbeaffected?Ignoreairtywillbelargerthanthepostedspeedresistance.limit.Thetimedoesnotchange—thetimeittakestohitthegrounddependsonlyApplyingConceptsonverticalvelocitiesandaccelera-pages164–165tion.40.ProjectileMotionAnalyzehowhorizontalb.Howwilltheincreasedspeedaffectthemotioncanbeuniformwhileverticaldistancefromwheretherockleftyourmotionisaccelerated.Howwillprojectilehandtowheretherockhitstheground?motionbeaffectedwhendragduetoairAhigherhorizontalspeedproducesresistanceistakenintoconsideration?alongerhorizontaldistance.Thehorizontalmotionisuniformbecausetherearenoforcesactingin44.FieldBiologyAzoologiststandingonathatdirection(ignoringfriction).Thecliffaimsatranquilizergunatamonkeyverticalmotionisacceleratedduetohangingfromadistanttreebranch.Thetheforceofgravity.Theprojectilebarrelofthegunishorizontal.Justasthemotionequationsinthisbookdonotzoologistpullsthetrigger,themonkeyletsholdwhenfrictionistakenintogoandbeginstofall.WillthedarthittheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.account.Projectilemotioninbothmonkey?Ignoreairresistance.directionswillbeimpactedwhenYes,infact,themonkeywouldbesafeifdragduetoairresistanceistakenitdidnotletgoofthebranch.Theverti-intoconsideration.Therewillbeacalaccelerationofthedartisthesamefrictionforceopposingthemotion.asthatofthemonkey.Therefore,thedartisatthesameverticalheightwhenPhysics:PrinciplesandProblemsSolutionsManual125

129Chapter6continueditreachesthemonkey.45.FootballAquarterbackthrowsafootballat24m/sata45°angle.Ifittakestheball3.0storeachthetopofitspathandtheballiscaughtatthesameheightatwhichitisthrown,howlongisitintheair?Ignoreairresistance.6.0s:3.0supand3.0sdown46.TrackandFieldYouareworkingonimprovingyourperformanceinthelongjumpandbelievethattheinformationinthischaptercanhelp.Doestheheightthatyoureachmakeanydifferencetoyourjump?Whatinfluencesthelengthofyourjump?Bothspeedandangleoflaunchmatter,soheightdoesmakeadifference.Maximumrangeisachievedwhentheresultantvelocityhasequalverticalandhorizontalcomponents—inotherwords,alaunchangleof45°.Forthisreason,heightandspeedaffecttherange.47.Imaginethatyouaresittinginacartossingaballstraightupintotheair.a.Ifthecarismovingataconstantvelocity,willtheballlandinfrontof,behind,orCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.inyourhand?Theballwilllandinyourhandbecauseyou,theball,andthecarallaremovingforwardwiththesamespeed.b.Ifthecarroundsacurveataconstantspeed,wherewilltheballland?Theballwilllandbesideyou,towardtheoutsideofthecurve.Atopviewwouldshowtheballmovingstraightwhileyouandthecarmovedoutfromundertheball.48.Youswingoneyo-yoaroundyourheadinahorizontalcircle.Thenyouswinganotheryo-yowithtwicethemassofthefirstone,butyoudon’tchangethelengthofthestringortheperiod.Howdothetensionsinthestringsdiffer?ThetensioninthestringisdoubledsinceFTmac.126SolutionsManualPhysics:PrinciplesandProblems

130Chapter6continued49.CarRacingThecurvesonaracetrackareshouldyoulookforthekeys?bankedtomakeiteasierforcarstogo12yvaroundthecurvesathighspeeds.Drawayt2gtfree-bodydiagramofacaronabankedSinceinitialverticalvelocityiszero,curve.Fromthemotiondiagram,findthe2y(2)(64m)directionoftheacceleration.tg9.80m/s2FrontViewTopView3.6sv2v1xvFNxt(8.0m/s)(3.6)28.8mv129mvv2a52.ThetoycarinFigure6-12runsofftheedgeFgofatablethatis1.225-mhigh.Thecarlands0.400mfromthebaseofthetable.vTheaccelerationisdirectedtowardthecenterofthetrack.a.Whatexertstheforceinthedirectionoftheacceleration?Thecomponentofthenormalforceactingtowardthecenterofthecurve,anddependingonthecar’s1.225mspeed,thecomponentofthefrictionforceactingtowardthecenter,bothcontributetothenetforceinthedirectionofacceleration.b.Canyouhavesuchaforcewithout0.400mfriction?■Figure6-12Yes,thecentripetalaccelerationneeda.Howlongdidittakethecartofall?onlybeduetothenormalforce.12yvy0t2gt50.DrivingontheHighwayExplainwhyitisSinceinitialverticalvelocityiszero,thatwhenyoupassacargoinginthesamedirectionasyouonthefreeway,ittakesa2y(2)(1.225m)tlongertimethanwhenyoupassacargoingg9.80m/s2intheoppositedirection.0.500sTherelativespeedoftwocarsgoinginb.Howfastwasthecargoingonthetable?thesamedirectionislessthantherela-x0.400mtivespeedoftwocarsgoinginthev0.800m/sxt0.500soppositedirection.Passingwiththelesserrelativespeedwilltakelonger.53.Adartplayerthrowsadarthorizontallyat12.4m/s.Thedarthitstheboard0.32mMasteringProblemsbelowtheheightfromwhichitwasthrown.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6.1ProjectileMotionHowfarawayistheplayerfromtheboard?page16512yvy0t2gtLevel151.Youaccidentallythrowyourcarkeysandbecauseinitialvelocityiszero,horizontallyat8.0m/sfromacliff64-m2ythigh.HowfarfromthebaseofthecliffgPhysics:PrinciplesandProblemsSolutionsManual127

131Chapter6continued(2)(0.32m)55.SwimmingYoutookarunningleapoffa9.80m/s2high-divingplatform.Youwererunningat2.8m/sandhitthewater2.6slater.How0.26shighwastheplatform,andhowfarfromNowxvxttheedgeoftheplatformdidyouhitthe(12.4m/s)(0.26s)water?Ignoreairresistance.3.2m12yvyit2gt54.ThetwobaseballsinFigure6-13werehit0(2.6s)1(9.80m/s2)(2.6s)2withthesamespeed,25m/s.Drawseparate2graphsofyversustandxversustforeach33m,sotheplatformis33mhighball.xvxt(2.8m/s)(2.6s)7.3mALevel256.ArcheryAnarrowisshotat30.0°aboveByAthehorizontal.Itsvelocityis49m/s,andit60°hitsthetarget.yB30°a.WhatisthemaximumheightthearrowxAxBwillattain?v2v22gdyyi■Figure6-13Atthehighpointvy0,soVerticalv.Time25(v)2y060d202g15(v2isin)102gCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Vertical(m)530((49m/s)(sin30.0°))2(2)(9.80m/s2)001423531mTime(s)b.ThetargetisattheheightfromwhichHorizontalv.Timethearrowwasshot.Howfarawayisit?8012yv60y0t2gt30butthearrowlandsatthesame4060height,soHorizontal(m)201y0and0vyi2gt0014235sot0orTime(s)2vytig2vinisg(2)(49m/s)(sin30.0°)9.80m/s25.0s128SolutionsManualPhysics:PrinciplesandProblems

132Chapter6continuedandxvbutvxtyi0,so(vicos)(t)t2yg(49m/s)(cos30.0°)(5.0s)(2)(1001m)2.1102m29.80m/s57.HittingaHomeRunApitchedballishit14.3sbyabatterata45°angleandjustclearstheb.Whatisthehorizontaldistancebetweenoutfieldfence,98maway.Ifthefenceisattheplaneandthevictimswhentheboxthesameheightasthepitch,findthevelocityisdropped?oftheballwhenitleftthebat.Ignoreairxvxtresistance.1h1000mThecomponentsoftheinitialvelocity(125km/h)3600s1kmarevxvicosiandvyivisini(14.3s)Nowxvxt(vicosi)t,so497mxtvsicoi59.DivingDiversinAcapulcodivefromacliffAndyv12,buty0,sothatis61mhigh.Iftherocksbelowtheyit2gtcliffextendoutwardfor23m,whatisthe10vyi2gttminimumhorizontalvelocityadivermust1havetocleartherocks?sot0orvyi2gt012yvyit2gtFromabovev1x0andsincevyi0,isini2gvsicoi2ytMultiplyingbyvgicosigivesv2sin1(2)(61m)iicosi2gx09.80m/s22gxsovi(2)(sin3.53si)(cosi)xvxtgxthus,vi(2)(sinxi)(cosi)vxt(9.80m/s2)(98m)23m(2)(sin45°)(cos45°)3.53s31m/sat45°6.5m/sLevel360.JumpShotAbasketballplayeristryingto58.At-SeaRescueAnairplanetraveling1001mmakeahalf-courtjumpshotandreleasestheabovetheoceanat125km/hisgoingtoballattheheightofthebasket.Assumingdropaboxofsuppliestoshipwreckedthattheballislaunchedat51.0°,14.0mvictimsbelow.fromthebasket,whatspeedmusttheplayergivetheball?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.HowmanysecondsbeforetheplaneisdirectlyoverheadshouldtheboxbeThecomponentsoftheinitialvelocitydropped?arevxivicosiandvyivisiniyv12Nowxvxit(vicosi)t,soyit2gtPhysics:PrinciplesandProblemsSolutionsManual129

133Chapter6continuedxtvsicoiAndyv12,buty0,soyit2gt10vyi2gttsot0orv120yi2gtFromabove1xv0sini2gvs0icoiMultiplyingbyvicosigivesv2(sin1ii)(cosi)2gx0gxsovi(2)(sini)(cosi)(9.80m/s2)(14.0m)(2)(sin51.0°)(cos51.0°)11.8m/s6.2CircularMotionpage166Level161.CarRacingA615-kgracingcarcompletesonelapin14.3saroundacirculartrackwitharadiusof50.0m.ThecarmovesataCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.constantspeed.a.Whatistheaccelerationofthecar?v2acr42rT242(50.0m)(14.3s)29.59m/s2b.Whatforcemustthetrackexertonthetirestoproducethisacceleration?F2)cmac(615kg)(9.59m/s5.90103N130SolutionsManualPhysics:PrinciplesandProblems

134Chapter6continued62.HammerThrowAnathletewhirlsa(42)(0.050m)20.61m/s7.00-kghammer1.8mfromtheaxisof(1.80s)2rotationinahorizontalcircle,asshowninr10.0cm:Figure6-14.Ifthehammermakesonerev-42r(42)(0.100m)olutionin1.0s,whatisthecentripetalacT2(1.80s)2accelerationofthehammer?Whatisthetensioninthechain?1.22m/s2r15.0cm:vtang42r(42)(0.150m)acT2(1.80s)21.83m/s2c.Whatforceacceleratesthecoin?frictionalforcebetweencoinandrecordd.Atwhichofthethreeradiiinpartbwouldthecoinbemostlikelytoflyoff■Figure6-14theturntable?Why?2r15.0cm,thelargestradius;the4acT2frictionforceneededtoholditisthegreatest.(42)(1.8m)(1.0s)264.Arotatingrodthatis15.3cmlongisspun71m/s2withitsaxisthroughoneendoftherodsoFthattheotherendoftherodhasaspeedofcmac2010m/s(4500mph).(7.00kg)(71m/s2)a.Whatisthecentripetalaccelerationof5.0102Ntheendoftherod?v2(2010m/s)2Level2acr0.153m63.Acoinisplacedonavinylstereorecordthatismaking331revolutionsperminute2.64107m/s23onaturntable.b.Ifyouweretoattacha1.0-gobjecttoa.Inwhatdirectionistheaccelerationoftheendoftherod,whatforcewouldbethecoin?neededtoholditontherod?TheaccelerationistowardtheFcmaccenteroftherecord.(0.0010kg)(2.64107m/s2)b.Findthemagnitudeoftheacceleration4N2.610whenthecoinisplaced5.0,10.0,and15.0cmfromthecenteroftherecord.65.Frictionprovidestheforceneededforacar11totravelaroundaflat,circularracetrack.Tf133revminWhatisthemaximumspeedatwhichacarCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3cansafelytraveliftheradiusofthetrackis60s(0.0300min)1.80s80.0mandthecoefficientoffrictionis0.40?1minFr5.0cm:cFfFNmg42rmv2mv2ac2ButFcr,thusrmg.TPhysics:PrinciplesandProblemsSolutionsManual131

135Chapter6continuedThemassofthecardividesouttogivev2agv2gr,socgorrv2vgrsorg(0.40)(9.80m/s2)(80.0m)2(120m/s)18m/s9.80m/s21.5103mLevel366.Acarnivalclownridesamotorcycledownarampandaroundaverticalloop.Ifthe6.3RelativeVelocityloophasaradiusof18m,whatisthepages166–167slowestspeedtheridercanhaveatthetopLevel1ofthelooptoavoidfalling?Hint:Atthis68.NavigatinganAirplaneAnairplanefliesslowestspeed,thetrackexertsnoforceontheat200.0km/hrelativetotheair.Whatismotorcycleatthetopoftheloop.thevelocityoftheplanerelativetotheFcmacFgmg,sogroundifitfliesduringthefollowingwindaconditions?cga.a50.0-km/htailwindv2g,sorTailwindisinthesamedirectionastheairplanevgr200.0km/h50.0km/h250.0km/h(9.80m/s2)(18m)b.a50.0-km/hheadwind13m/sHeadwindisintheoppositedirectionoftheairplane67.A75-kgpilotfliesaplaneinaloopasshown200.0km/h50.0km/h150.0km/hinFigure6-15.Atthetopoftheloop,whentheplaneiscompletelyupside-downforanCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.69.OdinaandLaToyaaresittingbyariverandinstant,thepilothangsfreelyintheseatdecidetohavearace.Odinawillrundownanddoesnotpushagainsttheseatbelt.Thetheshoretoadock,1.5kmaway,thenturnairspeedindicatorreads120m/s.Whatisaroundandrunback.LaToyawillalsoracetheradiusoftheplane’sloop?tothedockandback,butshewillrowaboatintheriver,whichhasacurrentofvtang120m/s2.0m/s.IfOdina’srunningspeedisequaltoLaToya’srowingspeedinstillwater,whichis4.0m/s,whowillwintherace?Assumethattheybothturninstantaneously.xxvt,sotvforOdina,3.0103mt4.0m/s■Figure6-157.5102sBecausethenetforceisequaltotheForLaToya(assumeagainstcurrentonweightofthepilot,thewaytothedock),xxFcmacFgmg,sot12vv12132SolutionsManualPhysics:PrinciplesandProblems

136Chapter6continued1.5103m1.5103m1vb/airtan4.0m/s2.0m/s4.0m/s2.0m/svair1.0103s115m/stan6.5m/sOdinawins.67°fromthehorizontowardthewestLevel270.CrossingaRiverYourowaboat,suchasLevel3theoneinFigure6-16,perpendicularto72.BoatingYouareboatingonariverthattheshoreofariverthatflowsat3.0m/s.flowstowardtheeast.BecauseofyourThevelocityofyourboatis4.0m/srelativeknowledgeofphysics,youheadyourboattothewater.53°westofnorthandhaveavelocityof6.0m/sduenorthrelativetotheshore.a.Whatisthevelocityofthecurrent?vtanw/s,sovbb/svvwvw/s(tan)(vb/s)(tan53°)(6.0m/s)8.0m/seast■Figure6-16b.Whatisthespeedofyourboatrelativea.Whatisthevelocityofyourboatrelativetothewater?totheshore?vv2(v2cosb/s,sob/s(vb/w)w/s)vb/w(4.0m/s)2(3.0m/s)2vb/svb/wcos5.0m/s6.0m/svtan1b/wcos53°vw/s1m/s1.01014.0m/stan3.0m/s73.AirTravelYouarepilotingasmallplane,53°fromshoreandyouwanttoreachanairport450kmb.Whatisthecomponentofyourvelocityduesouthin3.0h.Awindisblowingfromparalleltotheshore?Perpendicularthewestat50.0km/h.Whatheadingandtoit?airspeedshouldyouchoosetoreachyourdestinationintime?3.0m/s;4.0m/sds450kmv150km/h71.StudyingtheWeatherAweatherstationst3.0hreleasesaballoontomeasurecloudcondi-vp(150km/h)2(50.0km/h)2tionsthatrisesataconstant15m/srelativetotheair,butthereisalsoawindblowing1.6102km/hat6.5m/stowardthewest.Whatarethevtan1windmagnitudeanddirectionofthevelocityofvCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.stheballoon?150.0km/hv)2(v2tan150km/hb(vb/airair)18°westofsouth(15m/s)2(6.5m/s)216m/sMixedReviewPhysics:PrinciplesandProblemsSolutionsManual133

137Chapter6continuedpage167Level174.EarlyskepticsoftheideaofarotatingEarthsaidthatthefastspinofEarthwouldthrowpeopleattheequatorintospace.TheradiusofEarthisabout6.38103km.Showwhythisideaiswrongbycalculatingthefollowing.a.thespeedofa97-kgpersonattheequatord2r2(6.38106m)vtT3600s(24h)1h464m/sb.theforceneededtoacceleratethepersoninthecircleFcmacmv2r(97kg)(464m/s)26.38106m3.3Nc.theweightofthepersonFgmg(97)(9.80m/s2)9.5102Nd.thenormalforceofEarthontheperson,thatis,theperson’sapparentweightF2N3.3NN9.5109.5102NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.75.FiringaMissileAnairplane,movingat375m/srelativetotheground,firesamissileforwardataspeedof782m/srelativetotheplane.Whatisthespeedofthemissilerelativetotheground?vm/gvp/gvm/p375m/s782m/s1157m/s76.RocketryArocketinouterspacethatismovingataspeedof1.25km/srelativetoanobserverfiresitsmotor.Hotgasesareexpelledoutthebackat2.75km/srelativetotherocket.Whatisthespeedofthegasesrelativetotheobserver?vg/ovr/ovg/r1.25km/s(2.75km/s)1.50km/sLevel277.Twodogs,initiallyseparatedby500.0m,arerunningtowardseachother,eachmovingwithaconstantspeedof2.5m/s.Adragonfly,movingwithaconstantspeedof3.0m/s,fliesfromthenoseofonedogtotheother,thenturnsaround134SolutionsManualPhysics:PrinciplesandProblems

138Chapter6continuedinstantaneouslyandfliesbacktotheotherFor35mph:dog.ItcontinuestoflybackandforthuntilFcFgthedogsrunintoeachother.Whatdistancedoesthedragonflyflyduringthistime?mv2cosmgsinrThedogswillmeetinv2sin500.0m2srgcosgtan1.0105.0m/sv2Thedragonflyfliestan1gr(3.0m/s)(1.0102s)3.0102m.(15.7m/s)2tan1(9.80m/s2)(56.0m)78.A1.13-kgballisswungverticallyfroma0.50-mcordinuniformcircularmotion34.9°ataspeedof2.4m/s.WhatisthetensioninFor45mph:thecordatthebottomoftheball’smotion?21vtanFgrTFgFc21(20.1m/s)2mgmvtan2)(36.0m)r(9.80m/s(1.13kg)(9.80m/s2)48.9°(1.13kg)(2.4m/s)2Level30.50m80.The1.45-kgballinFigure6-17is24Nsuspended14.0°froma0.80-79.BankedRoadsCurvesonroadsoftenaremstringbankedtohelppreventcarsfromslippingandofftheroad.Ifthepostedspeedlimitforaswunginaparticularcurveofradius36.0mis15.7m/s■Figure6-17horizontal(35mph),atwhatangleshouldtheroadbecircleataconstantbankedsothatcarswillstayonacircularspeedsuchthatthestringmakesananglepatheveniftherewerenofrictionbetweenof14.0°withthevertical.theroadandthetires?Ifthespeedlimitwasa.Whatisthetensioninthestring?increasedto20.1m/s(45mph),atwhatFangleshouldtheroadbebanked?TcosmgmgsoFTcos(1.45m/s)(9.80m/s2)cos14.0°14.6Nb.Whatisthespeedoftheball?mv2FcFTsinrFgFTcosmgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FTsinmv2soFcosrmgTv2ortanrgPhysics:PrinciplesandProblemsSolutionsManual135

139Chapter6continuedsovrgtan(0.80m)(9.80m/s2)(tan14.0°)1.4m/s81.Abaseballishitdirectlyinlinewithanoutfielderatanangleof35.0°abovethehorizontalwithaninitialvelocityof22.0m/s.Theoutfielderstartsrunningassoonastheballishitataconstantvelocityof2.5m/sandbarelycatchestheball.Assumingthattheballiscaughtatthesameheightatwhichitwashit,whatwastheinitialseparationbetweenthehitterandoutfielder?Hint:Therearetwopossibleanswers.xvxitvptt(vxivp)Togett,yv12,y0yit2gtsov12,t0oryit2gt1vyi2gt2vyitg2vinisg2vinsoxis(vgxivp)2vinis(vCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.gicosvp)(2)(22.0m/s)(sin35.0°)29.80m/s((22.0m/s)(cos35.0°)2.5m/s)53mor4.0101m82.AJewelHeistYouareservingasatechnicalconsultantforalocallyproducedcartoon.Inoneepisode,twocrimi-nals,ShiftyandLefty,havestolensomejewels.Leftyhasthejewelswhenthepolicestarttochasehim,andherunstothetopofa60.0-mtallbuildinginhisattempttoescape.Meanwhile,Shiftyrunstotheconvenienthot-airballoon20.0mfromthebaseofthebuildingandunteth-ersit,soitbeginstoriseataconstantspeed.Leftytossesthebagofjewelshorizontallywithaspeedof7.3m/sjustastheballoon136SolutionsManualPhysics:PrinciplesandProblems

140Chapter6continuedbeginsitsascent.WhatmustthevelocityoftheballoonbeforShiftytoeasilycatchthebag?xxvxit,sotvxiy12,butvbagvyit2gtyi012soybag2gtyvballoonballoont60.0mybagt1260.0mgt2t60.0m1gtt260.0m1xgx2vxivxi(60.0m)vxigxx2vxi(60.0m)(7.3m/s)(20.0m)(9.80m/s)(20.0m)2(7.3m/s)8.5m/sThinkingCriticallypage16883.ApplyConceptsConsideraroller-coasterloopliketheoneinFigure6-18.Arethecarstravelingthroughtheloopinuniformcircularmotion?Explain.■Figure6-18Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Theverticalgravitationalforcechangesthespeedofthecars,sothemotionisnotuniformcircularmotion.84.UseNumbersA3-pointjumpshotisreleased2.2mabovethegroundand6.02mfromthebasket.Thebasketis3.05mabovethefloor.Forlaunchanglesof30.0°and60.0°,findthespeedtheballneedstobethrowntomakethebasket.Physics:PrinciplesandProblemsSolutionsManual137

141Chapter6continuedxxa.Isthehitahomerun?xvixt,sotvvosixicYes,thehitisahomerun;theball12clearsthewallby2.1m.yviyt2gtb.Whatistheminimumspeedatwhichx1x2theballcouldbehitandclearthewall?visinvos2gvosicicgxvgx2icos2((tan)xy)xtan2v2(cos)2i96.0xgmcos29.80m/s2sovicos2((tan)xy)6°(2)((tan26°)(96.0m)13m)For30.0°41m/s6.02c.Iftheinitialvelocityoftheballis42.0vmicos3m/s,forwhatrangeofangleswillthe9.80m/s20.0°ballgooverthewall?.02m)(3.05m2.2m))Fortheballtogooverthewall,the9.5m/srangeofanglesneedstobe25°70°.For60.0°87.AnalyzeAlbertEinsteinshowedthatthe6.02vicos62ruleyoulearnedfortheadditionofveloci-m9.80m/s0.0°2((tan60°(6.02m)(3.05m2.2m))tiesdoesnotworkforobjectsmovingnearthespeedoflight.Forexample,ifarocket8.6m/smovingatvelocityvAreleasesamissilethathasvelocityv85.AnalyzeForwhichangleinproblem84isBrelativetotherocket,thenthevelocityofthemissilerelativetoanitmoreimportantthattheplayerobserverthatisatrestisgivenbyv(vgetthespeedright?Toexplorethisques-Av2),wherecisthespeedoftion,varythespeedateachanglebyB)/(1vAvB/c8m/s.Thisformulagivesthelight,3.0010Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5percentandfindthechangeintherangecorrectvaluesforobjectsmovingatslowoftheattemptedshot.speedsaswell.SupposearocketmovingatVaryingspeedby5percentat30.0°11km/sshootsalaserbeamoutinfrontofchangesRbyabout0.90mineitherit.Whatspeedwouldanunmovingobserv-direction.At60.0°itchangesRbyonlyerfindforthelaserlight?Supposethataabout0.65m.Thus,thehighlaunchrocketmovesataspeedc/2,halfthespeedangleislesssensitivetospeedvaria-oflight,andshootsamissileforwardatations.speedofc/2relativetotherocket.Howfastwouldthemissilebemovingrelativetoa86.ApplyComputersandCalculatorsAbase-fixedobserver?ballplayerhitsabelt-high(1.0m)fastball(vdowntheleft-fieldline.Theballishitwithanvr/ovl/r)l/ovr/ov1/rinitialvelocityof42.0m/sat26°.Theleft-field1c2wallis96.0mfromhomeplateatthefoulpoleandis14-mhigh.Writetheequationfor1.1104m/s3.00108m/stheheightoftheball,y,asafunctionofits(1.1104m/s)(3.00108m/s)1distancefromhomeplate,x.Useacomputer(3.00108m/s)2orgraphingcalculatortoplotthepathofthe8m/s3.010ball.Tracealongthepathtofindhowhighvvabovethegroundtheballiswhenitisatther/om/rvm/ovvwall.1r/om/rc2138SolutionsManualPhysics:PrinciplesandProblems

142Chapter6continuedccisoutwardwhilethetensionisinward.22Thus,thetensionexertedbythestringccmustbeevenlarger.2212cWritinginPhysics4cpage168589.RollerCoastersIfyoutakealookatverti-88.AnalyzeandConcludeAballonalightcalloopsonrollercoasters,youwillnoticestringmovesinaverticalcircle.Analyzeandthatmostofthemarenotcircularinshape.describethemotionofthissystem.BesureResearchwhythisissoandexplainthetoconsidertheeffectsofgravityandten-physicsbehindthisdecisionbythecoastersion.Isthissysteminuniformcircularengineers.motion?Explainyouranswer.mv2Answerswillvary.SinceF,asvcrItisnotuniformcircularmotion.decreasesduetogravitywhengoingGravityincreasesthespeedoftheballuphill,risreducedtokeeptheforcewhenitmovesdownwardandreducesconstant.thespeedwhenitismovingupward.Therefore,thecentripetalacceleration90.Manyamusement-parkridesutilizecen-neededtokeepitmovinginacircletripetalaccelerationtocreatethrillsforwillbelargeratthebottomandsmall-thepark’scustomers.Choosetworideseratthetopofthecircle.Atthetop,otherthanrollercoastersthatinvolvetensionandgravityareinthesamecircularmotionandexplainhowthedirection,sothetensionneededwillphysicsofcircularmotioncreatesthebeevensmaller.Atthebottom,gravitysensationsfortheriders.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual139

143

144CHAPTER7GravitationPracticeProblems7.1PlanetaryMotionandGravitationpages171–178page1741.IfGanymede,oneofJupiter’smoons,hasaperiodof32days,howmanyunitsarethereinitsorbitalradius?UsetheinformationgiveninExampleProblem1.T2r3GGTrII3332days2rG(4.2units)1.8days33un323.410its29units2.AnasteroidrevolvesaroundtheSunwithameanorbitalradiustwicethatofEarth’s.PredicttheperiodoftheasteroidinEarthyears.T2r3aaTrwithra2rEEEr3TaT2arEE2r3E(1.0y)2rE2.8y3.FromTable7-1,onpage173,youcanfindthat,onaverage,Marsis1.52timesasfarfromtheSunasEarthis.PredictthetimerequiredforMarstoorbittheSuninEarthdays.T2r3MMTrwithrM1.52rEEEr31.52r3Thus,TMT2E(365days)2MrErEE4.68105days2684days4.TheMoonhasaperiodof27.3daysandameandistanceof3.90105kmfromCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.thecenterofEarth.a.UseKepler’slawstofindtheperiodofasatelliteinorbit6.70103kmfromthecenterofEarth.T2r3ssTrMMPhysics:PrinciplesandProblemsSolutionsManual141

145Chapter7continuedr3TsT2srMM6.70103km32(27.3days)3.90105km3.78103days26.15102days88.6minb.HowfaraboveEarth’ssurfaceisthissatellite?hrsrE6.70106m6.38106m3.2105m3.2102km5.UsingthedatainthepreviousproblemfortheperiodandradiusofrevolutionoftheMoon,predictwhatthemeandistancefromEarth'scenterwouldbeforanartificialsatellitethathasaperiodofexactly1.00day.T2r3ssTrMM33Ts235km31.00days2rsrMT(3.9010)27.3daysM313k37.9610m4.30104kmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.SectionReview7.1PlanetaryMotionandGravitationpages171–178page1786.Neptune’sOrbitalPeriodNeptuneorbitstheSunwithanorbitalradiusof4.4951012m,whichallowsgases,suchasmethane,tocondenseandformanatmosphere,asshowninFigure7-8.IfthemassoftheSunis1.991030kg,calculatetheperiodofNeptune’sorbit.r3T2GmS(4.4951012m)3■Figure7-82(6.671011Nm2/kg2)(1.991030kg)5.20109s6.02105days7.GravityIfEarthbegantoshrink,butitsmassremainedthesame,whatwouldhappentothevalueofgonEarth’ssurface?Thevalueofgwouldincrease.142SolutionsManualPhysics:PrinciplesandProblems

146Chapter7continued8.GravitationalForceWhatisthegravitationalforcebetweentwo15-kgpackagesthatare35cmapart?Whatfractionisthisoftheweightofonepackage?mmFEgGr2(6.671011Nm2kg2)(15kg)2(0.35m)21.2107NBecausetheweightismg147N,thegravitationalforceis8.21010or0.82partsperbillionoftheweight.9.UniversalGravitationalConstantCavendishdidhisexperimentusingleadspheres.Supposehehadreplacedtheleadsphereswithcopperspheresofequalmass.WouldhisvalueofGbethesameordifferent?Explain.Itwouldbethesame,becausethesamevalueofGhasbeenusedsuccessfullytodescribetheattractionofbodieshavingdiversechemicalcompositions:theSun(astar),theplanets,andsatellites.10.LawsorTheories?Kepler’sthreestatementsandNewton’sequationforgravitationalattractionarecalled“laws.”Weretheyevertheories?Willtheyeverbecometheories?No.Ascientificlawisastatementofwhathasbeenobservedtohappenmanytimes.Atheoryexplainsscientificresults.Noneofthesestatementsoffersexplanationsforwhythemotionofplanetsareastheyareorforwhygravitationalattractionactsasitdoes.11.CriticalThinkingPickinguparockrequireslesseffortontheMoonthanonEarth.a.HowwilltheweakergravitationalforceontheMoon’ssurfaceaffectthepathoftherockifitisthrownhorizontally?Horizontalthrowingrequiresthesameeffortbecausetheinertialcharacter,Fma,oftherockisinvolved.Themassoftherockdependsonlyontheamountofmatterintherock,notonitslocationintheuniverse.Thepathwouldstillbeaparabola,butitcouldbemuchwiderbecausetherockwouldgofartherbeforeithitstheground,giventhesmalleraccelerationrateandlongertimeofflight.b.Ifthethroweraccidentallydropstherockonhertoe,willithurtmoreorlessthanitwouldonEarth?Explain.AssumetherockswouldbedroppedfromthesameheightonEarthandontheMoon.ItwillhurtlessbecausethesmallervalueofgontheMoonmeansthattherockstrikesthetoewithasmallervelocitythanonEarth.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual143

147Chapter7continuedPracticeProblems7.2UsingtheLawofUniversalofGravitationpages179–185page181Forthefollowingproblems,assumeacircularorbitforallcalculations.12.SupposethatthesatelliteinExampleProblem2ismovedtoanorbitthatis24kmlargerinradiusthanitspreviousorbit.Whatwoulditsspeedbe?Isthisfasterorslowerthanitspreviousspeed?r(h2.40104m)rE(2.25105m2.40104m)6.38106m6.63106mGmvEr(6.671011Nm2/kg2)(5.971024kg)6.63106m7.75103m/s,slower13.UseNewton’sthoughtexperimentonthemotionofsatellitestosolvethefollowing.a.CalculatethespeedthatasatelliteshotfromacannonmusthavetoorbitEarth150kmaboveitssurface.GmE(6.671011Nm2/kg2)(5.971024kg)vr(6.38106m1.5105m)7.8103m/sb.Howlong,insecondsandminutes,wouldittakeforthesatellitetocompleteoneorbitandreturntothecannon?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.r3(6.38106m1.5105m)3T22GmE(6.671011Nm2/kg2)(5.971024kg)5.3103s88min14.UsethedataforMercuryinTable7-1onpage173tofindthefollowing.a.thespeedofasatellitethatisinorbit260kmaboveMercury’ssurfaceGmvMrrrM260km2.44106m0.26106m2.70106m(6.671011Nm2/kg2)(3.301023kg)v2.70106m2.86103m/sb.theperiodofthesatelliter3(2.70106m)3T22GmM(6.671011Nm2/kg2)(3.301023kg)5.94103s1.65h144SolutionsManualPhysics:PrinciplesandProblems

148Chapter7continuedSectionReview7.2UsingtheLawofUniversalofGravitationpages179–185page18515.GravitationalFieldsTheMoonis3.9105kmfromEarth’scenterand1.5108kmfromtheSun’scenter.ThemassesofEarthandtheSunare6.01024kgand2.01030kg,respectively.a.FindtheratioofthegravitationalfieldsduetoEarthandtheSunatthecenteroftheMoon.mGravitationalfieldduetotheSun:gSSGr2SmGravitationalfieldduetoEarth:gEEGr2Egmr2SSEgmr2EES(2.01030kg)(3.9105km)2(6.01024kg)(1.5108km)22.3b.WhentheMoonisinitsthirdquarterphase,asshowninFigure7-17,itsdirectionfromEarthisatrightanglestotheSun’sdirection.WhatisthenetgravitationalfieldduetotheSunandEarthatthecenteroftheMoon?MoonSunEarth■Figure7-17Becausethedirectionsareatrightangles,thenetfieldisthesquarerootofthesumofthesquaresofthetwofields.GmgSSr2(6.671011Nm2kg2)(2.01030kg)(1.51011m)25.9103N/kgSimilarly,g3N/kgE2.610gnet(5.9103N/kg)2(2.6103N/kg)26.4103N/kgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.16.GravitationalFieldThemassoftheMoonis7.31022kganditsradiusis1785km.WhatisthestrengthofthegravitationalfieldonthesurfaceoftheMoon?GM(6.671011Nm2kg2)(7.31022kg)gr2(1.785103m)21.5N/kg,aboutone-sixththatonEarthPhysics:PrinciplesandProblemsSolutionsManual145

149Chapter7continued17.ASatellite’sMassWhenthefirstartificialsatellitewaslaunchedintoorbitbytheformerSovietUnionin1957,U.S.presidentDwightD.Eisenhoweraskedhisscientificadvisorstocalculatethemassofthesatellite.Wouldtheyhavebeenabletomakethiscalculation?Explain.No.Becausethespeedandperiodoftheorbitdon’tdependatallonthemassofthesatellite,thescientificadvisorswouldnothavebeenabletocalculatethemassofthesatellite.18.OrbitalPeriodandSpeedTwosatellitesareincircularorbitsaboutEarth.Oneis150kmabovethesurface,theother160km.a.Whichsatellitehasthelargerorbitalperiod?Whentheorbitalradiusislarge,theperiodalsowillbelarge.Thus,theoneat160kmwillhavethelargerperiod.b.Whichonehasthegreaterspeed?Theoneat150km,becausethesmallertheorbitalradius,thegreaterthespeed.19.TheoriesandLawsWhyisEinstein’sdescriptionofgravitycalleda“theory,”whileNewton’sisa“law?”Newton’slawdescribeshowtocalculatetheforcebetweentwomassiveobjects.Einstein’stheoryexplainshowanobject,suchasEarth,attractstheMoon.20.WeightlessnessChairsinanorbitingspacecraftareweightless.Ifyouwereonboardsuchaspacecraftandyouwerebarefoot,wouldyoustubyourtoeifyoukickedachair?Explain.Yes.Thechairsareweightlessbutnotmassless.TheystillhaveinertiaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.andcanexertcontactforcesonyourtoe.21.CriticalThinkingItiseasiertolaunchasatellitefromEarthintoanorbitthatcircleseastwardthanitistolaunchonethatcircleswestward.Explain.Earthrotatestowardtheeast,anditsvelocityaddstothevelocitygiventothesatellitebytherocket,therebyreducingthevelocitythattherocketmustsupply.ChapterAssessmentConceptMappingpage19022.Createaconceptmapusingtheseterms:planets,stars,Newton’slawofuniversalgravitation,Kepler’sfirstlaw,Kepler’ssecondlaw,Kepler’sthirdlaw,Einstein’sgeneraltheoryofrelativity.146SolutionsManualPhysics:PrinciplesandProblems

150Chapter7continuedEinstein’sgeneraltheoryofrelativityExplainsNewton’slawofuniversalgravitationSupportsKepler’sfirstKepler’ssecondKepler’sthirdlawlawlawRelatesperiodsandDescribesmotionofradiiofplanetsmovingaroundstarsKepler’sfirstandsecondlawsdescribethemotionofasingleplanet.Kepler’sthirdlawdescribestheperiodsversustheorbitalradiiofallplanetsaroundastar.Newton’slawofuniversalgravitationsupportsKepler’slaws.Einstein’stheoryexplainsNewton’sandKepler’slaws.MasteringConceptspage19023.In1609,GalileolookedthroughhistelescopeatJupiterandsawfourmoons.ThenameofoneofthemoonsthathesawisIo.RestateKepler’sfirstlawforIoandJupiter.(7.1)ThepathofIoisanellipse,withJupiteratonefocus.24.EarthmovesmoreslowlyinitsorbitduringsummerinthenorthernhemisphereCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.thanitdoesduringwinter.IsitclosertotheSuninsummerorinwinter?(7.1)BecauseEarthmovesmoreslowlyinitsorbitduringsummer,byKepler’ssecondlaw,itmustbefartherfromtheSun.Therefore,EarthisclosertotheSuninthewintermonths.Physics:PrinciplesandProblemsSolutionsManual147

151Chapter7continued25.Istheareasweptoutperunitoftimeby32.WhatprovidestheforcethatcausestheEarthmovingaroundtheSunequaltocentripetalaccelerationofasatelliteintheareasweptoutperunitoftimebyMarsorbit?(7.2)movingaroundtheSun?(7.1)gravitationalattractiontothecentralNo.Theequalityoftheareasweptoutbodyperunitoftimeappliestoeachplanetindividually.33.Duringspaceflight,astronautsoftenrefertoforcesasmultiplesoftheforceofgravityon26.WhydidNewtonthinkthataforcemustactEarth’ssurface.Whatdoesaforceof5gontheMoon?(7.1)meantoanastronaut?(7.2)NewtonknewthattheMoonfollowedAforceof5gmeansthatanastronaut’sacurvedpath;therefore,itwasweightisfivetimesheavierthanitisonaccelerated.HealsoknewthataEarth.Theforceexertedontheastro-forceisrequiredforacceleration.nautisfivetimestheforceofEarth’sgravitationalforce.27.HowdidCavendishdemonstratethatagravitationalforceofattractionexists34.Newtonassumedthatagravitationalforcebetweentwosmallobjects?(7.1)actsdirectlybetweenEarthandtheMoon.Hecarefullymeasuredthemasses,theHowdoesEinstein’sviewoftheattractivedistancebetweenthemasses,andtheforcebetweenthetwobodiesdifferfromforceofattraction.HethencalculatedNewton’sview?(7.2)GusingNewton’slawofuniversalgrav-Einstein’sviewisthatgravityisanitation.effectofthecurvatureofspaceasaresultofthepresenceofmass,whereas28.WhathappenstothegravitationalforceNewton’sviewofgravityisthatitisabetweentwomasseswhenthedistanceforceactingdirectlybetweenobjects.betweenthemassesisdoubled?(7.1)Thus,accordingtoEinstein,theAccordingtoNewton,F1/r2.IftheattractionbetweenEarthandtheMoonCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.distanceisdoubled,theforceiscuttoistheeffectofcurvatureofspaceone-fourth.causedbytheircombinedmasses.29.AccordingtoNewton’sversionofKepler’s35.Showthatthedimensionsofginthethirdlaw,howwouldtheratioT2/r3changeequationgF/mareinm/s2.(7.2)ifthemassoftheSunweredoubled?(7.1)2FNkgm/s2BecauseT2/r342/GmTheunitsofarem/sS,ifthemassmkgkgoftheSun,mS,isdoubled,theratiowillbehalved.36.IfEarthweretwiceasmassivebutremainedthesamesize,whatwouldhappentothe30.Howdoyouanswerthequestion,“Whatvalueofg?(7.2)keepsasatelliteup?”(7.2)Thevalueofgwoulddouble.Itsspeed;itisfallingallthetime.ApplyingConcepts31.AsatelliteisorbitingEarth.Onwhichofpages190–191thefollowingdoesitsspeeddepend?(7.2)37.GolfBallTheforceofgravityactingonana.massofthesatelliteobjectnearEarth’ssurfaceisproportionalb.distancefromEarthtothemassoftheobject.Figure7-18c.massofEarthshowsatennisballandgolfballinfreefall.WhydoesatennisballnotfallfasterthanaSpeeddependsonlyonb,thedistancegolfball?fromtheEarth,andc,themassofEarth.148SolutionsManualPhysics:PrinciplesandProblems

152Chapter7continued40.DecidewhethereachoftheorbitsshowninFigure7-19isapossibleorbitforaplanet.ccSunSunccSunSun■Figure7-18mFG1m2r2■Figure7-19m1Earth’smassOnlyd(lowerright)ispossible.Fa(topleft)andb(topright)donotam2havetheSunatafocus,andinc(lowerleft),theplanetisnotinm2object’smassorbitaroundtheSun.GmThus,a1r241.TheMoonandEarthareattractedtoeachTheaccelerationisindependentoftheotherbygravitationalforce.Doesthemore-object’smass.ThisisbecausemoremassiveEarthattracttheMoonwithamassiveobjectsrequiremoreforcetogreaterforcethantheMoonattractsEarth?accelerateatthesamerate.Explain.No.Theforcesconstituteanaction-38.Whatinformationdoyouneedtofindthereactionpair,sounderNewton’sthirdmassofJupiterusingNewton’sversionoflaw,theyareequalandopposite.Kepler’sthirdlaw?Youmustknowtheperiodandorbital42.WhatwouldhappentothevalueofGifradiusofatleastoneofitssatellites.Earthweretwiceasmassive,butremainedthesamesize?39.ThemassofPlutowasnotknownuntilaNothing.Gisauniversalconstant,andsatelliteoftheplanetwasdiscovered.Why?itisindependentofEarth’smass.OrbitalmotionofaplanetdoesnotHowever,theforceofattractionwoulddependonitsmassandcannotbedouble.usedtofindthemass.Asatelliteorbitingtheplanetisnecessaryto43.Figure7-20showsasatelliteorbitingEarth.findtheplanet’smass.GmEExaminetheequationv,relatingrthespeedofanorbitingsatelliteanditsdistancefromthecenterofEarth.DoesaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.satellitewithalargeorsmallorbitalradiushavethegreatervelocity?Physics:PrinciplesandProblemsSolutionsManual149

153Chapter7continued300g3.Thus,gonJupiteristhree102SatellitetimesthatonEarth.47.AsatelliteisoneEarthradiusabovethesur-rfaceofEarth.HowdoestheaccelerationduetogravityatthatlocationcomparetoaccelerationatthesurfaceofEarth?EarthdrErE2rErE21so,agg2r4E■Figure7-20(Nottoscale)Asatellitewithasmallradiushasthe48.IfamassinEarth’sgravitationalfieldisgreatervelocity.doubled,whatwillhappentotheforceexertedbythefielduponthemass?44.SpaceShuttleIfaspaceshuttlegoesintoaItalsowilldouble.higherorbit,whathappenstotheshuttle’speriod?49.WeightSupposethatyesterdayyourbodyr3hadamassof50.0kg.ThismorningyouBecauseT2,iftheorbitalGmsteppedonascaleandfoundthatyouhadradiusincreases,sowilltheperiod.gainedweight.a.Whathappened,ifanything,toyour45.Marshasaboutone-ninththemassofmass?Earth.Figure7-21showssatelliteM,whichorbitsMarswiththesameorbitalradiusYourmassincreased.assatelliteE,whichorbitsEarth.Whichb.Whathappened,ifanything,totheratiosatellitehasasmallerperiod?ofyourweighttoyourmass?TheratioremainedconstantCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.becauseitisequaltothegravita-tionalfieldatthelocation,acon-MarsEarthstantg.rMrE50.Asanastronautinanorbitingspaceshuttle,EMhowwouldyougoabout“dropping”anobjectdowntoEarth?■Figure7-21(Nottoscale)To“drop”anobjectdowntoEarth,your3BecauseT2,asthemassofwouldhavetolaunchitbackwardattheGmsamespeedatwhichyouaretravelingtheplanetincreases,theperiodoftheinorbit.WithrespecttoEarth,thesatellitedecreases.BecauseEarthobject’sspeedperpendiculartoEarth’shasalargermassthanMars,Earth’sgravitywouldbezero,anditcouldthensatellitewillhaveasmallerperiod.“drop”downtoEarth.However,theobjectislikelytoburnupasaresultof46.Jupiterhasabout300timesthemassoffrictionwithEarth’satmosphereontheEarthandabouttentimesEarth’sradius.waydown.EstimatethesizeofgonthesurfaceofJupiter.51.WeatherSatellitesTheweatherpicturesmgE.IfJupiterhas300timesthethatyouseeeverydayonTVcomefromar2EspacecraftinastationarypositionrelativetomassandtentimestheradiusofEarth,150SolutionsManualPhysics:PrinciplesandProblems

154Chapter7continuedthesurfaceofEarth,35,700kmabovemmFGSJEarth’sequator.Explainhowitcanstayinr2exactlythesamepositiondayafterday.(6.671011Nm2/kg2)Whatwouldhappenifitwerecloser?Fartherout?Hint:Drawapictorialmodel.(5.9kg)(4.7102kg)(5.5102m)2Thesatelliteispositionedasclosetotheequatoraspossiblesoitdoesn’t6.1109Nappeartohavemuchnorth-southmove-ment.Becauseitisplacedatthatdis-54.UseTable7-1onp.173tocomputethetance,thesatellitehasaperiodof24.0gravitationalforcethattheSunexertsonh.Ifitwerepositionedanycloser,itsJupiter.periodwouldbelessthan24.0handitm1m2FGwouldappeartomovetowardtheeast.r2Ifitwerepositionedanyfarther,itsperi-(6.671011Nm2/kg2)odwouldbelongerthan24.0h.(1.991030kg)(1.901027kg)MasteringProblems(7.781011m)27.1PlanetaryMotionandGravitation4.171023Npages191–19255.Tomhasamassof70.0kgandSallyhasaLevel1massof50.0kg.TomandSallyarestanding52.Jupiteris5.2timesfartherfromtheSun20.0mapartonthedancefloor.SallylooksthanEarthis.FindJupiter’sorbitalperiodupandseesTom.Shefeelsanattraction.IfinEarthyears.theattractionisgravitational,finditssize.TJ2rJ3AssumethatbothTomandSallycanbeTErEreplacedbysphericalmasses.mr3FGTmSTJT22JrErE(6.671011Nm2/kg2)5.23(1.0y)2(70.0kg)(50.0kg)1.0(20.0m)2141y210N5.841012y56.Twoballshavetheircenters2.0mapart,as53.Figure7-22showsaCavendishapparatusshowninFigure7-23.OneballhasamassliketheoneusedtofindG.Ithasalargeof8.0kg.Theotherhasamassof6.0kg.leadspherethatis5.9kginmassandaWhatisthegravitationalforcebetweenthem?smallonewithamassof0.047kg.Theircentersareseparatedby0.055m.Findtheforceofattractionbetweenthem.2.0m8.0kg6.0kg5.9kg■Figure7-230.047kgmFG1m2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.r20.055m0.055m(6.671011Nm2/kg2)(8.0kg)(6.0kg)0.047kg(2.0m)25.9kg8.01010N■Figure7-22Physics:PrinciplesandProblemsSolutionsManual151

155Chapter7continued57.TwobowlingballseachhaveamassofFr2m6.8kg.TheyarelocatednexttoeachotherEGmwiththeircenters21.8cmapart.Whatgrav-(9.8N)(6.4106m)2itationalforcedotheyexertoneachother?(6.671011Nm2kg2)(1.0kg)mFG1m26.01024kgr2b.CalculatetheaveragedensityofEarth.(6.671011Nm2/kg2)(6.8kg)(6.8kg)43(4)(6.4106m)3(0.218m)2Vr336.5108N21m31.11058.Assumethatyouhaveamassof50.0kg.m6.01024kgD24kgandaV1.11021m3Earthhasamassof5.9710radiusof6.38106m.5.5103kg/m3a.Whatistheforceofgravitationalattrac-61.UranusUranusrequires84yearstocircletionbetweenyouandEarth?theSun.FindUranus’sorbitalradiusasaSampleanswer:multipleofEarth’sorbitalradius.mSmET2r3FG2UUrTrEE(6.671011Nm2/kg2)rU3TU2(50.0kg)(5.971024kg)rTEE(6.38106m)2384y2489N1.0yb.Whatisyourweight?19Sampleanswer:SorF2)U19rEgmg(50.0kg)(9.80m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4.90102N62.VenusVenushasaperiodofrevolutionof225Earthdays.Findthedistancebetween59.ThegravitationalforcebetweentwoelectronstheSunandVenusasamultipleofEarth’sthatare1.00mapartis5.541071N.Findorbitalradius.themassofanelectron.T2r3VVFGm1m2,wheremTErEr21m2merV3TV2Fr2SomerETEG3225days2(5.541071N)(1.00m)2365days6.671011Nm2/kg29.111031kg0.724SorV0.724rE60.A1.0-kgmassweighs9.8NonEarth’ssur-face,andtheradiusofEarthisroughlyLevel26.4106m.63.Ifasmallplanet,D,werelocated8.0timesasfarfromtheSunasEarthis,howmanyyearsa.CalculatethemassofEarth.wouldittaketheplanettoorbittheSun?mFG1m2T2r3r2DDTrEE152SolutionsManualPhysics:PrinciplesandProblems

156Chapter7continuedrD328.032TT(1.0y)23yearsDrE1.0E64.Twospheresareplacedsothattheircentersare2.6mapart.Theforcebetweenthetwospheresis2.751012N.Whatisthemassofeachsphereifonesphereistwicethemassoftheothersphere?mFG1m2,wheremr222m1(m1)(2m1)FG2rFr2m12G(2.751012N)(2.6m)2(2)(6.671011Nm2kg2)0.3733kgor0.37kgtotwosignificantdigitsm22m1(2)(0.3733kg)0.7466kgor0.75totwosignificantdigits65.TheMoonis3.9105kmfromEarth’scenterand1.5108kmfromtheSun’scenter.IfthemassesoftheMoon,Earth,andtheSunare7.31022kg,6.01024kg,and2.01030kg,respectively,findtheratioofthegravitationalforcesexertedbyEarthandtheSunontheMoon.mFG1m2r211Nm2/kg2)(6.01024kg)(7.31022kg)EarthonMoon:FE(6.6710(3.9108m)21.91020N11Nm2/kg2)(2.01030kg)(7.31022kg)SunonMoon:FE(6.6710(1.51011m)24.31020NFE1.91020N1.0RatioisFS4.31020N2.3TheSunpullsmorethantwiceashardontheMoonasdoesEarth.66.ToyBoatAforceof40.0Nisrequiredtopulla10.0-kgwoodentoyboatataconstantvelocityacrossasmoothglasssurfaceonEarth.WhatforcewouldberequiredtopullthesamewoodentoyboatacrossthesameglasssurfaceontheplanetJupiter?FFff,wheremFmgbisthemassofthetoyboat.NbOnJupiter,thenormalforceisequaltothegravitationalattractionCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.betweenthetoyboatandJupiter,ormmFbJNGr2JFmmNowf,soFbJFfJFNGr2NJPhysics:PrinciplesandProblemsSolutionsManual153

157Chapter7continuedFButf,soFmgfJbmbmJmJ(40.0N)(6.671011Nm2/kg2)(1.901027kg)FFfGmgr2fGgr2(9.80m/s2)(7.15107m)2bJJ101N67.Mimas,oneofSaturn’smoons,hasanorbitalradiusof1.87108mandanorbitalperiodofabout23.0h.UseNewton’sversionofKepler’sthirdlawtofindSaturn’smass.2423TrGm42r3mGT242(1.87108m)3(6.671011Nm2/kg2)(8.28104s)25.61026kgLevel368.TheMoonis3.9108mawayfromEarthandhasaperiodof27.33days.UseNewton’sversionofKepler’sthirdlawtofindthemassofEarth.Comparethismasstothemassfoundinproblem60.2423TrGm42r3m2GT42(3.9108m)3(6.671011Nm2/kg2)(2.361106s)2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6.31024kgThemassisconsiderablyclosetothatfoundinproblem60.69.Halley’sCometEvery74years,cometHalleyisvisiblefromEarth.FindtheaveragedistanceofthecometfromtheSuninastronomicalunits(AU).ForEarth,r1.0AUandT1.0yr3T2aarTbb3Ta2374y2rr3(1.0AU)3abT1.0yb18AU70.Areaismeasuredinm2,sotherateatwhichareaissweptoutbyaplanetorsatelliteismeasuredinm2/s.a.HowquicklyisanareasweptoutbyEarthinitsorbitabouttheSun?r1.501011mandT3.156107s,in365.25days1.00yr2(1.501011m)215m2/s2.2410T3.156107s154SolutionsManualPhysics:PrinciplesandProblems

158Chapter7continuedb.HowquicklyisanareasweptoutbytheMooninitsorbitaboutEarth?Use3.9108mastheaveragedistancebetweenEarthandtheMoon,and27.33daysastheperiodoftheMoon.(3.9108m)211m2/s2.0102.36106s7.2UsingtheLawofUniversalGravitationpages192–193Level171.SatelliteAgeosynchronoussatelliteisonethatappearstoremainoveronespotonEarth,asshowninFigure7-24.Assumethatageosynchronoussatellitehasanorbitalradiusof4.23107m.SatelliterEarth■Figure7-24(Nottoscale)a.Calculateitsspeedinorbit.GmvEr(6.671011Nm2/kg2)(5.97x1024kg)4.23107m3.07103m/sor3.07km/sb.Calculateitsperiod.r3T2GmE(4.23107m)32(6.671011Nm2/kg2)(5.971024kg)21.90108s28.66104sor24.1h72.AsteroidTheasteroidCereshasamassof71020kgandaradiusof500km.a.WhatisgonthesurfaceofCeres?Gmgr2(6.671011Nm2/kg2)(71020kg)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(500103m)20.2m/s2b.Howmuchwoulda90-kgastronautweighonCeres?F2)20Ngmg(90kg)(0.2m/sPhysics:PrinciplesandProblemsSolutionsManual155

159Chapter7continued73.BookA1.25-kgbookinspacehasaweightof8.35N.Whatisthevalueofthegravitationalfieldatthatlocation?F8.35Ng6.68N/kgm1.25kg74.TheMoon’smassis7.341022kg,anditis3.8108mawayfromEarth.Earth’smassis5.971024kg.a.CalculatethegravitationalforceofattractionbetweenEarthandtheMoon.mmFGEMr211Nm2/kg2)(5.981024kg)(7.341022kg)(6.6710(3.8108m)22.01020Nb.FindEarth’sgravitationalfieldattheMoon.F2.031020Ng0.0028N/kgm7.341022kgNotethat2.031020Ninsteadof2.01020Nisusedtopreventroundofferror.75.Two1.00-kgmasseshavetheircenters1.00mapart.Whatistheforceofattractionbetweenthem?mF1m2gGr2(6.671011Nm2/kg2)(1.00kg)(1.00kg)(1.00m)26.671011NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Level276.TheradiusofEarthisabout6.38103km.A7.20103-NspacecrafttravelsawayfromEarth.WhatistheweightofthespacecraftatthefollowingdistancesfromEarth’ssurface?a.6.38103kmdrErE2rETherefore,F1originalweight13N)1.80103Ng44(7.2010b.1.28104kmdrE2rE3rEF13N)8.00102Ng9(7.201077.RocketHowhighdoesarockethavetogoaboveEarth’ssurfacebeforeitsweightishalfofwhatitisonEarth?1Fgr21SorFg156SolutionsManualPhysics:PrinciplesandProblems

160Chapter7continued1Iftheweightis,thedistanceis2(r2E)orr2(6.38106m)9.02106m9.02106m6.38106m2.64106m2.64103km78.Twosatellitesofequalmassareputintoorbit30.0mapart.Thegravitationalforcebetweenthemis2.0107N.a.Whatisthemassofeachsatellite?mFG1m2,mr21m2mFr2(2.0107N)(30.0m)2mG6.671011Nm2kg21.6103kgb.Whatistheinitialaccelerationgiventoeachsatellitebygravitationalforce?FnetmaFnet2.0107N10m/s2a31.310m1.610kg79.Twolargespheresaresuspendedclosetoeachother.Theircentersare4.0mapart,asshowninFigure7-25.Onesphereweighs9.8102N.Theotherspherehasaweightof1.96102N.Whatisthegravitationalforcebetweenthem?4.0m9.8102N1.96102N■Figure7-25Fg9.8102N2kgm1.0101g9.80m/s2Fg1.96102N1kgm2.00102g9.80m/s2mFG1m2r2(6.671011Nm2/kg2)(1.0101kg)(2.00101kg)(4.0m)2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8.3109NPhysics:PrinciplesandProblemsSolutionsManual157

161Chapter7continued80.SupposethecentersofEarthandtheMoonare3.9108mapart,andthegravitationalforcebetweenthemisabout1.91020N.WhatistheapproximatemassoftheMoon?mmFGEMr2Fr2mMGmE(1.91020N)(3.9108m)2(6.671011Nm/kg2)(5.971024kg)7.31022kg81.OnthesurfaceoftheMoon,a91.0-kgphysicsteacherweighsonly145.6N.WhatisthevalueoftheMoon’sgravitationalfieldatitssurface?Fgmg,Fg145.6NSog1.60N/kgm91.0kgLevel382.Themassofanelectronis9.11031kg.Themassofaprotonis1.71027kg.Anelectronandaprotonareabout0.591010mapartinahydrogenatom.Whatgravitationalforceexistsbetweentheprotonandtheelectronofahydrogenatom?mempFG2r(6.671011Nm2/kg2)(9.11031kg)(1.71027kg)(1.01010m)2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.01047N83.Considertwospherical8.0-kgobjectsthatare5.0mapart.a.Whatisthegravitationalforcebetweenthetwoobjects?mFG1m2r2(6.671011Nm2/kg2)(8.0kg)(8.0kg)(5.0m)21.71010Nb.Whatisthegravitationalforcebetweenthemwhentheyare5.0101mapart?mFG1m2r2(6.671011Nm2/kg2)(8.0kg)(8.0kg)(5.0101m)21.71012N158SolutionsManualPhysics:PrinciplesandProblems

162Chapter7continued84.Ifyouweigh637NonEarth’ssurface,howmuchwouldyouweighontheplanetMars?Marshasamassof6.421023kgandaradiusof3.40106m.Fg637Nm65.0kgg9.80m/s2mFG1m2r2(6.671011Nm2/kg2)(65.0kg)(6.371023kg)(3.43106m)2235N85.UsingNewton’sversionofKepler’sthirdlawandinformationfromTable7-1onpage173,calculatetheperiodofEarth’sMooniftheorbitalradiusweretwicetheactualvalueof3.9108m.423)T(rMGmE428m3(7.810)(6.671011Nm2/kg2)(5.9791024kg)6.85106sor79days86.Findthevalueofg,accelerationduetogravity,inthefollowingsituations.a.Earth’smassistripleitsactualvalue,butitsradiusremainsthesame.GmgE9.80m/s2(r2E)2m2)19.6m/s2E→2g2(9.80m/sb.Earth’sradiusistripled,butitsmassremainsthesame.GmgE9.80m/s2(r2E)g9.80m/s222rE→442.45m/sc.BoththemassandradiusofEartharedoubled.GmgE(r2E)2g2(9.80m/s2)2m2Eand2rE→444.90m/s87.AstronautWhatwouldbethestrengthofEarth’sgravitationalfieldatapointwherean80.0-kgastronautwouldexperiencea25.0percentreductioninweight?F2)784Ngmg(80.0kg)(9.80m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Fg,reduced(784N)(0.750)588NFg,reduced588N2g7.35m/sreducedm80.0kgPhysics:PrinciplesandProblemsSolutionsManual159

163Chapter7continuedMixedReviewpages193–194Level188.UsetheinformationforEarthinTable7-1onpage173tocalculatethemassoftheSun,usingNewton’sversionofKepler’sthirdlaw.2423,TrGm2423andsomTrG42r3m2GT42(1.501011m)36.671011Nm2/kg2(3.156107s)22.011030kg89.Earth’sgravitationalfieldis7.83N/kgatthealtitudeofthespaceshuttle.Atthisaltitude,whatisthesizeoftheforceofattractionbetweenastudentwithamassof45.0kgandEarth?FgmFmg(45.0kg)(7.83N/kg)352N90.UsethedatafromTable7-1onpage173tofindthespeedandperiodofasatellitethatorbitsMars175kmaboveitssurface.rr6m0.175106mM175km3.40106m3.5810Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.GMvMr(6.671011Nm2/kg2)(6.421023kg)3.58106m3.46103m/sr3T2GMM(3.58106m)32(6.671011Nm2/kg2)(6.421023kg)6.45103sor1.79h160SolutionsManualPhysics:PrinciplesandProblems

164Chapter7continuedLevel291.SatelliteAsatelliteisplacedinorbit,asshowninFigure7-26,witharadiusthatishalftheradiusoftheMoon’sorbit.FindtheperiodofthesatelliteinunitsoftheperiodoftheMoon.MoonrMEarth1rM2Satellite■Figure7-26T2r3ssTrMMr30.50r3So,TsTMT2srMrMMM0.125T2M0.35TM92.CannonballTheMoon’smassis7.31022kganditsradiusis1785km.IfNewton’sthoughtexperimentoffiringacannonballfromahighmountainwereattemptedontheMoon,howfastwouldthecannonballhavetobefired?Howlongwouldittakethecannonballtoreturntothecannon?Gmvr(6.671011Nm2/kg2)(7.31022kg)1.785106m1.7103m/sr3T2Gm(1.785106m)32(6.671011Nm2/kg2)(7.31022kg)6.8103s93.TheperiodoftheMoonisonemonth.AnswerthefollowingquestionsassumingthatthemassofEarthisdoubled.a.WhatwouldtheperiodoftheMoonbe?Expressyourresultsinmonths.r3T2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.GmSo,ifEarth’smassweredoubled,buttheradiusremainedthesame,1thentheperiodwouldbereducedbyafactorof,or0.707months.2Physics:PrinciplesandProblemsSolutionsManual161

165Chapter7continuedb.Wherewouldasatellitewithanorbitalperiodofonemonthbelocated?T2r32GmT2r3(Gm)21Thus,r3wouldbedoubled,orrwouldbeincreasedby231.26timesthepresentradiusoftheMoon.c.HowwouldthelengthofayearonEarthbeaffected?ThelengthofayearonEarthwouldnotbeaffected.ItdoesnotdependonEarth’smass.Level394.HowfastwouldaplanetofEarth’smassandsizehavetospinsothatanobjectattheequatorwouldbeweightless?Givetheperiodofrotationoftheplanetinminutes.Thecentripetalaccelerationmustequaltheaccelerationduetogravitysothatthesurfaceoftheplanetwouldnothavetosupplyanyforce(otherwiseknownasweight).mv2mEmGrr2GmEvr2r2r2rBut,v,soTTvGmEr3rCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2GmE(6.38106m)32(6.671011Nm2/kg2)(5.971024kg)5.07103s84.5min95.CarRacesSupposethataMartianbasehasbeenestablishedandcarracesarebeingconsidered.Aflat,circularracetrackhasbeenbuiltfortherace.Ifacarcanachievespeedsofupto12m/s,whatisthesmallestradiusofatrackforwhichthecoefficientoffrictionis0.50?Theforcethatcausesthecentripetalaccelerationisthestaticfrictionforce:Fstaticsmgv2Thecentripetalaccelerationisacrmv2So,rsmgv2Thus,rgsGmNotethatg(r2planet)162SolutionsManualPhysics:PrinciplesandProblems

166Chapter7continuedTofindgonMars,thefollowingcalculationisused.m23kg,andR6mMars6.3710Mars3.4310So,g2Mars3.61m/s(12m/s)2Therefore,R2)(0.50)(3.61m/sR8.0101m96.Apollo11OnJuly19,1969,Apollo11’srevolutionaroundtheMoonwasadjustedtoanaverageorbitof111km.TheradiusoftheMoonis1785km,andthemassoftheMoonis7.31022kg.a.HowmanyminutesdidApollo11taketoorbittheMoononce?r3m111103morbit178510r3T2Gm(1896103m)32(6.671011Nm2/kg2)(7.31022kg)7.4103s1.2102minb.AtwhatvelocitydidApollo11orbittheMoon?Gmvrorbit(6.671011Nm2/kg2)(7.31022kg)1896103m1.6103m/sThinkingCriticallypage19497.AnalyzeandConcludeSomepeoplesaythatthetidesonEartharecausedbythepulloftheMoon.Isthisstatementtrue?a.DeterminetheforcesthattheMoonandtheSunexertonamass,m,ofwateronEarth.YouranswerwillbeintermsofmwithunitsofN.11Nm2/kg2(1.991030kg)(m)FS,m6.6710(1.501011m)2(5.90103N)m11Nm2/kg2(7.361022kg)(m)FM,m6.6710(3.80108m)2(3.40105N)mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Whichcelestialbody,theSunortheMoon,hasagreaterpullonthewatersofEarth?TheSunpullsapproximately100timesstrongeronthewatersofEarth.Physics:PrinciplesandProblemsSolutionsManual163

167Chapter7continuedc.DeterminethedifferenceinforceexertedbytheMoononthewateratthenearsurfaceandthewateratthefarsurface(ontheoppositeside)ofEarth,asillustratedinFigure7-27.Again,youranswerwillbeintermsofmwithunitsofN.FartidalbulgeNeartidalbulgeEarthMoon■Figure7-27(Nottoscale)F11Nm2kg2)(7.361022kg)(m)m,mAFm,mB(6.671011(3.80108m6.37106m)2(3.80108m6.37106m)2(2.28106N)md.DeterminethedifferenceinforceexertedbytheSunonwateratthenearsurfaceandonwateratthefarsurface(ontheoppositeside)ofEarth.F11Nm2kg2)(1.991030kg)(m)S,mAFS,mB(6.671011(1.501011m6.37106m)2(1.501011m6.37106m)2(1.00106N)me.WhichcelestialbodyhasagreaterdifferenceinpullfromonesideofEarthtoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.theother?theMoonf.WhyisthestatementthatthetidesresultfromthepulloftheMoonmisleading?MakeacorrectstatementtoexplainhowtheMooncausestidesonEarth.ThetidesareprimarilyduetothedifferencebetweenthepulloftheMoononEarth’snearsideandEarth’sfarside.98.MakeandUseGraphsUseNewton’slawofuniversalgravitationtofindanequationwherexisequaltoanobject’sdistancefromEarth’scenter,andyisitsaccelerationduetogravity.Useagraphingcalculatortographthisequation,using6400–6600kmastherangeforxand9–10m/s2astherangefory.Theequationshouldbeoftheformyc(1/x2).Tracealongthisgraphandfindyforthefollowinglocations.a.atsealevel,6400kmc(Gm6m2/km2)4.0108unitsE)(10acc9.77m/s2b.ontopofMt.Everest,6410km9.74m/s2164SolutionsManualPhysics:PrinciplesandProblems

168Chapter7continuedc.inatypicalsatelliteorbit,6500km9.47m/s2d.inamuchhigherorbit,6600km9.18m/s2WritinginPhysicspage19499.ResearchanddescribethehistoricaldevelopmentofthemeasurementofthedistancebetweentheSunandEarth.OneoftheearliestcrudemeasurementswasmadebyJamesBradleyin1732.TheanswersalsoshoulddiscussmeasurementsofthetransitsofVenusdoneinthe1690s.100.Explorethediscoveryofplanetsaroundotherstars.Whatmethodsdidtheastronomersuse?Whatmeasurementsdidtheytake?HowdidtheyuseKepler’sthirdlaw?Astronomersmeasurethestar’stinyvelocityduetothegravitationalforceexertedonitbyamassiveplanet.ThevelocityiscalculatedbymeasuringtheDopplershiftofthestar’slightthatresultsfromthatmotion.Thevelocityoscillatesbackandforthastheplanetsorbitthestar,allowingcalculationoftheplanet’speriod.Fromthesizeofthevelocitytheycanestimatetheplanet’sdistanceandmass.BycomparingthedistancesandperiodsofplanetsinsolarsystemswithmultipleplanetsandusingKepler’sthirdlaw,astronomerscanbetterseparatethedistancesandmassesofstarsandplanets.CumulativeReviewpage194101.AirplanesAjetairplanetookofffromPittsburghat2:20P.M.andlandedinWashington,DC,at3:15P.M.onthesameday.Ifthejet’saveragespeedwhileintheairwas441.0km/h,whatisthedistancebetweenthecities?(Chapter2)t55min0.917hdvtdvt(441.0km/h)(0.917h)404km102.CarolynwantstoknowhowmuchherbrotherJaredweighs.Heagreestostandonascaleforher,butonlyiftheyareridinginanelevator.Ifhestepsonthescalewhiletheelevatorisacceleratingupwardat1.75m/s2andthescalereadsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.716N,whatisJared’susualweightonEarth?(Chapter4)IdentifyJaredasthesystemandupwardaspositive.FnetFscaleonJaredFEarth’smassonJaredmaPhysics:PrinciplesandProblemsSolutionsManual165

169Chapter7continuedFEarth’smassonJaredFscaleonJaredFEarth’smassonJaredgaaFEarth’smassonJared1gFscaleonJaredFEarth’smassonJareda1g716N1.75m/s219.80m/s2608N103.PotatoBugA1.0-gpotatobugiswalkingaroundtheouterrimofanupside-downflyingdisk.Ifthediskhasadiameterof17.2cmandthebugmovesatarateof0.63cm/s,whatisthecentripetalforceactingonthebug?Whatagentprovidesthisforce?(Chapter6)mv2(0.0010kg)(0.0063cm)2F7N,cr0.086m5.010providedbythefrictionalforcebetweenthebugandtheflyingdiskChallengeProblempage176AstronomershavedetectedthreeplanetsthatorbitthestarUpsilonAndromedae.PlanetBhasanaverageorbitalradiusof0.059AUandaperiodof4.6170days.PlanetChasanaverageorbitalradiusof0.829AUandaperiodof241.5days.PlanetDhasanaverageorbitalradiusof2.53AUandaperiodof1284days.(DistancesaregiveninCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.astronomicalunits(AU)—Earth’saveragedistancefromtheSun.ThedistancefromEarthtotheSunis1.00AU.)DBrDrBCrCUpsilonAndromedae1.DotheseplanetsobeyKepler’sthirdlaw?r3Testbycalculatingthefollowingratio.T2r3(0.059AU)3ForplanetB,B9.6106AU3/days2T2(4.6170days)2B166SolutionsManualPhysics:PrinciplesandProblems

170Chapter7continuedr3(0.829AU)3ForplanetC,C9.77106AU3/days2T2(241.5days)2Cr3(2.53AU)3ForplanetD,D9.82106AU3/days2T2(1284days)2DThesevaluesarequiteclose,soKepler’sthirdlawisobeyed.2.FindthemassofthestarUpsilonAndromedaeinunitsoftheSun’smass.r3GmcentralbodyT242r3(1.000AU)33/y2FortheEarth-Sunsystem,1.000AUT2(1.000y)2FortheplanetC-Upsilonsystem,r39.77106AU3/days2T2(9.77106AU3/days2)(365days/y)21.30AU3/y2Gmstar42Theratioofthetwoshowsthestar’sslightlyheaviermasstobe1.30thatoftheSun.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual167

171

172CHAPTER8RotationalMotionBecausetheradiusofthewheelisPracticeProblemsreducedfrom35.4cmto24cm,the8.1DescribingRotationalangularaccelerationwillbeMotionincreased.pages197–200215.23rad/spage200a21.85m/s21.Whatistheangulardisplacementofeachof2r0.24mthefollowinghandsofaclockin1h?State27.7rad/syouranswerinthreesignificantdigits.a.thesecondhand4.Youwanttoreplacethetiresonyourcar(60)(2rad)withtiresthathavealargerdiameter.Afteryouchangethetires,fortripsatthesame120rador377radspeedandoverthesamedistance,howwillb.theminutehandtheangularvelocityandnumberofrevolu-2rador6.28radtionschange?c.thehourhandvBecause,ifrisincreased,will1r(2rad)decrease.Thenumberofrevolutions12willalsodecrease.rador0.524rad62.Ifatruckhasalinearaccelerationof1.85m/s2andthewheelshaveanangularSectionReviewaccelerationof5.23rad/s2,whatisthe8.1DescribingRotationaldiameterofthetruck’swheels?Motionrapages197–200page2001.85m/s25.AngularDisplacementAmovielasts2h.5.23rad/s2Duringthattime,whatistheangulardis-0.354mplacementofeachofthefollowing?Thus,thediameteris0.707m.a.thehourhand13.Thetruckinthepreviousproblemistowing(2rad)rad63atrailerwithwheelsthathaveadiameterofb.theminutehand48cm.(2)(2rad)4rada.Howdoesthelinearaccelerationofthetrailercomparewiththatofthetruck?6.AngularVelocityTheMoonrotatesThechangesinvelocityaretheonceonitsaxisin27.3days.ItsradiusisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.same,sothelinearaccelerations6m.1.7410arethesame.b.Howdotheangularaccelerationsofthewheelsofthetrailerandthewheelsofthetruckcompare?Physics:PrinciplesandProblemsSolutionsManual169

173Chapter8continueda.WhatistheperiodoftheMoon’srota-9.AngularAccelerationInthespincycletioninseconds?ofaclotheswasher,thedrumturnsatperiod(27.3day)(24h/day)635rev/min.Ifthelidofthewasheris(3600s/h)opened,themotoristurnedoff.Ifthedrumrequires8.0stoslowtoastop,what2.36106sistheangularaccelerationofthedrum?b.WhatisthefrequencyoftheMoon’si635rpm66.53rad/srotationinrad/s?1f0.0,so66.5rad/speriod66.5rad/s2and8.3rad/s1t8.0srev/s,or2.3610610.CriticalThinkingACD-ROMhasaspiral2.66106rad/strackthatstarts2.7cmfromthecenterofc.Whatisthelinearspeedofarockonthediskandends5.5cmfromthecenter.theMoon’sequatordueonlytotheThediskdrivemustturnthedisksothatMoon’srotationthelinearvelocityofthetrackisaconstant1.4m/s.Findthefollowing.vra.theangularvelocityofthedisk(inrad/s(1.74106m)(2.66106rad/s)andrev/min)forthestartofthetrack4.63m/svd.ComparethisspeedwiththespeedofarpersononEarth’sequatorduetoEarth’s1.4m/srotation.0.027mThespeedonEarth’sequatoris52rad/sor5.0102rev/min464m/s,orabout100timesfaster.b.thedisk’sangularvelocityattheendofthetrack7.AngularDisplacementTheballinacom-vputermouseis2.0cmindiameter.IfyouCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.rmovethemouse12cm,whatistheangulardisplacementoftheball?1.4m/s0.055mdr25rad/sor2.4102rev/mind12cmso12radr1.0cmc.thedisk’sangularaccelerationifthediskisplayedfor76min8.AngularDisplacementDoallpartsoftheminutehandonawatchhavethesamefiangulardisplacement?Dotheymovethettsamelineardistance?Explain.25rad/s52rad/sangulardisplacement—yes;lineardis-(76min)(60s/min)tance—no,becauselineardistanceisa5.9103rad/s2functionoftheradius170SolutionsManualPhysics:PrinciplesandProblems

174Chapter8continuedPracticeProblemsFrsinmgrsin8.2RotationalDynamics(65kg)(9.80m/s2)(0.18m)(sin55°)pages201–21094Nmpage20311.ConsiderthewrenchinExampleProblem15.Ifthepedalinproblem14ishorizontal,1.Whatforceisneededifitisappliedtohowmuchtorquewouldyouexert?Howthewrenchatapointperpendiculartothemuchtorquewouldyouexertwhenthewrench?pedalisvertical?FrsinHorizontal90.0°soFrsinsoFrsinmgrsin35Nm(0.25m)(sin90.0°)(65kg)(9.80m/s2)(0.18m)2N(sin90.0°)1.4101.1102Nm12.Ifatorqueof55.0NmisrequiredandtheVertical=0.0°largestforcethatcanbeexertedbyyouisSoFrsin135N,whatisthelengthoftheleverarmmgrsinthatmustbeused?(65kg)(9.80m/s2)(0.18m)Fortheshortestpossibleleverarm,(sin0.0°)90.0°.0.0NmFrsinsorFsinpage20555.0Nm16.Ashok,whosemassis43kg,sits1.8m(135N)(sin90.0°)fromthecenterofaseesaw.Steve,whose0.407mmassis52kg,wantstobalanceAshok.Howfarfromthecenteroftheseesaw13.Youhavea0.234-m-longwrench.AjobshouldStevesit?requiresatorqueof32.4Nm,andyoucanFexertaforceof232N.WhatisthesmallestArAFSrSangle,withrespecttothevertical,atwhichFArAsortheforcecanbeexerted?SFSFrsinmAgrA1mSgsosinFrmArAsin132.4NmmS(232N)(0.234m)(43kg)(1.8m)36.6°52kg1.5m14.Youstandonthepedalofabicycle.IfyouCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.haveamassof65kg,thepedalmakesan17.Abicycle-chainwheelhasaradiusofangleof35°abovethehorizontal,andthe7.70cm.Ifthechainexertsa35.0-Nforcepedalis18cmfromthecenterofthechainonthewheelintheclockwisedirection,ring,howmuchtorquewouldyouexert?whattorqueisneededtokeepthewheelTheanglebetweentheforceandthefromturning?radiusis90°35°55°.Physics:PrinciplesandProblemsSolutionsManual171

175Chapter8continuedchainFgr(35.0N)(0.0770m)2.70NmThus,atorqueof2.70Nmmustbeexertedtobalancethistorque.18.Twobasketsoffruithangfromstringsgoingaround4.5cm1.1cmpulleysofdifferentdiameters,asshowninFigure8-6.WhatisthemassofbasketA?12F1r1F2r2m1gr1m2gr2mm2r21r1(0.23kg)(1.1cm)A4.5cm0.056kg0.23kg■Figure8-619.Supposetheradiusofthelargerpulleyinproblem18wasincreasedto6.0cm.WhatisthemassofbasketAnow?mm2r21r1(0.23kg)(1.1cm)6.0cm0.042kg20.Abicyclist,ofmass65.0kg,standsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.45.0°onthepedalofabicycle.Thecrank,whichis0.170mlong,makesa45.0°anglewiththevertical,asshowninFigure8-7.Thecrankis9.70cmattachedtothechainwheel,whichhasaradiusof9.70cm.Whatforce0.170mmustthechainexerttokeepthewheelfromturning?crch■Figure8-7FcrrcrsinFchrchFsoFcrrcrsinchrchmgrcrsinrch(65.0kg)(9.80m/s2)(0.170m)(sin45.0°)0.097m789N172SolutionsManualPhysics:PrinciplesandProblems

176Chapter8continuedpage20824.Eachsphereinthepreviousproblemhasa21.Twochildrenofequalmassessit0.3mmassof0.10kg.Thedistancebetweenfromthecenterofaseesaw.AssumingthatspheresAandCis0.20m.Findthemomenttheirmassesaremuchgreaterthanthatofofinertiainthefollowinginstances:rotationtheseesaw,byhowmuchisthemomentofaboutsphereA,rotationaboutsphereC.inertiaincreasedwhentheysit0.6mfromAboutsphereA:thecenter?Imr2m(2r)2Forthemassofthetwochildren,Imr2mr22mr2.5mr2Whenrisdoubled,Iismultipliedbya(5)(0.10kg)(0.20m)2factorof4.0.020kgm222.SupposetherearetwoballswithequalAboutsphereC:diametersandmasses.Oneissolid,andtheImr2mr2otherishollow,withallitsmassdistributedatitssurface.Arethemomentsofinertiaof2mr2theballsequal?Ifnot,whichisgreater?(2)(0.10kg)(0.20m)2Themoreofthemassthatislocatedfar20.0080kgmfromthecenter,thegreaterthemomentofinertia.Thus,thehollowballhasagreatervalueofI.page21025.ConsiderthewheelinExampleProblem4.23.Figure8-9showsthreemassivespheresonIftheforceonthestrapweretwiceasgreat,arodofverysmallmass.Considerthewhatwouldbethespeedofrotationofthemomentofinertiaofthesystem,firstwhenwheelafter15s?itisrotatedaboutsphereA,andthenwhenTorqueisnowtwiceasgreat.Theangu-itisrotatedaboutsphereC.Arethelaraccelerationisalsotwiceasgreat,momentsofinertiathesameordifferent?sothechangeinangularvelocityisExplain.Ifthemomentsofinertiaaretwiceasgreat.Thus,thefinalangulardifferent,inwhichcaseisthemomentofvelocityis32rad/s,or16rev/s.inertiagreater?AC26.Asolidwheelacceleratesat3.25rad/s2whenaforceof4.5Nexertsatorqueonit.Ifthewheelisreplacedbyawheelwithallofitsmassontherim,themomentofiner-tiaisgivenbyImr2.Ifthesameangular■Figure8-9velocityweredesired,whatforcewouldThemomentsofinertiaaredifferent.Ifhavetobeexertedonthestrap?thespacingbetweenspheresisrandTheangularaccelerationhasnoteachspherehasmassm,thenrotationchanged,butthemomentofinertiaisaboutsphereAistwiceasgreat.Therefore,thetorqueImr2m(2r)25mr2.mustbetwiceasgreat.Theradiusofthewheelisthesame,sotheforceRotationaboutsphereCisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mustbetwiceasgreat,or9.0N.Imr2mr22mr2.ThemomentofinertiaisgreaterwhenrotatingaroundsphereA.Physics:PrinciplesandProblemsSolutionsManual173

177Chapter8continued27.Abicyclewheelcanbeacceleratedeitherbypullingonthechainthatisonthegearorbypullingonastringwrappedaroundthetire.Thewheel’sradiusis0.38m,whiletheradiusofthegearis0.14m.Ifyouobtainedtheneededaccelerationwithaforceof15Nonthechain,whatforcewouldyouneedtoexertonthestring?Thetorqueonthewheekcomesfromeitherthechainorthestring.chainIwheelwheelwheelIwheelwheelThus,chainwheelFchainrgearFstringrwheelFchainrgearFstringrwheel(15N)(0.14m)0.38m5.5N28.Thebicyclewheelinproblem27isusedwithasmallergearwhoseradiusis0.11m.Thewheelcanbeacceleratedeitherbypullingonthechainthatisonthegearorbypullingstringthatiswrappedaroundthetire.Ifyouobtainedtheneededaccelerationwithaforceof15Nonthechain,whatforcewouldyouneedtoexertonthestring?FchainrgearFstringrwheel(15N)(0.11m)0.38mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4.3N29.Adiskwithamomentofinertiaof0.26kgm2isattachedtoasmallerdiskmountedonthesameaxle.Thesmallerdiskhasadiameterof0.180mandamassof2.5kg.Astrapiswrappedaroundthesmallerdisk,asshowninFigure8-10.Findtheforceneededtogivethissystemanangularaccelerationof2.57rad/s2.FFrIIIsoFr(IsmallIlarge)■Figure8-10r1m2I2smallrsmalllarger120.26kgm22)2(2.5kg)(0.090m)(2.57rad/s0.090m7.7N174SolutionsManualPhysics:PrinciplesandProblems

178Chapter8continued33.MomentofInertiaRefertoTable8-2onSectionReviewpage206andrankthemomentsofinertia8.2RotationalDynamicsfromleasttogreatestofthefollowingpages201–210objects:asphere,awheelwithalmostallofitsmassattherim,andasoliddisk.Allpage210haveequalmassesanddiameters.Explain30.TorqueVijeshentersarevolvingdoorthattheadvantageofusingtheonewiththeisnotmoving.Explainwhereandhowleastmomentofinertia.Vijeshshouldpushtoproduceatorquewiththeleastamountofforce.fromleasttogreatest:sphere2mr2,5Toproduceatorquewiththeleastsoliddisk1mr2,wheel(mr2)force,youshouldpushasclosetothe2edgeaspossibleandatrightanglestoThelessthemomentofinertia,thelessthedoor.torqueneededtogiveanobjectthesameangularacceleration.31.LeverArmYoutrytoopenadoor,butyouareunabletopushatarightangletothe34.Newton’sSecondLawforRotationaldoor.So,youpushthedooratanangleofMotionAropeiswrappedaroundapulley55°fromtheperpendicular.Howmuchandpulledwithaforceof13.0N.Thepul-harderwouldyouhavetopushtoopentheley’sradiusis0.150m.Thepulley’srota-doorjustasfastasifyouweretopushitattionalspeedgoesfrom0.0to14.0rev/min90°?in4.50s.WhatisthemomentofinertiaofTheanglebetweentheforceandthethepulley?radiusis35°.TorqueisFrsin.Becausesin90°1,andsin35°0.57,IyouwouldhavetoincreasetheFr1forcebyaratioof1.75toobtain0.57thesametorque.tFrt32.NetTorqueTwopeoplearepullingonfiropeswrappedaroundtheedgeofalarge(13.0N)(0.150m)(4.50s)wheel.Thewheelhasamassof12kgandarevrev2radmin14.0min0.0minrev60sdiameterof2.4m.Onepersonpullsinaclockwisedirectionwitha43-Nforce,while25.99kgmtheotherpullsinacounterclockwisedirec-tionwitha67-Nforce.Whatisthenet35.CriticalThinkingAballonanextremelytorqueonthewheel?low-friction,tiltedsurface,willslidedown-net12hillwithoutrotating.Ifthesurfaceisrough,however,theballwillroll.Explainwhy,F1rF2rusingafree-bodydiagram.1(F1F2)2d1(43N67N)(2.4m)2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.29NmPhysics:PrinciplesandProblemsSolutionsManual175

179Chapter8continuedTorque:Frsin.Theforceisdueto1.92mfriction,andthetorquecausestheball0.30m0.66m0.51m0.45mtorotateclockwise.Ifthesurfaceisfric-tion-free,thenthereisnoforceparalleltothesurface,notorque,andthusnorotation.Remember,forcesactingonthepivotpoint(thecenteroftheball)ABareignored.■Figure8-15Extremelylow-frictionRougha.Whatarethetorquesactingonthesurfacesurfaceladder?FnormalFnormalclockwise:AFArAFfrictionFA(0.96m–0.30m)rrFfriction(0.66m)FASurfaceSurfacecounterclockwise:FgFgBFBrBFB(0.96m0.45m)PracticeProblems(0.51m)FB8.3Equilibriumb.Writetheequationforrotationalequi-pages211–217librium.page215netAB036.WhatwouldbetheforcesexertedbythesotwosawhorsesiftheladderinExampleBAProblem5hadamassof11.4kg?(0.51m)FB((0.66m)FA)Becausenodistanceshavechanged,(0.51m)FB(0.66m)FAtheequationsarestillvalid:Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.SolvetheequationforFrAintermsofFg.gmgFBrFBgFAFB(0.30m)(11.4kg)(9.80m/s2)thus,FFAFgFBB1.05m32N(0.66m)FrAgFFAmg1g0.51mrBFg(11.4kg)(9.80m/s2)0.30morF11.05mA0.66m10.51m8.0101Nma0.66m37.A7.3-kgladder,1.92mlong,restsontwo10.51msawhorses,asshowninFigure8-15.(7.3kg)(9.80m/s2)SawhorseA,ontheleft,islocated0.30m0.66m1fromtheend,andsawhorseB,ontheright,0.51mislocated0.45mfromtheotherend.31NChoosetheaxisofrotationtobethecenterd.Howwouldtheforcesexertedbytheofmassoftheladder.twosawhorseschangeifAweremovedverycloseto,butnotdirectlyunder,thecenterofmass?176SolutionsManualPhysics:PrinciplesandProblems

180Chapter8continuedFAwouldbecomegreater,andFBg(2mdivergmboard)wouldbeless.(9.80m/s2)38.A4.5-m-longwoodenplankwitha24-kg(2(85kg)14kg)massissupportedintwoplaces.Onesup-1.8103Nportisdirectlyunderthecenteroftheboard,andtheotherisatoneend.Whataretheforcesexertedbythetwosupports?SectionReviewPickthecenterofmassoftheboardasthepivot.Theunsupportedendexerts8.3Equilibriumnotorque,sothesupportedenddoespages211–217nothavetoexertanytorque.Therefore,page217alltheforceisexertedbythecenter40.CenterofMassCanthecenterofmassofsupport.Thatforceisequaltotheanobjectbelocatedinanareawheretheweightoftheboard:objecthasnomass?Explain.F2)Yes,anobjectmovesasifallitsmasscenterFg(24kg)(9.80m/s2.4102Nisconcentratedatthecenterofmass.ThereisnothinginthedefinitionthatFend0Nrequiresanyoralloftheobject’smass39.A85-kgdiverwalkstotheendofadivingtobeatthatlocation.board.Theboard,whichis3.5mlongwith41.StabilityofanObjectWhyisamodifiedamassof14kg,issupportedatthecentervehiclewithitsbodyraisedhighonrisersofmassoftheboardandatoneend.Whatlessstablethanasimilarvehiclewithitsaretheforcesonthetwosupports?bodyatnormalheight?ChoosethecenterofmassoftheboardThecenterofmassofthevehiclewillasthepivot.TheforceofEarth’sgravityberaised,butthesizeofitsbasewillontheboardisexertedtotallyonthenotbeincreased.Therefore,itneedstosupportunderthecenterofmass.betippedatasmallerangletogettheenddivercenterofmassoutsidethebaseoftheFendrendFdiverrdivervehicle.FThus,Fdiverrdiver42.ConditionsforEquilibriumGiveanendrendexampleofanobjectforeachofthefollow-mdivergrdiveringconditions.renda.rotationalequilibrium,butnottransla-(85kg)(9.80m/s2)(1.75m)tionalequilibrium1.75mabookthatisdroppedsoitfalls8.3102NwithoutrotatingTofindtheforceonthecentersupport,b.translationalequilibrium,butnotrota-noticethatbecausetheboardisnottionalequilibriummoving,aseesawthatisnotbalancedandFrotatesuntiloneperson’sfeethittheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.endFcenterFdiverFggroundThus,FcenterFdiverFgFend43.CenterofMassWhereisthecenterof2FdiverFgmassofarollofmaskingtape?2mdivergmboardgItisinthemiddleoftheroll,intheopenspace.Physics:PrinciplesandProblemsSolutionsManual177

181Chapter8continued44.LocatingtheCenterofMassDescribeChapterAssessmenthowyouwouldfindthecenterofmassofthistextbook.ConceptMappingObtainapieceofstringandattachapage222smallweighttoit.Suspendthestring47.Completethefollowingconceptmapusingandtheweightfromonecornerofthethefollowingterms:angularacceleration,book.Drawalinealongthestring.radius,tangentialacceleration,centripetalSuspendthetwofromanothercornerofacceleration.thebook.Again,drawalinealongthestring.ThepointwherethelinescrossAngularAngularvelocityaccelerationisthecenterofmass.Radius45.RotatingFramesofReferenceApennyisCentripetalTangentialplacedonarotating,old-fashionedrecordaccelerationaccelerationturntable.Atthehighestspeed,thepennystartsslidingoutward.Whataretheforcesactingonthepenny?MasteringConceptsEarth’smassexertsadownwardforce.page222Theturntable’ssurfaceexertsbothan48.Abicyclewheelrotatesataconstantupwardforcetobalancegravityandan25rev/min.Isitsangularvelocitydecreas-inwardforceduetofrictionthatgivesing,increasing,orconstant?(8.1)thepennyitscentripetalacceleration.Itisconstant.Thereisnooutwardforce.Ifitwerenotforfriction,thepennywouldmoveina49.Atoyrotatesataconstant5rev/min.Isitsstraightline.angularaccelerationpositive,negative,orzero?(8.1)46.CriticalThinkingYouhavelearnedwhyItiszero.thewindsaroundalow-pressureareamoveCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.inacounterclockwisedirection.Wouldthe50.DoallpartsofEarthrotateatthesamerate?windsmoveinthesameoroppositedirec-Explain.(8.1)tioninthesouthernhemisphere?Explain.Yes,becauseallpartsofarigidbodyTheywouldmoveintheoppositedirec-rotateatthesamerate.tion.Windsfromthenorthmovefromtheequator,wherethelinearspeeddue51.Aunicyclewheelrotatesataconstanttorotationishighest,tomid-latitudes,14rev/min.Isthetotalaccelerationofawhereitislower.Thus,thewindsbendpointonthetireinward,outward,tangen-totheeast.Windsfromthesouthblowtial,orzero?(8.1)fromregionswherethelinearspeedisItisinward(centripetal).lowtoregionswhereitishigher;thus,theybendtothewest.Thesetwofac-52.Thinkaboutsomepossiblerotationsoftorsresultinaclockwiserotationyourtextbook.Arethemomentsofinertiaaroundalow-pressurearea.aboutthesethreeaxesthesameordiffer-ent?Explain.(8.2)Theyarealldifferent.Theonewiththemostmass,farthestfromtheaxis,hasthegreatestmomentofinertia.178SolutionsManualPhysics:PrinciplesandProblems

182Chapter8continued53.Torqueisimportantwhentighteningbolts.58.Supposeyoustandflat-footed,thenyouriseWhyisforcenotimportant?(8.2)andbalanceontiptoe.IfyoustandwithAnangularaccelerationmustbepro-yourtoestouchingawall,youcannotducedtotightenabolt.Differentbalanceontiptoe.Explain.(8.3)torquescanbeexertedwithwrenchesYourcenterofmassmustbeabovetheofdifferentlengths.pointofsupport.Butyourcenterofmassisroughlyinthecenterofyour54.Rankthetorquesonthefivedoorsshownbody.Thus,whileyouareonyourtoes,inFigure8-18fromleasttogreatest.Noteabouthalfofyourbodymustbeinfrontthatthemagnitudeofalltheforcesistheofyourtoes,andhalfmustbebehind.Ifsame.(8.2)yourtoesareagainstthewall,nopartofyourbodycanbeinfrontofyourABtoes.59.WhydoesagymnastappeartobefloatingCDonairwhensheraisesherarmsaboveherheadinaleap?(8.3)Shemoveshercenterofmassclosertoherhead.E60.Whyisavehiclewithwheelsthathavea■Figure8-18largediametermorelikelytorolloverthan0EDCBAavehiclewithwheelsthathaveasmallerdiameter?(8.3)55.Explainhowyoucanchangeanobject’sThecenterofmassofthevehiclewithangularfrequency.(8.2)thelargerwheelsislocatedatahigherChangetheamountoftorqueappliedtopoint.Thusitdoesnothavetobetiltedtheobjectorchangethemomentofveryfarbeforeitrollsover.inertia.ApplyingConcepts56.Tobalanceacar’swheel,itisplacedonapages222–223verticalshaftandweightsareaddedto61.Twogearsareincontactandrotating.Onemakethewheelhorizontal.Whyisthisislargerthantheother,asshowninFigureequivalenttomovingthecenterofmass8-19.Comparetheirangularvelocities.Alsountilitisatthecenterofthewheel?(8.3)comparethelinearvelocitiesoftwoteethWhenthewheelisbalancedsoitdoesthatareincontact.nottilt(rotate)inanydirection,thenthereisnonettorqueonit.Thismeansthatthecenterofmassisatthepivotpoint.57.Astuntdrivermaneuversamonstertrucksothatitistravelingononlytwowheels.Whereisthecenterofmassofthetruck?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(8.3)■Figure8-19Itisdirectlyabovethelinebetweenthepointswherethetwowheelsaretouch-ingtheground.Thereisnonettorqueonthetruck,soitismomentarilystable.Physics:PrinciplesandProblemsSolutionsManual179

183Chapter8continuedTheteethhaveidenticallinearveloci-66.BowlingBallWhenabowlingballleavesaties.Becausetheradiiaredifferentbowler’shand,itdoesnotspin.Afterithasvgoneabouthalfthelengthofthelane,how-and,theangularvelocitiesarerever,itdoesspin.Explainhowitsrotationdifferent.rateincreasedandwhyitdoesnotcontinuetoincrease.62.VideotapeWhenavideotapeisrewound,Itsrotationratecanbeincreasedonlyifwhydoesitwindupfastesttowardstheend?atorqueisappliedtoit.ThefrictionalThemachineturnsthespoolatacon-forceofthealleyontheballprovidesstantangularvelocity.Towardstheend,thisforce.Oncetheballisrollingsothespoolhasthegreatestradius.thatthereisnovelocitydifferenceBecausevr,thevelocityofthetapebetweenthesurfaceoftheballandtheisfastestwhentheradiusisgreatest.alley,thenthereisnomorefrictionalforceandthusnomoretorque.63.SpinCycleWhatdoesaspincycleofawashingmachinedo?Explainintermsof67.FlatTireSupposeyourcarhasaflattire.theforcesontheclothesandwater.YougetoutyourtoolsandfindalugwrenchtoremovethenutsofftheboltInthespincycle,thewaterandclothesstuds.Youfinditimpossibletoturntheundergogreatcentripetalaccelerations.nuts.YourfriendsuggestswaysyoumightThedrumcanexertforcesontheproduceenoughtorquetoturnthem.Whatclothes,butwhenthewaterreachesthethreewaysmightyourfriendsuggest?holesinthedrum,noinwardforcecanbeexertedonitanditthereforemovesPutanextensionpipeontheendoftheinastraightline,outofthedrum.wrenchtoincreasetheleverarm,exertyourforceatrightanglestothewrench,64.Howcanyouexperimentallyfindtheorexertagreaterforce,perhapsbymomentofinertiaofanobject?standingontheendofthewrench.Youcanapplyaknowntorqueandmea-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.68.TightropeWalkersTightropewalkerssuretheresultingangularacceleration.oftencarrylongpolesthatsagsothatthe65.BicycleWheelsThreebicyclewheelshaveendsarelowerthanthecenterasshowninmassesthataredistributedinthreedifferentFigure8-20.Howdoessuchapoleincreaseways:mostlyattherim,uniformly,andthetightropewalker’sstability?Hint:mostlyatthehub.ThewheelsallhavetheConsiderbothcenterofmassandmomentofsamemass.Ifequaltorquesareappliedtoinertia.them,whichonewillhavethegreatestangularacceleration?Whichonewillhavetheleast?Themoremassthereisfarfromtheaxis,thegreaterthemomentofinertia.Iftorqueisfixed,thegreaterthemomentofinertia,thelesstheangularacceleration.Thus,thewheelwithmassmostlyatthehubhastheleastmoment■Figure8-20ofinertiaandthegreatestangularacceleration.Thewheelwithmassmostlyneartherimhasthegreatestmomentofinertiaandtheleastangularacceleration.180SolutionsManualPhysics:PrinciplesandProblems

184Chapter8continuedThepoleincreasesmomentofinertiaMasteringProblemsbecauseofitsmassandlength.The8.1DescribingRotationalMotiondroopingendsofthepolebringthepages223–224centerofmassclosertothewire,thusreducingthetorqueonthewalker.TheLevel1increasedmomentofinertiaand72.Awheelisrotatedsothatapointonthedecreasedtorquebothreducetheangu-edgemovesthrough1.50m.Theradiuslaraccelerationifthewalkerbecomesofthewheelis2.50m,asshowninunbalanced.ThewalkercanalsouseFigure8-21.Throughwhatangle(inradi-thepoletoeasilyshiftthecenterofans)isthewheelrotated?massoverthewiretocompensateforinstability.m0.51.50m69.Merry-Go-RoundWhileridingamerry-go-2round,youtossakeytoafriendstandingontheground.Foryourfriendtobeabletocatchthekey,shouldyoutossitasecondortwobeforeyoureachthespotwhereyourfriendisstandingorwaituntilyourfriendisdirectlybehindyou?Explain.■Figure8-21Youhaveforwardtangentialvelocity,sodrthekeywillleaveyourhandwiththatdsovelocity.Therefore,youshouldtossitrearly.1.50m2.50m70.Whycanyouignoreforcesthatactonthe0.600radaxisofrotationofanobjectinstaticequi-libriumwhendeterminingthenettorque?73.TheouteredgeofatrucktirethathasaradiusThetorquecausedbytheseforcesisof45cmhasavelocityof23m/s.Whatisthezerobecausetheleverarmiszero.angularvelocityofthetireinrad/s?vr,71.Insolvingproblemsaboutstaticequilibri-vum,whyistheaxisofrotationoftenplacedratapointwhereoneormoreforcesareact-23m/singontheobject?51rad/s0.45mThatmakesthetorquecausedbythat74.Asteeringwheelisrotatedthrough128°,asforceequaltozero,reducingthenum-showninFigure8-22.Itsradiusis22cm.beroftorquesthatmustbecalculated.Howfarwouldapointonthesteeringwheel’sedgemove?22cm128°Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.■Figure8-22Physics:PrinciplesandProblemsSolutionsManual181

185Chapter8continueddrLevel278.WashingMachineAwashingmachine’s2rad(0.22m)(128°)0.49mtwospincyclesare328rev/minand542360°rev/min.Thediameterofthedrumis0.43m.75.PropellerApropellerspinsat1880rev/min.a.Whatistheratioofthecentripetalaccel-a.Whatisitsangularvelocityinrad/s?erationsforthefastandslowspincycles?rev2radminv21880Recallthataandvrw.minrev60scrar2197rad/sfastfastaslowr2b.Whatistheangulardisplacementoftheslowpropellerin2.50s?(542rev/min)22t(328rev/min)(197rad/s)(2.50s)2.73492radb.Whatistheratioofthelinearvelocityofanobjectatthesurfaceofthedrumfor76.Thepropellerinthepreviousproblemthefastandslowspincycles?slowsfrom475rev/minto187rev/mininvfastfastr4.00s.Whatisitsangularacceleration?vslowslowrfasttslowfi542rev/mint328rev/min(187rev/min475rev/min)2rad1.654.00srev1min79.Findthemaximumcentripetalacceleration60sintermsofgforthewashingmachinein7.54rad/s2problem78.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2r1g77.Anautomobilewheelwitha9.00cmac9.80m/s2radius,asshowninFigure8-23,rotatesat2rad1min22.50rad/s.Howfastdoesapoint7.00cm542rev/minrev60sfromthecentertravel?0.43m1g229.80m/s71gLevel380.Alaboratoryultracentrifugeisdesignedtoproduceacentripetalaccelerationof0.35106gatadistanceof2.50cmfromtheaxis.Whatangularvelocityinrev/minisrequired?■Figure8-23vr(7.00cm)(2.50rad/s)17.5cm/s182SolutionsManualPhysics:PrinciplesandProblems

186Chapter8continueda2r83.Atoyconsistingoftwoballs,each0.45kg,cattheendsofa0.46-m-long,thin,light-asocweightrodisshowninFigure8-25.Findrthemomentofinertiaofthetoy.The(0.35106)(9.80m/s2)revmomentofinertiaistobefoundaboutthe0.025m2radcenteroftherod.60s0.45kg0.45kg1min1.1105rev/min8.2RotationalDynamics0.46mpage224■Figure8-25Level1Imr281.WrenchAboltistobetightenedwithatorqueof8.0Nm.Ifyouhaveawrench(0.45kg)(0.23m)2(0.45kg)(0.23m)2thatis0.35mlong,whatistheleast20.048kgmamountofforceyoumustexert?FrsinLevel2soFrsin84.Abicyclewheelwitharadiusof38cmisgivenanangularaccelerationof2.67rad/s2Fortheleastpossibleforce,theangleis90.0°,thenbyapplyingaforceof0.35Nontheedgeofthewheel.Whatisthewheel’smoment8.0NmFofinertia?(0.35m)(sin90.0°)23NI82.WhatisthetorqueonaboltproducedbyaI15-Nforceexertedperpendiculartoawrenchthatis25cmlong,asshowninFrsinFigure8-24?(0.35N)(0.38m)(sin90.0°)F2.67rad/s290°0.050kgm225cmLevel385.ToyTopAtoytopconsistsofarodwithadiameterof8.0-mmandadiskofmass0.0125kgandadiameterof3.5cm.Themomentofinertiaoftherodcanbeneglected.Thetopisspunbywrappingastringaroundtherodandpullingitwitha■Figure8-24velocitythatincreasesfromzeroto3.0m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Frsinover0.50s.(15N)(0.25m)(sin90.0°)3.8NmPhysics:PrinciplesandProblemsSolutionsManual183

187Chapter8continueda.Whatistheresultingangularvelocityofthetop?vffrrod3.0m/s1(0.0080m)27.5102rad/sb.Whatforcewasexertedonthestring?FrrodsinandIThus,FrrodsinIIFrrodsin12t2mrdiskrrodsinmr2disk2trrodsin(2fi)mrdisk2trrodsin0.035m2(7.5102rad/s0.0rad/s)(0.0125kg)20.0080m(2)(0.50s)(sin90.0°)20.72NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8.3Equilibriumpage224Level186.A12.5-kgboard,4.00mlong,isbeingheldupononeendbyAhmed.Hecallsforhelp,andJudiresponds.a.WhatistheleastforcethatJudicouldexerttolifttheboardtothehorizontalposition?Whatpartoftheboardshouldshelifttoexertthisforce?Attheoppositeend,shewillonlylifthalfthemass.Fleastmg12)(12.5kg)(9.80m/s261.2Nb.WhatisthegreatestforcethatJudicouldexerttolifttheboardtothehorizontalposition?Whatpartoftheboardshouldshelifttoexertthisforce?Attheboard’scenterofmass(middle),shewilllifttheentiremass.Fgreatestmg(12.5kg)(9.80m/s2)122N184SolutionsManualPhysics:PrinciplesandProblems

188Chapter8continuedLevel287.Twopeopleareholdinguptheendsofa4.25-kgwooden6.00kgboardthatis1.75mlong.A6.00-kgboxsitsontheboard,0.50mfromoneend,asshowninFigure8-26.Whatforcesdothetwopeopleexert?Inequilibrium,thesumofallforcesiszeroandthesumofthetorquesaboutanaxisofrotationiszero.1.25m0.50mFleftFrightFboardFbox0■Figure8-26leftrightboardbox0Wecanchoosetheaxisofrotationatthelocationofoneoftheunknownforces(Fleft)sothattorqueiseliminated,thussimplifyingthecalculations.FleftrleftFrightrrightFboardrboardFboxrbox0FleftrleftFrightrrightmboardgrboardmboxgrbox02)1.25m0.50mFleft(0)Fright(1.25m0.50m)(4.25kg)(9.80m/s2(6.00kg)(9.80m/s2)(1.25m)0Fright63NSubstitutingthisintotheforceequation,FleftFrightFboardFbox0FleftFrightFboardFboxFrightmboardgmboxg(63N)(4.25kg)(9.80m/s2)(6.00kg)(9.80m/s2)37NLevel388.Acar’sspecificationsstatethatitsweightdistributionis53percentonthefronttiresand47percentonthereartires.Thewheelbaseis2.46m.Whereisthecar’scenterofmass?Letthecenterofmassbexfromthefrontofthecar.LettheweightofthecarbeFg.frontrearFfrontrfrontFrearrrear(0.53Fg)x(0.47Fg)(2.46mx)x1.16mMixedReviewCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pages224–225Level189.Awoodendoorofmass,m,andlength,l,isheldhorizontallybyDanandAjit.Dansuddenlydropshisend.Physics:PrinciplesandProblemsSolutionsManual185

189Chapter8continueda.WhatistheangularaccelerationoftheThen,doorjustafterDanletsgo?leftbagsThetorqueisduetothegravitation-FleftrleftFbagsrbagsalforce.Theforceatthecenterofmassismg.FbagsrbagsFleftrleftThus,I(10)(175N)(0.50m)Frsin2.43m123.6102N3mlSubstitutethisintotheforceequation.1mg2l(sin90.0°)FleftFrightFbags012mlF3rightFleftFbags3gl3.6102N10(175N)21.4102Nb.Istheaccelerationconstant?Explain.No;theanglebetweenthedoorand91.BasketballAbasketballisrolleddownthetheweightischanging,andthere-court.Aregulationbasketballhasadiame-forethetorqueisalsochanging.terof24.1cm,amassof0.60kg,andaThus,theaccelerationchanges.momentofinertiaof5.8103kgm2.Thebasketball’sinitialvelocityis2.5m/s.90.TopsoilTenbagsoftopsoil,eachweighinga.Whatisitsinitialangularvelocity?175N,areplacedona2.43-m-longsheetofwood.Theyarestacked0.50mfromonevrendofthesheetofwood,asshowninFigure8-27.Twopeopleliftthesheetof2.5m/swood,oneateachend.Ignoringtheweight1(0.241m)2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofthewood,howmuchforcemusteachpersonexert?21rad/sb.Theballrollsatotalof12m.Howmanyrevolutionsdoesitmake?drdsor12mrev12rad(0.241m)21.93m16revc.Whatisitstotalangulardisplacement?0.50mdr■Figure8-27dInequilibrium,thesumoftheforcesissorzeroandthesumofthetorquesiszero.F12mleftFrightFbags01(0.241m)leftrightbags02ChoosethelocationofF1.0102radrightastheaxisofrotationtomakethattorquezero.186SolutionsManualPhysics:PrinciplesandProblems

190Chapter8continued92.Thebasketballinthepreviousproblemdstopsrollingaftertraveling12m.vta.Ifitsaccelerationwasconstant,what(2.50m)wasitsangularacceleration?(1.25s)v2v22adfi2.00m/sv2soaic.Whatistheangularvelocityofthe2dcylinder?av2Thus,ivr2rdr(2.5m/s)22.00m/s12(0.241m)(12m)12(50m)22.2rad/s28102rad/sb.Whattorquewasactingonitasitwasslowingdown?94.HardDriveAharddriveonamoderncomputerspinsat7200rpm(revolutionsIperminute).Ifthedriveisdesignedtostart(5.8103kgm2)(2.2rad/s2)fromrestandreachoperatingspeedin1.5s,1.3102Nmwhatistheangularaccelerationofthedisk?fi93.Acylinderwitha50mdiameter,asshownttinFigure8-28,isatrestonasurface.A(7200rpm0rpm)2rad1minropeiswrappedaroundthecylinderand1.5srev60spulled.Thecylinderrollswithoutslipping.5.0102rad/s295.SpeedometersMostspeedometersinauto-mobilesmeasuretheangularvelocityofthetransmissionandconvertittospeed.Howwillincreasingthediameterofthetiresaffectthereadingofthespeedometer?50mBecauseincreasingthediameterdecreasestheangularvelocity,itwillalsodecreasethereadingofthespeedometer.■Figure8-2896.Aboxisdraggedacrossthefloorusingaa.Aftertheropehasbeenpulledadis-ropethatisadistancehabovethefloor.tanceof2.50mataconstantspeed,Thecoefficientoffrictionis0.35.Theboxishowfarhasthecenterofmassofthe0.50mhighand0.25mwide.Findthecylindermoved?forcethatjusttipsthebox.ThecenterofmassisalwaysoverLetMequalthemassofthebox.Thethepointofcontactwiththesurfacecenterofmassoftheboxis0.25mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.forauniformcylinder.Thereforetheabovethefloor.Theboxjusttipswhencenterofmasshasmoved2.50m.thetorquesonitareequal.b.Iftheropewaspulledadistanceofropefriction2.50min1.25s,howfastwasthecen-Fterofmassofthecylindermoving?roperropeFfrictionrfrictionPhysics:PrinciplesandProblemsSolutionsManual187

191Chapter8continuedFThelongerpieceoflumberwouldbeFfrictionrfrictionropereasiertokeepfromrotatingbecauseropeithasagreatermomentofinertia.Mgrfrictionrrope99.SurfboardHarrisandPaulcarryasurf-(0.35)M(9.80m/s2)(0.25m)boardthatis2.43mlongandweighs143N.h0.25mPaulliftsoneendwithaforceof57N.(0.86m2/s2)Ma.WhatforcemustHarrisexert?h0.25mFHFgFPNotethatwhenyoupulltheboxattheheightofitscenterofmass,thedenom-143N57Ninatorbecomeszero.Thatis,youcan86Npullwithanyamountofforceandnotb.WhatpartoftheboardshouldHarrislift?tipthebox.ChoosethepointofrotationattheendwherePaullifts.97.Thesecondhandonawatchis12mmlong.Whatisthevelocityofitstip?HgvrFHrHFgrgmin(0.012m)(2rad/min)60sFgrgrHF1.3103m/sH2.43m(143N)2Level286N98.LumberYoubuya2.44-m-longpieceof2.0m10cm10cmlumber.YourfriendbuysapieceofthesamesizeandcutsitintotwoThus,Harrishastolift2.0mfromlengths,each1.22mlong,asshowninPaul’sendoftheboard.Figure8-29.YoueachcarryyourlumberCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.onyourshoulders.100.Asteelbeamthatis6.50mlongweighs325N.Itrestsontwosupports,3.00mapart,withequalamountsofthebeam2.44mextendingfromeachend.Suki,whoweighs575N,standsonthebeaminthecenterandthenwalkstowardoneend.Howclosetotheendcanshecomebeforethebeambeginstotip?1.22mEachsupportis1.75mfromtheendofthebeam.Choosethepointofrotation1.22mtobethesupportattheendcloserto■Figure8-29Suki.Thecenterofmassofthebeama.Whichloadiseasiertolift?Why?is1.50mfromthatsupport.ThebeamThemassesarethesame,sothewilljustbegintotipwhenSuki’sweightsarethesame.Thus,thetorque(S)equalsthetorqueofthesameupwardforceisrequiredtoliftbeam’scenterofmass(cm)andtheeachload.entireweightisonthesupportclosesttoSuki.b.BothyouandyourfriendapplyatorquewithyourhandstokeeptheScmlumberfromrotating.WhichloadisFSrSFcmrcmeasiertokeepfromrotating?Why?188SolutionsManualPhysics:PrinciplesandProblems

192Chapter8continuedFrcmrcmSFS3.00m(325N)2575NCable0.848mThatisSukicanmove0.848mfromAxisofCenterofthesupportor1.750.8480.90mrotationmass25.0°fromtheend.1.05m1.80mThinkingCritically2.10mpages225–226101.ApplyConceptsConsiderapointontheedgeofarotatingwheel.a.Underwhatconditionscanthecen-tripetalaccelerationbezero?when0.0b.Underwhatconditionscanthetangen-■Figure8-30tial(linear)accelerationbezero?Wecanusetorquestofindtheverticalwhen0.0component(FTy)ofthetension.Thecounterclockwisetorquesareinequi-c.Canthetangentialaccelerationbelibriumwiththeclockwisetorques.nonzerowhilethecentripetalaccelera-tioniszero?Explain.ccwcwWhen0.0instantaneously,butcablepolebannerisnotzero,willkeepchanging.FTyrcableFpolerpoleFbannerrbannerd.Canthecentripetalaccelerationbenonzerowhilethetangentialaccelera-FpolerpoleFbannerrbannerFtioniszero?Explain.TyrcableYes,aslongasisconstantbutThetotaltension,then,isnotzero.FFTypolerpoleFbannerrbannerFTsin25°r102.ApplyConceptsWhenyouapplythecablesin25°brakesinacar,thefrontenddips.Why?(175N)(1.05m)(105N)(1.80m)(2.10m)sin25°Theroadexertsaforceonthetiresthatbringsthecartorest.Thecenter420Nofmassisabovetheroad.Therefore,thereisanettorqueonthecar,caus-ingittorotateinthedirectionthatforcesthefrontdown.103.AnalyzeandConcludeAbannerissus-pendedfromahorizontal,pivotedpole,asCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.showninFigure8-30.Thepoleis2.10mlongandweighs175N.Thebanner,whichweighs105N,issuspended1.80mfromthepivotpointoraxisofrotation.Whatisthetensioninthecablesupportingthepole?Physics:PrinciplesandProblemsSolutionsManual189

193Chapter8continued104.AnalyzeandConcludeApivotedlampThetotaltension,then,ispoleisshowninFigure8-31.ThepoleFTyweighs27N,andthelampweighs64N.FTsin105°FpolerpoleFlamprlamprropesin105°Rope0.44my(27N)(64N)(0.33m)2(0.44m)(sin105°)105.0°Axisofx64Nrotation105.AnalyzeandConcludeGeraldand0.33mEvelyncarrythefollowingobjectsupa0.44mflightofstairs:alargemirror,adresser,andatelevision.Evelynisatthefrontend,andGeraldisatthebottomend.AssumeLampthatbothEvelynandGeraldexertonlyupwardforces.■Figure8-31a.Drawafree-bodydiagramshowinga.Whatisthetorquecausedbyeachforce?GeraldandEvelynexertingthesameforceonthemirror.gFgrsinMirror(27N)(0.22m)(sin90.0°)5.9NmFEFGrElampFlamprsincm(64N)(0.33m)(sin90.0°)rG21NmrErGEGCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Determinethetensionintheropesup-portingthelamppole.FgUsethetorquestofindtheverticalb.Drawafree-bodydiagramshowingcomponent(FTy)ofthetension.TheGeraldexertingmoreforceonthebot-counterclockwisetorquesareintomofthedresser.equilibriumwiththeclockwisetorques.DresserFGccwcwropepolelampFErEFcmTyrropeFpolerpoleFlamprlamprGFrErGEGpolerpoleFlamprlampFTyrropeFg190SolutionsManualPhysics:PrinciplesandProblems

194Chapter8continuedc.WherewouldthecenterofmassoftheCumulativeReviewtelevisionhavetobesothatGeraldpage226carriesalltheweight?108.Twoblocks,oneofmass2.0kgandthedirectlyabovewhereGeraldisotherofmass3.0kg,aretiedtogetherwithliftingamasslessrope.ThisropeisstrungoveraTVmassless,resistance-freepulley.Theblocksarereleasedfromrest.Findthefollowing.(Chapter4)FGcma.thetensionintheropeThetensionontheropeisTm2gm2aFgTm3gm3aThus,WritinginPhysicsma3m2gpage226m3m2106.AstronomersknowthatifasatelliteistooSubstituteintothefirstequation.closetoaplanet,itwillbetornapartby2mtidalforces.Thatis,thedifferenceintheT2m3gmgravitationalforceonthepartofthesatel-2m3litenearesttheplanetandthepartfarthest2(2kg)(3kg)2)(9.80m/sfromtheplanetisstrongerthantheforces2kg3kgholdingthesatellitetogether.Doresearch24NontheRochelimitanddeterminehowb.theaccelerationoftheblocks.closetheMoonwouldhavetoorbitEarthSubstituteintotheequationfora.tobeattheRochelimit.ma3m2gForaplanetandamoonwithidenticalm3m2densities,theRochelimitis2.446timestheradiusoftheplanet.Earth’s3kg2kg(9.80m/s2)3kg2kgRochelimitis18,470km.1.96m/s2107.Automobileenginesareratedbythetorquethattheyproduce.Researchand109.Ericsitsonasee-saw.Atwhatangle,rela-explainwhytorqueisanimportantquan-tivetothevertical,willthecomponentoftitytomeasure.hisweightparalleltotheplanebeequaltoone-thirdtheperpendicularcomponentofTheforceexertedbythegroundonthehisweight?(Chapter5)tireacceleratesthecar.Thisforceisproducedbytheengine.ItcreatestheFg,parallelFgsinforcebyrotatingtheaxle.ThetorqueisFequaltotheforceontheedgeoftheg,perpendicularFgcostiremultipliedbytheradiusofthetire.Fg,perpendicular3Fg,parallelGearsinthetransmissionmaycauseFg,perpendicularCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.theforcetochange,buttheydonot3Fg,parallelchangethetorque.Therefore,theFamountoftorquecreatedbythegcos13engineisdeliveredtothewheels.Fgsintantan1(3)71.6°Physics:PrinciplesandProblemsSolutionsManual191

195Chapter8continued110.Thepilotofaplanewantstoreachan111.A60.0-kgspeedskaterwithavelocityofairport325kmduenorthin2.75hours.18.0m/scomesintoacurveof20.0-mAwindisblowingfromthewestatradius.Howmuchfrictionmustbeexerted30.0km/h.Whatheadingandairspeedbetweentheskatesandicetonegotiatetheshouldbechosentoreachthedestinationcurve?(Chapter6)ontime?(Chapter6)2mvFLetgbethedistancenorthtotheair-fFnetrport,xbethewestwarddeflection,and(60.0kg)(18.0m/s)2972Nhbetheactualdistancetraveled.First,20.0mfindtheheadingastheangletraveledeastwardfromthenorthwardpath.ChallengeProblemvtanxvypage208Ranktheobjectsshowninthediagramaccordingvtan1xtotheirmomentsofinertiaabouttheindicatedvyaxes.Allsphereshaveequalmassesandallsepa-1vxtyrationsarethesame.tandybcdatan1(30.0km/h)(2.75h)A325km14.3°westofnorthBTheairspeed,then,shouldbev2v2v2hxyC1v2v2)2h(vxy21d2y2Dv2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.xty(325km)21(30.0km/h)22(2.75h)2122km/h192SolutionsManualPhysics:PrinciplesandProblems

196CHAPTERMomentumandIts9ConservationPracticeProblemsimpulseFt(5.0103N)(2.0s)9.1ImpulseandMomentumpages229–2351.0104NsTheimpulseisdirectedwestwardpage233andhasamagnitudeof1.0104Ns.1.Acompactcar,withmass725kg,ismovingat115km/htowardtheeast.Sketchtheb.Completethe“before”and“after”movingcar.sketches,anddeterminethemomentumandthevelocityofthecarnow.a.Findthemagnitudeanddirectionofitsmomentum.DrawanarrowonyourBeforeAftersketchshowingthemomentum.EpipfvpEp4kgm/seastwardi2.3210pmvFtppfpi(725kg)(115km/h)pfFtpi1000m1h1km3600s1.0104kgm/s2.32104kgm/seastward2.32104kgm/sb.Asecondcar,withamassof2175kg,1.3104kgm/seastwardhasthesamemomentum.Whatisitsvelocity?pfmvfvppf1.3104kgm/smvfm725kg(2.32104kgm/s)3600s1km18m/s1h1000m2175kg65km/heastward38.4km/heastward3.A7.0-kgbowlingballisrollingdownthe2.Thedriverofthecompactcarintheprevi-alleywithavelocityof2.0m/s.Foreachousproblemsuddenlyappliesthebrakesimpulse,showninFigures9-3aand9-3b,hardfor2.0s.Asaresult,anaverageforcefindtheresultingspeedanddirectionofof5.0103Nisexertedonthecartoslowitmotionofthebowlingball.down.a.b.t2.0s5512Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3N00F5.0101255Force(N)Force(N)a.Whatisthechangeinmomentum;thatis,themagnitudeanddirectionoftheTime(s)Time(s)impulse,onthecar?■Figure9-3Physics:PrinciplesandProblemsSolutionsManual193

197Chapter9continueda.Ftpfpimvfmvi141kmFtmv65.0kmvffm(5.0N)(1.0s)(7.0kg)(2.0m/s)125.0km7.0kga.Whatistheaverageforceexertedonthe2.7m/sinthesamedirectionperson?astheoriginalvelocityFtppfpiFtmvb.vfpfmFfpit(5.0N)(1.0s)(7.0kg)(2.0m/s)pfmvi7.0kgFt1.3m/sinthesamedirection(0.0kgm/s)(60.0kg)(94km/h)astheoriginalvelocity0.20s4.Thedriveracceleratesa240.0-kgsnowmo-1000m1h1km3600sbile,whichresultsinaforcebeingexertedthatspeedsupthesnowmobilefrom7.8103Noppositetothe6.00m/sto28.0m/soveratimeintervaldirectionofmotionof60.0s.b.Somepeoplethinkthattheycanstopa.Sketchtheevent,showingtheinitialtheirbodiesfromlurchingforwardinaandfinalsituations.vehiclethatissuddenlybrakingbyputtingtheirhandsonthedashboard.BeforeAfterFindthemassofanobjectthathasavivfweightequaltotheforceyoujustcalcu-lated.Couldyouliftsuchamass?Areyoustrongenoughtostopyourbodyxwithyourarms?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Whatisthesnowmobile’schangeinFgmgmomentum?WhatistheimpulseonF3Ng7.8102kgthesnowmobile?mg28.0109.80m/spFtSuchamassistooheavytolift.Youm(vfvi)cannotsafelystopyourselfwithyourarms.(240.0kg)(28.0m/s6.00m/s)5.28103kgm/sc.WhatisthemagnitudeoftheaverageSectionReviewforcethatisexertedonthesnowmobile?9.1ImpulseandMomentump5.28103kgm/spages229–235Ft60.0spage23588.0N6.MomentumIsthemomentumofacartravelingsouthdifferentfromthatofthe5.Supposea60.0-kgpersonwasinthevehiclesamecarwhenittravelsnorthatthesamethathittheconcretewallinExamplespeed?Drawthemomentumvectorstosup-Problem1.Thevelocityofthepersonequalsportyouranswer.thatofthecarbothbeforeandafterthecrash,andthevelocitychangesin0.20s.Yes,momentumisavectorquantity,Sketchtheproblem.andthemomentaofthetwocarsareinoppositedirections.194SolutionsManualPhysics:PrinciplesandProblems

198Chapter9continuedpN11.1kgm/s11.1NsWEd.IfthebatandsoftballareincontactforS0.80ms,whatistheaverageforcethatthebatexertsontheball?Ftm(vfvi)pm(vfvi)Ft7.ImpulseandMomentumWhenyoujumpfromaheighttotheground,youletyour(0.174kg)(38.0m/s(26.0m/s))legsbendatthekneesasyourfeethitthe1s(0.80ms)floor.Explainwhyyoudothisintermsof1000msthephysicsconceptsintroducedinthis1.4104Nchapter.Youreducetheforcebyincreasingthe10.MomentumThespeedofabasketballasitlengthoftimeittakestostoptheisdribbledisthesamewhentheballismotionofyourbody.goingtowardthefloorasitiswhentheballrisesfromthefloor.Isthebasketball’s8.MomentumWhichhasmoremomentum,changeinmomentumequaltozerowhenitasupertankertiedtoadockorafallinghitsthefloor?Ifnot,inwhichdirectionisraindrop?thechangeinmomentum?Drawthebas-ketball’smomentumvectorsbeforeandTheraindrophasmoremomentum,afterithitsthefloor.becauseasupertankeratresthaszeromomentum.No,thechangeinmomentumisupward.Beforetheballhitsthefloor,its9.ImpulseandMomentumA0.174-kgsoft-momentumvectorisdownward.Afterballispitchedhorizontallyat26.0m/s.Thetheballhitsthefloor,itsmomentumballmovesintheoppositedirectionatvectorisupward.38.0m/safteritishitbythebat.BeforeAftera.Drawarrowsshowingtheball’smomen-vftumbeforeandafterthebathitsit.pBeforeAfterpviyvivfx11.AngularMomentumAnice-skaterspinsb.Whatisthechangeinmomentumofwithhisarmsoutstretched.Whenhepullstheball?hisarmsinandraisesthemabovehishead,pm(vfvi)hespinsmuchfasterthanbefore.Dida(0.174kg)torqueactontheice-skater?Ifnot,how(38.0m/s(26.0m/s))couldhisangularvelocityhaveincreased?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11.1kgm/sNotorqueactedonhim.Bydrawinghisarmsinandkeepingthemclosetothec.Whatistheimpulsedeliveredbythebat?axisofrotation,hedecreasedhisFtpfpimomentofinertia.Becausetheangularpmomentumdidnotchange,theskater’sangularvelocityincreased.Physics:PrinciplesandProblemsSolutionsManual195

199Chapter9continued12.CriticalThinkingAnarchershootsarrowsmvPvPiatatarget.Someofthearrowsstickinthef(mPmG)target,whileothersbounceoff.Assuming(0.105kg)(24m/s)thatthemassesofthearrowsandtheveloc-0.034m/s(0.105kg75kg)itiesofthearrowsarethesame,whicharrowsproduceabiggerimpulseonthetar-15.A35.0-gbulletstrikesa5.0-kgstationaryget?Hint:Drawadiagramtoshowthepieceoflumberandembedsitselfinthemomentumofthearrowsbeforeandafterhit-wood.Thepieceoflumberandbulletflytingthetargetforthetwoinstances.offtogetherat8.6m/s.Whatwastheorigi-Theonesthatbounceoffgivemorenalspeedofthebullet?impulsebecausetheyendupwithmbvbimwvwi(mbmw)vfsomemomentuminthereversedirec-tion,meaningtheyhavealargerchangewherevfisthecommonfinalspeedofinmomentum.thebulletandpieceoflumber.Becausevwi0.0m/s,(mPracticeProblemsvbmw)vfbimb9.2Conservationof(0.0350kg5.0kg)(8.6m/s)Momentum0.0350kgpages236–2451.2103m/spage23813.Twofreightcars,eachwithamassof16.A35.0-gbulletmovingat475m/sstrikesa3.0105kg,collideandsticktogether.One2.5-kgbagofflourthatisonice,atrest.Thewasinitiallymovingat2.2m/s,andthebulletpassesthroughthebag,asshowninotherwasatrest.Whatistheirfinalspeed?Figure9-7,andexitsitat275m/s.Howfastisthebagmovingwhenthebulletexits?pipfCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mvAimvBi2mvfvvAivBif2275m/s2.2m/s0.0m/s21.1m/s14.A0.105-kghockeypuckmovingat24m/siscaughtandheldbya75-kggoalieatrest.Withwhatspeeddoesthegoalieslideontheice?■Figure9-7pPipGipPfpGfmmBvBimFvFimBvBfmFvFfPvPimGvGimPvPfmGvGfwherevBecausevFi0.0m/sGi0.0kgm/s,(mmvBvBimBvBf)PvPi(mPmG)vfFfmFwherevfvPfvGfisthecommonmB(vBivBf)finalspeedofthegoalieandthepuck.vFfmF196SolutionsManualPhysics:PrinciplesandProblems

200Chapter9continued(0.0350kg)(475m/s275m/s)2.5kg2.8m/s17.Thebulletinthepreviousproblemstrikesa2.5-kgsteelballthatisatrest.Thebulletbouncesbackwardafteritscollisionataspeedof5.0m/s.Howfastistheballmovingwhenthebulletbouncesbackward?Thesystemisthebulletandtheball.mbulletvbullet,imballvball,imbulletvbullet,fmballvball,fvball,i0.0m/sandvbullet,f5.0m/smbullet(vbullet,ivbullet,f)(0.0350kg)(475m/s(5.0m/s))sovball,fm2.5kgball6.7m/s18.A0.50-kgballthatistravelingat6.0m/scollideshead-onwitha1.00-kgballmovingintheoppositedirectionataspeedof12.0m/s.The0.50-kgballbouncesbackwardat14m/safterthecollision.Findthespeedofthesecondballafterthecollision.Saythatthefirstball(ballC)isinitiallymovinginthepositive(forward)direction.mCvCimDvDimCvCfmDvDfmsovCvCimDvDimCvCfDfmD(0.50kg)(6.0m/s)(1.00kg)(12.0m/s)(0.50kg)(14m/s)1.00kg2.0m/s,or2.0m/sintheoppositedirectionpage24019.A4.00-kgmodelrocketislaunched,expelling50.0gofburnedfuelfromitsexhaustataspeedof625m/s.Whatisthevelocityoftherocketafterthefuelhasburned?Hint:Ignoretheexternalforcesofgravityandairresistance.pripfuel,iprfpfuel,fwhereprfpfuel,f0.0kgm/sIftheinitialmassoftherocket(includingfuel)ismr4.00kg,thenthefinalmassoftherocketismrf4.00kg0.0500kg3.95kg0.0kgm/smrfvrfmfuelvfuel,fmvfuelvfuel,fCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.rfmrf(0.0500kg)(625m/s)3.95kg7.91m/sPhysics:PrinciplesandProblemsSolutionsManual197

201Chapter9continued20.Athreadholdsa1.5-kgcartanda4.5-kgcarttogether.Afterthethreadisburned,acompressedspringpushesthecartsapart,givingthe1.5-kgcartaspeedof27cm/stotheleft.Whatisthevelocityofthe4.5-kgcart?Letthe1.5-kgcartberepresentedby“C”andthe4.5-kgcartberepresentedby“D”.pCipDipCfpDfwherepCipDi0.0kgm/smDvDfmCvCfmsovCvCfDfmD(1.5kg)(27cm/s)4.5kg9.0cm/stotheright21.CarmenandJudidockacanoe.80.0-kgCarmenmovesforwardat4.0m/sassheleavesthecanoe.AtwhatspeedandinwhatdirectiondothecanoeandJudimoveiftheircombinedmassis115kg?pCipJipCfpJfwherepCipJi0.0kgm/smCvCfmJvJfmsovCvCfJfmJ(80.0kg)(4.0m/s)115kg2.8m/sintheoppositedirectionCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page24322.A925-kgcarmovingnorthat20.1m/scollideswitha1865-kgcarmovingwestat13.4m/s.Thetwocarsarestucktogether.Inwhatdirectionandatwhatspeeddotheymoveafterthecollision?Before:pi,ymyvi,y(925kg)(20.1m/s)1.86104kgm/spi,xmxvi,x(1865kg)(13.4m/s)2.50104kgm/spf,ypi,ypf,xpi,xpfpi(pf,x)2(pf,y)2198SolutionsManualPhysics:PrinciplesandProblems

202Chapter9continued(2.50104kgm/s)2(1.86104kgm/s)23.12104kgm/spvffm1m23.12104kgm/s11.2m/s(925kg1865kg)p1f,ytanpf,x11.86104kgm/stan2.50104kgm/s36.6°northofwest23.A1383-kgcarmovingsouthat11.2m/sisstruckbya1732-kgcarmovingeastat31.3m/s.Thecarsarestucktogether.Howfastandinwhatdirectiondotheymoveimmediatelyafterthecollision?Before:pi,xp1,xp2,x0m2v2ipi,yp1,yp2,ym1v1i0pfpip1,x2pi,y2(m2v2i)2(m1v1i)2pvffm1m2(m2v2i)2(m1v1i)2m1m2((1732kg)(31.3m/s))2((1383kg)(11.2m/s))21383kg1782kg18.1m/s1pi,y1m1v1i1(1383kg)(11.2m/s)tantantan15.9°pm(1732kg)(31.3m/s)i,x2v2isouthofeast24.Astationarybilliardball,withamassof0.17kg,isstruckbyanidenticalballmovingat4.0m/s.Afterthecollision,thesecondballmoves60.0°totheleftofitsoriginaldirection.Thestationaryballmoves30.0°totherightofthemovingCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ball’soriginaldirection.Whatisthevelocityofeachballafterthecollision?pCipDipCfpDfwherepCi0.0kgm/smCmDm0.17kgPhysics:PrinciplesandProblemsSolutionsManual199

203Chapter9continuedvDfVectorDiagramDf60.0pDiDiCivDi30.030.060.0CfpCfpDfvCfThevectordiagramprovidesfinalmomentumequationsfortheballthatisinitiallystationary,C,andtheballthatisinitiallymoving,D.pCfpDisin60.0°pDfpDicos60.0°Wecanusethemomentumequationforthestationaryballtofinditsfinalvelocity.pCfpDisin60.0°mvCfmvDisin60.0°vCfvDisin60.0°(4.0m/s)(sin60.0°)3.5m/s,30.0°totherightWecanusethemomentumequationforthemovingballtofinditsvelocity.pDfpDicos60.0°mvDfmvDicos60.0°vDfvDicos60.0°Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(4.0m/s)(cos60.0°)2.0m/s,60.0°totheleft25.A1345-kgcarmovingeastat15.7m/sisstruckbya1923-kgcarmovingnorth.Theyarestucktogetherandmovewithaninitialvelocityof14.5m/sat63.5°.Wasthenorth-movingcarexceedingthe20.1m/sspeedlimit?Before:pi,xm1v1,i(1345kg)(15.7m/s)2.11104kgm/spfpi(m1m2)vf(1345kg1923kg)(14.5m/s)4.74104kgm/spf,ypfsin(4.74104kgm/s)(sin63.5°)4.24104kgm/s200SolutionsManualPhysics:PrinciplesandProblems

204Chapter9continuedpwheref,ypi,ym2v2,ipf,y4.24104kgm/sFFgvaandm2,im1923kgmg2F22.1m/sgsinso,agsinFYes,itwasexceedingthespeedlimit.g/gThevelocityandaccelerationofthecartarerelatedbythemotionequa-SectionReviewtion,v2v22a(ddii)withv9.2Conservationofi0anddi0.Thus,Momentumv22adpages236–245v2adpage245(2)(gsin)(d)26.AngularMomentumTheouterrimofaplasticdiskisthickandheavy.Besidesmak-(2)(9.80m/s2)(sin30.0°)(1.00m)ingiteasiertocatch,howdoesthisaffect3.13m/stherotationalpropertiesoftheplasticdisk?b.Ifthetwocartssticktogether,withwhatMostofthemassofthediskislocatedinitialspeedwilltheymovealong?attherim,therebyincreasingitsmmomentofinertia.Therefore,whentheCvCi(mCmC)vfdiskisspinning,itsangularmomentummCvCiso,vfmislargerthanitwouldbeifmoremassCmDwerenearthecenter.WithalargerFCangularmomentum,thediskfliesgvCithroughtheairwithmorestability.FFCDgg27.SpeedAcart,weighing24.5N,isreleasedFCvCifromrestona1.00-mramp,inclinedatanFCFDangleof30.0°asshowninFigure9-14.Thecartrollsdowntheinclineandstrikesa(24.5N)(3.13m/s)24.5N36.8Nsecondcartweighing36.8N.1.25m/s24.5N28.ConservationofMomentumDuringatennisserve,theracketofatennisplayercontinuesforwardafterithitstheball.Ismomentumconservedinthecollision?1.0Explain,makingsurethatyoudefinethe0m36.8Nsystem.No,becausethemassoftheracketis30.0°muchlargerthanthatoftheball,onlya■Figure9-14smallchangeinitsvelocityisrequired.Inaddition,itisbeingheldbyamas-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Calculatethespeedofthefirstcartatsive,movingarmthatisattachedtoathebottomoftheincline.bodyincontactwithEarth.Thus,theTheforceparalleltothesurfaceofracketandballdonotcompriseantherampisisolatedsystem.FFgsin29.MomentumApole-vaulterrunstowardthePhysics:PrinciplesandProblemsSolutionsManual201

205Chapter9continuedlaunchpointwithhorizontalmomentum.Wheredoestheverticalmomentumcomeaverageforcetimeoverwhichtheforceisexertedfromastheathletevaultsoverthecrossbar?ProductistheTheverticalmomentumcomesfromtheimpulsiveforceofEarthagainstimpulsethepole.Earthacquiresanequalandoppositeverticalmomentum.Producesachangeinmomentum30.InitialMomentumDuringasoccergame,twoplayerscomefromoppositedirectionsandcollidewhentryingtoheadtheball.ProductisthemassvelocityTheycometorestinmidairandfalltotheground.Describetheirinitialmomenta.Becausetheirfinalmomentaarezero,MasteringConceptstheirinitialmomentawereequalandopposite.page25033.Canabullethavethesamemomentumasa31.CriticalThinkingYoucatchaheavyballtruck?Explain.(9.1)whileyouarestandingonaskateboard,Yes,forabullettohavethesameandthenyourollbackward.Ifyouweremomentumasatruck,itmusthaveastandingontheground,however,youhighervelocitybecausethetwomasseswouldbeabletoavoidmovingwhilecatch-arenotthesame.ingtheball.Explainbothsituationsusingmbulletvbulletmtruckvtruckthelawofconservationofmomentum.Explainwhichsystemyouuseineachcase.34.Apitcherthrowsacurveballtothecatcher.Inthecaseoftheskateboard,theball,Assumethatthespeedoftheballdoesn’ttheskateboard,andyouareanisolatedchangeinflight.(9.1)system,andthemomentumoftheballa.WhichplayerexertsthelargerimpulseCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.isshared.Inthesecondcase,unlessontheball?Earthisincluded,thereisanexternalThepitcherandthecatcherexertforce,somomentumisnotconserved.thesameamountofimpulseontheIfEarth’slargemassisincludedintheball,butthetwoimpulsesareinsystem,thechangeinitsvelocityisoppositedirections.negligible.b.Whichplayerexertsthelargerforceontheball?ChapterAssessmentThecatcherexertsthelargerforceontheballbecausethetimeintervalConceptMappingoverwhichtheforceisexertedispage250smaller.32.Completethefollowingconceptmapusingthefollowingterms:mass,momentum,aver-35.Newton’ssecondlawofmotionstatesthatageforce,timeoverwhichtheforceisexerted.ifnonetforceisexertedonasystem,noaccelerationispossible.Doesitfollowthatnochangeinmomentumcanoccur?(9.1)Nonetforceonthesystemmeansnonetimpulseonthesystemandnonetchangeinmomentum.However,individ-ualpartsofthesystemmayhavea202SolutionsManualPhysics:PrinciplesandProblems

206Chapter9continuedchangeinmomentumaslongastheThemomentumofafallingballisnetchangeinmomentumiszero.notconservedbecauseanetexter-nalforce,gravity,isactingonit.36.Whyarecarsmadewithbumpersthatcanb.Inwhatsystemthatincludesthefallingbepushedinduringacrash?(9.1)ballisthemomentumconserved?Carsaremadewithbumpersthatcom-Onesuchsysteminwhichtotalpressduringacrashtoincreasethemomentumisconservedincludestimeofacollision,therebyreducingthetheballplusEarth.force.42.Afallingbasketballhitsthefloor.Just37.Anice-skaterisdoingaspin.(9.1)beforeithits,themomentumisinthea.Howcantheskater’sangularmomen-downwarddirection,andafterithitsthetumbechanged?floor,themomentumisintheupwardbyapplyinganexternaltorquedirection.(9.2)b.Howcantheskater’sangularvelocitybea.Whyisn’tthemomentumofthebasket-changedwithoutchangingtheangularballconservedeventhoughthebouncemomentum?isacollision?bychangingthemomentofinertiaThefloorisoutsidethesystem,soitexertsanexternalforce,andthere-38.Whatismeantby“anisolatedsystem?”fore,animpulseontheball.(9.2)b.Inwhatsystemisthemomentumcon-Anisolatedsystemhasnoexternalserved?forcesactingonit.Momentumisconservedinthesys-temofballplusEarth.39.Aspacecraftinouterspaceincreasesitsvelocitybyfiringitsrockets.Howcanhot43.Onlyanexternalforcecanchangethegasesescapingfromitsrocketenginemomentumofasystem.Explainhowthechangethevelocityofthecraftwhenthereinternalforceofacar’sbrakesbringsthecarisnothinginspaceforthegasestopushtoastop.(9.2)against?(9.2)Theexternalforceofacar’sbrakescanMomentumisconserved.Thechangeinbringthecartoastopbystoppingthemomentumofgasesinonedirectionwheelsandallowingtheexternalfric-mustbebalancedbyanequalchangetionalforceoftheroadagainstthetiresinmomentumofthespacecraftinthetostopthecar.Ifthereisnofriction—oppositedirection.forexample,iftheroadisicy—thenthereisnoexternalforceandthecar40.Acueballtravelsacrossapooltableanddoesnotstop.collideswiththestationaryeightball.Thetwoballshaveequalmasses.Afterthecolli-44.Children’splaygroundsoftenhavecircular-sion,thecueballisatrest.Whatmustbemotionrides.Howcouldachildchangethetrueregardingthespeedoftheeightball?angularmomentumofsucharideasitis(9.2)turning?(9.2)TheeightballmustbemovingwiththeThechildwouldhavetoexertatorqueCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.samevelocitythatthecueballhadjustonit.Heorshecouldstandnexttoitbeforethecollision.andexertaforcetangentialtothecircleonthehandlesastheygopast.Heor41.ConsideraballfallingtowardEarth.(9.2)shealsocouldrunattherideandjumpa.Whyisthemomentumoftheballnotonboard.conserved?Physics:PrinciplesandProblemsSolutionsManual203

207Chapter9continuedApplyingConceptsAftertimeA,theobjectmoveswithaconstant,positivevelocity.AftertimeB,pages250–251theobjectisatrest.AftertimeC,the45.Explaintheconceptofimpulseusingphysi-objectmoveswithaconstant,negativecalideasratherthanmathematics.velocity.Aforce,F,exertedonanobjectoveratime,t,causesthemomentumofthe50.Duringaspacewalk,thetetherconnectingobjecttochangebythequantityFt.anastronauttothespaceshipbreaks.Usingagaspistol,theastronautmanagestogetback46.Isitpossibleforanobjecttoobtainalargertotheship.Usethelanguageoftheimpulse-impulsefromasmallerforcethanitdoesmomentumtheoremandadiagramtofromalargerforce?Explain.explainwhythismethodwaseffective.Yes,ifthesmallerforceactsforalongWhenthegaspistolisfiredintheoppo-enoughtime,itcanprovidealargersitedirection,itprovidestheimpulseimpulse.neededtomovetheastronauttowardthespaceship.47.FoulBallYouaresittingatabaseballgamewhenafoulballcomesinyourdirection.Youpreparetocatchitbare-handed.Tocatchitsafely,shouldyoumoveyourhandstowardtheball,holdthemstill,ormovetheminthesamedirectionasthemovingpball?Explain.Youshouldmoveyourhandsinthesamedirectiontheballistravelingtoincreasethetimeofthecollision,therebyreducingtheforce.p48.A0.11-gbulletleavesapistolat323m/s,51.TennisBallAsatennisballbouncesoffaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.whileasimilarbulletleavesarifleat396wall,itsmomentumisreversed.Explainthism/s.Explainthedifferenceinexitspeedsofactionintermsofthelawofconservationofthetwobullets,assumingthattheforcesmomentum.Definethesystemanddrawaexertedonthebulletsbytheexpandingdiagramasapartofyourexplanation.gaseshavethesamemagnitude.BeforeAfterThebulletisintheriflealongertime,ppsothemomentumitgainsislarger.vivf49.AnobjectinitiallyatrestexperiencestheimpulsesdescribedbythegraphinFigure9-15.Describetheobject’spEarthandwallmotionafterimpulsesA,B,andC.+x2Considerthesystemtobetheball,theAwall,andEarth.ThewallandEarthgain6810110somemomentuminthecollision.24BCForce(N)252.ImaginethatyoucommandspaceshipZeldon,whichismovingthroughinterplan-Time(s)etaryspaceathighspeed.Howcouldyou■Figure9-15204SolutionsManualPhysics:PrinciplesandProblems

208Chapter9continuedslowyourshipbyapplyingthelawofcon-MasteringProblemsservationofmomentum?9.1ImpulseandMomentumByshootingmassintheformofpages251–252exhaustgas,athighvelocityintheLevel1samedirectioninwhichyouaremov-56.GolfRocíostrikesa0.058-kggolfballwithing,itsmomentumwouldcausetheaforceof272Nandgivesitavelocityofship’smomentumtodecrease.62.0m/s.HowlongwasRocío’sclubincontactwiththeball?53.Twotrucksthatappeartobeidenticalcol-lideonanicyroad.Onewasoriginallyatmv(0.058kg)(62.0m/s)trest.ThetrucksarestucktogetherandmoveF272Natmorethanhalftheoriginalspeedofthe0.013smovingtruck.Whatcanyouconcludeaboutthecontentsofthetwotrucks?57.A0.145-kgbaseballispitchedat42m/s.ThebatterhitsithorizontallytothepitcherIfthetwotruckshadequalmasses,at58m/s.theywouldhavemovedoffathalfthespeedofthemovingtruck.Thus,thea.Findthechangeinmomentumofthemovingtruckmusthavehadamoreball.massiveload.Takethedirectionofthepitchtobepositive.54.Explain,intermsofimpulseandmomen-pmvfmvim(vfvi)tum,whyitisadvisabletoplacethebuttofarifleagainstyourshoulderwhenfirst(0.145kg)(58m/s(42m/s))learningtoshoot.14kgm/sWhenheldloosely,therecoilmomen-b.Iftheballandbatareincontactfortumoftherifleworksagainstonlythe4.6104s,whatistheaverageforcemassoftherifle,therebyproducingaduringcontact?largervelocityandstrikingyourshoul-Ftpder.Therecoilmomentummustworkpagainstthemassoftherifleandyou,Ftresultinginasmallervelocity.m(vfvi)55.BulletsTwobulletsofequalmassareshottatequalspeedsatblocksofwoodona(0.145kg)(58m/s(42m/s))44.610ssmoothicerink.Onebullet,madeofrub-ber,bouncesoffofthewood.Theother3.2104Nbullet,madeofaluminum,burrowsintothewood.Inwhichcasedoestheblockof58.BowlingAforceof186Nactsona7.3-kgwoodmovefaster?Explain.bowlingballfor0.40s.Whatisthebowlingball’schangeinmomentum?WhatisitsMomentumisconserved,sothechangeinvelocity?momentumoftheblockandbulletafterthecollisionequalsthemomentumofpFtthebulletbeforethecollision.Therub-(186N)(0.40s)berbullethasanegativemomentumCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.74Nsafterimpact,withrespecttotheblock,74kgm/ssotheblock’smomentummustbegreaterinthiscase.pvmFtmPhysics:PrinciplesandProblemsSolutionsManual205

209Chapter9continued62.HockeyAhockeyplayermakesaslapshot,(186N)(0.40s)7.3kgexertingaconstantforceof30.0Nonthe1m/shockeypuckfor0.16s.Whatisthemagni-1.010tudeoftheimpulsegiventothepuck?59.A5500-kgfreighttruckacceleratesFt(30.0N)(0.16s)from4.2m/sto7.8m/sin15.0sby4.8Nstheapplicationofaconstantforce.a.Whatchangeinmomentumoccurs?63.SkateboardingYourbrother’smassis35.6kg,andhehasa1.3-kgskateboard.pmvm(vfvi)Whatisthecombinedmomentumofyour(5500kg)(7.8m/s4.2m/s)brotherandhisskateboardiftheyaremov-2.0104kgm/singat9.50m/s?pmvb.Howlargeofaforceisexerted?(mpboymboard)vFt(35.6kg1.3kg)(9.50m/s)m(vfvi)3.5102kgm/st(5500kg)(7.8m/s4.2m/s)64.Ahockeypuckhasamassof0.115kgand15.0sisatrest.Ahockeyplayermakesashot,1.3103Nexertingaconstantforceof30.0Nonthepuckfor0.16s.Withwhatspeeddoesit60.Inaballisticstestatthepolicedepartment,headtowardthegoal?OfficerRiosfiresa6.0-gbulletat350m/sFtmvm(vfvi)intoacontainerthatstopsitin1.8ms.Whatistheaverageforcethatstopsthebullet?wherevi0pFtFtThusvfmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.m(vfvi)(30.0N)(0.16s)t0.115kg(0.0060kg)(0.0m/s350m/s)42m/s1.8103s1.2103N65.Beforeacollision,a25-kgobjectwasmov-ingat12m/s.Findtheimpulsethatacted61.VolleyballA0.24-kgvolleyballapproachesontheobjectif,afterthecollision,itmovedTinawithavelocityof3.8m/s.Tinabumpsatthefollowingvelocities.theball,givingitaspeedof2.4m/sbutina.8.0m/stheoppositedirection.WhataverageforceFtmvm(vfvi)didsheapplyiftheinteractiontimebetweenherhandsandtheballwas0.025s?(25kg)(8.0m/s12m/s)mv1.0102kgm/sFtb.8.0m/s(0.24kg)(2.4m/s3.8m/s)Ftmvm(v0.025sfvi)6.0101N(25kg)(8.0m/s12m/s)5.0102kgm/s206SolutionsManualPhysics:PrinciplesandProblems

210Chapter9continuedLevel266.A0.150-kgball,movinginthepositivedirection2at12m/s,isactedonbytheimpulseshowninthegraphinFigure9-16.Whatistheball’s0speedat4.0s?1234Force(N)Ftmv2AreaofgraphmvTime(s)12(2.0N)(2.0s)m(vfvi)■Figure9-162.0Ns(0.150kg)(vf12m/s)2.0kgm/svf0.150kg12m/s25m/s67.BaseballA0.145-kgbaseballismovingat35m/swhenitiscaughtbyaplayer.a.Findthechangeinmomentumoftheball.pm(vfvi)(0.145kg)(0.0m/s35m/s)5.1kgm/sb.Iftheballiscaughtwiththemittheldinastationarypositionsothattheballstopsin0.050s,whatistheaverageforceexertedontheball?pFaveragetpm(vfvi)so,Faveragett(0.145kg)(0.0m/s35m/s)0.500s1.0102Nc.If,instead,themittismovingbackwardsothattheballtakes0.500stostop,whatistheaverageforceexertedbythemittontheball?pFaveragetpm(vfvi)so,Faveragett(0.145kg)(0.0m/s35m/s)0.500s1.0101N68.HockeyAhockeypuckhasamassof0.115kgandstrikesthepoleofthenetat37m/s.Itbouncesoffintheoppositedirectionat25m/s,asshowninFigure9-17.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Whatistheimpulseonthepuck?Ftm(vfvi)0.115kg(0.115kg)(25m/s37m/s)25m/s7.1kgm/s■Figure9-17Physics:PrinciplesandProblemsSolutionsManual207

211Chapter9continuedb.Ifthecollisiontakes5.0104s,whatistheaverageforceonthepuck?Ftm(vfvi)m(vFfvi)t(0.115kg)(25m/s37m/s)45.010s1.4104N69.Anitrogenmoleculewithamassof4.71026kg,movingat550m/s,strikesthewallofacontainerandbouncesbackatthesamespeed.a.Whatistheimpulsethemoleculedeliverstothewall?Ftm(vfvi)(4.71026kg)(550m/s550m/s)5.21023kgm/sTheimpulsethewalldeliverstothemoleculeis5.21023kgm/s.Theimpulsethemoleculedeliverstothewallis5.21023kgm/s.b.Ifthereare1.51023collisionseachsecond,whatistheaverageforceonthewall?Ftm(vfvi)m(vFfvi)tForallthecollisions,theforceism(vF23)fvi)total(1.510t(4.71026kg)(550m/s550m/s)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(1.51023)1.0s7.8NLevel370.RocketsSmallrocketsareusedtomaketinyadjustmentsinthespeedsofsatellites.Onesuchrockethasathrustof35N.Ifitisfiredtochangethevelocityofa72,000-kgspacecraftby63cm/s,howlongshoulditbefired?Ftmvmtso,tF(72,000kg)(0.63m/s)35N1.3103s,or22min208SolutionsManualPhysics:PrinciplesandProblems

212Chapter9continued71.Ananimalrescueplaneflyingdueeastat36.0m/sdropsabale36.0m/sofhayfromanaltitudeof60.0m,asshowninFigure9-18.Ifthebaleofhayweighs175N,whatisthemomentumofthebalethemomentbeforeitstrikestheground?Givebothmagni-175Ntudeanddirection.Firstuseprojectilemotiontofindthevelocityofthebale.pmv60.0mTofindv,considerthehorizontalandverticalcomponents.vx36.0m/sv2v22dg2dg■Figure9-18yiyThus,vvx2vy2vx22dgThemomentum,then,isFgvFgvx22dgpgg(175N)(36.0m/s)2(2)(60.0m)(9.80m/s2)9.80m/s2888kgm/sTheanglefromthehorizontalisvytanvx2dgvx(2)(60.0m)(9.80m/s2)36.0m/s43.6°72.AccidentAcarmovingat10.0m/scrashesintoabarrierandstopsin0.050s.Thereisa20.0-kgchildinthecar.Assumethatthechild’svelocityischangedbythesameamountasthatofthecar,andinthesametimeperiod.a.Whatistheimpulseneededtostopthechild?Ftmvm(vfvi)(20.0kg)(0.0m/s10.0m/s)2.00102kgm/sb.Whatistheaverageforceonthechild?Ftmvm(vfvi)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.m(vso,Ffvi)t(20.0kg)(0.0m/s10.0m/s)0.050s4.0103NPhysics:PrinciplesandProblemsSolutionsManual209

213Chapter9continuedc.WhatistheapproximatemassofanBeforeobjectwhoseweightequalstheforceinvFBvDTpartb?FgmgFg4.0103Nso,mg9.80m/s24.1102kgd.Couldyouliftsuchaweightwithyour+xarm?No.Aftere.WhyisitadvisabletouseapropervFBvDT0restrainingseatratherthanholdachildonyourlap?Youwouldnotbeabletoprotectachildonyourlapintheeventofacollision.9.2ConservationofMomentum+xpages252–253b.Whatwasthefullback’smomentumLevel1beforethecollision?73.FootballA95-kgfullback,runningatp8.2m/s,collidesinmidairwitha128-kgFBmFBvFB(95kg)(8.2m/s)defensivetacklemovingintheopposite7.8102kgm/sdirection.Bothplayersendupwithzeroc.Whatwasthechangeinthefullback’sspeed.momentum?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Identifythe“before”and“after”situa-pFBpfpFBtionsanddrawadiagramofboth.0p2kgm/sBefore:mFB7.810FB95kgd.WhatwasthechangeinthedefensivevFB8.2m/stackle’smomentum?mDT128kg7.8102kgm/svDT?e.Whatwasthedefensivetackle’soriginalmomentum?After:m223kgv7.8102kgm/sf0m/sf.Howfastwasthedefensivetacklemov-ingoriginally?m2kgm/sDTvDT7.8107.8102kgm/sso,vDT128kg6.1m/s210SolutionsManualPhysics:PrinciplesandProblems

214Chapter9continued74.MarbleC,withmass5.0g,movesatad.CalculatethemomentumofmarbleDspeedof20.0cm/s.Itcollideswithasec-afterthecollision.ondmarble,D,withmass10.0g,movingatpCipDipCfpDf10.0cm/sinthesamedirection.Afterthepcollision,marbleCcontinueswithaspeedDfpCipDipCfof8.0cm/sinthesamedirection.1.00103kgm/sa.Sketchthesituationandidentifythesys-1.00103kgm/stem.Identifythe“before”and“after”sit-4.0104kgm/suationsandsetupacoordinatesystem.1.6103kgm/sBefore:mC5.0ge.WhatisthespeedofmarbleDafterthemD10.0gcollision?vCi20.0cm/spDfmDvDfvDi10.0cm/spDfso,vDfmDAfter:mC5.0g1.6103kgm/sm2D10.0g1.0010kgvCf8.0cm/s1.6101m/s0.16m/sv16cm/sDf?Before75.TwolabcartsarepushedtogetherwithaspringmechanismcompressedbetweenvCivDithem.Uponrelease,the5.0-kgcartrepelsCDonewaywithavelocityof0.12m/s,whilethe2.0-kgcartgoesintheoppositedirec-tion.Whatisthevelocityofthe2.0-kgcart?+xm1vim2vfmAfterv1vifm2(5.0kg)(0.12m/s)vCfvDfCD(2.0kg)0.30m/s+x76.A50.0-gprojectileislaunchedwithahori-zontalvelocityof647m/sfroma4.65-kgb.Calculatethemarbles’momentabeforelaunchermovinginthesamedirectionatthecollision.2.00m/s.Whatisthelauncher’svelocitym3kg)(0.200m/s)afterthelaunch?CvCi(5.010p1.0103kgm/sCipDipCfpDfmm2kg)(0.100m/s)CvCimDvDimCvCfmDvDfDvDi(1.0010mso,vCvCimDvDimCvCfCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.0103kgm/sDfmDc.CalculatethemomentumofmarbleCAssumingthattheprojectile,C,isafterthecollision.launchedinthedirectionofthem3kg)(0.080m/s)CvCf(5.010launcher,D,motion,4.0104kgm/sPhysics:PrinciplesandProblemsSolutionsManual211

215Chapter9continued(0.0500kg)(2.00m/s)(4.65kg)(2.00m/s)(0.0500kg)(647m/s)vDf4.65kg4.94m/s,or4.94m/sbackwardsLevel277.A12.0-grubberbullettravelsatavelocityof150m/s,hitsastationary8.5-kgconcreteblockrestingonafrictionlesssurface,andricochetsintheoppositedirectionwithavelocityof1.0102m/s,asshowninFigure9-19.Howfastwilltheconcrete21.010m/s8.5kgblockbemoving?mCvCimDvDimCvCfmDvDf12.0gmvCvCimDvDimCvCfDfm■Figure9-19Dsincetheblockisinitiallyatrest,thisbecomesmvC(vCivCf)DfmD(0.0120kg)(150m/s(1.0102m/s))8.5kg0.35m/s78.SkateboardingKofi,withmass42.00kg,isridingaskateboardwithamassof2.00kgandtravelingat1.20m/s.Kofijumpsoffandtheskateboardstopsdeadinitstracks.Inwhatdirectionandwithwhatvelocitydidhejump?(mLvLimsvsi)vimLvLfmsvsfCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.wherevsf0andvLivsivi(mThusvLms)viLfmL(42.00kg2.00kg)(1.20m/s)42.00kg1.26m/sinthesamedirectionasshewasriding79.BilliardsAcueball,withmass0.16kg,rollingat4.0m/s,hitsastationaryeightballofsimilarmass.Ifthecueballtravels45°aboveitsoriginalpathandtheeightballtravels45°belowthehorizontal,asshowninFigure9-20,whatisthevelocityofeachballafterthecollision?212SolutionsManualPhysics:PrinciplesandProblems

216Chapter9continued80.A2575-kgvanrunsintothebackofan825-kgcompactcaratrest.Theymoveofftogetherat8.5m/s.Assumingthatthefric-tionwiththeroadisnegligible,calculatetheinitialspeedofthevan.pCipDipCfpDfm45°CvCi(mCmD)vfmso,vCmD45°CimC(2575kg825kg)(8.5m/s)vf2575kg11m/sLevel381.In-lineSkatingDiegoandKeshiaareon■Figure9-20in-lineskatesandstandface-to-face,thenpusheachotherawaywiththeirhands.pCiDiegohasamassof90.0kgandKeshiahas45amassof60.0kg.p8fpa.Sketchtheevent,identifyingtheCf“before”and“after”situations,andsetupacoordinateaxis.Wecangetmomentumequationsfromthevectordiagram.Before:mK60.0kgpCfpCisin45°mD90.0kgmCvCfmCvCisin45°vi0.0m/svCfvCisin45°After:mK60.0kg(4.0m/s)(sin45°)mD90.0kg2.8m/svKf?Fortheeightball,vpDf?8fpcicos45°m8v8fmCvCi(cos45°)wherem8mC.Thus,v8fvCicos45°(4.0m/s)(cos45°)2.8m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual213

217Chapter9continuedBeforevKivDi0+xAftervKfvDf+xb.Findtheratiooftheskaters’velocitiesjustaftertheirhandslosecontact.pKipDi0.0kgm/spKfpDfso,mKvKfmDvDf0.0kgm/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.andmKvKfmDvDfThus,theratiosofthevelocitiesarevKfmD90.0kg1.50vm60.0kgDfKThenegativesignshowsthatthevelocitiesareinoppositedirections.c.Whichskaterhasthegreaterspeed?Keshia,whohasthesmallermass,hasthegreaterspeed.d.Whichskaterpushedharder?Theforceswereequalandopposite.82.A0.200-kgplasticballmoveswithavelocityof0.30m/s.Itcollideswithasecondplasticballofmass0.100kg,whichismovingalongthesamelineataspeedof0.10m/s.Afterthecollision,bothballscontinuemovinginthesame,originaldirection.Thespeedofthe0.100-kgballis0.26m/s.Whatisthenewvelocityofthe0.200-kgball?mCvCimDvDimCvCfmDvDfmso,vCvCimDvDimDvDfCfmC214SolutionsManualPhysics:PrinciplesandProblems

218Chapter9continued(0.200kg)(0.30m/s)(0.100kg)(0.10m/s)(0.100kg)(0.26m/s)0.200kg0.22m/sintheoriginaldirectionMixedReviewpages253–254Level183.Aconstantforceof6.00Nactsona3.00-kgobjectfor10.0s.Whatarethechangesintheobject’smomentumandvelocity?ThechangeinmomentumispFt(6.00N)(10.0s)60.0Ns60.0kgm/sThechangeinvelocityisfoundfromtheimpulse.FtmvFtvm(6.00N)(10.0s)3.00kg20.0m/s84.Thevelocityofa625-kgcarischangedfrom10.0m/sto44.0m/sin68.0sbyanexternal,constantforce.a.Whatistheresultingchangeinmomentumofthecar?pmvm(vfvi)(625kg)(44.0m/s10.0m/s)2.12104kgm/sb.Whatisthemagnitudeoftheforce?Ftmvmvso,Ftm(vfvi)t(625kg)(44.0m/s10.0m/s)68.0s313N85.DragsterAn845-kgdragsteracceleratesonaracetrackfromrestto100.0km/hin0.90s.a.Whatisthechangeinmomentumofthedragster?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pm(vfvi)1000m1h(845kg)(100.0km/h0.0km/h)1km3600s2.35104kgm/sPhysics:PrinciplesandProblemsSolutionsManual215

219Chapter9continuedb.Whatistheaverageforceexertedonthedragster?m(vFfvi)t(845kg)(100.0km/h0.0km/h)1000m1h0.90s1km3600s2.6104Nc.Whatexertsthatforce?Theforceisexertedbythetrackthroughfriction.Level286.IceHockeyA0.115-kghockeypuck,movingat35.0m/s,strikesa0.365-kgjacketthatisthrownontotheicebyafanofacertainhockeyteam.Thepuckandjacketslideofftogether.Findtheirvelocity.mPvPi(mPmJ)vfmvPmPifmPmJ(0.115kg)(35.0m/s)(0.115kg0.365kg)8.39m/s87.A50.0-kgwoman,ridingona10.0-kgcart,ismovingeastat5.0m/s.Thewomanjumpsoffthefrontofthecartandlandsonthegroundat7.0m/seastward,rela-tivetotheground.a.Sketchthe“before”and“after”situationsandassignacoordinateaxistothem.Before:mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.w50.0kgmc10.0kgvi5.0m/sAfter:mw50.0kgmc10.0kgvwf7.0m/svcf?BeforeAftervivwfvCf+x216SolutionsManualPhysics:PrinciplesandProblems

220Chapter9continuedb.Findthecart’svelocityafterthewomanjumpsoff.(mwmc)vimwvwfmcvcf(mso,vwmc)vimwvwfcfmc(50.0kg10.0kg)(5.0m/s)(50.0kg)(7.0m/s)10.0kg5.0m/s,or5.0m/swest88.GymnasticsFigure9-21showsagymnastperformingaroutine.First,shedoesgiantswingsonthehighbar,holdingherbodystraightandpivotingaroundherhands.Then,sheletsgoofthehighbarandgrabsherkneeswithherhandsinthetuckposition.Finally,shestraightensupandlandsonherfeet.a.Inthesecondandfinalpartsofthegymnast’sroutine,aroundwhataxisdoesshespin?Shespinsaroundthecenterofmassofherbody,firstinthetuckposi-tionandthenalsoasshestraightensout.b.Rankinorder,fromgreatesttoleast,hermomentsofinertiaforthethreepositions.giantswing(greatest),straight,tuck(least)c.Rankinorder,fromgreatesttoleast,herangularvelocitiesinthethreepositions.tuck(greatest),straight,giantswing(least)■Figure9-21Level389.A60.0-kgmaledancerleaps0.32mhigh.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Withwhatmomentumdoeshereachtheground?vv22dg0Thus,thevelocityofthedancerisv2dgPhysics:PrinciplesandProblemsSolutionsManual217

221Chapter9continuedandhismomentumispmvm2dg(60.0kg)(2)(0.32m)(9.80m/s2)1.5102kgm/sdownb.Whatimpulseisneededtostopthedancer?Ftmvm(vfvi)Tostopthedancer,vf0.Thus,Ftmv2kgm/supfp1.510c.Asthedancerlands,hiskneesbend,lengtheningthestoppingtimeto0.050s.Findtheaverageforceexertedonthedancer’sbody.Ftmvm2dgm2dgso,Ft(60.0kg)(2)(0.32m)(9.80m/s2)0.050s3.0103Nd.Comparethestoppingforcewithhisweight.F2)5.98102Ngmg(60.0kg)(9.80m/sTheforceisaboutfivetimestheweight.ThinkingCriticallypage25490.ApplyConceptsA92-kgfullback,runningat5.0m/s,attemptstodivedirectlyCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.acrossthegoallineforatouchdown.Justashereachestheline,heismethead-oninmidairbytwo75-kglinebackers,bothmovinginthedirectionoppositethefull-back.Oneismovingat2.0m/sandtheotherat4.0m/s.Theyallbecomeentan-gledasonemass.a.Sketchtheevent,identifyingthe“before”and“after”situations.Before:mA92kgmB75kgmC75kgvAi5.0m/svBi2.0m/svCi4.0m/sAfter:mA92kgmB75kgmC75kgvf?218SolutionsManualPhysics:PrinciplesandProblems

222Chapter9continuedBeforeAftervAivBivCivf+xb.Whatisthevelocityofthefootballplayersafterthecollision?pAipBipCipAfpBfpCfmAvAimBvBimCvCimAvAfmBvBfmCvCf(mAmBmC)vfmvAvAimBvBimCivCifmAmBmC(92kg)(5.0m/s)(75kg)(2.0m/s)(75kg)(4.0m/s)92kg75kg75kg0.041m/sc.Doesthefullbackscoreatouchdown?Yes.Thevelocityispositive,sothefootballcrossesthegoallineforatouchdown.91.AnalyzeandConcludeAstudent,holdingabicyclewheelwithitsaxisvertical,sitsonastoolthatcanrotatewithoutfriction.Sheusesherhandtogetthewheelspinning.Wouldyouexpectthestudentandstooltoturn?Ifso,inwhichdirection?Explain.Thestudentandthestoolwouldspinslowlyinthedirectionoppositetothatofthewheel.Withoutfrictiontherearenoexternaltorques.Thus,theangularmomentumofthesystemisnotchanged.Theangularmomentumofthestudentandstoolmustbeequalandoppositetotheangularmomentumofthespinningwheel.92.AnalyzeandConcludeTwoballsduringacollisionareshowninFigure9-22,whichisdrawntoscale.Theballsenterfromtheleft,collide,andthenbounceaway.Theheavierball,atthebottomofthediagram,hasamassof0.600kg,andtheotherhasamassof0.400kg.Usingavectordiagram,determinewhethermomentumisconservedinthiscollision.Explainanydifferenceinthemomentumofthesystembeforeandafterthecollision.■Figure9-22Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Dottedlinesshowthatthechangesofmomentumforeachballareequalandopposite:(mAvA)(mBvB).Becausethemasseshavea3:2ratio,a2:3ratioofvelocitychangeswillcompensate.Physics:PrinciplesandProblemsSolutionsManual219

223Chapter9continuedvB1Systemsarebeingdevelopedthatwilladjusttherateatwhichgasesfilltheairbagstomatchthesizeoftheperson.mB0.400kgvBvvB2B1CumulativeReviewpage254v95.A0.72-kgballisswungverticallyfromavA1A20.60-mstringinuniformcircularmotionatvAaspeedof3.3m/s.Whatisthetensioninthecordatthetopoftheball’smotion?vA1mA0.600kg(Chapter6)WritinginPhysicsThetensionisthegravitationalforcepage254minusthecentripetalforce.93.HowcanhighwaybarriersbedesignedtoFTFgFcbemoreeffectiveinsavingpeople’slives?mv2v2Researchthisissueanddescribehowmgmgrrimpulseandchangeinmomentumcanbeusedtoanalyzebarrierdesigns.(3.3m/s)2(0.72kg)(9.80m/s2)0.60mThechangeinacar’smomentumdoes6.0Nnotdependonhowitisbroughttoastop.Thus,theimpulsealsodoesnot96.Youwishtolaunchasatellitethatwillchange.Toreducetheforce,thetimeremainabovethesamespotonEarth’ssur-overwhichacarisstoppedmustbeface.Thismeansthesatellitemusthaveaincreased.Usingbarriersthatcanperiodofexactlyoneday.Calculatetheextendthetimeittakestostopacarradiusofthecircularorbitthissatellitewillreducetheforce.Flexible,plasticmusthave.Hint:TheMoonalsocirclesEarthcontainersfilledwithsandoftenareandboththeMoonandthesatellitewillobeyused.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Kepler’sthirdlaw.TheMoonis3.9108mfromEarthanditsperiodis27.33days.94.Whileairbagssavemanylives,theyalso(Chapter7)havecausedinjuriesandevendeath.ResearchtheargumentsandresponsesofTs2rs3automobilemakerstothisstatement.TmrmDeterminewhethertheproblemsinvolve21Ts33impulseandmomentumorotherissues.sorsTrmmTherearetwowaysanairbagreduces21injury.First,anairbagextendsthetime1.000day(3.9108m)3327.33daysoverwhichtheimpulseacts,therebyreducingtheforce.Second,anairbag4.3107mspreadstheforceoveralargerarea,therebyreducingthepressure.Thus,97.Aropeiswrappedaroundadrumthatistheinjuriesduetoforcesfromsmall0.600mindiameter.Amachinepullswithobjectsarereduced.Thedangersofairaconstant40.0Nforceforatotalof2.00s.bagsmostlycenteronthefactthatairInthattime,5.00mofropeisunwound.bagsmustbeinflatedveryrapidly.TheFind,at2.00s,andI.(Chapter8)surfaceofanairbagcanapproachtheTheangularaccelerationistheratioofpassengeratspeedsofupto322km/hthelinearaccelerationofthedrum’s(200mph).Injuriescanoccurwhentheedgeanddrum’sradius.movingbagcollideswiththeperson.220SolutionsManualPhysics:PrinciplesandProblems

224Chapter9continuedarThelinearaccelerationisfoundfromtheequationofmotion.12xat22xa2tThus,theangularaccelerationisa2x2rrt(2)(5.00m)0.600m(2.00s)228.33rad/s2Attheendof2.00s,theangularvelocityist2xtrt22xrt(2)(5.00m)0.600m(2.00s)216.7rad/sThemomentofinertiaisIFrsinFr2t2sin2x2x2rt(40.0N)0.600m2(2.00s)2(sin90.0°)2(2)(5.00m)1.44kgm2ChallengeProblempage244Yourfriendwasdrivingher1265-kgcarnorthonOakStreetwhenshewashitbya925-kgcompactcargoingwestonMapleStreet.Thecarsstucktogetherandslid23.1mat42°northofwest.Thespeedlimitonbothstreetsis22m/s(50mph).AssumethatmomentumwasconservedduringthecollisionandthataccelerationCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.wasconstantduringtheskid.Thecoefficientofkineticfrictionbetweenthetiresandthepavementis0.65.Physics:PrinciplesandProblemsSolutionsManual221

225Chapter9continuedy925kgx42°1265kg1.Yourfriendclaimsthatshewasn’tspeeding,butthatthedriverofothercarwas.Howfastwasyourfrienddrivingbeforethecrash?Thevectordiagramprovidesamomentumequationforthefriend’scar.pCipfsin42°Thefriend’sinitialvelocity,then,isp(mvCiCmD)vfsin42°CimmCCWecanfindvffirstbyfindingtheaccelerationandtimeoftheskid.TheaccelerationisFFg(mCmD)gagmmmCmDThetimecanbederivedfromthedistanceequation.12dat2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2d2dtagThefinalvelocity,then,is2dv2dgfatggUsingthis,wenowcanfindthefriend’sinitialvelocity.(mvCmD)vfsin42°CimC(mCmD)2dgsin42°mC(1265kg925kg)(2)(23.1m)(0.65)(9.80m/s2)(sin42°)1265kg2.0101m/s222SolutionsManualPhysics:PrinciplesandProblems

226Chapter9continued2.Howfastwastheothercarmovingbeforethecrash?Canyousupportyourfriend’scaseincourt?Fromthevectordiagram,themomentumequationfortheothercarispDipfcos42°(mCmD)vfcos42°(mCmD)2dg(cos42°)Theothercar’sinitialvelocity,then,is,pvDiDimD(mCmD)2dg(sin42°)mD(1265kg925kg)(2)(23.1m)(0.65)(9.80m/s2)(cos42°)925kg3.0101m/sThefriendwasnotexceedingthe22m/sspeedlimit.Theothercarwasexceedingthespeedlimit.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual223

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228CHAPTEREnergy,Work,and10SimpleMachines3.Arockclimberwearsa7.5-kgbackpackPracticeProblemswhilescalingacliff.After30.0min,the10.1EnergyandWorkclimberis8.2mabovethestartingpoint.pages257–265a.Howmuchworkdoestheclimberdopage261onthebackpack?1.RefertoExampleProblem1tosolvetheWFdfollowingproblem.mgda.Ifthehockeyplayerexertedtwiceas(7.5kg)(9.80m/s2)(8.2m)muchforce,9.00N,onthepuck,how2J6.010wouldthepuck’schangeinkineticb.Iftheclimberweighs645N,howmuchenergybeaffected?workdoesshedoliftingherselfandtheBecauseWFdandKEW,backpack?doublingtheforcewoulddoubleWFd6.0102Jthework,whichwoulddoublethechangeinkineticenergyto1.35J.(645N)(8.2m)6.0102Jb.Iftheplayerexerteda9.00N-force,but5.9103Jthestickwasincontactwiththepuckc.Whatistheaveragepowerdevelopedbyforonlyhalfthedistance,0.075m,theclimber?whatwouldbethechangeinkinetic3W5.910J1minenergy?Pt30.0min60sBecauseWFd,halvingthe3.3Wdistancewouldcuttheworkinhalf,whichalsowouldcutthechangeinkineticenergyinhalf,to0.68J.page2624.IfthesailorinExampleProblem2pulled2.Together,twostudentsexertaforceof825Nwiththesameforce,andalongthesameinpushingacaradistanceof35m.distance,butatanangleof50.0°,howa.Howmuchworkdothestudentsdoonmuchworkwouldhedo?thecar?WFdcosWFd(825N)(35m)(255N)(30.0m)(cos50.0°)2.9104J4.92103Jb.Iftheforcewasdoubled,howmuchworkwouldtheydopushingthecarthe5.Twopeopleliftaheavyboxadistanceofsamedistance?15m.Theyuseropes,eachofwhichmakesanangleof15°withthevertical.EachWFdpersonexertsaforceof225N.Howmuch(2)(825N)(35m)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.workdotheydo?5.8104JwhichistwiceasWFdcosmuchwork(2)(225N)(15m)(cos15°)6.5103JPhysics:PrinciplesandProblemsSolutionsManual225

229Chapter10continued6.Anairplanepassengercarriesa215-Nsuit-WFdcaseupthestairs,adisplacementof4.20m(25N)(275m)vertically,and4.60mhorizontally.3J6.910a.Howmuchworkdoesthepassengerdo?b.HowmuchworkisdonebytheforceofSincegravityactsvertically,onlythegravityonthebike?verticaldisplacementneedstobeTheforceisdownward(90°),andconsidered.thedisplacementis25°abovetheWFd(215N)(4.20m)903Jhorizontalor115°fromtheforce.b.ThesamepassengercarriesthesameWFdcossuitcasebackdownthesamesetofmgdcosstairs.Howmuchworkdoesthe(13kg)(9.80m/s2)(275m)passengerdonow?(cos115°)Forceisupward,butvertical1.5104Jdisplacementisdownward,soWFdcospage264(215N)(4.20m)(cos180.0°)9.Aboxthatweighs575Nisliftedadistance903Jof20.0mstraightupbyacableattachedtoamotor.Thejobisdonein10.0s.Whatpower7.AropeisusedtopullametalboxadistanceisdevelopedbythemotorinWandkW?of15.0macrossthefloor.TheropeisheldWFd(575N)(20.0m)atanangleof46.0°withthefloor,andaPtt10.0sforceof628Nisappliedtotherope.How1.15103W1.15kWmuchworkdoestheforceontheropedo?WFdcos10.Youpushawheelbarrowadistanceof(628N)(15.0m)(cos46.0°)60.0mataconstantspeedfor25.0s,by6.54103Jexertinga145-Nforcehorizontally.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Whatpowerdoyoudevelop?8.AbicycleriderpushesabicyclethathasaWFd(145N)(60.0m)massof13kgupasteephill.TheinclineisP348Wtt25.0s25°andtheroadis275mlong,asshownb.IfyoumovethewheelbarrowtwiceasinFigure10-4.Theriderpushesthebikefast,howmuchpowerisdeveloped?paralleltotheroadwithaforceof25N.tishalved,soPisdoubledto696W.11.Whatpowerdoesapumpdeveloptolift5m35Lofwaterperminutefromadepthof27110m?(1Lofwaterhasamassof1.00kg.)WmgdmPgdtttm25Nwhere(35L/min)(1.00kg/L)25°tThus,■Figure10-4(Nottoscale)a.HowmuchworkdoestheriderdoonPmtgdthebike?(35L/min)(1.00kg/L)(9.80m/s2)Forceanddisplacementareinthesamedirection.(110m)(1min/60s)0.63kW226SolutionsManualPhysics:PrinciplesandProblems

230Chapter10continued12.Anelectricmotordevelops65kWofpower1(15m)(210.0N40.0N)asitliftsaloadedelevator17.5min35s.2Howmuchforcedoesthemotorexert?1.9103JWFdPttPt(65103W)(35s)210.0Fd17.5m1.3105NForce(N)40.013.Awinchdesignedtobemountedona0truck,asshowninFigure10-7,isadver-15tisedasbeingabletoexerta6.8103-NDisplacement(m)forceandtodevelopapowerof0.30kW.Howlongwouldittakethetruckandthewinchtopullanobject15m?SectionReview10.1EnergyandWorkpages257–265page26515.WorkMurimipushesa20-kgmass10macrossafloorwithahorizontalforceof80N.CalculatetheamountofworkdonebyMurimi.WFd(80N)(10m)8102JThemassisnotimportanttothisproblem.■Figure10-716.WorkAmoverloadsa185-kgrefrigeratorWFdintoamovingvanbypushingitupaPtt10.0-m,friction-freerampatanangleofFdinclinationof11.0°.HowmuchworkistPdonebythemover?(6.8103N)(15m)y(10.0m)(sin11.0°)340s(0.30103W)1.91m5.7minWFdmgdsin(185kg)(9.80m/s2)(10.0m)(sin11.0°)14.Yourcarhasstalledandyouneedtopushit.Younoticeasthecargetsgoingthatyou3.46103Jneedlessandlessforcetokeepitgoing.Supposethatforthefirst15m,yourforce17.WorkandPowerDoestheworkrequireddecreasedataconstantratefrom210.0Ntotoliftabooktoahighshelfdependon40.0N.Howmuchworkdidyoudoonthehowfastyouraiseit?Doesthepowercar?Drawaforce-displacementgraphtorequiredtoliftthebookdependonhowCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.representtheworkdoneduringthisperiod.fastyouraiseit?Explain.TheworkdoneistheareaoftheNo,workisnotafunctionoftime.trapezoidunderthesolidline:However,powerisafunctionoftime,sothepowerrequiredtoliftthebookdoes1Wd(F21F2)dependonhowfastyouraiseit.Physics:PrinciplesandProblemsSolutionsManual227

231Chapter10continued18.PowerAnelevatorliftsatotalmassof23.CriticalThinkingExplainhowtofindthe1.1103kgadistanceof40.0min12.5s.changeinenergyofasystemifthreeagentsHowmuchpowerdoestheelevatorgenerate?exertforcesonthesystematonce.WFdmgdSinceworkisthechangeinkineticPtttenergy,calculatetheworkdonebyeach(1.1103kg)(9.80m/s2)(40.0m)force.Theworkcanbepositive,nega-12.5stive,orzero,dependingontherelative4Wanglesoftheforceanddisplacementof3.410theobject.Thesumofthethreeworksisthechangeinenergyofthesystem.19.WorkA0.180-kgballfalls2.5m.Howmuchworkdoestheforceofgravitydoontheball?PracticeProblemsWFgdmgd10.2Machines(0.180kg)(9.80m/s2)(2.5m)pages266–2734.4Jpage27224.IfthegearradiusinthebicycleinExample20.MassAforkliftraisesabox1.2mandProblem4isdoubled,whiletheforceexerteddoes7.0kJofworkonit.Whatisthemassonthechainandthedistancethewheelrimofthebox?movesremainthesame,whatquantitiesWFdmgdchange,andbyhowmuch?W7.0103Jrsome8.00cmgd(9.80m/s2)(1.2m)IMA0.225(doubled)r35.6cmr6.0102kge95.0MAIMA(0.225)10010021.WorkYouandafriendeachcarryidentical0.214(doubled)boxesfromthefirstfloorofabuildingtoaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.roomlocatedonthesecondfloor,fartherFrMAsoFdownthehall.YouchoosetocarrytheboxFr(MA)(Fe)efirstupthestairs,andthendownthehallto(0.214)(155N)theroom.Yourfriendcarriesitdownthehallonthefirstfloor,thenupadifferentstairwell33.2Ntothesecondfloor.Whodoesmorework?deIMABothdothesameamountofwork.drOnlytheheightliftedandtheverticalsodforceexertedcount.e(IMA)(dr)(0.225)(14.0cm)22.WorkandKineticEnergyIfthework3.15cmdoneonanobjectdoublesitskineticenergy,doesitdoubleitsvelocity?Ifnot,25.Asledgehammerisusedtodriveawedgebywhatratiodoesitchangethevelocity?intoalogtosplitit.Whenthewedgeisdri-Kineticenergyisproportionaltotheven0.20mintothelog,thelogisseparatedsquareofthevelocity,sodoublingtheadistanceof5.0cm.Aforceof1.7104Nenergydoublesthesquareoftheveloc-isneededtosplitthelog,andthesledge-ity.Thevelocityincreasesbyafactorofhammerexertsaforceof1.1104N.thesquarerootof2,or1.4.a.WhatistheIMAofthewedge?de(0.20m)IMA4.0d(0.050m)r228SolutionsManualPhysics:PrinciplesandProblems

232Chapter10continuedb.WhatistheMAofthewedge?27.Youexertaforceof225NonalevertoFraisea1.25103-NrockadistanceofrMA13cm.IftheefficiencyoftheleverisFe88.7percent,howfardidyoumoveyour(1.7104N)1.5endofthelever?(1.1104N)Woc.CalculatetheefficiencyofthewedgeasefficiencyW100iamachine.FrdrMA100e100FdIMAee1.510038%Frdr(100)4.0SodeFe(efficiency)26.Aworkerusesapulleysystemtoraise(1.25103N)(0.13m)(100)a24.0-kgcarton16.5m,asshownin(225N)(88.7)Figure10-14.Aforceof129Nisexerted,0.81mandtheropeispulled33.0m.28.Awinchhasacrankwitha45-cmradius.Aropeiswrappedaroundadrumwitha7.5-cmradius.Onerevolutionofthecrankturnsthedrumonerevolution.33.0ma.Whatistheidealmechanicaladvantage24.0kgofthismachine?Compareeffortandresistancedistancesfor1rev:16.5mde(2)45cmIMA6.0129Ndr(2)7.5cmb.If,duetofriction,themachineisonly■Figure10-1475percentefficient,howmuchforcea.WhatistheMAofthepulleysystem?wouldhavetobeexertedonthehandleFrmgofthecranktoexert750NofforceonMAtherope?FFeeMA(24.0kg)(9.80m/s2)efficiency100IMA129NFr1.82100(F(IMA)e)b.Whatistheefficiencyofthesystem?(Fr)(100)MAsoFefficiency100e(efficiency)(IMA)IMA(750N)(100)(MA)(100)(75)(6.0)dedr1.7102NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(MA)(dr)(100)de(1.82)(16.5m)(100)33.0m91.0%Physics:PrinciplesandProblemsSolutionsManual229

233Chapter10continuedde(3)(2r)SectionReviewIMAd2rr10.2Machines(3)(2)(45cm)pages266–273(2)(7.5cm)page2731829.SimpleMachinesClassifythetoolsbelowasalever,awheelandaxle,aninclined32.EfficiencySupposeyouincreasetheeffi-plane,awedge,orapulley.ciencyofasimplemachine.DotheMAanda.screwdriverIMAincrease,decrease,orremainthesame?wheelandaxleEitherMAincreaseswhileIMAremainsb.pliersthesame,orIMAdecreaseswhileMAremainsthesame,orMAincreasesleverwhileIMAdecreases.c.chiselwedge33.CriticalThinkingThemechanicaladvan-d.nailpullertageofamulti-gearbicycleischangedbymovingthechaintoasuitablereargear.levera.Tostartout,youmustacceleratethe30.IMAAworkeristestingamultiplepulleybicycle,soyouwanttohavethebicyclesystemtoestimatetheheaviestobjectthatexertthegreatestpossibleforce.Shouldhecouldlift.Thelargestdownwardforceheyouchooseasmallorlargegear?couldexertisequaltohisweight,875N.rgearlarge,toincreaseIMAWhentheworkermovestherope1.5m,therwheelobjectmoves0.25m.Whatistheheaviestb.Asyoureachyourtravelingspeed,youobjectthathecouldlift?wanttorotatethepedalsasfewtimesasFrpossible.ShouldyouchooseasmallorMAFelargegear?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.soFSmall,becauselesschaintravel,r(MA)(Fe)hencefewpedalrevolutions,willbeAssumingtheefficiencyis100%,requiredperwheelrevolution.dec.ManybicyclesalsoletyouchoosetheMAIMAd(Fe)rsizeofthefrontgear.Ifyouwanteven(1.5m)moreforcetoacceleratewhileclimbing(875N)(0.25m)ahill,wouldyoumovetoalargerorsmallerfrontgear?5.2103Nsmaller,toincreasepedal-frontgear31.CompoundMachinesAwinchhasacrankIMAbecauseona45-cmarmthatturnsadrumwitharpedalIMA7.5-cmradiusthroughasetofgears.Ittakesrfrontgearthreerevolutionsofthecranktorotatethedrumthroughonerevolution.WhatistheIMAofthiscompoundmachine?TheIMAofthesystemistheproductoftheIMAofeachmachine.Forthecrankanddrum,theratioofdistancesis2(45cm)6.0.2(7.5cm)230SolutionsManualPhysics:PrinciplesandProblems

234Chapter10continued39.WhatisawattequivalenttointermsofChapterAssessmentkilograms,meters,andseconds?(10.1)ConceptMappingWJ/spage278Nm/s34.Createaconceptmapusingthefollowing(kgm/s2)m/sterms:force,displacement,directionofmotion,kgm2/s3work,changeinkineticenergy.40.IsitpossibletogetmoreworkoutofaDirectionForceDisplacementmachinethanyouputintoit?(10.2)ofmotionno,e100%Work41.Explainhowthepedalsofabicycleareasimplemachine.(10.2)PedalstransferforcefromtheridertoChangeinthebikethroughawheelandaxle.kineticenergyApplyingConceptsMasteringConceptspage278page27842.Whichrequiresmorework,carrying35.Inwhatunitsisworkmeasured?(10.1)a420-Nbackpackupa200-m-highhilljoulesorcarryinga210-Nbackpackupa400-m-highhill?Why?36.SupposeasatelliterevolvesaroundEarthinEachrequiresthesameamountofacircularorbit.DoesEarth’sgravitydoanyworkbecauseforcetimesdistanceisworkonthesatellite?(10.1)thesame.No,theforceofgravityisdirectedtowardEarthandisperpendiculartothe43.LiftingYouslowlyliftaboxofbooksfromdirectionofdisplacementofthesatellite.thefloorandputitonatable.Earth’sgravityexertsaforce,magnitudemg,downward,and37.Anobjectslidesatconstantspeedonayouexertaforce,magnitudemg,upward.frictionlesssurface.WhatforcesactontheThetwoforceshaveequalmagnitudesandobject?Whatworkisdonebyeachforce?oppositedirections.Itappearsthatnowork(10.1)isdone,butyouknowthatyoudidwork.Explainwhatworkwasdone.Onlygravityandanupward,normalforceactontheobject.NoworkisYoudopositiveworkontheboxdonebecausethedisplacementisbecausetheforceandmotionareintheperpendiculartotheseforces.Theresamedirection.Gravitydoesnegativeisnoforceinthedirectionofdisplace-workontheboxbecausetheforceofmentbecausetheobjectisslidingatagravityisoppositetothedirectionofconstantspeed.motion.Theworkdonebyyouandbygravityareseparateanddonotcancel38.Defineworkandpower.(10.1)eachother.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Workistheproductofforceandthe44.Youhaveanafter-schooljobcarryingcar-distanceoverwhichanobjectismovedtonsofnewcopypaperupaflightofstairs,inthedirectionoftheforce.Powerisandthencarryingrecycledpaperbackdownthetimerateatwhichworkisdone.thestairs.ThemassofthepaperdoesnotPhysics:PrinciplesandProblemsSolutionsManual231

235Chapter10continuedchange.Yourphysicsteachersaysthatyou48.Howcanyouincreasetheidealmechanicaldonotworkallday,soyoushouldnotbeadvantageofamachine?paid.InwhatsenseisthephysicsteacherIncreasetheratioofde/drtoincreasecorrect?WhatarrangementofpaymentstheIMAofamachine.mightyoumaketoensurethatyouareproperlycompensated?49.WedgeHowcanyouincreasethemechani-Thenetworkiszero.Carryingthecartoncaladvantageofawedgewithoutchangingupstairsrequirespositivework;carryingitsidealmechanicaladvantage?itbackdownisnegativework.TheworkReducefrictionasmuchaspossibletodoneinbothcasesisequalandoppo-reducetheresistanceforce.sitebecausethedistancesareequalandopposite.Thestudentmightarrangethe50.OrbitsExplainwhyaplanetorbitingtheSunpaymentsonthebasisofthetimeitdoesnotviolatethework-energytheorem.takestocarrypaper,whetheruporAssumingacircularorbit,theforcedown,notonthebasisofworkdone.duetogravityisperpendiculartothedirectionofmotion.Thismeansthe45.Youcarrythecartonsofcopypaperdownworkdoneiszero.Hence,thereisnothestairs,andthenalonga15-m-longchangeinkineticenergyoftheplanet,hallway.Areyouworkingnow?Explain.soitdoesnotspeeduporslowdown.No,theforceontheboxisupandtheThisistrueforacircularorbit.displacementisdownthehall.Theyareperpendicularandnoworkisdone.51.ClawHammerAclawhammerisusedtopullanailfromapieceofwood,asshown46.ClimbingStairsTwopeopleofthesameinFigure10-16.Whereshouldyouplacemassclimbthesameflightofstairs.Theyourhandonthehandleandwhereshouldfirstpersonclimbsthestairsin25s;thethenailbelocatedintheclawtomakethesecondpersondoessoin35s.effortforceassmallaspossible?a.Whichpersondoesmorework?ExplainCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.youranswer.Bothpeoplearedoingthesameamountofworkbecausetheybothareclimbingthesameflightofstairsandtheyhavethesamemass.b.Whichpersonproducesmorepower?Explainyouranswer.Thepersonwhoclimbsin25s■Figure10-16expendsmorepower,aslesstimeisYourhandshouldbeasfarfromtheneededtocoverthedistance.headaspossibletomakedeaslargeaspossible.Thenailshouldbeascloseto47.ShowthatpowerdeliveredcanbewrittenastheheadaspossibletomakedPFvcos.rassmallaspossible.WP,butWFdcostFdcosso,Ptdbecausev,tPFvcos232SolutionsManualPhysics:PrinciplesandProblems

236Chapter10continuedMasteringProblemsWFdPtt10.1EnergyandWork(15.0N)(2.51m)pages278–28030.0sLevel1126W52.Thethirdfloorofahouseis8mabovestreetlevel.Howmuchworkisneededtomovea58.Astudentlibrarianliftsa2.2-kgbookfrom150-kgrefrigeratortothethirdfloor?thefloortoaheightof1.25m.HecarriestheWFdmgdbook8.0mtothestacksandplacesthebook(150kg)(9.80m/s2)(8m)onashelfthatis0.35mabovethefloor.Howmuchworkdoeshedoonthebook?1104JOnlythenetverticaldisplacementcounts.53.Halokedoes176Jofworkliftinghimself0.300m.WhatisHaloke’smass?WFdmgdWFdmgd;therefore,(2.2kg)(9.80m/s2)(0.35m)W176J7.5Jmgd(9.80m/s2)(0.300m)59.9kg59.Aforceof300.0Nisusedtopusha145-kgmass30.0mhorizontallyin3.00s.54.FootballAfterscoringatouchdown,ana.Calculatetheworkdoneonthemass.84.0-kgwidereceivercelebratesbyleapingWFd(300.0N)(30.0m)1.20mofftheground.Howmuchworkwasdonebythewidereceiverinthe9.00103Jcelebration?9.00kJWFdmgdb.Calculatethepowerdeveloped.(84.0kg)(9.80m/s2)(1.20m)W9.00103JPt3.00s988J3.00103W55.Tug-of-WarDuringatug-of-war,teamA3.00kWdoes2.20105JofworkinpullingteamB8.00m.WhatforcewasteamAexerting?Level2WFd,so60.WagonAwagonispulledbyaforceofW2.20105J4N38.0NexertedonthehandleatanangleofF2.7510d8.00m42.0°withthehorizontal.Ifthewagonispulledinacircleofradius25.0m,how56.Tokeepacartravelingataconstantvelocity,muchworkisdone?a551-NforceisneededtobalancefrictionalWFdcosforces.Howmuchworkisdoneagainstfric-(F)(2r)costionbythecarasittravelsfromColumbustoCincinnati,adistanceof161km?(38.0N)(2)(25.0m)(cos42.0°)WFd(551N)(1.61105m)4.44103JCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8.87107J61.LawnMowerShaniispushingalawnmowerwithaforceof88.0Nalongahan-57.CyclingAcyclistexertsaforceof15.0Nasdlethatmakesanangleof41.0°withtheheridesabike251min30.0s.Howmuchhorizontal.Howmuchworkisdonebypowerdoesthecyclistdevelop?Physics:PrinciplesandProblemsSolutionsManual233

237Chapter10continuedShaniinmovingthelawnmower1.2km(225N)(1.15m)tomowtheyard?sin30.0°WFdcos518J(88.0N)(1.2103m)(cos41.0°)65.PianoA4.2103-Npianoistobeslidupa8.0104J3.5-mfrictionlessplankataconstantspeed.Theplankmakesanangleof30.0°withthe62.A17.0-kgcrateistobepulledadistanceofhorizontal.Calculatetheworkdonebythe20.0m,requiring1210Jofworktobepersonslidingthepianouptheplank.done.IfthejobisdonebyattachingaropeTheforceparalleltotheplaneisgivenbyandpullingwithaforceof75.0N,atwhatFangleistheropeheld?FsinsoWFWFdcosdFdsinW(4200N)(3.5m)(sin30.0°)1WcosFd7.4103J11210Jcos(75.0N)(20.0m)66.SledDiegopullsa4.5-kgsledacrosslevelsnowwithaforceof225Nonaropethat36.2°is35.0°abovethehorizontal,asshowninFigure10-18.Ifthesledmovesadistance63.LawnTractorA120-kglawntractor,of65.3m,howmuchworkdoesDiegodo?showninFigure10-17,goesupa21°inclinethatis12.0mlongin2.5s.CalculatethepowerthatisdevelopedN25bythetractor.2kg4.5kg35.0°120.0mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc..012■Figure10-18WFdcos21°(225N)(65.3m)(cos35.0°)■Figure10-174J1.2010WFdsinmgdsinPttt67.EscalatorSau-Lanhasamassof52kg.(120kg)(9.80m/s2)(12.0m)(sin21°)SheridesuptheescalatoratOceanParkinHongKong.Thisistheworld’slongest2.5sescalator,withalengthof227mandan2.0103W2.0kWaverageinclinationof31°.HowmuchworkdoestheescalatordoonSau-Lan?64.Youslideacrateuparampatanangleof30.0°byexertinga225-NforceparalleltoWFdsinmgdsintheramp.Thecratemovesataconstant(52kg)(9.80m/s2)(227m)(sin31°)speed.Thecoefficientoffrictionis0.28.6.0104JHowmuchworkdidyoudoonthecrateasitwasraisedaverticaldistanceof1.15m?68.LawnRollerAlawnrollerispushedFanddareparallelsoacrossalawnbyaforceof115Nalonghthedirectionofthehandle,whichis22.5°WFdFsin234SolutionsManualPhysics:PrinciplesandProblems

238Chapter10continuedabovethehorizontal.If64.6Wofpowerisdevelopedfor90.0s,whatdistanceistherollerpushed?WFdcoskgPso,.0tt60m1.0m2.0PtdNFcos0.040(64.6W)(90.0s)(115N)(cos22.5°)■Figure10-1954.7ma.HowmuchworkdoesMaricruzdoinslidingthecrateuptheramp?69.Johnpushesacrateacrossthefloorofafactorywithahorizontalforce.Therough-WFd(400.0N)(2.0m)8.0102Jnessofthefloorchanges,andJohnmustb.Howmuchworkwouldbedoneifexertaforceof20Nfor5m,then35NforMaricruzsimplyliftedthecratestraight12m,andthen10Nfor8m.upfromthefloortotheplatform?a.DrawagraphofforceasafunctionofWFdmgddistance.(60.0kg)(9.80m/s2)(1.0m)405.9102J302071.BoatEngineAnenginemovesaboatForce(N)throughthewaterataconstantspeedof1015m/s.Theenginemustexertaforce0510152025of6.0kNtobalancetheforcethattheDisplacement(m)waterexertsagainstthehull.Whatpowerdoestheenginedevelop?b.FindtheworkJohndoespushingtheWFdPFvcrate.ttWF(6.0103N)(15m/s)1d1F2d2F3d3(20N)(5m)(35N)(12m)9.0104W9.0101kW(10N)(8m)Level3600J72.InFigure10-20,themagnitudeoftheforcenecessarytostretchaspringisplotted70.Maricruzslidesa60.0-kgcrateupanagainstthedistancethespringisstretched.inclinedrampthatis2.0-mlongandattachedtoaplatform1.0mabovefloorlevel,asshowninFigure10-19.A400.0-N8.00force,paralleltotheramp,isneededtoslidethecrateuptherampataconstant6.00speed.4.00Force(N)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.000.000.100.200.30■Figure10-20Physics:PrinciplesandProblemsSolutionsManual235

239Chapter10continueda.Calculatetheslopeofthegraph,k,and74.Aworkerpushesacrateweighing93NupshowthatFkd,wherek25N/m.aninclinedplane.Theworkerpushesthey5.00N0.00Ncratehorizontally,paralleltotheground,askx0.20m0.00millustratedinFigure10-21.F1kd1Letd10.20m85NFromthegraph,Fm3.0m1is5.00N..05F1Sokd14.0m5.00N25N/m0.20m93Nb.Findtheamountofworkdonein■Figure10-21stretchingthespringfrom0.00mtoa.Theworkerexertsaforceof85N.0.20mbycalculatingtheareaunderHowmuchworkdoeshedo?thegraphfrom0.00mto0.20m.Displacementindirectionofforce1A(base)(height)is4.0m,21soWFd(85N)(4.0m)(0.20m)(5.00N)23.4102J0.50Jb.Howmuchworkisdonebygravity?c.Showthattheanswertopartb(Becarefulwiththesignsyouuse.)canbecalculatedusingtheformula1DisplacementindirectionofforceWkd2,whereWisthework,2is3.0m,k25N/m(theslopeofthegraph),soWFd(93N)(3.0m)anddisthedistancethespringisstretched(0.20m).2.8102JCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.Thecoefficientoffrictionis0.20.W1kd21222(25N/m)(0.20m)Howmuchworkisdonebyfriction?(Becarefulwiththesignsyouuse.)0.50JWFNd(Fyou,Fg,)d73.UsethegraphinFigure10-20tofindthe0.20(85N)(sin)workneededtostretchthespringfrom0.12mto0.28m.(93N)(cos)(5.0m)Addtheareasofthetriangleand3.00.20(85N)rectangle.Theareaofthetriangleis:5.01bh1(0.28m0.12m)(7.00N3.00N)4.022(93N)5.0(5.0m)0.32J1.3102J(workdoneagainstTheareaoftherectangleis:friction)bh(0.28m0.12m)(3.00N0.00N)75.OilPumpIn35.0s,apumpdelivers0.48J0.550m3ofoilintobarrelsonaplatformTotalworkis0.32J0.48J0.80J25.0mabovetheintakepipe.Theoil’sdensityis0.820g/cm3.236SolutionsManualPhysics:PrinciplesandProblems

240Chapter10continueda.Calculatetheworkdonebythepump.78.ThegraphinFigure10-22showstheforceTheworkdoneisanddisplacementofanobjectbeingpulled.WFgdmgh(volume)(density)gh(0.550m3)(0.820g/cm3)1kg40.01000g(1.00106cm3/m3)(9.80m/s2)Force(N)20.0(25.0m)1.10105Jb.Calculatethepowerproducedbythe0.02.04.06.0pump.Displacement(m)W1.10105J■Figure10-22Pt35.0sa.Calculatetheworkdonetopullthe3.14103W3.14kWobject7.0m.76.ConveyorBeltA12.0-m-longconveyorFindtheareaunderthecurve(seebelt,inclinedat30.0°,isusedtograph):transportbundlesofnewspapersfromthe0.0to2.0m:mailroomuptothecargobaytobe11J(20.0N)(2.0m)2.010loadedontodeliverytrucks.Eachnewspa-2perhasamassof1.0kg,andthereare2.0mto3.0m:25newspapersperbundle.Determinethe1powerthattheconveyordevelopsifit(30.0N)(1.0m)(20N)(1.0m)35J2delivers15bundlesperminute.3.0mto7.0m:WFdmgdPttt(50.0N)(4.0m)2.0102J(25newspapers)(15bundles/min)Totalwork:(1.0kg/newspaper)(9.80m/s2)2.0101J35J2.0102J(12.0m)(sin30.0°)(1min/60s)2.6102J3.7102Wb.Calculatethepowerthatwouldbedevelopediftheworkwasdonein2.0s.77.AcarisdrivenataconstantspeedofW2.6102J2WP1.31076km/hdownaroad.Thecar’senginet2.0sdelivers48kWofpower.Calculatetheaverageforcethatisresistingthemotionofthecar.WFdPFvttPCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.soFv48,000W76km1000m1h1h1km3600s2.3103NPhysics:PrinciplesandProblemsSolutionsManual237

241Chapter10continued10.2Machines81.Apulleysystemliftsa1345-Nweightapages280–281distanceof0.975m.PaulpullstheropeaLevel1distanceof3.90m,exertingaforceof375N.79.PianoTakeshiraisesa1200-Npianoaa.Whatistheidealmechanicaladvantagedistanceof5.00musingasetofpulleys.ofthesystem?Hepullsin20.0mofrope.de3.90mIMA4.00a.HowmucheffortforcewouldTakeshid0.975mrapplyifthiswereanidealmachine?b.Whatisthemechanicaladvantage?FdreFdFr1345NerMA3.59F375NeFrdr(1200N)(5.00m)soFc.Howefficientisthesystem?ed20.0meMA2Nefficiency1003.010IMAb.Whatforceisusedtobalancethefriction3.59100forceiftheactualeffortis340N?4.00F89.8%eFfFe,idealF2N82.Aforceof1.4NisexertedthroughafFeFe,ideal340N3.010distanceof40.0cmonaropeinapulley40Nsystemtolifta0.50-kgmass10.0cm.c.Whatistheoutputwork?Calculatethefollowing.WoFrdr(1200N)(5.00m)a.theMA6.0103JFmgrMAd.Whatistheinputwork?FeFeWiFede(340N)(20.0m)(0.50kg)(9.80m/s2)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3J1.4N6.8103.5e.Whatisthemechanicaladvantage?b.theIMAFr1200NMA3.5dF340Ne40.0cmeIMA4.00d10.0cmr80.LeverBecausethereisverylittlefriction,c.theefficiencytheleverisanextremelyefficientsimpleMAmachine.Usinga90.0-percent-efficientefficiency100IMAlever,whatinputworkisrequiredtoliftan3.518.0-kgmassthroughadistanceof0.50m?10088%4.00WoefficiencyW10083.Astudentexertsaforceof250Nonalever,ithroughadistanceof1.6m,asheliftsa(Wo)(100)(mgd)(100)150-kgcrate.IftheefficiencyoftheleverisWiefficiency90.090.0percent,howfaristhecratelifted?(18.0kg)(9.80m/s2)(0.50m)(100)Fr90.0FMAee9010010098JIMAdedFrrdr100Fdee238SolutionsManualPhysics:PrinciplesandProblems

242Chapter10continuedeFedeeFedeb.WhatistheIMAoftheramp?so,dr100F100mgdre18mIMA4.0(90.0)(250N)(1.6m)df4.5m(100)(150kg)(9.80m/s2)c.WhataretherealMAandtheefficiency0.24moftherampifaparallelforceof75Nisactuallyrequired?FrLevel2MAF84.Whatworkisrequiredtolifta215-kgmasseadistanceof5.65m,usingamachinethatmg(25kg)(9.80m/s2)3.3is72.5percentefficient?F75NeWoMAe100efficiency100WIMAi3.3Frdr10082%1004.0Wi86.BicycleLuisapedalsabicyclewithagearmgdrradiusof5.00cmandawheelradiusof100Wi38.6cm,asshowninFigure10-24.Ifthemgdwheelrevolvesonce,whatisthelengthofrW100iethechainthatwasused?(215kg)(9.80m/s2)(5.65m)(100)72.51.64104J38.6cm85.TherampinFigure10-23is18mlongand4.5mhigh.5.00cm18mFA4.5m■Figure10-24d2r2(5.00cm)31.4cmFg■Figure10-23Level3a.Whatforce,paralleltotheramp(FA),is87.CraneAmotorwithanefficiencyofrequiredtoslidea25-kgboxatconstant88percentoperatesacranewithanspeedtothetopoftherampiffrictionefficiencyof42percent.Ifthepowerisdisregarded?suppliedtothemotoris5.5kW,withWFgdmghwhatconstantspeeddoesthecraneliftmgha410-kgcrateofmachineparts?soFFgdTotalefficiency(88%)(42%)37%Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(25kg)(9.80m/s2)(4.5m)UsefulPower(5.5kW)(37%)18m2.0kW61N3W2.010Physics:PrinciplesandProblemsSolutionsManual239

243Chapter10continuedWFddc.IfyoumovetheeffortsideoftheleverPFFvttt12.0cm,howfaristheboxlifted?PP2.0103Wde1vFmg2)IMAcg(410kg)(9.80m/sdr20.50m/sde112.0cmd2.0cmr2IMA6.0c88.Acompoundmachineisconstructedbyattachingalevertoapulleysystem.MixedReviewConsideranidealcompoundmachinecon-pages281–282sistingofaleverwithanIMAof3.0andapulleysystemwithanIMAof2.0.Level189.RampsIsrahastogetapianoontoaa.ShowthattheIMAofthiscompound2.0-m-highplatform.Shecanuseamachineis6.0.3.0-m-longfrictionlessrampora4.0-m-longWi1Wo1Wi2Wo2frictionlessramp.WhichrampshouldIsrauseWifshewantstodotheleastamountofwork?i1Wo2Eitherramp:onlytheverticaldistanceFe1de1Fr2dr2isimportant.IfIsrausedalongerramp,Forthecompoundmachineshewouldrequirelessforce.Theworkdonewouldbethesame.de1IMAcdr290.Brutus,achampionweightlifter,raisesde1de2240kgofweightsadistanceof2.35m.IMAIMAd1andd2a.HowmuchworkisdonebyBrutusr1r2dliftingtheweights?r1de2WFdmgdde1dr1de2(IMA2)(dr2)(240kg)(9.80m/s2)(2.35m)IMA1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.d5.5103Je1(IMA1)(IMA2)(dr2)b.HowmuchworkisdonebyBrutusde1IMAholdingtheweightsabovehishead?dc(IMA1)(IMA2)r2d0,sonowork(3.0)(2.0)6.0c.HowmuchworkisdonebyBrutusb.Ifthecompoundmachineis60.0percentloweringthembacktotheground?efficient,howmucheffortmustbedisoppositeofmotioninparta,soappliedtothelevertolifta540-Nbox?Wisalsotheopposite,5.5103J.Frd.DoesBrutusdoworkifheletsgooftheMAFeweightsandtheyfallbacktotheground?e100100IMAIMANo.Heexertsnoforce,sohedoes(F)(100)rnowork,positiveornegative.(F)(IMA)ee.IfBrutuscompletestheliftin2.5s,how(Fr)(100)muchpowerisdeveloped?soFe(e)(IMA)3JW5.510P2.2kW(540N)(100)t2.5s150N(60.0)(6.0)240SolutionsManualPhysics:PrinciplesandProblems

244Chapter10continuedLevel293.Sallydoes11.4kJofworkdraggingawooden91.Ahorizontalforceof805Nisneededtocrate25.0macrossafloorataconstantdragacrateacrossahorizontalfloorwithaspeed.Theropemakesanangleof48.0°constantspeed.Youdragthecrateusingawiththehorizontal.ropeheldatanangleof32°.a.Howmuchforcedoestheropeexertona.Whatforcedoyouexertontherope?thecrate?FxFcosWFdcosW11,400JFx805NsoFsoFdcos(25.0m)(cos48.0°)coscos32°2N681N9.510b.Whatistheforceoffrictionactingonb.Howmuchworkdoyoudoonthecratethecrate?ifyoumoveit22m?ThecratemoveswithconstantWFxd(805N)(22m)speed,sotheforceoffrictionequals1.8104Jthehorizontalcomponentofthec.Ifyoucompletethejobin8.0s,whatforceoftherope.powerisdeveloped?FfFxFcosW1.8104J(681N)(cos48.0°)P2.2kWt8.0s456N,oppositetothedirectionofmotion92.DollyandRampAmover’sdollyisusedtotransportarefrigeratoruparampintoc.Whatworkisdonebythefloorthroughahouse.Therefrigeratorhasamassoftheforceoffrictionbetweenthefloor115kg.Therampis2.10mlongandrisesandthecrate?0.850m.ThemoverpullsthedollywithaForceanddisplacementareinforceof496Nuptheramp.Thedollyandoppositedirections,sorampconstituteamachine.WFd(456N)(25.0m)a.Whatworkdoesthemoverdo?1.14104JWiFd(496N)(2.10m)(Becausenonetforcesactonthe1.04103Jcrate,theworkdoneonthecratemustbeequalinmagnitudebutb.WhatistheworkdoneontheoppositeinsigntotheenergySallyrefrigeratorbythemachine?expands:1.14104J)dheightraised0.850mW94.SleddingAn845-NsledispulledaoFgdmgddistanceof185m.Thetaskrequires(115kg)(9.80m/s2)(0.850m)1.20104Jofworkandisdonebypullingonaropewithaforceof125N.Atwhat958Jangleistheropeheld?c.Whatistheefficiencyofthemachine?WFdcos,soWo1.20104Jefficiency100cos1W1WcosiFd(125N)(185m)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.958J10058.7°1.04103J92.1%Physics:PrinciplesandProblemsSolutionsManual241

245Chapter10continuedLevel3onlyoneboxatatime,mostoftheenergy95.Anelectricwinchpullsa875-Ncrateupawillgointoraisingyourownbody.The15°inclineat0.25m/s.Thecoefficientofpower(inwatts)thatyourbodycandevelopfrictionbetweenthecrateandinclineis0.45.overalongtimedependsonthemassthata.Whatpowerdoesthewinchdevelop?youcarry,asshowninFigure10-25.ThisisanexampleofapowercurvethatappliestoWorkisdoneonthecratebythemachinesaswellastopeople.Findthewinch,gravity,andfriction.Becausenumberofboxestocarryoneachtripthatthekineticenergyofthecratedoeswouldminimizethetimerequired.Whatnotchange,thesumofthethreetimewouldyouspenddoingthejob?Ignoreworksisequaltozero.thetimeneededtogobackdownthestairsTherefore,andtoliftandlowereachbox.WwinchWfrictionWgravityPowerv.Massor,PwinchPfrictionPgravity25FFdNdg20tt15ddFNtFgt10Power(W)FNvFgv5(F0g)(cos)(v)Fgv51015202530(0.45)(875N)(cos15°)Mass(kg)■Figure10-25(0.25m/s)Theworkhastobedonethesame,(875N)(0.25m/s)WFgdmgd3.1102W(150kg)(9.80m/s2)(12m)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Ifthewinchis85percentefficient,whatistheelectricalpowerthatmust1.76104J.bedeliveredtothewinch?Fromthegraph,themaximumpowerisWoPo25Wat15kg.SincethemassperboxiseWiPi150kg5kg,thisrepresentsthree30boxesPoso,Pieboxes.3.1102WPWsotWtt0.853.6102W1.76104J25WThinkingCritically7.0102spage28212min96.AnalyzeandConcludeYouworkatastore,carryingboxestoastorageloftthatis12m97.ApplyConceptsAsprinterofmass75kgabovetheground.Youhave30boxeswitharunsthe50.0-mdashin8.50s.Assumethattotalmassof150kgthatmustbemovedasthesprinter’saccelerationisconstantquicklyaspossible,soyouconsidercarryingthroughouttherace.morethanoneupatatime.Ifyoutrytoa.Whatistheaveragepowerofthemovetoomanyatonce,youknowthatyousprinteroverthe50.0m?willgoveryslowly,restingoften.Ifyoucarry242SolutionsManualPhysics:PrinciplesandProblems

246Chapter10continuedAssumeconstantacceleration,Finalvelocity:thereforeconstantforcevfviat12ddivit2atvi0sobutdivi0vfata(t1)2dm2dWFdmadtTherefore,Ptttt1dat2at2md2(2)(75kg)(50.0m)f211t2t3(8.50s)31at2t2W211t26.110db.Whatisthemaximumpowergeneratedfa1t2tbythesprinter?211t2Powerincreaseslinearlyfromzero,50.0msincethevelocityincreaseslinearly1(1.00s)2(1.00s)(7.50s)asshownby2WFddPttFtFv.6.25m/s2ThereforeForthefirstsecond:P3Wd1at212)(1.00s)2max2Pave1.210(6.25m/s22c.Makeaquantitativegraphofpower3.12mversustimefortheentirerace.FromProblem97,madPt1.2x103(75kg)(6.25m/s2)(3.12m)Power(W)Pave1.00s1.5103W08.50Time(s)b.Whatisthemaximumpowerthatthesprinternowgenerates?98.ApplyConceptsThesprinterintheP3Wmax2Pave3.010previousproblemrunsthe50.0-mdashinthesametime,8.50s.However,thisWritinginPhysicstimethesprinteracceleratesinthefirstpage282secondandrunstherestoftheraceata99.Justasabicycleisacompoundmachine,soconstantvelocity.isanautomobile.Findtheefficienciesofa.Calculatetheaveragepowerproducedthecomponentpartsofthepowertrainforthatfirstsecond.(engine,transmission,wheels,andtires).DistancefirstsecondExplorepossibleimprovementsineachofDistancerestofrace50.0mtheseefficiencies.12Theoverallefficiencyis15–30percent.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.dfdivitat2Thetransmission’sefficiencyisaboutd90percent.Rollingfrictioninthetiresisivi0soabout1percent(ratioofpushingforced1a(t2vf21)f(t2)50.0mtoweightmoved).Thelargestgainispossibleintheengine.Physics:PrinciplesandProblemsSolutionsManual243

247Chapter10continued100.Thetermsforce,work,power,andenergy12dgtfydiyviyt2oftenmeanthesamethingineverydayuse.Obtainexamplesfromadvertisements,diyviy0printmedia,radio,andtelevisionthat12illustratemeaningsforthesetermsthatdif-sodfy2gtferfromthoseusedinphysics.1(9.80m/s2)(0.457s)2Answerswillvary.Someexamples2include,thecompanyConsumers’1.02mPowerchangeditsnametoConsumers’Energywithoutchanging103.PeoplesometimessaythattheMoonstaysitsproduct,naturalgas.“It’snotjustinitsorbitbecausethe“centrifugalforceenergy,it’spower!”hasappearedinjustbalancesthecentripetalforce,givingthepopularpress.nonetforce.”Explainwhythisideaiswrong.(Chapter8)CumulativeReviewThereisonlyoneforceonthemoon,page282thegravitationalforceofEarth’smass101.Youarehelpingyourgrandmotherwithonit.Thisnetforcegivesitanacceler-somegardeningandhavefilledagarbageationwhichisitscentripetalaccelera-canwithweedsandsoil.NowyouhavetotiontowardEarth’scenter.movethegarbagecanacrosstheyardandrealizeitissoheavythatyouwillneedtopushit,ratherthanliftit.IfthecanhasaChallengeProblemmassof24kg,thecoefficientofkineticfrictionbetweenthecan’sbottomandthepage268muddygrassis0.27,andthestaticcoeffi-Anelectricpumppullswateratarateof0.25cientoffrictionbetweenthosesamesur-m3/sfromawellthatis25mdeep.Thewaterfacesis0.35,howharddoyouhavetoleavesthepumpataspeedof8.5m/s.pushhorizontallytogetthecantojustCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.startmoving?(Chapter5)8.5m/sFyouoncanFfrictionsFNsmg(0.35)(24kg)(9.80m/s2)82N102.BaseballIfamajorleaguepitcherthrowsafastballhorizontallyataspeedof40.3m/s(90mph)andittravels18.4m(60ft,6in),25mhowfarhasitdroppedbythetimeitcrosseshomeplate?(Chapter6)dfxdixvxtddfxixsotvx18.4m0.0m0.457s40.3m/s244SolutionsManualPhysics:PrinciplesandProblems

248Chapter10continued1.Whatpowerisneededtoliftthewatertothesurface?TheworkdoneinliftingisFgdmgd.Therefore,thepowerisWFgdmgdPliftttt(0.25m3)(1.00103kg/m3)(9.80m/s2)(25m)1.0s6.1104W61kW2.Whatpowerisneededtoincreasethepump’skineticenergy?12.Theworkdoneinincreasingthepump’skineticenergyismv212mvWKE2mv2Therefore,Pttt2t(0.25m3)(1.00103kg/m3)(8.5m/s)2(2)(1.0s)9.0103W9.0kW3.Ifthepump’sefficiencyis80percent,howmuchpowermustbedeliveredtothepump?WoWtPooe100100100so,WWiPiitPo9.0103WP100100ie801.1104W11kWCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual245

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250CHAPTER11EnergyandItsConservationThefinalkineticenergyisPracticeProblemsKE121(875.0kg)(44.0m/s)211.1TheManyFormsofEnergyf2mv2pages285–2928.47105Jpage287Theworkdoneis1.Askaterwithamassof52.0kgmovingatKE5J2.12105J2.5m/sglidestoastopoveradistanceoffKEi8.471024.0m.Howmuchworkdidthefrictionof6.35105Jtheicedotobringtheskatertoastop?Howmuchworkwouldtheskaterhaveto3.Acometwithamassof7.851011kgdotospeedupto2.5m/sagain?strikesEarthataspeedof25.0km/s.Findthekineticenergyofthecometinjoules,Tobringtheskatertoastop:andcomparetheworkthatisdonebyEarthWKEfKEiinstoppingthecomettothe4.21015Jof1mv21mv2energythatwasreleasedbythelargest2f2inuclearweaponeverbuilt.1(52.0kg)(0.00m/s)212KEmv221(52.0kg)(2.5m/s)21(7.851011kg)(2.50104m/s)222160J2.451020JTospeedupagain:KEcomet2.451020J45.810Thisisthereverseofthefirstquestion.KEbomb4.21015JWKEfKEi5.8104bombswouldberequiredtoproducethesameamountofenergy1mv21mv22f2iusedbyEarthinstoppingthecomet.12(52.0kg)(2.5m/s)2page2911(52.0kg)(0.00m/s)24.InExampleProblem1,whatisthepoten-2tialenergyofthebowlingballrelativeto160Jtherackwhenitisonthefloor?PEmgh2.An875.0-kgcompactcarspeedsupfrom22.0m/sto44.0m/swhilepassinganother(7.30kg)(9.80m/s2)(0.610m)car.Whatareitsinitialandfinalenergies,43.6Jandhowmuchworkisdoneonthecarto5.Ifyouslowlylowera20.0-kgbagofsandincreaseitsspeed?1.20mfromthetrunkofacartothedrive-TheinitialkineticenergyofthecarisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.way,howmuchworkdoyoudo?KE121(875.0kg)(22.0m/s)2i2mv2WFd2.12105Jmg(hfhi)(20.0kg)(9.80m/s2)(0.00m1.20m)2.35102JPhysics:PrinciplesandProblemsSolutionsManual247

251Chapter11continued6.Aboyliftsa2.2-kgbookfromhisdesk,SectionReviewwhichis0.80mhigh,toabookshelfthatis2.10mhigh.Whatisthepotentialenergyof11.1TheManyFormsofEnergythebookrelativetothedesk?pages285–292PEmg(hfhi)page292(2.2kg)(9.80m/s2)(2.10m0.80m)9.ElasticPotentialEnergyYougetaspring-28Jloadedtoypistolreadytofirebycompress-ingthespring.Theelasticpotentialenergy7.Ifa1.8-kgbrickfallstothegroundfromaofthespringpushestherubberdartoutofchimneythatis6.7mhigh,whatisthethepistol.Youusethetoypistoltoshootchangeinitspotentialenergy?thedartstraightup.DrawbargraphsthatdescribetheformsofenergypresentintheChoosethegroundasthereferencefollowinginstances.level.a.Thedartispushedintothegunbarrel,PEmg(hfhi)therebycompressingthespring.(1.8kg)(9.80m/s2)(0.0m6.7m)1.2102J8.Awarehouseworkerpicksupa10.1-kgboxfromthefloorandsetsitonalong,SpringDartDart1.1-m-hightable.Heslidesthebox5.0melasticgravitationalkineticalongthetableandthenlowersitbacktopotentialpotentialenergythefloor.Whatwerethechangesintheenergyenergyenergyofthebox,andhowdidthetotalThereshouldbethreebars:oneforenergyoftheboxchange?(Ignorefriction.)thespring’spotentialenergy,oneforTolifttheboxtothetable:gravitationalpotentialenergy,andWFdoneforkineticenergy.Thespring’spotentialenergyisatthemaximumCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mg(hfhi)level,andtheothertwoarezero.PEb.Thespringexpandsandthedartleaves(10.1kg)(9.80m/s2)(1.1m0.0m)thegunbarrelafterthetriggerispulled.1.1102JToslidetheboxacrossthetable,W0.0becausetheheightdidnotchangeandweignoredfriction.SpringDartDartTolowertheboxtothefloor:elasticgravitationalkineticpotentialpotentialenergyWFdenergyenergymg(hfhi)Thekineticenergyisatthemaxi-PEmumlevel,andtheothertwoare(10.1kg)(9.80m/s2)(0.0m1.1m)zero.1.1102JThesumofthethreeenergychangesis1.1102J0.0J(1.1102J)0.0J248SolutionsManualPhysics:PrinciplesandProblems

252Chapter11continuedc.Thedartreachesthetopofitsflight.Thebowlingballhaszerokineticenergywhenitisrestingontherackorwhenitisheldnearyourshoulder.Therefore,thetotalworkdoneontheballbyyouandbygravitymustequalzero.SpringDartDartelasticgravitationalkinetic13.PotentialEnergyA90.0-kgrockclimberpotentialpotentialenergyfirstclimbs45.0muptothetopofaquarry,energyenergythendescends85.0mfromthetoptotheThegravitationalpotentialenergyisbottomofthequarry.Iftheinitialheightisatthemaximumlevel,andtheotherthereferencelevel,findthepotentialenergytwoarezero.ofthesystem(theclimberandEarth)atthetopandatthebottom.Drawbargraphsfor10.PotentialEnergyA25.0-kgshellisshotbothsituations.fromacannonatEarth’ssurface.Therefer-3.97104JencelevelisEarth’ssurface.WhatisthegravitationalpotentialenergyofthesystemAtthebottomwhentheshellisat425m?WhatisthechangeinpotentialenergywhentheshellAtthefallstoaheightof225m?topa.PEmgh3.53104J(25.0kg)(9.80m/s2)(425m)PEmgh1.04105JAtthetop,b.PEmghPE(90.0kg)(9.80m/s2)(45.0m)(25.0kg)(9.80m/s2)(225m)3.97104J5.51104JAtthebottom,ThechangeinenergyisPE(90.0kg)(9.80m/s2)(45.0m85.0m)(1.04105J)(5.51104J)4.89104J3.53104J11.RotationalKineticEnergySupposesome14.CriticalThinkingKarlusesanairhosetochildrenpushamerry-go-roundsothatitexertaconstanthorizontalforceonapuck,turnstwiceasfastasitdidbeforetheypushedwhichisonafrictionlessairtable.Hekeepsit.Whataretherelativechangesinangularthehoseaimedatthepuck,therebycreatingmomentumandrotationalkineticenergy?aconstantforceasthepuckmovesafixeddistance.Theangularmomentumisdoubledbecauseitisproportionaltotheangulara.Explainwhathappensintermsofworkvelocity.Therotationalkineticenergyisandenergy.Drawbargraphs.quadrupledbecauseitisproportionaltothesquareoftheangularvelocity.KEinitialWKEfinalThechildrendidworkinrotatingtheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.merry-go-round.KarlexertedaconstantforceFoveradistancedanddidanamountof12.Work-EnergyTheoremHowcanyouworkWFdonthepuck.Thisworkapplythework-energytheoremtoliftingachangedthekineticenergyofthebowlingballfromastorageracktoyourpuck.shoulder?Physics:PrinciplesandProblemsSolutionsManual249

253Chapter11continuedW(KEThesystemisthebikeriderEarth.fKEi)Therearenoexternalforces,sototal1mv21mv22f2ienergyisconserved.12KE1mv22mvf2b.SupposeKarlusesadifferentpuckwith1(85.0kg)(8.5m/s)22halfthemassofthefirstone.Allotherconditionsremainthesame.Howwill3.1103JthekineticenergyandworkdifferfromKEiPEiKEfPEfthoseinthefirstsituation?1mv200mghIfthepuckhashalfthemass,itstill2receivesthesameamountofworkv2(8.5m/s)2andhasthesamechangeinkinetich2g(2)(9.80m/s2)energy.However,thesmallermass3.7mwillmovefasterbyafactorof1.414.c.Explainwhathappenedinpartsaandb16.Supposethatthebikeriderinproblem15intermsofimpulseandmomentum.pedaledupthehillandnevercametoaThetwopucksdonothavethesamestop.Inwhatsystemisenergyconserved?finalmomentum.FromwhatformofenergydidthebikegainMomentumofthefirstpuck:mechanicalenergy?pThesystemofEarth,bike,andrider1m1v1remainsthesame,butnowtheenergyMomentumofthesecondpuck:involvedisnotmechanicalenergyp2m2v2alone.Theridermustbeconsideredas1havingstoredenergy,someofwhichism21(1.414v1)convertedtomechanicalenergy.0.707p1Energycamefromthechemicalpoten-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tialenergystoredintherider’sbody.Thus,thesecondpuckhaslessmomentumthanthefirstpuckdoes.17.AskierstartsfromrestatthetopofaBecausethechangeinmomentum45.0-m-highhill,skisdowna30°inclineisequaltotheimpulseprovidedbyintoavalley,andcontinuesupa40.0-m-theairhose,thesecondpuckhighhill.Theheightsofbothhillsaremea-receivesasmallerimpulse.suredfromthevalleyfloor.Assumethatyoucanneglectfrictionandtheeffectoftheskipoles.HowfastistheskiermovingatPracticeProblemsthebottomofthevalley?Whatistheskier’s11.2ConservationofEnergyspeedatthetopofthenexthill?Dotheanglesofthehillsaffectyouranswers?pages293–301Bottomofvalley:page297KE15.AbikeriderapproachesahillataspeedofiPEiKEfPEf8.5m/s.Thecombinedmassofthebike1200mghmvandtherideris85.0kg.Chooseasuitable2system.Findtheinitialkineticenergyofthev22ghsystem.Theridercoastsupthehill.Assumingthereisnofriction,atwhatv2ghheightwillthebikecometorest?2)(45.0(2)(9.80m/sm)250SolutionsManualPhysics:PrinciplesandProblems

254Chapter11continued29.7m/sConservationofmomentum:Topofnexthill:mv(mM)V,orKEiPEiKEfPEf(mM)Vvm0mgh12mghi2mvf(0.00800kg9.00kg)(0.100m/s)0.00800kgv22g(hihf)1.13102m/s2g(hihf)20.A0.73-kgmagnetictargetissuspendedona(2)(9.80ms2)(45.0m40.0m)string.A0.025-kgmagneticdart,shothori-9.90m/szontally,strikesthetargethead-on.ThedartNo,theanglesdonothaveanyimpact.andthetargettogether,actinglikeapendu-lum,swing12.0cmabovetheinitiallevel18.Inabelly-flopdivingcontest,thewinnerisbeforeinstantaneouslycomingtorest.thediverwhomakesthebiggestsplashupona.Sketchthesituationandchooseasystem.hittingthewater.Thesizeofthesplashdependsnotonlyonthediver’sstyle,butSystemalsoontheamountofkineticenergythatthediverhas.Consideracontestinwhicheachdiverjumpsfroma3.00-mplatform.Onediverhasamassof136kgandsimplystepsofftheplatform.Anotherdiverhasamassof102kgandleapsupwardfromtheplatform.12cmHowhighwouldtheseconddiverhavetoPE0leaptomakeacompetitivesplash?ThesystemincludesthesuspendedUsingthewaterasareferencelevel,thetargetandthedart.kineticenergyonentryisequaltotheb.Decidewhatisconservedineachpartpotentialenergyofthediveratthetopandexplainyourdecision.ofhisflight.ThelargediverhasPEmgh(136kg)(9.80m/s2)(3.00m)Onlymomentumisconservedinthe3Jinelasticdart-targetcollision,so4.0010mvToequalthis,thesmallerdiverwouldiMVi(mM)VfhavetojumptowhereVi0sincethetargetisini-tiallyatrestandV4.00103Jfisthecommonh4.00mvelocityjustafterimpact.Asthe(102kg)(9.80m/s2)dart-targetcombinationswingsThus,thesmallerdiverwouldhavetoupward,energyisconserved,soleap1.00mabovetheplatform.PEKEor,atthetopoftheswing,(mM)gh12f2(mM)(Vf)page30019.An8.00-gbulletisfiredhorizontallyintoa9.00-kgblockofwoodonanairtableandisembeddedinit.Afterthecollision,theCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.blockandbulletslidealongthefrictionlesssurfacetogetherwithaspeedof10.0cm/s.Whatwastheinitialspeedofthebullet?Physics:PrinciplesandProblemsSolutionsManual251

255Chapter11continuedc.Whatwastheinitialvelocityofthedart?SolveforVf.Vf2ghfSubstitutevfintothemomentumequationandsolveforvi.mMvim2ghf0.025kg0.73kg2)(0.12(2)(9.80m/s0m)0.025kg46m/s21.A91.0-kghockeyplayerisskatingoniceat5.50m/s.Anotherhockeyplayerofequalmass,movingat8.1m/sinthesamedirection,hitshimfrombehind.Theyslideofftogether.a.Whatarethetotalenergyandmomentuminthesystembeforethecollision?KE121m2i2m1v122v21(91.0kg)(5.50m/s)21(91.0kg)(8.1m/s)2224.4103Jpim1v1m2v2(91.0kg)(5.5m/s)(91.0kg)(8.1m/s)1.2103kgm/sb.Whatisthevelocityofthetwohockeyplayersafterthecollision?Afterthecollision:Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pipfm1v1m2v2(m1m2)vfm1v1m2v2vfm1m2(91.0kg)(5.50ms)(91.0kg)(8.1ms)91.0kg91.0kg6.8m/sc.Howmuchenergywaslostinthecollision?Thefinalkineticenergyis12KEf2(mimf)vf12(91.0kg91.0kg)(6.8m/s)24.2103JThus,theenergylostinthecollisionisKE3J4.2103JiKEf4.4102102J252SolutionsManualPhysics:PrinciplesandProblems

256Chapter11continued25.KineticEnergyIntabletennis,averylightSectionReviewbuthardballishitwithahardrubberor11.2ConservationofEnergywoodenpaddle.Intennis,amuchsofterpages293–301ballishitwitharacket.Whyarethetwosetsofequipmentdesignedinthisway?page301Canyouthinkofotherball-paddlepairsin22.ClosedSystemsIsEarthaclosed,isolatedsports?Howaretheydesigned?system?Supportyouranswer.TheballsandthepaddleandracketareTosimplifyproblemsthattakeplacedesignedtomatchsothatthemaximumoverashorttime,Earthisconsideredaamountofkineticenergyispassedfromclosedsystem.Itisnotactuallyisolated,thepaddleorrackettotheball.Asofterhowever,becauseitisacteduponbytheballreceivesenergywithlesslossfromagravitationalforcesfromtheplanets,thesofterpaddleorracket.Othercombina-Sun,andotherstars.Inaddition,Earthistionsareagolfballandclub(bothhard)therecipientofcontinuouselectromag-andabaseballandbat(alsobothhard).neticenergy,primarilyfromtheSun.26.PotentialEnergyArubberballisdropped23.EnergyAchildjumpsonatrampoline.fromaheightof8.0montoahardconcreteDrawbargraphstoshowtheformsoffloor.Ithitsthefloorandbouncesrepeat-energypresentinthefollowingsituations.1edly.Eachtimeithitsthefloor,itlosesa.Thechildisatthehighestpoint.5ofitstotalenergy.Howmanytimeswillitbouncebeforeitbouncesbackuptoaheightofonlyabout4m?EtotalmghTrampolineChildChildelasticgravitationalkineticSincethereboundheightisproportionalpotentialpotentialenergytoenergy,eachbouncewillreboundtoenergyenergy4theheightofthepreviousbounce.5b.Thechildisatthelowestpoint.4Afteronebounce:h(8m)6.4m54Aftertwobounces:h(6.4m)55.12mTrampolineChildChild4elasticgravitationalkineticAfterthreebounces:h5(5.12m)potentialpotentialenergy4.1menergyenergy24.KineticEnergySupposeaglobofchew-inggumandasmall,rubberballcollidehead-oninmidairandthenreboundapart.Wouldyouexpectkineticenergytobecon-served?Ifnot,whathappenstotheenergy?EventhoughtherubberballreboundsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.withlittleenergyloss,kineticenergywouldnotbeconservedinthiscasebecausetheglobofchewinggumprob-ablywasdeformedinthecollision.Physics:PrinciplesandProblemsSolutionsManual253

257Chapter11continued27.EnergyAsshowninFigure11-15,aChapterAssessment36.0-kgchildslidesdownaplaygroundslidethatis2.5mhigh.AtthebottomoftheConceptMappingslide,sheismovingat3.0m/s.Howmuchpage306energywaslostasshesliddowntheslide?29.Completetheconceptmapusingthefol-lowingterms:gravitationalpotentialenergy,elasticpotentialenergy,kineticenergy.36.0kgEnergykineticpotential2.5mlinearrotationalgravitationalelasticMasteringConcepts■Figure11-15page306EUnlessotherwisedirected,assumethatairresis-imghtanceisnegligible.(36.0kg)(9.80m/s2)(2.5m)30.Explainhowworkandachangeinenergy880Jarerelated.(11.1)12Ef2mvTheworkdoneonanobjectcausesachangeintheobject’senergy.Thisis12(36.0kg)(3.0m/s)thework-energytheorem.2160J31.Whatformofenergydoesawound-upCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Energyloss880J–160Jwatchspringhave?Whatformofenergy720Jdoesafunctioningmechanicalwatchhave?Whenawatchrunsdown,whathashap-28.CriticalThinkingAballdrops20m.penedtotheenergy?(11.1)Whenithasfallenhalfthedistance,orThewound-upwatchspringhaselastic10m,halfofitsenergyispotentialandhalfpotentialenergy.Thefunctioningwatchiskinetic.Whentheballhasfallenforhalfhaselasticpotentialenergyandrota-theamountoftimeittakestofall,willtionalkineticenergy.Thewatchrunsmore,less,orexactlyhalfofitsenergybedownwhenalloftheenergyhasbeenpotentialenergy?convertedtoheatbyfrictionintheTheballfallsmoreslowlyduringthegearsandbearings.beginningpartofitsdrop.Therefore,inthefirsthalfofthetimethatitfalls,it32.Explainhowenergychangeandforcearewillnothavetraveledhalfofthedis-related.(11.1)tancethatitwillfall.Therefore,theballAforceexertedoveradistancedoeswillhavemorepotentialenergythanwork,whichproducesachangeinenergy.kineticenergy.33.Aballisdroppedfromthetopofabuild-ing.Youchoosethetopofthebuildingtobethereferencelevel,whileyourfriendchoosesthebottom.Explainwhetherthe254SolutionsManualPhysics:PrinciplesandProblems

258Chapter11continuedenergycalculatedusingthesetworeference38.Thesportofpole-vaultingwasdrasticallylevelsisthesameordifferentforthefol-changedwhenthestiff,woodenpoleswerelowingsituations.(11.1)replacedbyflexible,fiberglasspoles.a.theball’spotentialenergyatanypointExplainwhy.(11.2)ThepotentialenergiesaredifferentAflexible,fiberglasspolecanstoreelas-duetothedifferentreferencelevels.ticpotentialenergybecauseitcanbebenteasily.Thisenergycanbereleasedb.thechangeintheball’spotentialenergytopushthepole-vaulterhighervertically.asaresultofthefallBycontrast,thewoodenpoledoesnotThechangesinthepotentialener-storeelasticpotentialenergy,andthegiesasaresultofthefallareequalpole-vaulter’smaximumheightislimitedbecausethechangeinhisthesamebythedirectconversionofkineticenergyforbothreferencelevels.togravitationalpotentialenergy.c.thekineticenergyoftheballatanypointThekineticenergiesoftheballat39.Youthrowaclayballatahockeypuckonanypointareequalbecausetheice.Thesmashedclayballandthehockeyvelocitiesarethesame.pucksticktogetherandmoveslowly.(11.2)a.Ismomentumconservedinthecollision?34.CanthekineticenergyofabaseballeverbeExplain.negative?(11.1)ThetotalmomentumoftheballandThekineticenergyofabaseballcanthepucktogetherisconservedinneverbenegativebecausethekineticthecollisionbecausetherearenoenergydependsonthesquareoftheunbalancedforcesonthissystem.velocity,whichisalwayspositive.b.Iskineticenergyconserved?Explain.35.CanthegravitationalpotentialenergyofaThetotalkineticenergyisnotbaseballeverbenegative?Explainwithoutconserved.Partofitislostintheusingaformula.(11.1)smashingoftheclayballandtheadhesionoftheballtothepuck.Thegravitationalpotentialenergyofabaseballcanbenegativeiftheheightof40.Drawenergybargraphsforthefollowingtheballislowerthanthereferencelevel.processes.(11.2)36.Ifasprinter’svelocityincreasestothreea.Anicecube,initiallyatrest,slidesdowntimestheoriginalvelocity,bywhatfactorafrictionlessslope.doesthekineticenergyincrease?(11.1)Thesprinter’skineticenergyincreasesbyafactorof9,becausethevelocityissquared.PEiKEiPEfKEf37.Whatenergytransformationstakeplaceb.Anicecube,initiallymoving,slidesupawhenanathleteispole-vaulting?(11.2)frictionlessslopeandinstantaneouslyThepole-vaulterruns(kineticenergy)comestorest.andbendsthepole,therebyaddingelasticpotentialenergytothepole.AsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.he/sheliftshis/herbody,thatkineticandelasticpotentialenergyistrans-ferredintokineticandgravitationalPEiKEiPEfKEfpotentialenergy.Whenhe/shereleasesthepole,allofhis/herenergyiskineticandgravitationalpotentialenergy.Physics:PrinciplesandProblemsSolutionsManual255

259Chapter11continued41.Describethetransformationsfromkinetic44.Acompactcarandatrailertruckarebothenergytopotentialenergyandviceversafortravelingatthesamevelocity.Didthecararoller-coasterride.(11.2)engineorthetruckenginedomoreworkinOnaroller-coasterride,thecarhasacceleratingitsvehicle?mostlypotentialenergyatthetopsofThetrailertruckhasmorekineticenergy,thehillsandmostlykineticenergyat12,becauseithasgreatermassKEmvthebottomsofthehills.2thanthecompactcar.Thus,according42.Describehowthekineticenergyandelastictothework-energytheorem,thetruck’spotentialenergyarelostinabouncingenginemusthavedonemorework.rubberball.Describewhathappenstothemotionoftheball.(11.2)45.CatapultsMedievalwarriorsusedcatapultstoassaultcastles.SomecatapultsworkedOneachbounce,some,butnotall,ofthebyusingatightlywoundropetoturntheball’skineticenergyisstoredaselasticcatapultarm.Whatformsofenergyarepotentialenergy;theball’sdeformationinvolvedincatapultingarocktothecastledissipatestherestoftheenergyasther-wall?malenergyandsound.Afterthebounce,thestoredelasticpotentialenergyisElasticpotentialenergyisstoredinthereleasedaskineticenergy.Duetothewoundrope,whichdoesworkontheenergylossesinthedeformation,eachrock.Therockhaskineticandpotentialsubsequentbouncebeginswithasmallerenergyasitfliesthroughtheair.Whenamountofkineticenergy,andresultsinithitsthewall,theinelasticcollisiontheballreachingalowerheight.causesmostofthemechanicalenergyEventually,alloftheball’senergyisdissi-tobeconvertedtothermalandsoundpated,andtheballcomestorest.energyandtodoworkbreakingapartthewallstructure.Someofthemechan-ApplyingConceptsicalenergyappearsinthefragmentsthrownfromthecollision.pages306–307Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.43.Thedriverofaspeedingcarappliesthe46.Twocarscollideandcometoacompletebrakesandthecarcomestoastop.Thestop.Wheredidalloftheirenergygo?systemincludesthecarbutnottheroad.TheenergywentintobendingsheetApplythework-energytheoremtothefol-metalonthecars.Energyalsowaslostlowingsituations.duetofrictionalforcesbetweenthea.Thecar’swheelsdonotskid.carsandthetires,andintheformofIfthecarwheelsdonotskid,thethermalenergyandsound.brakesurfacesrubagainsteachotheranddoworkthatstopsthecar.47.Duringaprocess,positiveworkisdoneonaTheworkthatthebrakesdoisequalsystem,andthepotentialenergydecreases.tothechangeinkineticenergyofCanyoudetermineanythingaboutthechangethecar.Thebrakesurfacesheatupinkineticenergyofthesystem?Explain.becausethekineticenergyistrans-Theworkequalsthechangeinthetotalformedtothermalenergy.mechanicalenergy,W(KEPE).IfWb.Thebrakeslockandthecar’swheelsskid.ispositiveandPEisnegative,thenKEIfthebrakeslockandthecarwheelsmustbepositiveandgreaterthanW.skid,thewheelsrubbingontheroadaredoingtheworkthatstopsthecar.48.Duringaprocess,positiveworkisdoneonaThetiresurfacesheatup,notthesystem,andthepotentialenergyincreases.Canbrakes.Thisisnotanefficientwaytoyoutellwhetherthekineticenergyincreased,stopacar,anditruinsthetires.decreased,orremainedthesame?Explain.256SolutionsManualPhysics:PrinciplesandProblems

260Chapter11continuedTheworkequalsthechangeinthetotal51.Givespecificexamplesthatillustratethemechanicalenergy,W(KEPE).Iffollowingprocesses.WispositiveandPEispositive,thena.Workisdoneonasystem,therebyyoucannotsayanythingconclusiveincreasingkineticenergywithnoaboutKE.changeinpotentialenergy.pushingahockeypuckhorizontally49.SkatingTwoskatersofunequalmasshaveacrossice;systemconsistsofthesamespeedandaremovinginthesamehockeypuckonlydirection.Iftheiceexertsthesamefrictionalforceoneachskater,howwillthestoppingb.Potentialenergyischangedtokineticdistancesoftheirbodiescompare?energywithnoworkdoneonthesystem.Thelargerskaterwillhavemorekineticdroppingaball;systemconsistsofenergy.ThekineticenergyofeachballandEarthskaterwillbedissipatedbythenegativec.Workisdoneonasystem,increasingwork,WFd,donebythefrictionofpotentialenergywithnochangeintheice.Sincethefrictionalforcesarekineticenergy.equal,thelargerskaterwillgofarthercompressingthespringinatoypis-beforestopping.tol;systemconsistsofspringonlyd.Kineticenergyisreduced,butpotential50.Youswinga55-gmassontheendofaenergyisunchanged.Workisdoneby0.75-mstringaroundyourheadinanearlythesystem.horizontalcircleatconstantspeed,asshowninFigure11-16.Acar,speedingonaleveltrack,brakesandreducesitsspeed.52.RollerCoasterYouhavebeenhiredtomakearollercoastermoreexciting.Theownerswantthespeedatthebottomofthe0.75mfirsthilldoubled.Howmuchhighermust55gthefirsthillbebuilt?Thehillmustbemadehigherbyafactorof4.53.Twoidenticalballsarethrownfromthetop■Figure11-16ofacliff,eachwiththesamespeed.Oneisa.Howmuchworkisdoneonthemassthrownstraightup,theotherstraightdown.bythetensionofthestringinoneHowdothekineticenergiesandspeedsofrevolution?theballscompareastheystriketheground?NoworkisdonebythetensionforceEventhoughtheballsaremovinginonthemassbecausethetensionisoppositedirections,theyhavethesamepullingperpendiculartothemotionkineticenergyandpotentialenergyofthemass.whentheyarethrown.Therefore,theyb.Isyouranswertopartainagreementwillhavethesamemechanicalenergywiththework-energytheorem?Explain.andspeedwhentheyhittheground.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thisdoesnotviolatethework-energytheorembecausethekineticenergyofthemassisconstant;itismovingataconstantspeed.Physics:PrinciplesandProblemsSolutionsManual257

261Chapter11continuedMasteringProblemsc.Whatistheratioofthekineticenergiesinpartsaandb?Explain.Unlessotherwisedirected,assumethatairresistanceisnegligible.1(mv2)21v2(10.0)24111.1TheManyFormsofEnergy12)v2(5.0)21(mv222pages307–308Level1Twicethevelocitygivesfourtimes54.A1600-kgcartravelsataspeedof12.5m/s.thekineticenergy.ThekineticenergyWhatisitskineticenergy?isproportionaltothesquareofthevelocity.KE1mv21(1600kg)(12.5m/s)22258.KatiaandAngelaeachhaveamassof45kg,1.3105Jandtheyaremovingtogetherwithaspeedof10.0m/s.55.Aracingcarhasamassof1525kg.Whatisa.Whatistheircombinedkineticenergy?itskineticenergyifithasaspeedof108km/h?KE121(m2c2mv2KmA)v12KEmv12(45kg45kg)(10.0m/s)221(108kmh)(1000mkm)22(1525kg)3600sh4.5103J6.86105Jb.WhatistheratiooftheircombinedmasstoKatia’smass?56.ShawnandhisbikehaveacombinedmassmKmA45kg45kgof45.0kg.Shawnrideshisbike1.80kminmK45kg10.0minataconstantvelocity.Whatis2Shawn’skineticenergy?1121md2c.WhatistheratiooftheircombinedKEmv22tkineticenergytoKatia’skineticenergy?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2Explain.1(1.80km)(1000mkm)(45kg)2(10.0min)(60smin)KE121(45kg)(10.0m/s)2K2mKv2203J2.3103J57.Tonyhasamassof45kgandismoving12(mwithaspeedof10.0m/s.KEC2KmA)vmKmAa.FindTony’skineticenergy.KEK1m2mK2KvKE1mv21(45kg)(10.0m/s)222212.3103JTheratiooftheircombinedkineticb.Tony’sspeedchangesto5.0m/s.NowenergytoKatia’skineticenergyiswhatishiskineticenergy?thesameastheratiooftheircom-KE1mv21(45kg)(5.0m/s)2binedmasstoKatia’smass.Kinetic22energyisproportionaltomass.5.6102J258SolutionsManualPhysics:PrinciplesandProblems

262Chapter11continued59.TrainInthe1950s,anexperimentaltrain,Fg4kg,wasNowmwhichhadamassof2.5010gpoweredacrossaleveltrackbyajetengine12thatproducedathrustof5.00105Nforamv2Soddistanceof509m.Fa.Findtheworkdoneonthetrain.1Fg22gvWFd(5.00105N)(509m)F2.55108J114,700N222(25.0m/s)b.Findthechangeinkineticenergy.9.80m/s7100NKEW2.55108J66mc.Findthefinalkineticenergyofthetrainifitstartedfromrest.61.A15.0-kgcartismovingwithavelocityofKEKEfKEi7.50m/sdownalevelhallway.Aconstantforceof10.0Nactsonthecart,anditssoKEfKEKEivelocitybecomes3.20m/s.2.55108J0.00Ja.Whatisthechangeinkineticenergyof2.55108Jthecart?KEKE12v2)d.FindthefinalspeedofthetrainiftherefKEi2m(vfihadbeennofriction.12(15.0kg)((3.20m/s)122KEf2mvf(7.50m/s)2)KESov2f345Jf1m2b.Howmuchworkwasdoneonthecart?2.55108JWKE345J1(2.50104kg)c.Howfardidthecartmovewhilethe2forceacted?Sovf2.04104m2/s2143m/sWFdW345J60.CarBrakesA14,700-Ncaristravelingatsod34.5mF10.0N25m/s.Thebrakesareappliedsuddenly,andthecarslidestoastop,asshownin62.HowmuchpotentialenergydoesDeAnnaFigure11-17.Theaveragebrakingforcewithamassof60.0kg,gainwhenshebetweenthetiresandtheroadis7100N.climbsagymnasiumropeadistanceofHowfarwillthecarslideoncethebrakes3.5m?areapplied?PEmghBeforeAfter(60.0kg)(9.80m/s2)(3.5m)(initial)(final)2.1103Jv25m/sv0.0m/s63.BowlingA6.4-kgbowlingballisliftedCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.1mintoastoragerack.Calculatetheincreaseintheball’spotentialenergy.m14,700NPEmgh■Figure11-17(6.4kg)(9.80m/s2)(2.1m)12WFdmv1.3102J2Physics:PrinciplesandProblemsSolutionsManual259

263Chapter11continued64.Maryweighs505N.ShewalksdownaLevel2flightofstairstoalevel5.50mbelowher69.TennisItisnotuncommonduringtheservestartingpoint.WhatisthechangeinMary’sofaprofessionaltennisplayerfortheracketpotentialenergy?toexertanaverageforceof150.0NonthePEmghFball.Iftheballhasamassof0.060kgandisghincontactwiththestringsoftheracket,as(505N)(5.50m)showninFigure11-18,for0.030s,whatis2.78103Jthekineticenergyoftheballasitleavestheracket?Assumethattheballstartsfromrest.65.WeightliftingAweightlifterraisesa180-kgbarbelltoaheightof1.95m.Whatistheincreaseinthepotentialenergyofthebarbell?PEmgh(180kg)(9.80m/s2)(1.95m)150.0N3.4103J66.A10.0-kgtestrocketisfiredverticallyfromCapeCanaveral.Itsfuelgivesitakinetic■Figure11-18energyof1960JbythetimetherocketFtmvmvfmviandvi0engineburnsallofthefuel.Whatadditionalheightwilltherocketrise?Ft(150.0N)(3.0102s)sovfm6.0102kgPEmghKE75m/sKE1960hmg(10.0kg)(9.80m/s2)12KEmv220.0m1(6.0102kg)(75m/s)2267.Antwanraiseda12.0-NphysicsbookfromCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.atable75cmabovethefloortoashelf1.7102J2.15mabovethefloor.Whatwasthechangeinthepotentialenergyofthe70.Pam,wearingarocketpack,standsonfric-system?tionlessice.Shehasamassof45kg.Therocketsuppliesaconstantforcefor22.0m,PEmghFghFg(hfhi)andPamacquiresaspeedof62.0m/s.(12.0N)(2.15m0.75m)a.Whatisthemagnitudeoftheforce?17J12KEf2mvf68.Ahallwaydisplayofenergyisconstructedinwhichseveralpeoplepullonaropethat1(45kg)(62.0ms)22liftsablock1.00m.Thedisplayindicatesthat1.00Jofworkisdone.Whatisthe8.6104Jmassoftheblock?b.WhatisPam’sfinalkineticenergy?WPEmghWorkdoneonPamequalsherW1.00Jchangeinkineticenergy.mgh(9.80m/s2)(1.00m)WFdKEKEfKEi0.102kgKEi0JKEf8.6104JSo,Fd22.0m3.9103N260SolutionsManualPhysics:PrinciplesandProblems

264Chapter11continued71.CollisionA2.00103-kgcarhasaspeedof11.2ConservationofEnergy12.0m/s.Thecarthenhitsatree.Thetreepages308–309doesn’tmove,andthecarcomestorest,asLevel1showninFigure11-19.73.A98.0-Nsackofgrainishoistedtoastor-ageroom50.0mabovethegroundfloorofBeforeagrainelevator.(initial)After(final)a.Howmuchworkwasdone?vi12.0m/svf0.0m/sWPEmghFgh(98.0N)(50.0m)4.90103J3m2.0010kgb.Whatistheincreaseinpotentialenergy■Figure11-19ofthesackofgrainatthisheight?a.Findthechangeinkineticenergyof3JPEW4.9010thecar.1c.TheropebeingusedtoliftthesackofKEKE2v2)fKEi2m(vfigrainbreaksjustasthesackreachesthe1(2.00103kg)((0.0m/s)2storageroom.Whatkineticenergydoes2thesackhavejustbeforeitstrikesthe(12.0m/s)2)groundfloor?1.44105JKEPE4.90103Jb.Findtheamountofworkdoneasthefrontofthecarcrashesintothetree.74.A20-kgrockisontheedgeofa100-mcliff,5JasshowninFigure11-20.WKE1.4410c.Findthesizeoftheforcethatpushedin20kgthefrontofthecarby50.0cm.WFdWsoF100md1.44105J0.500m2.88105N■Figure11-20a.Whatpotentialenergydoestherock72.Aconstantnetforceof410Nisappliedpossessrelativetothebaseofthecliff?upwardtoastonethatweighs32N.ThePEmgh(20kg)(9.80m/s2)(100m)upwardforceisappliedthroughadistance4J210of2.0m,andthestoneisthenreleased.Tob.Therockfallsfromthecliff.Whatisitswhatheight,fromthepointofrelease,willkineticenergyjustbeforeitstrikesthethestonerise?ground?WFd(410N)(2.0m)8.2102JKEPE2104JButWPEmgh,soc.WhatspeeddoestherockhaveasitW8.2102Jh26mstrikestheground?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mg32N12KEmv22KE(2)(2104J)vm20kg40m/sPhysics:PrinciplesandProblemsSolutionsManual261

265Chapter11continued75.ArcheryAnarcherputsa0.30-kgarrowto77.Aphysicsbookofunknownmassisthebowstring.Anaverageforceof201Nisdropped4.50m.Whatspeeddoesthebookexertedtodrawthestringback1.3m.havejustbeforeithitstheground?a.AssumingthatalltheenergygoesintoKEPEthearrow,withwhatspeeddoesthe12mghmvarrowleavethebow?2WorkdoneonthestringincreasesThemassofthebookdividesout,sothestring’selasticpotentialenergy.1v2ghWPEFd2Allofthestoredpotentialenergyisv2gh2(9.80m/s2)(4.50m)transformedtothearrow’skinetic9.39m/senergy.KE1mv2PEFd78.RailroadCarArailroadcarwithamassof25.0105kgcollideswithastationaryrail-22Fdroadcarofequalmass.Afterthecollision,vmthetwocarslocktogetherandmoveoffat4.0m/s,asshowninFigure11-21.2Fd(2)(201N)(1.3m)vm0.30kgm5.0105kg42m/sv4.0m/sb.Ifthearrowisshotstraightup,howhighdoesitrise?Thechangeinthearrow’spotentialenergyequalstheworkdonetopullthestring.■Figure11-21PEmghFda.Beforethecollision,thefirstrailroadcarFd(201N)(1.3m)h2)wasmovingat8.0m/s.WhatwasitsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mg(0.30kg)(9.80msmomentum?89mmv(5.0105kg)(8.0m/s)76.A2.0-kgrockthatisinitiallyatrestloses4.0106kgm/s407Jofpotentialenergywhilefallingtob.Whatwasthetotalmomentumofthetheground.Calculatethekineticenergytwocarsafterthecollision?thattherockgainswhilefalling.Whatistherock’sspeedjustbeforeitstrikestheBecausemomentumisconserved,itground?mustbe4.0106kgm/sPEc.WhatwerethekineticenergiesoftheiKEiPEfKEftwocarsbeforeandafterthecollision?KEi0Beforethecollision:So,KE12fPEiPEf407JKEi2mvKE1215kg)(8.0m/s)2f2mvf(5.01022KEv2f1.6107JfmAfterthecollision:v2KEf(2)(407J)KE12fm(2.0kg)f2mv2.0101m/s262SolutionsManualPhysics:PrinciplesandProblems

266Chapter11continued1(5.0105kg5.0105kg)b.IfKellimovesthroughthelowestpoint2at2.0m/s,howmuchworkwasdone(4.0m/s)2ontheswingbyfriction?8.0106JTheworkdonebyfrictionequalsthechangeinmechanicalenergy.d.Accountforthelossofkineticenergy.WPEKEWhilemomentumwasconservedduringthecollision,kineticenergymg(h12fhi)2mvfwasnot.Theamountnotconservedwasturnedintothermalenergyand(420N)(0.40m1.00m)soundenergy.1420N229.80ms2(2.0ms)79.Fromwhatheightwouldacompactcar1.7102Jhavetobedroppedtohavethesamekineticenergythatithaswhenbeingdriven81.Hakeemthrowsa10.0-gballstraightdownat1.00102km/h?fromaheightof2.0m.Theballstrikesthe2km1000m1hfloorataspeedof7.5m/s.Whatwasthev1.0010h1km3600sinitialspeedoftheball?27.8m/sKEfKEiPEiKEPE121mv2mghmv12f2imv2mgh2themassoftheballdividesout,so1v2ghv2v22gh,2ifv2(27.8m/s)2vv22ghhif2g2(9.80m/s2)39.4m(7.5m/s)2(2)(9.80m/s2)(2.0m)4.1m/sLevel280.Kelliweighs420N,andsheissittingona82.SlideLorena’smassis28kg.Sheclimbstheplaygroundswingthathangs0.40mabove4.8-mladderofaslideandreachesavelocitytheground.Hermompullstheswingbackof3.2m/satthebottomoftheslide.Howandreleasesitwhentheseatis1.00mmuchworkwasdonebyfrictiononLorena?abovetheground.TheworkdonebyfrictiononLorenaa.HowfastisKellimovingwhentheswingequalsthechangeinhermechanicalpassesthroughitslowestposition?energy.PEmghmg(hfhi)WPEKEKE1m(v2v2)1mv2mg(h12v2)2fi2ffhi)2m(vfiByconservationofmechanicalenergy:(28kg)(9.80ms2)(0.0m4.8m)PEKE01(28kg)((3.2ms)2(0.0ms)2)2mg(h120fhi)2mvf3JCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.210vf2g(hihf)83.Apersonweighing635Nclimbsupa(2)(9.80ms2)(1.00m0.40m)laddertoaheightof5.0m.UsethepersonandEarthasthesystem.3.4m/sPhysics:PrinciplesandProblemsSolutionsManual263

267Chapter11continueda.Drawenergybargraphsofthesystem86.HighJumpTheworldrecordforthemen’sbeforethepersonstartstoclimbthehighjumpisabout2.45m.Toreachthatladderandafterthepersonstopsattheheight,whatistheminimumamountoftop.Hasthemechanicalenergyworkthata73.0-kgjumpermustexertinchanged?Ifso,byhowmuch?pushingofftheground?WEmgh3200J2)(2.45m)(73.0kg)(9.80m/s1.75kJ87.AstuntwomanfindsthatshecansafelyKEiPEiKEfPEfbreakherfallfromaone-storybuildingbylandinginaboxfilledtoa1-mdepthwithYes.Themechanicalenergyhasfoampeanuts.Inhernextmovie,thescriptchanged,increaseinpotentialcallsforhertojumpfromafive-storybuild-energyof(635N)(5.0m)3200J.ing.Howdeepaboxoffoampeanutsb.Wheredidthisenergycomefrom?shouldsheprepare?fromtheinternalenergyofthepersonAssumethatthefoampeanutsexertaconstantforcetoslowhimdown,MixedReviewWFdEmgh.Iftheheightispages309–310increasedfivetimes,thenthedepthofthefoampeanutsalsoshouldbeLevel1increasedfivetimesto5m.84.Supposeachimpanzeeswingsthroughthejungleonvines.IfitswingsfromatreeonaLevel213-m-longvinethatstartsatanangleof88.FootballA110-kgfootballlinebackerhasa45°,whatisthechimp’svelocitywhenithead-oncollisionwitha150-kgdefensivereachestheground?end.Aftertheycollide,theycometoaThechimpanzee’sinitialheightiscompletestop.Beforethecollision,whichCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.h(13m)(1cos45°)3.8mplayerhadthegreatermomentumandConservationofmechanicalenergy:whichplayerhadthegreaterkineticenergy?PEKE0Themomentumafterthecollisioniszero;therefore,thetwoplayershadmg(h12v2)0fhi)2m(vfiequalandoppositemomentabeforethecollision.Thatis,mgh120i2mvfplinebackermlinebackervlinebackerpend2)mvf2ghi2(9.80m/s(3.8m)endvend.Afterthecollision,eachhad8.6m/szeroenergy.Theenergylossforeach121m2v2p2playerwasmv.85.An0.80-kgcartrollsdownafrictionlesshill22m2mofheight0.32m.Atthebottomofthehill,Becausethemomentawereequalbutthecartrollsonaflatsurface,whichexertsmlinebackermendthelinebackerlostafrictionalforceof2.0Nonthecart.Howfardoesthecartrollontheflatsurfacemoreenergy.beforeitcomestoastop?89.A2.0-kglabcartanda1.0-kglabcartareEmghWFdheldtogetherbyacompressedspring.Themgh(0.80kg)(9.80m/s2)(0.32m)labcartsmoveat2.1m/sinonedirection.dF2.0NThespringsuddenlybecomesuncompressed1.3mandpushesthetwolabcartsapart.The2-kg264SolutionsManualPhysics:PrinciplesandProblems

268Chapter11continuedlabcartcomestoastop,andthe1.0-kglab91.An0.80-kgcartrollsdowna30.0°hillfromcartmovesahead.Howmuchenergydidtheaverticalheightof0.50masshowninspringaddtothelabcarts?Figure11-22.ThedistancethatthecartE121(2.0kg1.0kg)(2.1m/s)2mustrolltothebottomofthehillisi2mv20.50m/sin30.0°1.0m.Thesurfaceofthe6.6Jhillexertsafrictionalforceof5.0Nonthecart.Doesthecartrolltothebottomofthehill?pimv(2.0kg1.0kg)(2.1m/s)m0.80kg6.3kgm/spf(1.0kg)vfF5.0Nso,vf6.3m/sE121(1.0kg)(6.3m/s)219.8Jf2mvf20.50mE19.8J6.6J13.2J30.0°13.2Jwasaddedbythespring.■Figure11-22E2)(0.50m)90.A55.0-kgscientistropingthroughthetopofimgh(0.80kg)(9.80m/satreeinthejungleseesalionaboutto3.9Jattackatinyantelope.ShequicklyswingsTheworkdonebyfrictionover1.0mdownfromher12.0-m-highperchandgrabswouldbetheantelope(21.0kg)assheswings.TheyWFd(5.0N)(1.0m)5.0J.barelyswingbackuptoatreelimboutofreachofthelion.Howhighisthistreelimb?Theworkdonebyfrictionisgreaterthantheenergyofthecart.ThecartEimBghwouldnotreachthebottomofthehill.ThevelocityofthebotanistwhenshereachesthegroundisLevel3E12m92.ObjectA,slidingonafrictionlesssurfaceati2mBvBgh3.2m/s,hitsa2.0-kgobject,B,whichis2Ei2mBghmotionless.ThecollisionofAandBiscom-v2ghmmBpletelyelastic.Afterthecollision,AandBmoveawayfromeachotheratequalandMomentumisconservedwhentheoppositespeeds.WhatisthemassofobjectA?botanistgrabstheantelope.pmimAv10Bv(mBmA)vfmBvmBpfmA(v2)mBv2so,vf(mm2ghBmA)BmApipf(conservationofmomentum)Thefinalenergyofthetwoistherefore,m2Av1mA(v2)mBv12Ef2(mBmA)vf(mBmA)v2mAv11mB2(mBmA)(2gh)(mAv1)2mBmAv2(mBmA)(mBmA)ghf12Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mEi2mAv1B2So,hhfm1BmAE21m2f2mAv22Bv255.0kg2(12.0m)55.0kg21.0kg6.28mPhysics:PrinciplesandProblemsSolutionsManual265

269Chapter11continued12Fromtheconservationofmomentum,Ef2(mAmB)v2mcvc1mcvc2mbvb21(mAv1)22(mAmB)(mmbvb2BmA)Solveforvc2,vc2vc1mcEiEf(conservationofenergyinFromconservationofenergy,elasticcollision)therefore,1m21m21m22cvc12cvc22bvb21m21(m2Av12AmB)Multiplybytwoandsubstitutetoget:(mAv1)22mmbvb222mcvc1cvc1mmbvb2(mBmA)cAftercancellingoutcommonfactors,orm2m22mcvc1cvc1bvc2vc1(m2m2v2AmB)mA(mBmA)bb22mm22m2mcbvb2BAmBmAmSimplifyandfactor:B2.00kgmA330.67kgm2vbb20(mvbvb2)2vc1mb293.HockeyA90.0-kghockeyplayermovingatc5.0m/scollideshead-onwitha110-kgmbvb20orhockeyplayermovingat3.0m/sinthem2vboppositedirection.Afterthecollision,theyc1m1vb20cmoveofftogetherat1.0m/s.Howmuchenergywaslostinthecollision?Ignoringthesolutionvb20,thenBefore:E1mv21mv22vc12122vb2mb1m1(90.0kg)(5.0m/s)2c2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2(44m/s)73m/s1(110kg)(3.0m/s)20.046kg120.220kg1.6103J195.ApplyConceptsAflyhittingthewind-After:E(mm)v22fshieldofamovingpickuptruckisanexam-1pleofacollisioninwhichthemassofone(200.0kg)(1.0m/s)22oftheobjectsismanytimeslargerthanthe1.0102Jother.Ontheotherhand,thecollisionoftwobilliardballsisoneinwhichthemassesEnergyloss1.6103J1.0102Jofbothobjectsarethesame.Howisenergy1.5103Jtransferredinthesecollisions?Consideranelasticcollisioninwhichbilliardballm1hasvelocityvThinkingCritically1andballm2ismotionless.a.Ifmpage3101m2,whatfractionoftheinitialenergyistransferredtom94.ApplyConceptsAgolfballwithamassof2?0.046kgrestsonatee.ItisstruckbyagolfIfm1m2,weknowthatm1willbeclubwithaneffectivemassof0.220kgandatrestafterthecollisionandm2willaspeedof44m/s.Assumingthatthecolli-movewithvelocityv1.Allofthesioniselastic,findthespeedoftheballenergywillbetransferredtom2.whenitleavesthetee.266SolutionsManualPhysics:PrinciplesandProblems

270Chapter11continuedb.Ifm1m2,whatfractionoftheinitialenergyistransferredtom2?Ifm1m2,weknowthatthemotionofm1willbeunaffectedbythecollisionandthattheenergytransfertom2willbeminimal.c.Inanuclearreactor,neutronsmustbesloweddownbycausingthemtocollidewithatoms.(Aneutronisaboutasmassiveasaproton.)Wouldhydrogen,carbon,orironatomsbemoredesirabletouseforthispurpose?Thebestwaytostopaneutronistohaveithitahydrogenatom,whichhasaboutthesamemassastheneutron.96.AnalyzeandConcludeInaperfectlyelasticcollision,bothmomentumandmechanicalenergyareconserved.Twoballs,withmassesmAandmB,aremovingtowardeachotherwithspeedsvAandvB,respectively.Solvetheappropriateequationstofindthespeedsofthetwoballsafterthecollision.conservationofmomentum(1)mAvA1mBvB1mAvA2mBvB2mAvA1mAvA2mBvB1mBvB2(2)mA(vA1vA2)mB(vB1vB2)conservationofenergy1m21m21m21m22AvA12BvB12AvA22BvB2m2m2m2m2AvA1AvA2BvB1BvB2m2v2)m2v2)A(vA1A2B(vB1B2(3)mA(vA1vA2)(vA1vA2)mB(vB1vB2)(vB1vB2)Divideequation(3)by(2)toobtain(4)vA1vA2vB1vB2Solveequation(1)forvA2andvB2mBvA2vA1m(vB1vB2)AmAv(vB2vB1mA1vA2)BSubstituteinto(4)andsolveforvB2andvA2mBvA1vA1m(vB1vB2)vB1vB2A2mAvA1mBvB1mBvB2mAvB1mAvB22mAmBmAvB2mvA1mvB1AmBAmBCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mAv(vA1vA2vB1vB1mA1vA2)BmBvA1mBvA22mBvB1mAvA1mAvA2mAmB2mBvA2mvA1mvB1AmBAmBPhysics:PrinciplesandProblemsSolutionsManual267

271Chapter11continued97.AnalyzeandConcludeA25-gballisfiredwithaninitialspeedofv1towarda125-gballthatishangingmotionlessfroma1.25-mstring.Theballshaveaperfectlyelasticcollision.Asaresult,the125-gballswingsoutuntilthestringmakesanangleof37.0°withthevertical.Whatisv1?Object1istheincomingball.Object2istheoneattachedtothestring.Inthecollision,momentumisconserved.p1ip1fp2form1v1im1v1fm2v2fInthecollision,kineticenergyisconserved.1m21m21m221v1i21v1f22v2fm2m2m21v1i1v1f2v2fmmm(m2)12)12)21v1im(m1v1fm(m2v2fm112m2v2m2v2m2v211i11f22fm1m1m2p2p2p21i1f2fmmm112mp2p2121i1fmp2f2Wedon’tcareaboutv1f,sogetridofp1fusingp1fp1ip2fmp2(p21p21i1ip2f)m2f2m2p22p21p2p1i1i1ip2fp2f2fCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.m2m122p1ip2f1mp2f21m1p1i21mp2f21m1v1i2(m2m1)v2f1m2v1i2m1v2f1Nowconsiderthependulum.1m2m22v2f2ghorv2f2ghwherehL(1cos)Thus,v2f2gL(1cos)v2f(2)(9.80m/s2)(1.25m)(1cos37.0°)2.22m/s268SolutionsManualPhysics:PrinciplesandProblems

272Chapter11continued1125gmakemoleculesandreleasedwhenthev1i225g1(2.22m/s)moleculesarebrokenuporrearranged.Separationofelectricchargesproduces6.7m/selectricpotentialenergy,asinabattery.ElectricpotentialenergyisconvertedtoWritinginPhysicskineticenergyinthemotionofelectricpage310chargesinanelectriccurrentwhenacon-98.AllenergycomesfromtheSun.Inwhatformsductivepath,orcircuit,isprovided.hasthissolarenergycometoustoallowustoBiologicalprocessesareallchemical,andliveandtooperateoursociety?Researchthethus,biologicalenergyisjustaformofwaysthattheSun’senergyisturnedintoachemicalenergy.Solarenergyisfusionformthatwecanuse.AfterweusetheSun’senergyconvertedtotoelectromagneticenergy,wheredoesitgo?Explain.radiation.(SeetheanswertothepreviousTheSunproducesenergythroughquestion.)Lightisawaveformofelectro-nuclearfusionandreleasesthatenergymagneticenergywhosefrequencyisinaintheformofelectromagneticradiation,rangedetectiblebythehumaneye.whichistransferredthroughthevacu-umofspacetoEarth.EarthabsorbsCumulativeReviewthatelectromagneticradiationinitspage310atmosphere,land,andoceansinthe100.Asatelliteisplacedinacircularorbitwithformofthermalenergyorheat.Partof7mandaperiodofaradiusof1.010thevisibleradiationalsoisconverted3s.CalculatethemassofEarth.9.910byplantsintochemicalenergythroughHint:Gravityisthenetforceonsuchasatel-photosynthesis.Thereareseveralotherlite.Scientistshaveactuallymeasuredthechemicalreactionsmediatedbysun-massofEarththisway.(Chapter7)light,suchasozoneproduction.Them2GmenergythenistransferredintovarioussvsmeFnetrr2forms,someofwhicharethechemicalprocessesthatallowustodigestfood2rSince,vTandturnitintochemicalenergytobuildtissues,tomove,andtothink.Inthems42r2GmsmerT2r2end,afterwehaveusedtheenergy,theremainderisdispersedaselectromag-42r2neticradiationbackintotheuniverse.meGT242(1.0107m)399.Allformsofenergycanbeclassifiedas(6.671011Nm2/kg2)(9.910s)2eitherkineticorpotentialenergy.How6.01024kgwouldyoudescribenuclear,electric,chemi-cal,biological,solar,andlightenergy,and101.A5.00-gbulletisfiredwithavelocityofwhy?Foreachofthesetypesofenergy,100.0m/stowarda10.00-kgstationaryresearchwhatobjectsaremovingandhowsolidblockrestingonafrictionlesssurface.energyisstoredinthoseobjects.(Chapter9)Potentialenergyisstoredinthebindinga.Whatisthechangeinmomentumofoftheprotonsandneutronsinthenucle-thebulletifitisembeddedintheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.us.Theenergyisreleasedwhenaheavyblock?nucleusisbrokenintosmallerpieces(fis-sion)orwhenverysmallnucleiarecom-mbvb1mbv2mwv2binedtomakebiggernuclei(fusion).In(mbmw)v2thesameway,chemicalpotentialenergyisstoredwhenatomsarecombinedtoPhysics:PrinciplesandProblemsSolutionsManual269

273Chapter11continuedmsovbvb1IMA2)(33cm)2mded(2.010bmwrThen,66mpvmb(v2vb1)mbvb1mbmvb1ChallengeProblembmwmpage300bmbvb1m1Abulletofmassm,movingatspeedv1,goesbmwthroughamotionlesswoodenblockandexitsmbmwvwithspeedvmb12.Afterthecollision,theblock,bmwwhichhasmassmB,ismoving.(5.00103kg)(10.00kg)5.00103kg10.00kgInitialFinal(100.0ms)v1v20.500kgm/svBb.Whatisthechangeinmomentumofthebulletifitricochetsintheoppositedirectionwithaspeedof99m/s?pvmb(v2vb1)(5.00103kg)1.WhatisthefinalspeedvB,oftheblock?Conservationofmomentum:(99.0m/s100.0m/s)mv0.995kgm/s1mv2mBvBc.InwhichcasedoestheblockendupmBvBm(v1v2)withagreaterspeed?m(v1v2)vWhenthebulletricochets,itsBmBCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.changeinmomentumislargerinmagnitude,andsoistheblock’s2.Howmuchenergywaslosttothebullet?changeinmomentum,sotheblockForthebulletalone:endsupwithagreaterspeed.12KE12mv1102.Anautomobilejackmustexertalifting12forceofatleast15kN.KE22mv2a.IfyouwanttolimittheeffortforcetoKE1m(v2v2)0.10kN,whatmechanicaladvantageis212needed?3.Howmuchenergywaslosttofrictioninside15kNMA150theblock?0.10kNEnergylosttofrictionKEb.Ifthejackis75%efficient,overwhat1distancemusttheeffortforcebeexertedKE2KEblockinordertoraisetheauto33cm?E121mv21m2IMAMA2.0102.lost2mv1222BvBedSinceeIMA,dr270SolutionsManualPhysics:PrinciplesandProblems

274CHAPTER12ThermalEnergypage319PracticeProblems3.Whenyouturnonthehotwatertowash12.1Temperatureanddishes,thewaterpipeshavetoheatup.ThermalEnergyHowmuchheatisabsorbedbyacopperwaterpipewithamassof2.3kgwhenitspages313–322temperatureisraisedfrom20.0°Ctopage31780.0°C?1.ConvertthefollowingKelvintemperaturesQmCTtoCelsiustemperatures.(2.3kg)(385J/kgK)a.115K(80.0°C20.0°C)TCTK273115273158°C5.3104Jb.172K4.ThecoolingsystemofacarenginecontainsTCTK273172273101°C20.0Lofwater(1Lofwaterhasamassofc.125K1kg).TCTK273125273148°Ca.Whatisthechangeinthetemperatured.402Kofthewateriftheengineoperatesuntil836.0kJofheatisadded?TCTK273402273129°CQmCTe.425KQ(8.36105J)TCTK273425273152°CTmC(20.0kg)(4180J/kgK)f.212K10.0KTCTK27321227361°Cb.Supposethatitiswinter,andthecar’s2.FindtheCelsiusandKelvintemperaturescoolingsystemisfilledwithmethanol.forthefollowing.Thedensityofmethanolis0.80g/cm3.a.roomtemperatureWhatwouldbetheincreaseintempera-tureofthemethanolifitabsorbedRoomtemperatureisabout72°F,836.0kJofheat?22°C.ThemassofmethanolwouldbeTKTC27322273295K0.80timesthemassof20.0Lofb.atypicalrefrigeratorwater,or16kg.Arefrigeratoriskeptatabout4°C.QmCTTKTC2734273277K5JQ8.3610c.ahotsummerdayinNorthCarolinaTmC(16kg)(2450J/kgK)Ahotsummerdayisabout95°F,21K35°C.c.Whichisthebettercoolant,waterorCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.TKTC27335273308Kmethanol?Explain.d.awinternightinMinnesotaFortemperaturesabove0°C,waterAtypicalwinternightinMinnesotaisthebettercoolantbecauseitcanisabout14°F,10°C.absorbheatwithoutchangingitstem-TKTC27310273263Kperatureasmuchasmethanoldoes.Physics:PrinciplesandProblemsSolutionsManual271

275Chapter12continued5.ElectricpowercompaniessellelectricitybythekWh,where1kWh3.6106J.Supposethatitcosts$0.08perkWhtorunanelectricwaterheaterinyourneigh-borhood.Howmuchdoesitcosttoheat75kgofwaterfrom15°Cto43°Ctofillabathtub?QmCT(75kg)(4180J/kgK)(43°C15°C)8.8106J8.8106J2.4kWh3.6106J/kWh(2.4kWh)($0.15perkWh)$0.36page3216.A2.00102-gsampleofwaterat80.0°Cismixedwith2.00102gofwaterat10.0°C.Assumethatthereisnoheatlosstothesurroundings.Whatisthefinaltemperatureofthemixture?mACA(TfTAi)mBCB(TfTBi)0SincemAmBandCACB,thereiscancellationinthisparticularcasesothatTT80.0°C10.0°CTAiBi45.0°Cf227.A4.00102-gsampleofmethanolat16.0°Cismixedwith4.00102gofwaterat85.0°C.Assumethatthereisnoheatlosstothesurroundings.Whatisthefinaltemperatureofthemixture?mACA(TfTAi)mWCW(TfTWi)0Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Sinceinthisparticularcase,mAmW,themassescancelandCATAiCWTWiTfCACW(2450J/kgK)(16.0°C)(4180J/kgK)(85.0°C)59.5°C2450J/kgK4180J/kgK8.Threeleadfishingweights,eachwithamassof1.00102gandatatemperatureof100.0°C,areplacedin1.00102gofwaterat35.0°C.Thefinaltemperatureofthemixtureis45.0°C.Whatisthespecificheatoftheleadintheweights?Heatgainedbythewater:QmCT(0.100kg)(4180J/kg°C)(10.0°C)4.18kJThus,heatlostbytheweights4.18kJmweightsCweightsT(4.184kJ)(1000J/kJ)hence,Cweights(0.100kg)(55.0°C)2.53102J/kg°C272SolutionsManualPhysics:PrinciplesandProblems

276Chapter12continued9.A1.00102-galuminumblockat100.0°Cd.55°Cisplacedin1.00102gofwaterat10.0°C.218KThefinaltemperatureofthemixtureise.184°C25.0°C.Whatisthespecificheatofthe89Kaluminum?Heatgainedbythewater:12.ThermalEnergyCouldthethermalenergyQmCTofabowlofhotwaterequalthatofabowl(0.100kg)(4180J/kg°C)(15.0°C)ofcoldwater?Explainyouranswer.Thermalenergyisthemeasureofthe6.27kJtotalenergyofallthemoleculesinanThus,heatlostbythealuminumblockobject.Thetemperature(hotorcold)6.27kJmAluminumCAluminumTmeasurestheamountofenergyperQmolecule.Ifthebowlsareidenticalandhence,CAluminummAluminumTcontainthesameamountofwater,theyhavethesamenumberofmolecules,6.27kJbutthebowlofhotwaterhasmore(0.100kg)(75.0°C)totalthermalenergy.However,ifthe8.36102J/kg°Ccoldwatermassisslightlymorethanthatofthehotwater,thetwoenergiescouldbeequal.SectionReview12.1Temperatureand13.HeatFlowOnadinnerplate,abakedpotatoalwaysstayshotlongerthananyThermalEnergyotherfood.Why?pages313–322Apotatohasalargespecificheatandpage322conductsheatpoorly,soitlosesits10.TemperatureMakethefollowingheatenergyslowly.conversions.a.5°Ctokelvins14.HeatThehardtilefloorofabathroom278Kalwaysfeelscoldtobarefeeteventhoughtherestoftheroomiswarm.Isthefloorb.34KtodegreesCelsiuscolderthantherestoftheroom?239°CThefloorisusuallyatthesametempera-c.212°Ctokelvinstureastherestoftheroom,butthetile485Kconductsheatmoreefficientlythanmostd.316KtodegreesCelsiusmaterials,soitconductsheatfromaperson’sfeet,makingthemfeelcold.43°C15.SpecificHeatIfyoutakeaplasticspoon11.ConversionsConvertthefollowingoutofacupofhotcocoaandputitinyourCelsiustemperaturestoKelvinmouth,youarenotlikelytoburnyourtemperatures.tongue.However,youcouldveryeasilya.28°CburnyourtongueifyouputthehotcocoaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.301Kinyourmouth.Why?b.154°CTheplasticspoonhasalowerspecific427Kheatthanthecocoa,soitdoesnotc.568°Ctransmitmuchheattoyourtongueasitcools.841KPhysics:PrinciplesandProblemsSolutionsManual273

277Chapter12continued16.HeatChefsoftenusecookingpansmadeofthickaluminum.Whyisthickalu-minumbetterthanthinaluminumforcooking?Thickaluminumconductsheatbetteranddoesnothaveany“hotspots.”17.HeatandFoodIttakesmuchlongertobakeawholepotatothantocookfrenchfries.Why?Potatoesdonotconductheatwell.Increasingsurfaceareabycuttingapotatointosmallpartsincreasesheatflowintothepotato.Heatflowfromhotoiltothepotatoisalsomoreefficientthanfromhotair.18.CriticalThinkingAswaterheatsinapotonastove,thewatermightproducesomemistaboveitssurfacerightbeforethewaterbeginstoroll.Whatishappen-ing,andwhereisthecoolestpartofthewaterinthepot?Theheatflowsfromtheburner(thehottestpart)tothetopsurfaceofthewater(coldest).Thewaterfirsttransfersheatfrombottomtotopthroughconduction,andthenconvectionbeginstomovehotwaterincurrentstothetop.PracticeProblems12.2ChangesofStateandtheLawsofThermodynamicspages323–331page32519.Howmuchheatisabsorbedby1.00102goficeat20.0°Ctobecomewaterat0.0°C?QmCTmHfCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(0.100kg)(2060J/kg°C)(20.0°C)(0.100kg)(3.34105J/kg)3.75104J20.A2.00102-gsampleofwaterat60.0°Cisheatedtosteamat140.0°C.Howmuchheatisabsorbed?QmCwaterTmHvmCsteamT(0.200kg)(4180J/kg°C)(100.0°C60.0°C)(0.200kg)(2.26106J/kg)(0.200kg)(2020J/kg°C)(140.0°C100.0°C)502kJ21.Howmuchheatisneededtochange3.00102goficeat30.0°Ctosteamat130.0°C?QmCiceTmHfmCwaterTmHvmCsteamT(0.300kg)(2060J/kg°C)(0.0°C(30.0°C))(0.300kg)(3.34105J/kg)(0.300kg)(4180J/kg°C)(100.0°C0.0°C)(0.300kg)(2.26106J/kg)(0.300kg)(2020J/kg°C)(130.0°C100.0°C)9.40102kJ274SolutionsManualPhysics:PrinciplesandProblems

278Chapter12continuedpage328(0.15kg)(4180J/kg°C)(2.0°C)22.Agasballoonabsorbs75Jofheat.Thebal-1.3103J.loonexpandsbutstaysatthesametemper-ature.HowmuchworkdidtheballoondoThenumberofstirsisinexpanding?1.3103J4stirs2.610UQW0.050JSincetheballoondidnotchange26.Howcanthefirstlawofthermodynamicstemperature,U0.beusedtoexplainhowtoreducetheTherefore,QW.temperatureofanobject?Thus,theballoondid75JofworkinSinceUQW,itispossibletoexpanding.haveanegativeUandtherefore,coolanobjectifQ0andtheobjectdoes23.Adrillboresasmallholeina0.40-kgblockwork,forinstance,byexpanding.ofaluminumandheatsthealuminumbyAlternatively,haveW0andQ5.0°C.Howmuchworkdidthedrilldoinnegativebyhavingittransferheattoboringthehole?itssurroundings.AnycombinationofUQWblock;sinceWdrillWblockthesewillworkwell.andassumenoheataddedtodrill:0WdrillmCTSectionReview(0.40kg)(897J/kg°C)(5.0°C)12.2ChangesofStateandthe1.8103JLawsofThermodynamicspages323–33124.Howmanytimeswouldyouhavetodropapage3310.50-kgbagofleadshotfromaheightof27.HeatofVaporizationOld-fashionedheat-1.5mtoheattheshotby1.0°C?ingsystemssentsteamintoradiatorsineachUmCTroomofahouse.Intheradiators,thesteam(0.50kg)(130J/kg°C)(1.0°C)condensedbacktowater.Analyzethis65Jprocessandexplainhowitheatedaroom.EachtimethebagisraiseditspotentialThecondensingsteamreleaseditsheatenergyisofvaporizationintotheroomandwasthencirculatedbacktotheboilertoPEmghreceivetheheatofvaporizationagain.(0.50kg)(9.80m/s2)(1.5m)7.4J28.HeatofVaporizationHowmuchheatisWhenthebaghitsthegroundthisneededtochange50.0gofwaterat80.0°Cenergyis(mostly)transmittedasworktosteamat110.0°C?ontheleadshot.ThenumberofdropsQmCwaterTmHvmCsteamT65Jis9drops7.4J(0.500kg)(4180J/kg°C)(100.0°C80.0°C)(0.500kg)25.Whenyoustiracupoftea,youdoaboutCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.050Jofworkeachtimeyoucirclethe(2.26106J/kg)(0.500kg)spooninthecup.Howmanytimeswould(2020J/kg°C)(110.0°C100.0°C)youhavetostirthespoontoheata0.15-kgcupofteaby2.0°C?1.18105JUmCTPhysics:PrinciplesandProblemsSolutionsManual275

279Chapter12continued29.HeatofVaporizationThespecificheat32.MechanicalEnergyandThermalEnergyofmercuryis140J/kg°C.Itsheatofvapor-Waterflowsoverafallthatis125.0mhigh,izationis3.06105J/kg.HowmuchenergyasshowninFigure12-17.Ifthepotentialisneededtoheat1.0kgofmercurymetalenergyofthewaterisallconvertedtother-from10.0°Ctoitsboilingpointandmalenergy,calculatethetemperaturediffer-vaporizeitcompletely?Theboilingpointencebetweenthewateratthetopandtheofmercuryis357°C.bottomofthefall.QmCHgTmHv(1.0kg)(140J/kg°C)(357°C10.0°C)125.0m(1.0kg)(3.06105J/kg)3.5105J30.MechanicalEnergyandThermalEnergyJamesJoulecarefullymeasuredthediffer-■Figure12-17enceintemperatureofwateratthetopandPEgravityQabsorbedbywaterbottomofawaterfall.WhydidheexpectamghmCTdifference?ThewateratthetophasgravitationalghTpotentialenergythatisdissipatedintoCthermalenergywhenthewatersplash-(9.80m/s2)(125.0m)esatthebottom.Thewatershouldbe4180J/kg°Chotteratthebottom,butnotbymuch.0.293°Criseintemperatureatthebottom31.MechanicalEnergyandThermalEnergyAmanusesa320-kghammermovingat33.EntropyEvaluatewhyheatingahomeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5.0m/stosmasha3.0-kgblockofleadwithnaturalgasresultsinanincreasedagainsta450-kgrock.Whenhemeasuredamountofdisorder.thetemperaturehefoundthatithadThegasreleasesheat,Q,atitscombus-increasedby5.0°C.Explainhowthistiontemperature,T.Thenaturalgashappened.moleculesbreakupandcombustwithPartofthekineticenergyofthehammeroxygen.Theheatisdistributedinmanyisabsorbedasthermalenergybythenewways,andthenaturalgasmoleculesleadblock.Thehammer’senergyiscannotreadilybereassembled.1mv21(320kg)(5.0m/s)24.0kJ.22Thechangeinthermalenergyofthe34.CriticalThinkingAnewdeckofcardshasblockisallthesuits(clubs,diamonds,hearts,andspades)inorder,andthecardsareorderedUmCTbynumberwithinthesuits.Ifyoushuffle(3.0kg)(130J/kgK)(5.0°C)thecardsmanytimes,areyoulikelyto2.0kJreturnthecardstotheiroriginalorder?Hence,abouthalfofthehammer’sExplain.Ofwhatphysicallawisthisanenergywenttotheleadblock.example?Thecardsareveryunlikelytoreturntotheiroriginalorder.Thisisanexampleofthesecondlawofthermodynamics,inwhichdisorderincreases.276SolutionsManualPhysics:PrinciplesandProblems

280Chapter12continued40.WhenheatflowsfromawarmerobjectinChapterAssessmentcontactwithacolderobject,dothetwoConceptMappinghavethesametemperaturechanges?(12.1)page336Thetwoobjectswillchangetempera-35.Completethefollowingconceptmapusingturesdependingontheirmassesandthefollowingterms:heat,work,internalenergy.specificheats.Thetemperaturechangesarenotnecessarilythesameforeach.Firstlawofthermodynamics41.Canyouaddthermalenergytoanobjectwithoutincreasingitstemperature?Explain.(12.2)heatworkinternalWhenyoumeltasolidorboilaenergyliquid,youaddthermalenergywithoutchangingthetemperature.externalentropytemperatureforces42.Whenwaxfreezes,doesitabsorborreleaseenergy?(12.2)MasteringConceptsWhenwaxfreezes,itreleasesenergy.page33636.Explainthedifferencesamongthemechani-43.Explainwhywaterinacanteenthatissur-calenergyofaball,itsthermalenergy,androundedbydryairstayscoolerifithasaitstemperature.(12.1)canvascoverthatiskeptwet.(12.2)ThemechanicalenergyisthesumofWhenthewaterinthecoverevaporatesthepotentialandkineticenergiesoftheintothedryair,itmustabsorbanballconsideredasonemass.Thether-amountofenergyproportionaltoitsmalenergyisthesumofthepotentialheatoffusion.Indoingso,itcoolsoffandkineticenergiesoftheindividualthecanteen.Thisworksonlyiftheairparticlesthatmakeupthemassoftheisdry;iftheairishumid,thenthewaterball.Thetemperatureisameasureofwillnotevaporate.theinternalenergyoftheball.44.Whichprocessoccursatthecoilsofarun-37.Cantemperaturebeassignedtoavacuum?ningairconditionerinsideahouse,vapor-Explain.(12.1)izationorcondensation?Explain.(12.2)No,becausetherearenoparticlesthatInsidethehouse,thecoolantisevapo-haveenergyinavacuum.ratinginthecoilstoabsorbenergyfromtherooms.38.Doallofthemoleculesoratomsinaliquidhavethesamespeed?(12.1)ApplyingConceptsNo.Thereisadistributionofvelocitiespage336oftheatomsormolecules.45.CookingSallyiscookingpastainapotofboilingwater.Willthepastacookfasterif39.Isyourbodyagoodjudgeoftemperature?thewaterisboilingvigorouslyorifitisOnacoldwinterday,ametaldoorknobfeelsboilinggently?muchcoldertoyourhandthanawoodenCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Itshouldmakenodifference.Eitherway,doordoes.Explainwhythisistrue.(12.1)thewaterisatthesametemperature.Yourskinmeasuresheatflowtoorfromitself.Themetaldoorknobabsorbsheatfromyourskinfasterthanthewoodendoor,soitfeelscolder.Physics:PrinciplesandProblemsSolutionsManual277

281Chapter12continued46.Whichliquidwouldanicecubecoolfaster,MasteringProblemswaterormethanol?Explain.12.1TemperatureandThermalEnergyMethanol,becauseithasalowerpages336–337specificheat;foragivenmassandheatLevel1transfer,itgeneratesabiggerTsince52.HowmuchheatisneededtoraisetheQmCT.temperatureof50.0gofwaterfrom4.5°Cto83.0°C?47.Equalmassesofaluminumandleadareheatedtothesametemperature.ThepiecesQmCTofmetalareplacedonablockofice.Which(0.0500kg)(4180J/kg°C)metalmeltsmoreice?Explain.(83.0°C4.5°C)Thespecificheatofaluminumismuchgreaterthanthatoflead;therefore,it1.64104Jmeltsmoreice.53.A5.00102-gblockofmetalabsorbs5016J48.Whydoeasilyvaporizedliquids,suchasofheatwhenitstemperaturechangesfromacetoneandmethanol,feelcooltotheskin?20.0°Cto30.0°C.Calculatethespecificheatofthemetal.Astheyevaporate,theyabsorbtheirheatofvaporizationfromtheskin.QmCTQsoC49.ExplainwhyfruitgrowersspraytheirtreesmTwithwaterwhenfrostisexpectedtoprotect5016Jthefruitfromfreezing.(5.00101kg)(30.0°C20.0°C)Thewaterontheleaveswillnotfreeze1.00103J/kg°Cuntilitcanreleaseitsheatoffusion.Thisprocesskeepstheleaveswarmer1.00103J/kgKlonger.Theheatcapacityoftheiceslowsdownthecoolingbelow0°C.54.CoffeeCupA4.00102-gglasscoffeecupCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.is20.0°Catroomtemperature.Itisthen50.Twoblocksofleadhavethesametempera-plungedintohotdishwateratatempera-ture.BlockAhastwicethemassofblockB.tureof80.0°C,asshowninFigure12-18.TheyaredroppedintoidenticalcupsofIfthetemperatureofthecupreachesthatofwaterofequaltemperatures.Willthetwothedishwater,howmuchheatdoesthecupcupsofwaterhaveequaltemperaturesafterabsorb?Assumethatthemassofthedish-equilibriumisachieved?Explain.waterislargeenoughsothatitstempera-ThecupwithblockAwillbehotterturedoesnotchangeappreciably.becauseblockAcontainsmorethermal20.0°C80.0°Cenergy.51.WindowsOften,architectsdesignmostofthewindowsofahouseonthenorthside.2Howdoesputtingwindowsonthesouthside4.0010gaffecttheheatingandcoolingofthehouse?■Figure12-18Inthenorthernhemisphere,thesun-QmCTlightcomesfromthesouth.TheSun’s(4.00101kg)(840J/kg°C)lightwouldhelpheatthehouseinthewinterbutalsowouldalsoheatthe(80.0°C20.0°C)houseinthesummer.2.02104J278SolutionsManualPhysics:PrinciplesandProblems

282Chapter12continued55.A1.00102-gmassoftungstenat100.0°Cisplacedin2.00102gofwaterat20.0°C.Themixturereachesequilibriumat21.6°C.Calculatethespecificheatoftungsten.QTQW0ormTCTTTmWCWTWmWCWTW(0.200kg)(4180J/kgK)(21.6°C20.0°C)CTmTTT(0.100kg)(21.6°C100.0°C)171J/kgK56.A6.0102-gsampleofwaterat90.0°Cismixedwith4.00102gofwaterat22.0°C.Assumethatthereisnoheatlosstothesurroundings.Whatisthefinaltemperatureofthemixture?mACATAimBCBTBiTfmACAmBCBbutCACBbecausebothliquidsarewater,andtheC’swilldivideout.m(6.0102g)(90.0°C)(4.00102g)(22.0°C)TATAimBTBifmAmB6.0102g4.00102g63°C57.A10.0-kgpieceofzincat71.0°Cisplacedinacontainerofwater,asshowninFigure12-19.Thewaterhasamassof20.0kganda20.0kgtemperatureof10.0°Cbeforethezincisadded.Whatisthefinalgktemperatureofthewaterand.00thezinc?110.0°CmTACATAimBCBTBifmACAmBCB■Figure12-19(10.0kg)(388J/kgK)(71.0°C)(20.0kg)(4180J/kgK)(10.0°C)(10.0kg)(388J/kgK)(20.0kg)(4180J/kgK)12.7°CLevel258.Thekineticenergyofacompactcarmovingat100km/his2.9105J.Togetafeelingfortheamountofenergyneededtoheatwater,whatvolumeofwater(inliters)would2.9105Jofenergywarmfromroomtemperature(20.0°C)toboiling(100.0°C)?QmCTVCTwhereisthedensityofthematerialCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Q2.9105Jso,VCT(1.00kg/L)(4180J/kg°C)(100.0°C20.0°C)0.87LPhysics:PrinciplesandProblemsSolutionsManual279

283Chapter12continued59.WaterHeaterA3.0102-Welectric2immersionheaterisusedtoheatacup3.0010Wofwater,asshowninFigure12-20.Thecupismadeofglass,anditsmassis3.00102g.Itcontains250gofwaterat15°C.Howmuchtimeisneededtobringthewatertotheboilingpoint?Assume15°Cthatthetemperatureofthecupisthesameasthetemperatureofthewaterat250galltimesandthatnoheatislosttotheair.QmGCGTGmWCWTW3.00102gbutTGTW,so■Figure12-20Q(mGCGmWCW)T((0.300kg)(840J/kg°C)(0.250kg)(4180J/kg°C))(100.0°C15°C)1.1105JEQNowP,sottQ1.1105JtP3.00102J/s370s6.1min60.CarEngineA2.50102-kgcast-ironcarenginecontainswaterasacoolant.Supposethattheengine’stemperatureis35.0°Cwhenitisshutoff,andtheairtemperatureis10.0°C.Theheatgivenoffbytheengineandwaterinitastheycooltoairtemperatureis4.40106J.Whatmassofwaterisusedtocooltheengine?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.QmWCWTmiCiTQmiCiT(4.4106J)((2.50102kg)(450J/kg°C)(35.0°C10.0°C))mWCT(4180J/kg°C)(35.0°C10.0°C)W15kg12.2ChangesofStateandtheLawsofThermodynamicspage337Level161.Yearsago,ablockoficewithamassofabout20.0kgwasuseddailyinahomeicebox.Thetemperatureoftheicewas0.0°Cwhenitwasdelivered.Asitmelted,howmuchheatdidtheblockoficeabsorb?QmH5J/kg)6.68106Jf(20.0kg)(3.341062.A40.0-gsampleofchloroformiscondensedfromavaporat61.6°Ctoaliquidat61.6°C.Itliberates9870Jofheat.Whatistheheatofvaporizationofchloroform?QmHvHQ9870J5J/kgvm0.0400kg2.4710280SolutionsManualPhysics:PrinciplesandProblems

284Chapter12continued63.A750-kgcarmovingat23m/sbrakestoaQmCTstop.Thebrakescontainabout15kgofiron,(0.0100kg)(2060J/kg°C)whichabsorbstheenergy.Whatistheincreaseintemperatureofthebrakes?(120.0°C100.0°C)Duringbraking,thekineticenergyof404JthecarisconvertedintoheatThetotalheatisenergy.So,412J3.34103J4.18103JKECQB0.0,andKEC2.26104J404J3.09104JmBCBT0.0so,65.A4.2-gleadbulletmovingat275m/s1m2v2)KEC2C(vfistrikesasteelplateandcomestoastop.IfTmBCBmBCBallitskineticenergyisconvertedtothermalenergyandnoneleavesthebullet,whatis1(750kg)(0.02(23m/s)2)itstemperaturechange?2(15kg)(450J/kg°C)Becausethekineticenergyisconvertedtothermalenergy,KEQ0.So29°CKEmBCBTandLevel264.Howmuchheatisaddedto10.0goficeat1m2v2)KE2B(vfi20.0°Ctoconvertittosteamat120.0°C?TmBCBmBCBAmountofheatneededtoheaticetoandthemassofthebulletdividesoutso0.0°C:QmCT1(v2v2)2fiT(0.0100kg)(2060J/kg°C)CB(0.0°C(20.0°C))1((0.0m/s)2(275m/s)2)2412J130J/kg°CAmountofheattomeltice:290°CQmHf(0.0100kg)(3.34105J/kg)66.SoftDrinkAsoftdrinkfromAustraliaislabeled“Low-JouleCola.”Thelabelsays3.34103J“100mLyields1.7kJ.”ThecancontainsAmountofheattoheatwaterto375mLofcola.Chandradrinksthecola100.0°C:andthenwantstooffsetthisinputoffoodQmCTenergybyclimbingstairs.HowhighwouldChandrahavetoclimbifshehasamassof(0.0100kg)(4180J/kg°C)65.0kg?(100.0°C0.0°C)Chandragained(3.75)(1.7kJ)6.4103J4.18103Jofenergyfromthedrink.Amountofheattoboilwater:Toconserveenergy,EPE0or6.4103Jmghso,QmHCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.v6.4103J6.4103J(0.0100kg)(2.26106J/kg)hmg(65.0kg)(9.80m/s2)2.26104J1.0101m,oraboutthreeflightsAmountofheattoheatsteamtoofstairs120.0°C:Physics:PrinciplesandProblemsSolutionsManual281

285Chapter12continuedMixedReviewKETherefore,TmCpages337–338Level14.7105J(45kg)(897J/kg°C)67.Whatistheefficiencyofanenginethatpro-duces2200J/swhileburningenoughgaso-12°Clinetoproduce5300J/s?Howmuchwasteheatdoestheengineproducepersecond?70.IcedTeaTomakeicedtea,youstartbybrewingtheteawithhotwater.ThenyouW2200JEfficiency100100addice.Ifyoustartwith1.0Lof90°Ctea,Q5300JHwhatistheminimumamountoficeneeded42%tocoolitto0°C?WoulditbebettertoletTheheatlossistheteacooltoroomtemperaturebefore5300J2200J2900Jaddingtheice?Heatlostbythetea68.StampingPressAmetalstampingmachineQmCTinafactorydoes2100Jofworkeachtimeitstampsoutapieceofmetal.Eachstamped(1.0kg)(4180J/kgK)(90°C)pieceisthendippedina32.0-kgvatofwater376kJforcooling.ByhowmanydegreesdoestheAmountoficemeltedvatheatupeachtimeapieceofstampedQmetalisdippedintoit?mHfIfweassumethe2100Jofworkfrom376kJthemachineisabsorbedasthermal1.1kg334kJenergyinthestampedpiece,thentheThus,youneedslightlymoreicethanvatmustabsorb2100Jintheformoftea,butthisratiowouldmakewateryheatfromeachpiece.Noworkisdonetea.Lettheteacooltoroomtempera-onthewater,onlyheatistransferred.turebeforeaddingtheice.ThechangeintemperatureofthewaterCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.isgivenbyLevel2UmCT,71.Ablockofcopperat100.0°Ccomesincon-Utactwithablockofaluminumat20.0°C,asthereforeTmCshowninFigure12-21.Thefinaltempera-2100Jtureoftheblocksis60.0°C.Whatarethe(32.0kg)(4180J/kg°C)relativemassesoftheblocks?0.016°C.69.A1500-kgautomobilecomestoastop100.0°C20.0°Cfrom25m/s.Alloftheenergyoftheautomobileisdepositedinthebrakes.CopperAluminumAssumingthatthebrakesareabout45kgofaluminum,whatwouldbethechangeintemperatureofthebrakes?60.0°C60.0°CTheenergychangeinthecarisKE1(1500kg)(25m/s)24.7105J.2CopperAluminumIfallofthisenergyistransferredasworktothebrakes,then■Figure12-21UKEmCT.282SolutionsManualPhysics:PrinciplesandProblems

286Chapter12continuedTheheatlostfromthecopperequalsTherefore,6.6Jisaddedtotheice.theheatgainedbythealuminum.TheTheamountoficemeltedisgivenbyTforthecopperis40.0°CandtheKEaluminumheatsby40.0°C.mHftherefore,6.6JmcopperCcoppermaluminumCaluminum3.34105J/kgmC2.0105kgcopperaluminumandmCaluminumcopperThinkingCritically897J/kgK2.3page338385J/kgK74.AnalyzeandConcludeAcertainheatThecopperblockhas2.3timesasengineremoves50.0Jofthermalenergymuchmassasthealuminumblock.fromahotreservoirattemperatureTH545Kandexpels40.0Jofheattoa72.A0.35-kgblockofcopperslidingonthecolderreservoirattemperatureTL325K.floorhitsanidenticalblockmovingattheIntheprocess,italsotransfersentropyfromsamespeedfromtheoppositedirection.onereservoirtotheother.Thetwoblockscometoastoptogethera.Howdoestheoperationoftheengineafterthecollision.Theirtemperatureschangethetotalentropyofthereservoirs?increaseby0.20°Casaresultofthecolli-Astheengineoperates,itremovession.Whatwastheirvelocitybeforetheenergyfromthehotreservoir.collision?QHThechangeininternalenergyoftheTherefore,SHTsothattheHblocksisentropyofthehotreservoirUmCTdecreases.(0.70kg)(385J/kg°C)(0.20°C)TheentropyofthecoldreservoirQ54JSLincreases.ThenetLTLTherefore,54Jequalsthekineticincreaseinentropyofthereservoirsenergyoftheblocksbeforethetogetheris12collision.54J(2)mv2STSLSH54Jv0.35kgQLQHTT12m/sLH40.0J50.0JSLevel3T325K545K73.A2.2-kgblockoficeslidesacrossarough0.0313J/Kfloor.Itsinitialvelocityis2.5m/sanditsb.Whatwouldbethetotalentropychangefinalvelocityis0.50m/s.HowmuchoftheinthereservoirsifTiceblockmeltedasaresultoftheworkL205K?donebyfriction?40.0J50.0JS0.103J/KT205K545KTheworkdonebyfrictionequalstheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thetotalentropychangeinthenegativeofthechangeinkineticenergyoftheblock,assumingnottoomuchofreservoirs,andintheuniverse,hastheblockmelted.increasedapproximatelybyafactorofthree.12KE(2.2kg)(0.50m/s)21(2.2kg)(2.5m/s)26.6J2Physics:PrinciplesandProblemsSolutionsManual283

287Chapter12continued75.AnalyzeandConcludeDuringagame,WritinginPhysicsthemetabolismofbasketballplayersoftenpage338increasesbyasmuchas30.0W.Howmuch78.Ourunderstandingoftherelationshipperspirationmustaplayervaporizeperbetweenheatandenergywasinfluencedbyahourtodissipatethisextrathermalenergy?soldiernamedBenjaminThompson,CountTheamountofthermalenergytoRumford;andabrewernamedJamesPrescottbedissipatedin1.00hisJoule.BothreliedonexperimentalresultstoU(30.0J/s)(3600s/h)1.08105J.developtheirideas.Investigatewhatexperi-Theamountofwaterthisenergy,mentstheydidandevaluatewhetherornottransmittedasheat,wouldvaporizeisitisfairthattheunitofenergyiscalledtheQJouleandnottheThompson.mHVIn1799,heatwasthoughttobealiquid1.08105Jthatflowedfromoneobjecttoanother.2.26106J/kgHowever,CountRumfordthoughtthatheatwascausedbythemotionofparti-0.0478kgclesinthemetalcannon.Hedidnotdo76.AnalyzeandConcludeChemistsuseanyquantitativemeasurementsandhiscalorimeterstomeasuretheheatproducedideaswerenotwidelyaccepted.In1843,bychemicalreactions.Forinstance,aJoule,doingcarefulmeasurements,chemistdissolves1.01022moleculesofameasuredthechangeintemperaturepowderedsubstanceintoacalorimetercon-causedbyaddingheatordoingworktaining0.50kgofwater.Themoleculesonaquantityofwater.Heprovedthatbreakupandreleasetheirbindingenergytoheatisaflavorofenergyandthatener-thewater.Thewatertemperatureincreasesgyisconserved.Jouledeservestheby2.3°C.Whatisthebindingenergypercreditandtheeponymicunit.moleculeforthissubstance?79.WaterhasanunusuallylargespecificheatTheamountofenergyaddedtotheandlargeheatsoffusionandvaporization.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.waterisOurweatherandecosystemsdependuponUmCTwaterinallthreestates.Howwouldour(0.50kg)(4180J/kg°C)(2.3°C)worldbedifferentifwater’sthermodynamicpropertieswerelikeothermaterials,suchas4.8kJmethanol?Theenergypermoleculeistherefore,Thelargespecificheatandlargeheats4.8kJ4.81019J/moleculeoffusionandvaporizationmeanthat1022moleculeswater,ice,andwatervaporcanstorealotofthermalenergywithoutchanging77.ApplyConceptsAlloftheenergyonEarththeirtemperaturestoomuch.Theimpli-comesfromtheSun.Thesurfacetempera-4K.cationsaremany.TheoceansandlargetureoftheSunisapproximately10lakesmoderatethetemperatureWhatwouldbetheeffectonourworldif3K?changesinnearbyregionsonadailytheSun’ssurfacetemperaturewere10andseasonalbasis.Theday-to-nightStudentanswerswillvary.Answerstemperaturevariationnearalakeisshouldreflectchangingaveragemuchsmallerthantheday-to-nighttemperatureonEarth,differentweathertemperaturevariationinthedesert.Thepatterns,plantandanimalspecieslargeheatoffusionofwatercontrolsdyingout,etc.thechangeofseasonsinthefarnorthandsouth.Theabsorptionofenergybyfreezingwaterinthefallanditsrelease284SolutionsManualPhysics:PrinciplesandProblems

288Chapter12continuedinthespringslowsthetemperature81.Aweightlifterraisesa180-kgbarbelltoachangesintheatmosphere.Waterheightof1.95m.Howmuchworkisdoneabsorbsandstoresalotofenergyasitbytheweightlifterinliftingthebarbell?vaporizes.Thisenergycanbeusedto(Chapter10)drivemeteorologicalevents,suchasWmghthunderstormsandhurricanes.(180kg)(9.80m/s2)(1.95m)CumulativeReview3.4103Jpage33880.Aropeiswoundaroundadrumwitha82.InaGreekmyth,themanSisyphusisradiusof0.250mandamomentofinertiacondemnedbythegodstoforeverrollanof2.25kgm2.Theropeisconnectedtoaenormousrockupahill.Eachtimehe4.00-kgblock.(Chapter8)reachesthetop,therockrollsbackdowna.Findthelinearaccelerationoftheblock.tothebottom.Iftherockhasamassof215kg,thehillis33minheight,andSolveNewton’ssecondlawfortheSisyphuscanproduceanaveragepowerblock:mgFTma,wherethepos-of0.2kW,howmanytimesin1hcanitivedirectionisdownwardandherolltherockupthehill?(Chapter11)whereFTistheforceoftheropeonthedrum.Newton’ssecondlawforTheamountofworkneededtorollthethedrumisFrockuponceisTrIorFTrIa/r.Thatis,F2.Therefore,Wmgh(215kg)(9.80m/s2)(33m)TIa/rmg(I/r2)ama.Thatis,70,000Jamg/(mI/r2)g/10.0InonehourSisyphusdoesanamount0.980m/s2.ofworkb.Findtheangularaccelerationofthe(0.2103J)(3600s)720,000Jdrum.Hepushestherockupthehill2(720,000)/(70,000)10timesinonehoura0.980m/sr0.250m3.92rad/s2ChallengeProblemc.Findthetension,FT,intherope.page329IEntropyhassomeinterestingproperties.FTrComparethefollowingsituations.Explainhow(2.25kgm2)(3.92rad/s2)andwhythesechangesinentropyaredifferent.0.250m35.3Nd.FindtheangularvelocityofthedrumQaftertheblockhasfallen5.00m.1kg1kgx1at2,sot2x3.19s2aTiTfTherefore,t(3.92rad/s2)(3.19s)1.Heating1.0kgofwaterfrom273Kto274K.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12.5rad/sSQmCTTT(1.0kg)(4180J/kgK)(274K273K)273K15J/KPhysics:PrinciplesandProblemsSolutionsManual285

289Chapter12continued2.Heating1.0kgofwaterfrom353Kto354K.QmCTSTT(1.0kg)(4180J/kgK)(354K353K)353K12J/K3.Completelymelting1.0kgoficeat273K.QmHfSTT(1.0kg)(3.34105J/kg)273K1.2103J/K4.Heating1.0kgofleadfrom273Kto274K.QmCTSTT(1.0kg)(130J/kgK)(274K273K)273K0.48J/KCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.286SolutionsManualPhysics:PrinciplesandProblems

290CHAPTER13StatesofMatter4.Inatornado,thepressurecanbe15percentPracticeProblemsbelownormalatmosphericpressure.13.1PropertiesofFluidsSupposethatatornadooccurredoutsideapages341–348doorthatis195cmhighand91cmwide.Whatnetforcewouldbeexertedonthepage344doorbyasudden15percentdropin1.Theatmosphericpressureatsealevelis5Pa.Whatistheforceatseanormalatmosphericpressure?Inwhatabout1.010directionwouldtheforcebeexerted?levelthatairexertsonthetopofadeskthatis152cmlongand76cmwide?ThepressuredifferenceacrossthedoorisFPAPlw(1.0105Pa)(1.52m)(0.76m)Pdiff(15%)(Patm)1.2105N(0.15)(1.0105Pa)1.5104PaFPdiffAPdifflw2.Acartiremakescontactwiththegroundonarectangularareaof12cmby18cm.Ifthe(1.5104Pa)(1.95m)(0.91m)car’smassis925kg,whatpressuredoesthe2.7104Ndirectedfromtheinsidecarexertonthegroundasitrestsonallofthehouseoutwardfourtires?FFg,carmcarg5.Inindustrialbuildings,largepiecesofequip-PAA4lwmentmustbeplacedonwidesteelplates2)thatspreadtheweightoftheequipmentover(925kg)(9.80m/slargerareas.Ifanengineerplanstoinstalla(4)(0.12m)(0.18m)454-kgdeviceonafloorthatisratedto1.0102kPa4Pa,withstandadditionalpressureof5.010howlargeshouldthesteelsupportplatebe?3.Aleadbrick,5.0cm10.0cm20.0cm,restsonthegroundonitssmallestface.FgmgThemaximumpressurePLeadhasadensityof11.8g/cm3.WhatAApressuredoesthebrickexertontheground?mgTherefore,APmbrickVlwh(454kg)(9.80m/s2)(11.8g/cm3)(5.0cm)4Pa5.010(10.0cm)(20.0cm)8.9102m21.18104g11.8kg6.AtankofheliumgasusedtoinflatetoyFg,brickmbrickg6PaPballoonsisatapressureof15.510Alwandatemperatureof293K.Thetank’sVglwhgvolumeis0.020m3.Howlargeaballoonhglwlwwoulditfillat1.00atmosphereand323K?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(11.8g/cm3)(20.0cm)PPT1V12V2,soV2P1V12)1kg(100cm)2T1T22P2T1(9.80m/s1000g2(1m)1.00atm1.013105Pa23kPaPhysics:PrinciplesandProblemsSolutionsManual287

291Chapter13continued(323K)(15.5106Pa)(0.020m3)VSectionReview2(1.013105Pa)(293K)313.1PropertiesofFluids3.4mpages341–3487.Whatisthemassoftheheliumgasinthepage348previousproblem?Themolarmassof10.PressureandForceSupposethatyouhaveheliumgasis4.00g/mol.twoboxes.Oneis20cm20cm20cm.PVnRTTheotheris20cm20cm40cm.PV(15.5106Pa)(0.020m3)a.HowdoesthepressureoftheaironthenRT(8.31Pam3/molK)(293K)outsideofthetwoboxescompare?127.3molThepressureoftheairisthesameonthetwoboxes.m(127.3mol)(4.00g/mol)5.1102gb.Howdoesthemagnitudeofthetotal8.Atankcontaining200.0Lofhydrogengasforceoftheaironthetwoboxesat0.0°Ciskeptat156kPa.Thetempera-compare?tureisraisedto95°C,andthevolumeisBecauseFPAthetotalforceofdecreasedto175L.Whatisthenewtheairisgreaterontheboxwiththepressureofthegas?greaterarea.ThesecondboxhasPPtwicethesurfacearea,soithas1V12V2TTwithT1273Kandtwicethetotalforceofthefirstbox.12T295°C273°C368K11.MeteorologyAweatherballoonusedbyTmeteorologistsismadeofaflexiblebagthat2P1V1P2Vallowsthegasinsidetofreelyexpand.Ifa2T1weatherballooncontaining25.0m3of(368K)(156kPa)(200.0L)heliumgasisreleasedfromsealevel,what(175L)(273K)isthevolumeofgaswhentheballoon2.4102kPareachesaheightof2100m,wheretheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pressureis0.82105Pa?Assumethat9.Theaveragemolarmassofthecomponentsthetemperatureisunchanged.ofair(mainlydiatomicoxygengasandP1V1P2V2diatomicnitrogengas)isabout29g/mol.Whatisthevolumeof1.0kgofairatP1V1V2Patmosphericpressureand20.0°C?2PVnRT(1.013105Pa)(25.0m3)0.82105PanRTVP3.1101m3m1.0103gwherenM29g/mol12.GasCompressionInacertaininternal-combustionengine,0.0021m3ofairatandT20.0°C273293Katmosphericpressureand303Kisrapidly1.0103g(8.31Pam3/molK)(293K)compressedtoapressureof20.1105Pa29g/molV5Pa)andavolumeof0.0003m3.Whatisthe(1.01310finaltemperatureofthecompressedgas?0.83m3PP1V12V2TT12TT1P2V22P1V1288SolutionsManualPhysics:PrinciplesandProblems

292Chapter13continued(303K)(20.1105Pa)(0.0003m3)Thereareanequalnumberofparticles(1.013105Pa)(0.0021m3)inthetwosamples.Inanidealgas,thesizeoftheparticlesisnotrelevantto9102Kthevolumeofthegasorthepressureexertedbythegas.13.DensityandTemperatureStartingat0°C,howwillthedensityofwaterchangeifitisheatedto4°C?To8°C?SectionReviewAsthewaterisheatedfrom0°C,thedensitywillincreaseuntilitreachesa13.2ForcesWithinLiquidsmaximumat4°C.Onfurtherheatingtopages349–3518°C,thedensityofthewaterwillpage351decrease.17.EvaporationandCoolingInthepast,whenababyhadahighfever,thedoctor14.TheStandardMolarVolumeWhatisthemighthavesuggestedgentlyspongingoffvolumeof1.00molofagasatatmosphericthebabywithrubbingalcohol.Whywouldpressureandatemperatureof273K?thishelp?nRTSincealcoholevaporateseasily,thereisVPaverynoticeableevaporativecooling(1.00mol)(8.31Pam3/molK)(273K)effect.1.013105Pa0.0224m318.SurfaceTensionApaperclip,whichhasadensitygreaterthanthatofwater,canbe15.TheAirinaRefrigeratorHowmanymadetostayonthesurfaceofwater.Whatmolesofairareinarefrigeratorwithavol-proceduresmustyoufollowforthistoumeof0.635m3atatemperatureofhappen?Explain.2.00°C?IftheaveragemolarmassofairisThepaperclipshouldbeplacedcare-29g/mol,whatisthemassoftheairinthefullyandflatlyontothesurfaceoftherefrigerator?water.ThiswillreducetheweightperPVunitareaofwatersurfaceonwhichitnRTwillrest.Thesurfacetensionofthe(1.013105Pa)(0.635m3)waterthenissufficienttosupportthe(8.31Pam3molK)(275K)reducedweightperunitareaofthepaperclip.28.1molmnM19.LanguageandPhysicsTheEnglishlan-guageincludesthetermsadhesivetapeand(28.1mol)(29g/mol)workingasacohesivegroup.Intheseterms,0.81kgareadhesiveandcohesivebeingusedinthesamecontextastheirmeaningsinphysics?16.CriticalThinkingComparedtotheparti-Yes,adhesivetapeisstickingtosome-clesthatmakeupcarbondioxidegas,thethingdifferentthantape.Acohesiveparticlesthatmakeupheliumgasareverygroupisacollectionofpeopleworkingsmall.Whatcanyouconcludeaboutthetogether.numberofparticlesina2.0-LsampleofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.carbondioxidegascomparedtothenum-20.AdhesionandCohesionIntermsofberofparticlesina2.0-Lsampleofheliumadhesionandcohesion,explainwhygasifbothsamplesareatthesametemper-alcoholclingstothesurfaceofaglassrodatureandpressure?butmercurydoesnot.Physics:PrinciplesandProblemsSolutionsManual289

293Chapter13continuedAdhesionistheforcebetweenunlikeF(1600N)(72cm2)F1A2materials.Alcoholhasagreater2A1440cm21adhesiveattractiontoglassthanmer-1N8.010curyhas.Thecohesiveforcesofmer-curyarestrongenoughtoovercomeits24.Amechanicexertsaforceof55Nonaadhesiveforcewithglass.0.015m2hydraulicpistontoliftasmallautomobile.Thepistonthattheautomobile21.FloatingHowcanyoutellthatthepaper2.Whatisthesitsonhasanareaof2.4mclipinproblem18wasnotfloating?weightoftheautomobile?Ifthepaperclipbrokethroughthesur-F(55N)(2.4m2)faceofthewater,itsank.AnobjectthatF1A23N2A(0.015m2)8.810floatswouldsimplybobbacktothe1surface.25.Bymultiplyingaforce,ahydraulicsystem22.CriticalThinkingOnahot,humidday,servesthesamepurposeasaleverorBethsatonthepatiowithaglassofcoldseesaw.Ifa400-Nchildstandingononewater.Theoutsideoftheglasswascoatedpistonisbalancedbya1100-Nadultwithwater.Heryoungersister,Jo,suggestedstandingonanotherpiston,whatisthethatthewaterhadleakedthroughtheglassratiooftheareasoftheirpistons?fromtheinsidetotheoutside.SuggestanF1A2FexperimentthatBethcoulddotoshowJo2A1wherethewatercamefrom.A2F2400NBethcouldweightheglassbefore0.4A1F11100NputtingitintherefrigeratorforawhileTheadultstandsonthelargerpiston.tocoolitdown.Thenshecouldremoveitfromtherefrigeratorandallowmois-26.Inamachineshop,ahydraulicliftisusedturetocollectontheoutside.Finally,toraiseheavyequipmentforrepairs.Theshewouldweightheglassasecondsystemhasasmallpistonwithacross-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.time.Ifwatersimplyleaksfromthesectionalareaof7.0102m2andalargeinsidetotheoutside,themassofthepistonwithacross-sectionalareaofglassandwaterwillbeunchanged.2.1101m2.AnengineweighingHowever,ifthemoistureiscondensa-2.7103Nrestsonthelargepiston.tion,therewillbeanincreaseinthemassatthesecondweighing.a.Whatforcemustbeappliedtothesmallpistontolifttheengine?F1A2FPracticeProblems2A113.3FluidsatRestand(2.7103N)(7.0102m2)2.1101m2inMotionpages352–3589.0102Nb.Iftheenginerises0.20m,howfardoespage353thesmallerpistonmove?23.Dentists’chairsareexamplesofhydraulic-liftsystems.Ifachairweighs1600NandV1V2andA1h1A2h2restsonapistonwithacross-sectionalareaA1h1(2.1101m2)(0.20m)of1440cm2,whatforcemustbeappliedtoh22m2A27.010thesmallerpiston,withacross-sectionalareaof72cm2,toliftthechair?0.60m290SolutionsManualPhysics:PrinciplesandProblems

294Chapter13continuedpage356waterandisbarelysubmerged,leavingthe27.Commonbrickisabout1.8timesdenserbricksdry?thanwater.WhatistheapparentweightofThefoamwoulddisplacea0.20m3blockofbricksunderwater?3V(1.0m)(1.0m)(0.10m)0.10mFofwater.TheweightofthefoamisapparentFgFbuoyantbrickVgwaterVgFg,foamfoamVg((1.0102kg/m3)(0.10m3)brickwater)Vg(9.80m/s2)(1.8103kg/m398N1.00103kg/m3)Thebuoyantforceis(0.20m3)(9.80m/s2)FbuoyantwaterVg1.6103N(1.00103kg/m3)(0.10m3)(9.80m/s2)28.Agirlisfloatinginafreshwaterlakewith980Nherheadjustabovethewater.Ifsheweighs610N,whatisthevolumeofthesubmergedTheweightofbrickthatyoucouldstackpartofherbody?isSheisfloatingsoshedisplacesaFg,brickFbuoyantFgfoamvolumeofwaterthatweighsasmuch980N98Nasshedoes.2N8.810FgFbuoyantwaterVgF31.CanoesoftenhaveplasticfoamblocksgVmountedundertheseatsforflotationinwatergcasethecanoefillswithwater.Whatisthe610Napproximateminimumvolumeoffoam(1.00103kg/m3)(9.80m/s2)neededforflotationfora480-Ncanoe?6.2102m3Thebuoyantforceonthefoammustequal480N.Weareassumingthe29.Whatisthetensioninawiresupportingacanoeismadeofdensematerial.1250-Ncamerasubmergedinwater?TheFvolumeofthecamerais16.5103m3.buoyantwaterVgFToholdthecamerainplacethetensionbuoyantVinthewiremustequaltheapparentwatergweightofthecamera.480N(1.00103kg/m3)(9.80m/s2)TFapparentF4.9102m3gFbuoyantFgwaterVg1250N(1.00103kg/m3)SectionReview(16.5103m3)(9.80m/s2)13.3FluidsatRestandCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.09103NinMotionpages352–35830.Plasticfoamisabout0.10timesasdensepage358aswater.Whatweightofbrickscouldyou32.FloatingandSinkingDoesafullsodapopstackona1.0m1.0m0.10mslabcanfloatorsinkinwater?Tryit.Doesitmat-offoamsothattheslaboffoamfloatsinterwhetherornotthedrinkisdiet?AllsodaPhysics:PrinciplesandProblemsSolutionsManual291

295Chapter13continuedpopcanscontainthesamevolumeofliquid,354mL,anddisplacethesamevol-umeofwater.Whatisthedifferencebetweenacanthatsinksandonethatfloats?Thedifferenceisalotofsugar.Aboutone-fourthcupofsugarisdissolvedintheregulardrink,makingitdenserthanwater.Thedietdrinkhasasmallamountofanartificialsweetener.Thedietdrinkislessdensethanthesugar-ladenregularsoftdrink.33.FloatingandDensityAfishingbobbermadeofcorkfloatswithone-tenthofitsvolumebelowthewater’ssurface.Whatisthedensityofcork?Theweightofthewaterdisplacedequalstheweightofthebobber.FgwaterVwatergcorkVcorkgVTherefore,corkwaterVwatercork110Thecorkisaboutone-tenthasdenseaswater.34.FloatinginAirAheliumballoonrisesbecauseofthebuoyantforceoftheairliftingit.Thedensityofheliumis0.18kg/m3,andthedensityofairis1.3kg/m3.Howlargeavolumewouldaheliumballoonneedtolifta10-Nleadbrick?Fapparentmustequal10Ntocounteracttheweightoftheleadbrick.FapparentFgFbuoyantheliumVballoongairVballoong(heliumair)VballoongCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thus,FapparentVballoon(heliumair)g10N(0.18kg/m31.3kgm3)(9.80ms2)0.9m335.TransmissionofPressureAtoyrocketlauncherisdesignedsothatachildstompsonarubbercylinder,whichincreasestheairpressureinalaunchingtubeandpushesafoamrocketintothesky.Ifthechildstompswithaforceof150Nona2.5103m2areapiston,whatistheadditionalforcetransmittedtothe4.0104m2launchtube?F1A2F2A1(150N)(4.0104m2)2.5103m224N292SolutionsManualPhysics:PrinciplesandProblems

296Chapter13continued36.PressureandForceAnautomobileweigh-Thefast-movingairofthetornadohasaing2.3104Nisliftedbyahydrauliccylin-lowerpressurethanthestillairinsidederwithanareaof0.15m2.thehouse.Therefore,theairinsidethea.Whatisthepressureinthehydraulichouseisatahigherpressureandpro-cylinder?ducesanenormousforceonthewin-dows,doors,andwallsofthehouse.FPThispressuredifferenceisreducedbyAopeningdoorsandwindowstoletthe2.3104Nairflowfreelytotheoutside.0.15m21.5105Pab.ThepressureintheliftingcylinderisPracticeProblemsproducedbypushingona0.0082m213.4Solidscylinder.Whatforcemustbeexertedonpages359–363thissmallcylindertolifttheautomobile?page362F1A239.Apieceofaluminumhousesidingis3.66mF2A1longonacoldwinterdayof28°C.How(2.3104N)(0.0082m2)muchlongerisitonaveryhotsummerday2at39°C?0.15m1.3103NL2L1L1(T2T1),soLL1(T2T1)37.DisplacementWhichofthefollowing(25106°C1)(3.66m)displacesmorewaterwhenitisplacedin(39°C(28°C))anaquarium?6.1103m6.1mma.A1.0-kgblockofaluminumora1.0-kgblockoflead?40.Apieceofsteelis11.5cmlongat22°C.ItisBothaluminumandironwillsinktoheatedto1221°C,closetoitsmeltingtem-thebottomoftheaquarium.Becauseperature.Howlongisit?aluminumislessdensethaniron,L1kgofaluminumhasagreatervol-2L1L1(T2T1)umethan1kgofiron.Therefore,the(0.115m)(12106°C1)blockofaluminumwilldisplace(0.115m)(1221°C22°C)morewater.1.2101m12cmb.A10-cm3blockofaluminumora10-cm3blockoflead?41.A400-mLglassbeakeratroomtemperatureisfilledtothebrimwithcoldwaterat4.4°C.Bothblockswillsink,andeachwillWhenthewaterwarmsupto30.0°C,howdisplacethesamevolumeofwater,10cm3.muchwaterwillspillfromthebeaker?Atthebeginning,400mLof4.4°Cwater38.CriticalThinkingAsyoudiscoveredinisinthebeaker.FindthechangeinPracticeProblem4,atornadopassingoveravolumeat30.0°C.housesometimesmakesthehouseexplodeVVTCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fromtheinsideout.HowmightBernoulli’s(210106°C1)(400106m3)principleexplainthisphenomenon?What(30.0°C4.4°C)couldbedonetoreducethedangerofadoororwindowexplodingoutward?2106m32mLPhysics:PrinciplesandProblemsSolutionsManual293

297Chapter13continued42.Atanktrucktakesonaloadof45,725LofBywhatpercentagewouldtherulerbegasolineinHouston,wherethetemperatureincorrectat30.0°C?is28.0°C.ThetruckdeliversitsloadinBecausethesteelshrinks,thedistancesMinneapolis,wherethetemperatureisbetweenthemillimetermarksdecrease12.0°C.whencooled.a.HowmanylitersofgasolinedoestheLtruckdeliver?%incorrect(100)LVV2V1(100)(TfTi)V1TV1T6°C1)(100)(1210V2V1TV1(30.0°C30.0°C)(950106°C1)(45,725L)0.072%(12.0°C28.0°C)45,725L4.4104Lb.Whathappenedtothegasoline?SectionReviewThegasolinevolumedecreased13.4Solidsbecausethetemperaturedecreased.pages359–363Themassofthegasolineremainedpage363thesame.45.RelativeThermalContractionOnahotday,youareinstallinganaluminumscreen43.Aholewithadiameterof0.85cmisdrilleddoorinaconcretedoorframe.Youwantintoasteelplate.At30.0°C,theholeexactlythedoortofitwellonacoldwinterday.accommodatesanaluminumrodoftheShouldyoumakethedoorfittightlyinthesamediameter.Whatisthespacingbetweenframeorleaveextraroom?theplateandtherodwhentheyarecooledto0.0°C?Fitthedoortightly.AluminumshrinkswhencooledmuchmorethanconcreteThealuminumshrinksmorethanthedoes.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.steel.LetLbethediameteroftherod.LaluminumLT46.StatesofMatterWhycouldcandlewaxbe(25106°C1)consideredasolid?Whymightitalsobe(0.85cm)(0.0°C30.0°C)consideredaviscousliquid?6.38104cmThewaxcouldbeconsideredasolidbecauseithasadefinitevolumeandForthesteel,thediameteroftheholeshape.ItcouldbeconsideredaviscousshrinksbyliquidbecausetheparticlesdonotformLsteelLTafixedcrystallinepattern.(12106°C1)(0.85cm)(0.0°C30.0°C)47.ThermalExpansionCanyouheatapiece3.06104cmofcopperenoughtodoubleitslength?ThethermalexpansioncoefficientforThespacingbetweentherodandthecopperis16106/°C.TodoubleitsholewillbelengthLLLT,whichmeansthat1(6.4104cm3.1104cm)2T11.6104cmT144.Asteelrulerismarkedinmillimetersso1thattherulerisabsolutelycorrectat30.0°C.16106°C163,000°C294SolutionsManualPhysics:PrinciplesandProblems

298Chapter13continuedThecopperwouldbevaporizedatthattemperature.StatesofMatter48.StatesofMatterDoesTable13-2provideawaytodistinguishbetweensolidsandliquids?fluidssolidsThecoefficientsofvolumeexpansionaremuchgreaterforliquidsthanforsolids.densityviscositydensityelasticity49.SolidsandLiquidsAsolidcanbedefinedasamaterialthatcanbebentandwillresistpressurebending.Explainhowthesepropertiesrelatetothebindingofatomsinasolid,MasteringConceptsbutdonotapplytoaliquid.page368Particlesinasolidarecloserand,there-52.Howareforceandpressuredifferent?(13.1)fore,moretightlybound.Theyvibrateaboutafixedposition.ThisallowstheForcedependsonlyonthepushorpullsolidtobebent,butitalsoresistsonanobject.Pressuredependsonthebending.Particlesinaliquidarefartherforceaswellastheareaoverwhichtheapartandlesstightlybound.Becauseforceisapplied.theparticlesarefreetoflowpastone53.Agasisplacedinasealedcontainer,andanother,aliquidcannotbebent.someliquidisplacedinacontainerofthe50.CriticalThinkingTheironringinsamesize.ThegasandliquidbothhaveFigure13-23wasmadebycuttingasmalldefinitevolume.Howdotheydiffer?(13.1)piecefromasolidring.IftheringintheTheliquid’svolumewillremainfigureisheated,willthegapbecomewiderunchanged.Thegaswillexpandtofillornarrower?Explainyouranswer.thevolumeofthecontainer.54.Inwhatwayaregasesandplasmassimilar?Inwhatwayaretheydifferent?(13.1)Bothgasesandplasmashavenodefi-nitevolumeandnodefiniteshape.Agasismadeofatoms.Aplasmaismadeofpositivelychargedionsandnegatively■Figure13-23chargedelectrons.TheparticlesofaThegapwillbecomewider.Alloftheplasmaaremoreenergeticthanthemeasurementsoftheringincreaseparticlesofagas.Plasmascanconductwhenheated.electricity.Gasescannot.55.TheSunismadeofplasma.HowisthisChapterAssessmentplasmadifferentfromtheplasmasonEarth?(13.1)ConceptMappingTheSun’splasmaisextremelyhot,butpage368moreimportantly,itisverydense—Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.51.CompletetheconceptmapbelowusingthedenserthanmostsolidsonEarth.followingterms:density,viscosity,elasticity,pressure.Atermmaybeusedmorethanonce.56.LakesAfrozenlakemeltsinthespring.Whateffectdoesthishaveonthetempera-tureoftheairabovethelake?(13.2)Physics:PrinciplesandProblemsSolutionsManual295

299Chapter13continuedTomelt,theicemustabsorbenergyin61.DoesArchimedes’principleapplytoantheamountofitsheatoffusionfromobjectinsideaflaskthatisinsideaspace-theairandwater.Itwillcooltheairshipinorbit?(13.3)aboveit.No,itdoesnot.Theapparentweightofthedisplacedfluidiszerobecausethe57.HikingCanteensusedbyhikersoftenarefluidisinfree-fall.Thus,thereisnocoveredwithcanvasbags.Ifyouwetthebuoyantforce.canvasbagcoveringacanteen,thewaterinthecanteenwillbecooled.Explain.(13.2)62.AstreamofwatergoesthroughagardenThewaterevaporatesintotheair,hoseintoanozzle.Asthewaterspeedsup,absorbingenergyfromthecanteenandwhathappenstothewaterpressure?(13.3)thewaterinside.ThewaterpressuredecreasesbecauseofBernoulli’sprinciple.58.WhatdotheequilibriumtubesinFigure13-24tellyouaboutthepressure63.Howdoesthearrangementofatomsinaexertedbyaliquid?(13.3)crystallinesubstancedifferfromthatinanamorphoussubstance?(13.4)Theatomsinacrystallinesubstancearearrangedinanorderedpattern.Intheamorphoussubstance,theatomsarerandomlyarranged.64.Doesthecoefficientoflinearexpansiondependontheunitoflengthused?Explain.(13.4)No.Thecoefficientisameasureoftheexpansionofanobjectrelativetoitstotallength.UnitsandtotallengthdoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.notchangethecoefficientoflinear■Figure13-24expansion.TheequilibriumtubesillustratethatpressureisindependentoftheshapeofApplyingConceptsthecontainer.page368–36965.Arectangularboxwithitslargestsurface59.AccordingtoPascal’sprinciple,whathap-restingonatableisrotatedsothatitspenstothepressureatthetopofacontainersmallestsurfaceisnowonthetable.Hasifthepressureatthebottomisincreased?thepressureonthetableincreased,(13.3)decreased,orremainedthesame?ChangesinpressurearedistributedThepressureincreased.Theweightequallytoallpartsofthecontainer.Thestayedthesame,buttheweightperpressureatthetopincreases.areaincreased.60.Howdoesthewaterpressure1mbelowthe66.Showthatapascalisequivalenttoasurfaceofasmallpondcomparewiththekg/ms2.waterpressurethesamedistancebelowthePaN/m2(kgm/s2)/m2surfaceofalake?(13.3)kg/ms2Sizeorshapeofthebodyofwaterdoesnotmatter,onlythedepth.Thepressureisthesameineachcase.296SolutionsManualPhysics:PrinciplesandProblems

300Chapter13continued67.ShippingCargoComparedtoanidenticalthefoil,willthesizeoftheholedecreaseoremptyship,wouldashipfilledwithtable-increase?Explain.tennisballssinkdeeperintothewaterorAsyouheatthefoil,theholebecomesriseinthewater?Explain.larger.HeatingtransfersmoreenergytoItwouldsinkdeeperintothewaterparticlesofthealuminum,causingthebecauseitwouldhaveagreaterweight.volumeofthealuminumtoincrease.68.Dropsofmercury,water,ethanol,and72.Equalvolumesofwaterareheatedintwoacetoneareplacedonasmooth,flatnarrowtubesthatareidentical,exceptthatsurface,asshowninFigure13-25.FromtubeAismadeofsoftglassandtubeBisthisfigure,whatcanyouconcludeaboutmadeofovenproofglass.Asthetemperaturethecohesiveforcesintheseliquids?increases,thewaterlevelriseshigherintubeBthanintubeA.Giveapossibleexplanation.Theovenproofglassexpandslessthanthesoftglasswhenheated.ThewaterdoesnotriseashighinAbecausethesoftglasstubeisexpandinginvolume.73.Aplatinumwireeasilycanbesealedinaglasstube,butacopperwiredoesnotformatightsealwiththeglass.Explain.Platinum’scoefficientofthermalexpan-■Figure13-25sionissimilartothatofglass,soitThecohesiveforcesarestrongestinexpandsandcontractsastheglassmercuryandweakestinacetone.Thedoes.Copperhasamuchlargercoeffi-strongerthecohesiveforce,themorecientthanglass.sphericalthedropwillbe.74.Fiveobjectswiththefollowingdensitiesare69.Howdeepwouldawatercontainerhavetoputintoatankofwater.betohavethesamepressureatthebottoma.0.85g/cm3d.1.15g/cm3asthatfoundatthebottomofa10.0-cmb.0.95g/cm3e.1.25g/cm3deepbeakerofmercury,whichis13.55c.1.05g/cm3timesasdenseaswater?Thedensityofwateris1.00g/cm3.ThediagraminFigure13-26showssixpossiblePwaterPmercurypositionsoftheseobjects.Selectaposition,waterhwatergmercuryhmercurygfrom1to6,foreachofthefiveobjects.Notallpositionsneedtobeselected.mercuryhwaterwaterhmercury(13.55)(10.0cm)123136cm4570.Alcoholevaporatesmorequicklythanwaterdoesatthesametemperature.WhatdoesthisobservationallowyoutoconcludeabouttheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.propertiesoftheparticlesinthetwoliquids?6Thecohesiveforcesofwateraregreaterthanthoseofalcohol.■Figure13-2671.SupposeyouuseaholepunchtomakeaThepositionsoftheobjectsshouldbecircularholeinaluminumfoil.Ifyouheata–1,b–2,c–6,d–6,e–6Physics:PrinciplesandProblemsSolutionsManual297

301Chapter13continuedMasteringProblemsdissolvedina2-Lbottleofsoda.ThemolarmassofCO13.1PropertiesofFluids2is44g/mol.a.Howmanymolesofcarbondioxidearepages369–370inthe2-Lbottle?(1L0.001m3)Level1Fromtheidealgaslaw75.TextbooksA0.85-kgphysicsbookwithdimensionsof24.0cm20.0cmisatrestPVnonatable.RT(1.01105Pa)(0.0080m3)a.Whatforcedoesthebookapplytothe(8.31Pam3molK)(300.0K)table?0.32molesTheforceonthetableistheweightofthebook.b.Whatisthemassofthecarbondioxide2)inthe2-Lbottleofsoda?Wmg(0.85kg)(9.80m/sThemolecularweightofcarbon8.3Ndioxideisb.Whatpressuredoesthebookapply?M122(16)Thepressureappliedbythebookis44g/molFPTherefore,themassisAmgmnMlw(0.32mol)(44g/mol)(0.85kg)(9.80ms2)14g(2.40101m)(2.00101m)2Pa79.AsshowninFigure13-27,aconstant-1.710pressurethermometerismadewitha76.A75-kgsolidcylinderthatis2.5mlongcylindercontainingapistonthatcanmoveandhasanendradiusof7.0cmstandsonfreelyinsidethecylinder.Thepressureandoneend.Howmuchpressuredoesitexert?theamountofgasenclosedinthecylinderCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.arekeptconstant.AsthetemperatureFPincreasesordecreases,thepistonmovesupAordowninthecylinder.At0°C,theheightmg2ofthepistonis20cm.Whatistheheightofrthepistonat100°C?(75kg)(9.80ms2)(0.070m)2Cylinder4.8104PaPiston20cm77.WhatisthetotaldownwardforceoftheGasatmosphereonthetopofyourheadrightnow?Assumethatthetopofyourheadhas2.■Figure13-27anareaofabout0.025mFPA(1.01105Pa)(0.025m2)Becausethepressureiskeptconstant,V3N1/T1V2/T2.Theheightofthepiston2.510isdirectlyproportionaltothevolumeofthecylinder.Therefore,78.SoftDrinksSodasaremadefizzybythecarbondioxide(COh1h22)dissolvedintheliquid.AnamountofcarbondioxideequalT1T2toabout8.0Lofcarbondioxidegasatatmosphericpressureand300.0Kcanbe298SolutionsManualPhysics:PrinciplesandProblems

302Chapter13continuedhT(Onepoundpersquareinchequals12h2T6.90103Pa.)Thetermgaugepressure1meansthepressureaboveatmosphericpres-(20cm)(373K)sure.Thus,theactualpressureinthetireis273K1.01105Pa(30.0psi)(6.90103Pa/psi)3101cm3.08105Pa.Asthecarisdriven,thetire’stemperatureincreases,andthevolumeLevel2andpressureincrease.Supposeyoufilleda80.Apistonwithanareaof0.015m2enclosescar’stiretoavolumeof0.55m3atatem-aconstantamountofgasinacylinderwithperatureof280K.Theinitialpressurewasavolumeof0.23m3.Theinitialpressureof30.0psi,butduringthedrive,thetire’sthegasis1.5105Pa.A150-kgmassistemperatureincreasedto310Kandthetire’svolumeincreasedto0.58m3.thenplacedonthepiston,andthepistonmovesdownwardtoanewposition,asa.Whatisthenewpressureinthetire?showninFigure13-28.IfthetemperaturePP1V12V2isconstant,whatisthenewvolumeoftheTT12gasinthecylinder?PP1V1T2Volume0.23m3Volume?2T1V2Pistonarea0.015m2(3.08105Pa)(0.55m3)(310K)(280K)(0.58m3)3.2105Pa150kgb.Whatisthenewgaugepressure?(30.0psi)(0.55m3)(310K)Pgauge3)(280K)(0.58m31psiGasGas13.3FluidsatRestandinMotionpage370Level1■Figure13-2882.ReservoirsAreservoirbehindadamisP1V1P2V217-mdeep.Whatisthepressureofthewateratthefollowinglocations?P1V1V2Pa.thebaseofthedam2PhgP1V1(1.00103kg/m3)(17m)mgP1A(9.80m/s2)(1.5105Pa)(0.23m3)1.7105Pa(150kg)(9.80m/s2)b.4.0mfromthetopofthedam1.5105Pa0.015m2PhgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.14m3(1.00103kg/m3)(4.0m)(9.80m/s2)Level33.9104Pa81.AutomobilesAcertainautomobiletireisspecifiedtobeusedatagaugepressureof30.0psi,or30.0poundspersquareinch.Physics:PrinciplesandProblemsSolutionsManual299

303Chapter13continued83.Atesttubestandingverticallyinatest-tubeFg195N8N203Nrackcontains2.5cmofoil(0.81g/cm3)and6.5cmofwater.Whatisthepressureb.Therockisremovedfromtheaquarium,exertedbythetwoliquidsonthebottomofandtheamountofwaterisadjustedthetesttube?untilthescaleagainreads195N.Afishweighing2Nisaddedtotheaquarium.PPoilPwaterWhatisthescalereadingwiththefishintheaquarium?oilhoilgwaterhwaterg(810kg/m3)(0.025m)(9.80m/s2)Fg195N2N197N(1.00103kg/m3)(0.065m)Ineachcasethebuoyantforceisequaltotheweightofthewater(9.80m/s2)displaced.8.4102Pa86.Whatisthesizeofthebuoyantforceona26.0-Nballthatisfloatinginfreshwater?84.AntiquesAnantiqueyellowmetalstatuetteofabirdissuspendedfromaspringscale.IftheballisfloatingThescalereads11.81NwhenthestatuetteFbuoyantFg26.0Nissuspendedinair,anditreads11.19Nwhenthestatuetteiscompletelysubmerged87.Whatistheapparentweightofarocksub-inwater.mergedinwateriftherockweighs45Nina.Findthevolumeofthestatuette.airandhasavolumeof2.1103m3?FbuoyantwaterVgFgFapparentFapparentFgFbuoyantThus,FgwaterVgFgFapparent45N(1.00103kg/m3)Vwaterg(2.1103m3)(9.80m/s2)11.81N11.19N(1.00103kgm3)(9.80ms2)24NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6.33105m388.Whatisthemaximumweightthataballoonb.Isthebirdmadeofgoldfilledwith1.00m3ofheliumcanliftinair?(19.3103kg/m3)orgold-platedAssumethatthedensityofairis1.20kg/m3aluminum(2.7103kg/m3)?andthatofheliumis0.177kg/m3.Neglectthemassoftheballoon.mVFapparentFgFbuoyantFgheliumVgairVgVg(heliumair)Vg11.81N(6.33105m3)(9.80ms2)(0.177kg/m31.20kg/m3)19.0103kg/m3(1.00m3)(9.80m/s2)Thestatuetteismadeofgold.10.0N85.Duringanecologyexperiment,anaquariumLevel2half-filledwithwaterisplacedonascale.89.Ifarockweighs54NinairandhasanThescaleshowsaweightof195N.apparentweightof46Nwhensubmergeda.Arockweighing8Nisaddedtotheinaliquidwithadensitytwicethatofaquarium.Iftherocksinkstothebottomwater,whatwillbeitsapparentweightoftheaquarium,whatwillthescaleread?whenitissubmergedinwater?300SolutionsManualPhysics:PrinciplesandProblems

304Chapter13continuedFapparent,waterFgwaterVgandFapparent,liquidFg2waterVgFgFapparent,liquidorV2watergSubstitutethisintothefirstequation.FgFapparent,liquidFapparent,waterFgwaterg2waterg1Fg2(FgFapparent,liquid)1(FgFapparent,liquid)21(54N46N)25.0101NLevel390.OceanographyAsshowninFigure13-29,alargebuoyusedtosupportanoceanographicresearchinstrumentismadeofacylindrical,?mhollowirontank.Thetankis2.1minheightand0.33mindiameter.Thetotalmassofthebuoyandtheresearchinstrumentisabout120kg.2.1mThebuoymustfloatsothatoneendisabovethewatertosupportaradiotransmitter.Assumingthatthemassofthebuoyisevenlydistributed,how0.33mmuchofthebuoywillbeabovethewaterlinewhenitisfloating?■Figure13-29TheheightofthebuoyabovewaterisVl1wateraboveVltotalbuoymwater1r2hltotalm1r2hltotalwater120kgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1(2.1m)12(2.1m)(1.00103kgm3)2(0.33m)0.70mPhysics:PrinciplesandProblemsSolutionsManual301

305Chapter13continued13.4SolidsVV1Tpages370–371(36106°C1)(1.0m3)(45°C)Level191.Abarofanunknownmetalhasalengthof1.6103m30.975mat45°Candalengthof0.972mat23°C.Whatisitscoefficientoflinear96.BridgesBridgebuildersoftenuserivetsexpansion?thatarelargerthantherivetholetomakethejointtighter.TherivetiscooledbeforeitL2L1isputintothehole.SupposethatabuilderL1(T2T1)drillsahole1.2230cmindiameterfora0.972m0.975msteelrivet1.2250cmindiameter.Towhat(0.975m)(23°C45°C)temperaturemusttherivetbecooledifitis1.4104°C1tofitintotherivethole,whichisat20.0°C?L2L1L1(T2T1)92.Aninventorconstructsathermometerfrom(Lanaluminumbarthatis0.500minlengthT2L1)2T1Lat273K.Hemeasuresthetemperatureby1measuringthelengthofthealuminumbar.1.2230cm1.2250cm20.0°CIftheinventorwantstomeasurea1.0-K(12106°C1)(1.2250cm)changeintemperature,howpreciselymust1.2102°Chemeasurethelengthofthebar?T1.0K1.0°CLevel2LL1T97.Asteeltankfilledwithmethanolis2.000m(25106°C1)(0.500m)(1.0°C)indiameterand5.000minheight.Itis1.3105mcompletelyfilledat10.0°C.Ifthetempera-turerisesto40.0°C,howmuchmethanol(inliters)willflowoutofthetank,given93.BridgesHowmuchlongerwilla300-mthatboththetankandthemethanolwillCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.steelbridgebeona30°CdayinAugustexpand?thanona10°CnightinJanuary?VVLL1T1TL1(T2T1)6°C1)(300m)(2h)(T(1210methanolsteel)(r2T1)(30°C(10°C))(1200106°C135106°C1)0.1m()(1.000m)2(5.000m)(40.0°C10.0°C)94.Whatisthechangeinlengthofa2.00-m30.55mcopperpipeifitstemperatureisraisedfrom23°Cto978°C?98.Analuminumsphereisheatedfrom11°CtoLL1TL1(T2T1)580°C.Ifthevolumeofthesphereis1.78cm3at11°C,whatistheincreaseinvolume(16106°C1)(2.00m)ofthesphereat580°C?(978°C23°C)VV1T3.1102m(75106°C1)(1.78cm3)95.Whatisthechangeinvolumeofa1.0-m3(580°C11°C)concreteblockifitstemperatureisraised7.6102cm345°C?302SolutionsManualPhysics:PrinciplesandProblems

306Chapter13continued99.Thevolumeofacoppersphereis2.56cm3afterbeingheatedfrom12°Cto984°C.Whatwasthevolumeofthecoppersphereat12°C?V2V1V1TV1(1T)V2V11T2.56cm3(1(48106°C1)(984°C12°C))2.4cm3Level3100.Asquareofironplatethatis0.3300moneachsideisheatedfrom0°Cto95°C.a.Whatisthechangeinthelengthofthesidesofthesquare?LL1TL1(T2T1)(12106°C1)(0.3300m)(95°C0°C)3.8104mb.Whatistherelativechangeinareaofthesquare?ArelativechangeA1AA21A1L2L221L21(L2L21L)1L21(0.3300m3.8104m)2(0.3300m)2(0.3300m)22.3103101.Analuminumcubewithavolumeof0.350m3at350.0Kiscooledto270.0K.a.Whatisitsvolumeat270.0K?V2V1V1TV1(1T)V1(1(T2T1))(0.350m3)(1(75106°C1)(270.0K350.0K))0.348m3b.Whatisthelengthofasideofthecubeat270.0K?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13L(V2)1(0.348m3)30.703mPhysics:PrinciplesandProblemsSolutionsManual303

307Chapter13continued102.IndustryAmachinistbuildsarectangularmechanicalpartforaspecialrefrig-eratorsystemfromtworectangularpiecesofsteelandtworectangularpiecesofaluminum.At293K,thepartisaperfectsquare,butat170K,thepartbecomeswarped,asshowninFigure13-30.Whichpartsweremadeofsteelandwhichweremadeofaluminum?2211T293K33T170K44■Figure13-30Parts1and2experiencedagreaterreductioninlengththanparts3and4;therefore,parts1and2musthavebeenmadeofaluminum,whichhasalargercoefficientofexpansionthansteel.MixedReviewpage371Level1103.Whatisthepressureonthehullofasubmarineatadepthof65m?PPatmospherewatergh(1.01105Pa)(1.00103kg/m3)(9.80m/s2)(65m)7.4105Pa104.ScubaDivingAscubadiverswimmingatadepthof5.0munderwaterexhalesCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a4.2106m3bubbleofair.Whatisthevolumeofthatbubblejustbeforeitreachesthesurfaceofthewater?P1V1P2V2P1V1V2P2(Patmospherewatergh)V1Patmosphere(1.01105Pa(1.00103kgm3)(9.80ms2)(5.0m))(4.2106m3)1.01105Pa6.2106m3105.An18-Nbowlingballfloatswithabouthalfoftheballsubmerged.a.Whatisthediameterofthebowlingball?VballFgVwaterg2g,434d3d3whereVball3r3261d3ThenFg26g,304SolutionsManualPhysics:PrinciplesandProblems

308Chapter13continued312Fgsothatdg3(12)(18N)(1.00103kgm3)(9.80ms2)0.19mb.Whatwouldbetheapproximateapparentweightofa36-Nbowlingball?Halftheballsankwhentheweightwas18N.Theapparentweightofthe36-Nballshouldbenearzero.Level2106.Analuminumbarisfloatinginabowlofmercury.Whenthetemperatureisincreased,doesthealuminumfloathigherorsinkdeeperintothemercury?Thevolumecoefficientofexpansionofmercuryisgreaterthanthevolumecoefficientofexpansionofaluminum.Therefore,astheyareheated,thealuminumbecomesdenserrelativetothemercuryandwouldsinkdeeperintothemercury.107.Thereis100.0mLofwaterinan800.0-mLsoft-glassbeakerat15.0°C.Howmuchwillthewaterlevelhavedroppedorrisenwhenthebottleandwaterareheatedto50.0°C?Thewaterexpands:VVT(210106°C1)(100.0mL)(35.0°C)0.735mLThebottleexpands:VVT(27106°C1)(800.0mL)(35.0°C)0.756mLThewaterlevelwillgodownslightly,butnotenoughtonotice.108.AutoMaintenanceAhydraulicjackusedtoliftcarsforrepairsiscalledathree-tonjack.Thelargepistonis22mmindiameter,andthesmalloneis6.3mmindiameter.Assumethataforceof3tonsis3.0104N.a.Whatforcemustbeexertedonthesmallpistontolifta3-tonweight?F1A2F2A1F21r2r21Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.d2F21d214N)6.3mm2(3.01022mm2.5103NPhysics:PrinciplesandProblemsSolutionsManual305

309Chapter13continuedb.MostjacksusealevertoreducetheThepressureinthewaterisforceneededonthesmallpiston.If(8.7107Pa)/(1.01105Pa)theresistancearmis3.0cm,howlong860timesgreaterthanstandardairmusttheeffortarmofanidealleverbepressure.Therefore,thedensityofairtoreducetheforceto100.0N?wouldbe860timesgreaterthantheFdensityoftheaironthesurfaceofrLrFeLetheocean.FrLrLeFeThinkingCritically(2.5103N)(3.0cm)page372100.0N111.ApplyConceptsYouarewashingdishes75cminthesink.Aservingbowlhasbeenfloat-inginthesink.Youfillthebowlwithwater109.BallooningAhot-airballooncontainsafromthesink,anditsinkstothebottom.fixedvolumeofgas.Whenthegasisheat-Didthewaterlevelinthesinkgoupored,itexpandsandpushessomegasoutatdownwhenthebowlwassubmerged?thelower,openend.Asaresult,themassWhenitwasfloating,thebowldis-ofthegasintheballoonisreduced.Whyplacedavolumeofwaterthatweighedwouldtheairinaballoonhavetobehot-asmuchasitdid.Whenitsank,itdis-tertoliftthesamenumberofpeopleaboveplacedavolumeofwaterthatweighedVail,Colorado,whichhasanaltitudeoflessthanthebowl,becausethebuoy-2400m,thanabovethetidewaterflatsofancyforceistheweightofthewaterVirginia,whichhaveanaltitudeof6m?displaced.Inthesecondcase,theAtmosphericpressureislowerathigh-bowldisplaceslesswaterandtheleveleraltitudes.Therefore,themassoftheinthesinkgoesdown.volumeoffluiddisplacedbyaballoonofthesamevolumeislessathigher112.ApplyConceptsPersonsconfinedtobedaltitudes.ToobtainthesamebuoyantarelesslikelytodevelopbedsoresiftheyCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.forceathigheraltitudes,aballoonuseawaterbedratherthananordinarymustexpelmoregas,requiringhighermattress.Explain.temperatures.Thesurfaceofthewaterbedconformsmorethanthesurfaceofamattressto110.TheLivingWorldSomeplantsandani-thecontoursofyourbody.One“sinks”malsareabletoliveinconditionsofmoreeasilyintoawaterbed.Becauseextremepressure.HOmattress,thebuoyantforcefrom2a.Whatisthepressureexertedbytheawaterbedisless.waterontheskinofafishorwormthatlivesnearthebottomofthe113.AnalyzeandConcludeOnemethodofPuertoRicoTrench,8600mbelowthemeasuringthepercentageofbodyfatissurfaceoftheAtlanticOcean?Usebasedonthefactthatfattytissueisless1030kg/m3forthedensityofseawater.densethanmuscletissue.Howcanaper-son’saveragedensitybeassessedwithaThepressureisscaleandaswimmingpool?Whatmea-Pghsurementsdoesaphysicianneedtorecord(1030kg/m3)(9.80m/s2)(8600m)tofindaperson’saveragepercentageof8.7107Pabodyfat?Thephysicianweighsthepersonnor-b.Whatwouldbethedensityofairatmallyandthenweighsthepersonthatpressure,relativetoitsdensitytotallysubmerged.Weighthastobeabovethesurfaceoftheocean?306SolutionsManualPhysics:PrinciplesandProblems

310Chapter13continuedaddedtotheweighingdevicebecausethedensityofahumanisnormal-lylessthanthedensityofwater.Thevolumeofwaterdisplacedbythepersonalsoshouldbemeasured.Theaveragedensityofthepersoncanbecalculatedfromthebalanceofforcesthatholdthepersoninequilibri-umunderwater.114.AnalyzeandConcludeAdownwardforceof700Nisrequiredtofullysubmergeaplasticfoamsphere,asshowninFigure13-31.Thedensityofthefoamis95kg/m3.a.Whatpercentageofthespherewouldbesubmergedifthespherewerereleasedtofloatfreely?Thedensityofthefoamrelativetothe95kgm3wateris0.095,so1.00103kgm39.5percentofthefloatingspherewouldbe700Nsubmerged.b.Whatistheweightofthesphereinair?■Figure13-31Theweightofwaterdisplacedby9.5percentofthesphere’svolumebalancestheentireweightofthesphere,Fg.Anadditional700Nisneededtosubmergetheremaining90.5percentofthesphere’svolume.Thus,Fg700N0.0950.905F1Ng710c.Whatisthevolumeofthesphere?FbuoyantFgFdownwaterVgfoamVgFdownFdownV(waterdown)g700N(1.00103kgm395kgm3)(9.80ms2)8102m3115.ApplyConceptsTropicalfishforaquariumsareoftentransportedhomefrompetshopsintransparentplasticbagsfilledmostlywithwater.Ifyouplacedafishinitsunopenedtransportbaginahomeaquarium,whichofthecasesinFigure13-32bestrepresentswhatwouldhappen?Explainyourreasoning.BagBagAquariumwaterwaterwaterlevelCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.levellevelBagwaterlevelAquariumAquariumwaterlevelwaterlevel■Figure13-32Physics:PrinciplesandProblemsSolutionsManual307

311Chapter13continuedThedensityofthewaterinthebag,theofoxygengas.Avogadrobuiltonfish,andtheplasticareallneartheGay-Lussac’sworktodevelopthedensityofthewaterintheaquarium.relationshipbetweenmolesofagasTherefore,thebagshouldfloatwithandvolume.thewaterlevelinthebagatthesameheightasthewaterlevelintheCumulativeReviewaquarium.page372118.TwoblocksareconnectedbyastringoverWritinginPhysicsafrictionless,masslesspulleysuchpage372thatoneisrestingonaninclinedplane116.Somesolidmaterialsexpandwhentheyandtheotherishangingoverthetopedgearecooled.Waterbetween4°and0°Cisoftheplane,asshowninFigure13-33.themostcommonexample,butrubberThehangingblockhasamassof3.0kgbandsalsoexpandinlengthwhencooled.andtheblockontheplanehasamassofResearchwhatcausesthisexpansion.2.0kg.ThecoefficientofkineticfrictionRubberbandsaremadeoflongrubberbetweentheblockandtheinclinedplanemoleculescalledpolymersthatactlikeis0.19.Answerthefollowingquestionschainswithmanylonglinks.Theprop-assumingtheblocksarereleasedfromrest.ertiesofrubbercomefromtheability(Chapter5)ofthechainlinkstotwistandturn.a.Whatistheaccelerationoftheblocks?Whentherubberiscolder,thepolymerlinksarestretchedoutinastraightline,likethelinksinasteelchainthatyouholdatoneendandlethang2.0kgfreely.Becausethelinksareordered3.0kgthatway,thepolymershaverelativelylittledisorder,orentropy.Addingheat45°tothepolymersincreasestheirther-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.malmotion.Thelinksbegintoshake■Figure13-33aboutandtheirdisorderincreases.IfForthe2-kgblock,youshakeachainlikethis,youwillTseethatitsaveragelengthbecomeskFNmgsinmalessthanifthechainwerehangingwhereFNmgcosisthenormalmotionless.force.Thus,117.ResearchJosephLouisGay-LussacandhisTcontributionstothegaslaws.HowdidkmgcosmgsinmaGay-Lussac’sworkcontributetothediscov-T(mg)(kcossin)maeryoftheformulaforwater?Forthe3-kgblock,Gay-LussacwasaFrenchscientistMgTMawhoalsowasinterestedinhigh-altitudeballoonascents.Hediscov-or,TMgMaeredthatwhengasesareatthesameSubstitutethisforTinthe2-kgtemperatureandpressure,theirvol-blockequation.umesreactinratiosofsmall,wholeMgMa(mg)(kcossin)numbers.Gay-Lussac’sworkcon-matributedtothediscoveryofwater’sformulabyshowingthattwovolumesofhydrogengasreactwithonevolume308SolutionsManualPhysics:PrinciplesandProblems

312Chapter13continuedSolveforacceleration.Mg(mg)(akcossin)mM(3.0kg)(9.80ms2)(2.0kg)(9.80ms2)((0.19)(cos45°)sin45°)2.0kg3.0kg2.6m/s2b.Whatisthetensioninthestringconnectingtheblocks?TofindT,usetheequationforthe3-kgblock.TMgMaM(ga)(3.0kg)(9.80m/s22.6m/s2)22N119.Acompactcarwithamassof875kg,movingsouthat15m/s,isstruckbyafull-sizedcarwithamassof1584kg,movingeastat12m/s.Thetwocarssticktogether,andmomentumisconserved.(Chapter9)a.Sketchthesituation,assigningcoordinateaxesandidentifying“before”and“after.”y-axisAmA875kgBeforevA15m/s(State1)BAfter(State2)x-axismB1584kgv2?vB12m/sp2?m22459kgb.Findthedirectionandspeedofthewreckimmediatelyafterthecollision,rememberingthatmomentumisavectorquantity.pA1mAvA(875kg)(15m/s)1.31104kgm/ssouthpB1mBvB(1584kg)(12m/s)1.90104kgm/seastp2pA12pB12(1.31104kgm/s)2(1.90104kgm/s)2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.3104kgm/sptanB1pA1Physics:PrinciplesandProblemsSolutionsManual309

313Chapter13continuedptan1B1ChallengeProblempA1page36311.90104kgm/stan1.31104kgm/sYouneedtomakea1.00-m-longbarthatexpandswithtemperatureinthesamewayasa55°eastofsouth1.00-m-longbarofcopperwould.Asshowninp22.3104kgm/sthefigureattheright,yourbarmustbemadev2m2459kg2fromabarofironandabarofaluminum9.4m/sattachedendtoend.Howlongshouldeachofthembe?c.ThewreckskidsalongthegroundandLcomestoastop.ThecoefficientofcopperLaluminumLironkineticfrictionwhilethewreckisskid-anddingis0.55.Assumethattheaccelera-copperLcopperTtionisconstant.Howfardoesthe(aluminumLaluminumironLiron)Twreckskidafterimpact?Tofindthedistance,usetheequa-SubstitutingLaluminumtionofmotion:LcopperLiron,thisgivesv2v22a(ddii)(copperaluminum)LcopperwherethefinalvelocityiszeroandLironironaluminumdi0.Solveford:(16106°C125106°C1)(1.00m)v212106°C125106°C1di2a0.69mTofindacceleration,noticethattheLaluminumLcopperLironforcethatslowsthecarsequalsthefrictionalforce.1.00m0.69m(m0.31mamb)ak(mamb)gCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.akgThedistanceisthenv2d02kg(9.4ms)2(2)(0.55)(9.80ms2)8.2m120.A188-Wmotorwillliftaloadattherate(speed)of6.50cm/s.Howgreataloadcanthemotorliftatthisrate?(Chapter10)v6.50cm/s0.0650m/sWFddPFFvtttPFgvP188W3NF2.8910gv0.0650m/s310SolutionsManualPhysics:PrinciplesandProblems

314CHAPTER14VibrationsandWaves12PEkxPracticeProblemssp214.1PeriodicMotion2PEsp(2)(48J)x0.61mpages375–380k256N/mpage3781.Howmuchforceisnecessarytostretchapage379spring0.25mwhenthespringconstant6.WhatistheperiodonEarthofapendulumis95N/m?withalengthof1.0m?FkxT2l21.0m2.0sg9.80m/s2(95N/m)(0.25m)24N7.HowlongmustapendulumbeontheMoon,whereg1.6m/s2,tohavea2.Aspringhasaspringconstantof56N/m.periodof2.0s?Howfarwillitstretchwhenablocklweighing18Nishungfromitsend?T2gFkx22T2)2.0sF18Nlg(1.6m/s0.16m22x0.32mk56N/m8.Onaplanetwithanunknownvalueofg,3.Whatisthespringconstantofaspringthattheperiodofa0.75-m-longpendulumstretches12cmwhenanobjectweighingis1.8s.Whatisgforthisplanet?24Nishungfromit?lFkxT2gF22kgl2(0.75m)29.1m/s2xT1.8s24N0.12m2.0102N/mSectionReview14.1PeriodicMotion4.Aspringwithaspringconstantof144N/mpages375–380iscompressedbyadistanceof16.5cm.page380Howmuchelasticpotentialenergyisstored9.Hooke’sLawTwospringslookalikebutinthespring?havedifferentspringconstants.Howcould12PEsp2kxyoudeterminewhichonehasthegreaterspringconstant?1(144N/m)(0.165m)21.96J2Hangthesameobjectfrombothsprings.TheonethatstretcheslessCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5.Aspringhasaspringconstantof256N/m.hasthegreaterspringconstant.Howfarmustitbestretchedtogiveitanelasticpotentialenergyof48J?10.Hooke’sLawObjectsofvariousweightsarehungfromarubberbandthatissus-pendedfromahook.TheweightsoftheobjectsareplottedonagraphagainstthePhysics:PrinciplesandProblemsSolutionsManual311

315Chapter14continuedstretchoftherubberband.Howcanyou14.CriticalThinkingHowisuniformcirculartellfromthegraphwhetherornotthemotionsimilartosimpleharmonicrubberbandobeysHooke’slaw?motion?Howaretheydifferent?Ifthegraphisastraightline,therubberBothareperiodicmotions.InuniformbandobeysHooke’slaw.Ifthegraphiscircularmotion,theacceleratingforcecurved,itdoesnot.isnotproportionaltothedisplacement.Also,simpleharmonicmotionisone-11.PendulumHowmustthelengthofapen-dimensionalanduniformcirculardulumbechangedtodoubleitsperiod?motionistwo-dimensional.Howmustthelengthbechangedtohalvetheperiod?PracticeProblems12,soPEsp2kx14.2WavePropertiesPEx2pages381–38611PEx222page386(0.40m)215.Asoundwaveproducedbyaclockchimeis(0.20m)2heard515maway1.50slater.4.0a.Whatisthespeedofsoundoftheclock’schimeinair?Theenergyofthefirstspringis4.0timesgreaterthantheenergyofvdtthesecondspring.515m12.EnergyofaSpringWhatisthedifference1.50sbetweentheenergystoredinaspringthatis343m/sstretched0.40mandtheenergystoredinb.Thesoundwavehasafrequencyofthesamespringwhenitisstretched0.20m?436Hz.Whatistheperiodofthewave?lT2l21Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.T2,soTgTlf11Todoubletheperiod:1436HzTll222,so243sTll2.2910111c.Whatisthewave’swavelength?Thelengthmustbequadrupled.vTohalvetheperiod:fT2l21l21343m/s,soT1l12l14436HzThelengthisreducedtoone-fourthits0.787moriginallength.16.Ahikershoutstowardaverticalcliff465m13.ResonanceIfacar’swheelisoutofbal-away.Theechoisheard2.75slater.ance,thecarwillshakestronglyataspecifica.Whatisthespeedofsoundofthespeed,butnotwhenitismovingfasterorhiker’svoiceinair?slowerthanthatspeed.Explain.d(2)(465m)Atthatspeed,thetire’srotationv338m/st2.75sfrequencymatchestheresonantfrequencyofthecar.312SolutionsManualPhysics:PrinciplesandProblems

316Chapter14continuedb.Thewavelengthofthesoundis0.750m.SectionReviewWhatisitsfrequency?v338m/s14.2WavePropertiesvf,sof451Hz0.750mpages381–386c.Whatistheperiodofthewave?page386T112.22103s22.SpeedinDifferentMediaIfyoupullonf451Hzoneendofacoiled-springtoy,doesthepulsereachtheotherendinstantaneously?17.IfyouwanttoincreasethewavelengthofWhathappensifyoupullonarope?Whatwavesinarope,shouldyoushakeitatahappensifyouhittheendofametalrod?higherorlowerfrequency?Compareandcontrastthepulsestravelingatalowerfrequency,becausewave-throughthesethreematerials.lengthvariesinverselywithfrequencyIttakestimeforthepulsetoreachtheotherendineachcase.Ittravelsfaster18.Whatisthespeedofaperiodicwavedistur-ontheropethanonthespring,andbancethathasafrequencyof3.50Hzandafastestinthemetalrod.wavelengthof0.700m?vf(0.700m)(3.50Hz)2.45m/s23.WaveCharacteristicsYouarecreatingtransversewavesinaropebyshakingyour19.Thespeedofatransversewaveinastringishandfromsidetoside.Withoutchanging15.0m/s.Ifasourceproducesadisturbancethedistancethatyourhandmoves,youthathasafrequencyof6.00Hz,whatisitsbegintoshakeitfasterandfaster.Whatwavelength?happenstotheamplitude,wavelength,v15.0m/sfrequency,period,andvelocityofthewave?vf,so2.50mf6.00HzTheamplitudeandvelocityremainunchanged,butthefrequencyincreases20.Fivepulsesaregeneratedevery0.100sinawhiletheperiodandthewavelengthtankofwater.Whatisthespeedofpropaga-decrease.tionofthewaveifthewavelengthofthesurfacewaveis1.20cm?24.WavesMovingEnergySupposethat0.100s0.0200s/pulse,soyouandyourlabpartnerareaskedto5pulsesdemonstratethatatransversewavetrans-T0.0200sportsenergywithouttransferringmatter.vT,soHowcouldyoudoit?TieapieceofyarnsomewherenearthevTmiddleofarope.Withyourpartner1.20cmholdingoneendoftherope,shakethe0.0200sotherendupanddowntocreatea60.0cm/s0.600m/stransversewave.Notethatwhilethewavemovesdowntherope,theyarn21.Aperiodiclongitudinalwavethathasamovesupanddownbutstaysinthefrequencyof20.0Hztravelsalongacoilsameplaceontherope.spring.IfthedistancebetweensuccessiveCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.compressionsis0.600m,whatisthespeed25.LongitudinalWavesDescribelongitudinalofthewave?waves.Whattypesofmediatransmitvf(0.600m)(20.0Hz)12.0m/slongitudinalwaves?Inlongitudinalwaves,theparticlesofthemediumvibrateinadirectionparalleltothemotionofthewave.Physics:PrinciplesandProblemsSolutionsManual313

317Chapter14continuedNearlyallmedia—solids,liquids,and30.CriticalThinkingAsanotherwaytogases—transmitlongitudinalwaves.understandwavereflection,covertheright-handsideofeachdrawinginFigure14-13a26.CriticalThinkingIfaraindropfallsintoawithapieceofpaper.Theedgeofthepaperpool,itcreateswaveswithsmallamplitudes.shouldbeatpointN,thenode.Now,con-Ifaswimmerjumpsintoapool,waveswithcentrateontheresultantwave,showninlargeamplitudesareproduced.Whydoesn’tdarkerblue.Notethatitactslikeawavetheheavyraininathunderstormproducereflectedfromaboundary.Istheboundaryalargewaves?rigidwall,orisitopen-ended?RepeatthisTheenergyoftheswimmeristrans-exerciseforFigure14-13b.ferredtothewaveinasmallspaceFigure14-14abehaveslikearigidwalloverashorttime,whereastheenergybecausethereflectedwaveisinverted;oftheraindropsisspreadoutin14-14bbehaveslikeanopenendareaandtime.becausetheboundaryisanantinodeandthereflectedwaveisnotinverted.SectionReview14.3WaveBehaviorChapterAssessmentpages387–391ConceptMappingpage391page39627.WavesatBoundariesWhichofthe31.Completetheconceptmapusingthefol-followingwavecharacteristicsremainlowingtermsandsymbols:amplitude,unchangedwhenawavecrossesaboundaryfrequency,v,,T.intoadifferentmedium:frequency,ampli-tude,wavelength,velocity,and/ordirection?WavesFrequencyremainsunchanged.Ingeneral,amplitude,wavelength,andCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.velocitywillchangewhenawaveentersspeedamplitudeperiodfrequencywavelengthanewmedium.Directionmayormaynotchange,dependingontheoriginaldirectionofthewave.vATf28.RefractionofWavesNoticeinFigure14-17ahowthewavechangesMasteringConceptsdirectionasitpassesfromonemediumtopage396another.Cantwo-dimensionalwavescross32.Whatisperiodicmotion?Givethreeexam-aboundarybetweentwomediawithoutplesofperiodicmotion.(14.1)changingdirection?Explain.PeriodicmotionismotionthatrepeatsYes,iftheystriketheboundarywhileinaregularcycle.Examplesincludetravelingnormaltoitssurface,oriftheyoscillationofaspring,swingofasimplehavethesamespeedinbothmedia.pendulum,anduniformcircularmotion.29.StandingWavesInastandingwaveona33.Whatisthedifferencebetweenfrequencystringfixedatbothends,howisthenumberandperiod?Howaretheyrelated?(14.1)ofnodesrelatedtothenumberofantinodes?FrequencyisthenumberofcyclesorThenumberofnodesisalwaysonerepetitionspersecond,andperiodisthegreaterthanthenumberofantinodes.timerequiredforonecycle.Frequencyistheinverseoftheperiod.314SolutionsManualPhysics:PrinciplesandProblems

318Chapter14continued34.Whatissimpleharmonicmotion?Givean41.Whatistheprimarydifferencebetweenaexampleofsimpleharmonicmotion.mechanicalwaveandanelectromagnetic(14.1)wave?(14.2)SimpleharmonicmotionisperiodicTheprimarydifferenceisthatmechani-motionthatresultswhentherestoringcalwavesrequireamediumtotravelforceonanobjectisdirectlypropor-throughandelectromagneticwavesdotionaltoitsdisplacement.Ablocknotneedamedium.bouncingontheendofaspringisoneexample.42.Whatarethedifferencesamongtransverse,longitudinal,andsurfacewaves?(14.2)35.IfaspringobeysHooke’slaw,howdoesitAtransversewavecausestheparticlesbehave?(14.1)ofthemediumtovibrateinadirectionThespringstretchesadistancethatisthatisperpendiculartothedirectionindirectlyproportionaltotheforcewhichthewaveismoving.Alongitudi-appliedtoit.nalwavecausestheparticlesofthemediumtovibrateinadirectionparallel36.Howcanthespringconstantofaspringbewiththedirectionofthewave.Surfacedeterminedfromagraphofforceversuswaveshavecharacteristicsofboth.displacement?(14.1)Thespringconstantistheslopeofthe43.WavesaresentalongaspringoffixedgraphofFversusx.length.(14.2)a.Canthespeedofthewavesinthespring37.Howcanthepotentialenergyinaspringbebechanged?Explain.determinedfromthegraphofforceversusSpeedofthewavesdependsonlydisplacement?(14.1)onthemediumandcannotbeThepotentialenergyistheareaunderchanged.thecurveofthegraphofFversusx.b.Canthefrequencyofawaveinthespringbechanged?Explain.38.DoestheperiodofapendulumdependonFrequencycanbechangedbythemassofthebob?Thelengthofthechangingthefrequencyatwhichthestring?Uponwhatelsedoestheperiodwavesaregenerated.depend?(14.1)no;yes;theaccelerationofgravity,g44.Whatisthewavelengthofawave?(14.2)Wavelengthisthedistancebetweentwo39.Whatconditionsarenecessaryforadjacentpointsonawavethatareinresonancetooccur?(14.1)phase.Resonancewilloccurwhenaforceisappliedtoanoscillatingsystematthe45.Supposeyousendapulsealongarope.samefrequencyasthenaturalfrequencyHowdoesthepositionofapointontheofthesystem.ropebeforethepulsearrivescomparetothepoint’spositionafterthepulsehaspassed?40.Howmanygeneralmethodsofenergy(14.2)transferarethere?GivetwoexamplesOncethepulsehaspassed,thepointisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofeach.(14.2)exactlyasitwaspriortotheadventofTwo.Energyistransferredbyparticlethepulse.transferandbywaves.Therearemanyexamplesthatcanbegivenofeach:abaseballandabulletforparticletransfer;soundwavesandlightwaves.Physics:PrinciplesandProblemsSolutionsManual315

319Chapter14continued46.Whatisthedifferencebetweenawavepulse53.Whenawavecrossesaboundarybetweenandaperiodicwave?(14.2)athinandathickrope,asshowninApulseisasingledisturbanceinaFigure14-18,itswavelengthandspeedmedium,whereasaperiodicwavecon-change,butitsfrequencydoesnot.Explainsistsofseveraladjacentdisturbances.whythefrequencyisconstant.(14.3)47.Describethedifferencebetweenwavefrequencyandwavevelocity.(14.2)■Figure14-18FrequencyisthenumberofvibrationsThefrequencydependsonlyontheratepersecondofapartofthemedium.atwhichthethinropeisshakenandtheVelocitydescribesthemotionofthethinropecausesthevibrationsinthewavethroughthemedium.thickrope.48.Supposeyouproduceatransversewaveby54.Howdoesaspringpulsereflectedfromashakingoneendofaspringfromsidetorigidwalldifferfromtheincidentpulse?side.Howdoesthefrequencyofyourhand(14.3)comparewiththefrequencyofthewave?Thereflectedpulsewillbeinverted.(14.2)Theyarethesame.55.Describeinterference.Isinterferenceaprop-ertyofonlysometypesofwavesoralltypes49.Whenarepointsonawaveinphasewithofwaves?(14.3)eachother?Whenaretheyoutofphase?ThesuperpositionoftwoormoreGiveanexampleofeach.(14.2)wavesisinterference.Thesuperposi-Pointsareinphasewhentheyhavetionoftwowaveswithequalbutoppo-thesamedisplacementandthesamesiteamplitudesresultsindestructivevelocity.Otherwise,thepointsareoutinterference.Thesuperpositionoftwoofphase.Twocrestsareinphasewithwaveswithamplitudesinthesameeachother.Acrestandatroughareoutdirectionresultsinconstructiveinter-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofphasewitheachother.ference;allwaves;itisaprimetestforwavenature.50.Whatistheamplitudeofawaveandwhatdoesitrepresent?(14.2)56.WhathappenstoaspringatthenodesofaAmplitudeisthemaximumdisplace-standingwave?(14.3)mentofawavefromtherestorequilib-Nothing,thespringdoesnotmove.riumposition.Theamplitudeofthewaverepresentstheamountofenergy57.ViolinsAmetalplateisheldfixedinthetransferred.centerandsprinkledwithsugar.Withaviolinbow,theplateisstrokedalongone51.Describetherelationshipbetweentheedgeandmadetovibrate.Thesugarbeginsamplitudeofawaveandtheenergyittocollectincertainareasandmoveawaycarries.(14.2)fromothers.DescribetheseregionsintermsTheenergycarriedbyawaveispropor-ofstandingwaves.(14.3)tionaltothesquareofitsamplitude.Bareareasareantinodalregionswherethereismaximumvibration.Sugar-52.Whenawavereachestheboundaryofacoveredareasarenodalregionswherenewmedium,whathappenstoit?(14.3)thereisnovibration.Partofthewavecanbereflectedandpartofthewavecanbetransmittedintothenewmedium.316SolutionsManualPhysics:PrinciplesandProblems

320Chapter14continued58.Ifastringisvibratinginfourparts,thereare62.Supposeyouholda1-mmetalbarinyourpointswhereitcanbetouchedwithouthandandhititsendwithahammer,first,disturbingitsmotion.Explain.Howmanyinadirectionparalleltoitslength,andofthesepointsexist?(14.3)second,inadirectionatrightanglestoitsAstandingwaveexistsandthestringlength.Describethewavesproducedinthecanbetouchedatanyofitsfivenodaltwocases.points.Inthefirstcase,longitudinalwaves;inthesecondcase,transversewaves.59.Wavefrontspassatananglefromonemediumintoasecondmedium,wherethey63.Supposeyourepeatedlydipyourfingerintotravelwithadifferentspeed.Describetwoasinkfullofwatertomakecircularwaves.changesinthewavefronts.WhatdoesnotWhathappenstothewavelengthasyouchange?(14.3)moveyourfingerfaster?ThewavelengthanddirectionoftheThefrequencyofthewaveswillwavefrontschange.Thefrequencyincrease;thespeedwillremainthedoesnotchange.same;thewavelengthwilldecrease.ApplyingConcepts64.Whathappenstotheperiodofawaveasthefrequencyincreases?page39760.AballbouncesupanddownontheendofAsthefrequencyincreases,theperiodaspring.Describetheenergychangesthatdecreases.takeplaceduringonecompletecycle.Doesthetotalmechanicalenergychange?65.Whathappenstothewavelengthofawaveasthefrequencyincreases?Atthebottomofthemotion,theelasticpotentialenergyisatamaximum,whileAsthefrequencyincreases,thewave-gravitationalpotentialenergyisatalengthdecreases.minimumandthekineticenergyiszero.66.SupposeyoumakeasinglepulseonaAttheequilibriumposition,theKEisatstretchedspring.Howmuchenergyisamaximumandtheelasticpotentialrequiredtomakeapulsewithtwicetheenergyiszero.Atthetopofthebounce,amplitude?theKEiszero,thegravitationalpotentialenergyisatamaximum,andtheelasticapproximatelytwosquared,orfourpotentialenergyisatamaximum.Thetimestheenergytotalmechanicalenergyisconserved.67.Youcanmakewatersloshbackandforthin61.Canapendulumclockbeusedintheorbit-ashallowpanonlyifyoushakethepaningInternationalSpaceStation?Explain.withthecorrectfrequency.Explain.No,thespacestationisinfree-fall,andTheperiodofthevibrationmustequaltherefore,theapparentvalueofgisthetimeforthewavetogobackandzero.Thependulumwillnotswing.forthacrossthepantocreateconstruc-tiveinterference.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual317

321Chapter14continued68.IneachofthefourwavesinFigure14-19,70.CarShocksEachofthecoilspringsofathepulseontheleftistheoriginalpulsecarhasaspringconstantof25,000N/m.movingtowardtheright.ThecenterpulseisHowmuchiseachspringcompressedifitareflectedpulse;thepulseontherightisasupportsone-fourthofthecar’s12,000-Ntransmittedpulse.Describetherigidityofweight?theboundariesatA,B,C,andD.Fkx,FAsoxk1(12,000N)425,000N/mB0.12m71.HowmuchpotentialenergyisstoredinaCspringwithaspringconstantof27N/mifitisstretchedby16cm?12PEsp2kxD1(27N/m)(0.16m)20.35J2■Figure14-19Level2BoundaryAismorerigid;boundaryB72.RocketLauncherAtoyrocket-launcherislessrigid;boundaryCislessrigid;containsaspringwithaspringconstantofboundaryDismorerigid.35N/m.HowfarmustthespringbeMasteringProblemscompressedtostore1.5Jofenergy?14.1PeriodicMotion12,PEsp2kxpages397–398Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2PEsp(2)(1.5J)Level1soxk35N/m69.Aspringstretchesby0.12mwhensomeapplesweighing3.2Naresuspendedfrom0.29mit,asshowninFigure14-20.Whatisthespringconstantofthespring?Level373.Force-versus-lengthdataforaspringareplottedonthegraphinFigure14-21.12.08.0Force(N)4.03.2N0.00.200.400.60■Figure14-20Length(m)Fkx,■Figure14-21F3.2Nsok27N/mx0.12m318SolutionsManualPhysics:PrinciplesandProblems

322Chapter14continueda.Whatisthespringconstantofthe1fspring?T11kslopeT8.3sf0.12HzF12.0N4.0Nx0.6m0.2m76.OceanWavesAnoceanwavehasalength20N/mof12.0m.Awavepassesafixedlocationb.Whatistheenergystoredinthespringevery3.0s.Whatisthespeedofthewave?whenitisstretchedtoalengthof11vf(12.0m)50.0cm?T3.0s14.0m/sPEsparea2bh177.Waterwavesinashallowdishare6.0-cm(0.500m)(10.0N)2.50J2long.Atonepoint,thewatermovesupanddownatarateof4.8oscillations/s.74.Howlongmustapendulumbetohaveaa.Whatisthespeedofthewaterwaves?periodof2.3sontheMoon,where2?vfg1.6m/slT2g(0.060m)(4.8Hz)0.29m/sT2,solg42b.Whatistheperiodofthewaterwaves?(2.3s)2(1.6m/s2)110.21mTf4.8Hz0.21s4278.Waterwavesinalaketravel3.4min1.8s.14.2WavePropertiesTheperiodofoscillationis1.1s.page398a.Whatisthespeedofthewaterwaves?Level175.BuildingMotionTheSearsTowerind3.4mv1.9m/st1.8sChicago,showninFigure14-22,swaysbackandforthinthewindwithafrequencyb.Whatistheirwavelength?ofabout0.12Hz.WhatisitsperiodofvvTvibration?f(1.9m/s)(1.1s)2.1mLevel279.SonarAsonarsignaloffrequency1.00106Hzhasawavelengthof1.50mminwater.a.Whatisthespeedofthesignalinwater?vf(1.50103m)(1.00106Hz)1.50103m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Whatisitsperiodinwater?11Tf1.00106Hz1.00106s■Figure14-22Physics:PrinciplesandProblemsSolutionsManual319

323Chapter14continuedc.Whatisitsperiodinair?vf1.00106s(6.0m)(0.60Hz)Theperiodandfrequencyremain3.6m/sunchanged.83.EarthquakesThevelocityofthetrans-80.Asoundwaveofwavelength0.60mandaversewavesproducedbyanearthquakeisvelocityof330m/sisproducedfor0.50s.8.9km/s,andthatofthelongitudinalwavesis5.1km/s.Aseismographrecordsa.Whatisthefrequencyofthewave?thearrivalofthetransversewaves68svfbeforethearrivalofthelongitudinalv330m/swaves.Howfarawayistheearthquake?f0.60mdvt.Wedonotknowt,onlythe550Hzdifferenceintime,t.Thetransverseb.Howmanycompletewavesareemitteddistance,dTvTt,isthesameastheinthistimeinterval?longitudinaldistance,dLvL(tt).ft(550Hz)(0.50s)UsevTtvL(tt),andsolvefort:280completewavesvLtc.After0.50s,howfaristhefrontofthetvTvLwavefromthesourceofthesound?(5.1km/s)(68s)dvtt91s8.9km/s5.1km/s(330m/s)(0.50s)Thenputtingtbackinto1.6102mdTvTt(8.9km/s)(91s)81.Thespeedofsoundinwateris1498m/s.A8.1102kmsonarsignalissentstraightdownfromashipatapointjustbelowthewatersurface,14.3WaveBehaviorand1.80slater,thereflectedsignalisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pages398–399detected.Howdeepisthewater?Level1Thetimeforthewavetotraveldown84.Sketchtheresultforeachofthethreecasesandbackupis1.80s.ThetimeonewayshowninFigure14-23,whenthecentersishalf1.80sor0.900s.ofthetwoapproachingwavepulseslieondvtthedashedlinesothatthepulsesexactly(1498m/s)(0.900s)overlap.1350m1Level382.PepeandAlfredoarerestingonanoffshore2raftafteraswim.Theyestimatethat3.0mseparatesatroughandanadjacentcrestofeachsurfacewaveonthelake.Theycount12creststhatpassbytheraftin20.0s.3Calculatehowfastthewavesaremoving.(2)(3.0m)6.0m12wavesf0.60Hz20.0s■Figure14-23320SolutionsManualPhysics:PrinciplesandProblems

324Chapter14continued1.Theamplitudeisdoubled.15cmfromtheotherend,wherethedistancestraveledarethesame.87.SketchtheresultforeachofthefourcasesshowninFigure14-24,whenthecentersof2.Theamplitudescanceleachother.eachofthetwowavepulseslieonthedashedlinesothatthepulsesexactlyoverlap.3.Iftheamplitudeofthefirstpulseis1one-halfofthesecond,theresultantpulseisone-halftheamplitudeofthesecond.2385.Ifyousloshthewaterinabathtubatthecorrectfrequency,thewaterrisesfirstatoneendandthenattheother.Supposeyoucan4makeastandingwaveina150-cm-longtubwithafrequencyof0.30Hz.Whatisthevelocityofthewaterwave?■Figure14-242(1.5m)3.0mvf(3.0m)(0.30Hz)10.90m/sLevel2286.GuitarsThewavespeedinaguitarstringis265m/s.Thelengthofthestringis63cm.You3pluckthecenterofthestringbypullingitupandlettinggo.Pulsesmoveinbothdirectionsandarereflectedofftheendsofthestring.4a.Howlongdoesittakeforthepulsetomovetothestringendandreturntothecenter?(2)(63cm)d63cmMixedReview2page399–400sotd0.63m2.4103sLevel1v265m/s88.Whatistheperiodofapendulumwithab.Whenthepulsesreturn,isthestringlengthof1.4m?aboveorbelowitsrestinglocation?lPulsesareinvertedwhenreflectedT2gCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fromamoredensemedium,soreturningpulseisdown(below).1.4m22.4s9.80m/s2c.Ifyoupluckedthestring15cmfromoneendofthestring,wherewouldthe89.Thefrequencyofyellowlightis5.11014Hz.twopulsesmeet?Findthewavelengthofyellowlight.Thespeedoflightis3.00108m/s.Physics:PrinciplesandProblemsSolutionsManual321

325Chapter14continuedcc.Inthecourseof15syoucounttenfwavesthatpassyou.Whatistheperiod3.00108m/softhewaves?5.11014Hz15sT1.5s10waves5.9107md.Whatisthefrequencyofthewaves?90.RadioWaveAM-radiosignalsarebroad-11f0.67Hzcastatfrequenciesbetween550kHzT1.5s(kilohertz)and1600kHzandtravele.Youestimatethatthewavecrestsare3m3.0108m/s.apart.Whatisthevelocityofthewaves?a.Whatistherangeofwavelengthsforvf(3m)(0.67Hz)2m/sthesesignals?f.Afterreturningtothebeach,youlearnvfthatthewavesaremovingat1.8m/s.Whatistheactualwavelengthofthev3.0108m/s1f5.5105Hzwaves?1v1.8m/s550m2.7mf0.67Hzv3.0108m/s2f1.6106HzLevel2292.BungeeJumperAhigh-altitudebungee190mjumperjumpsfromahot-airballoonusingRangeis190mto550m.a540-m-bungeecord.Whenthejumpisb.FMfrequenciesrangebetween88MHzcompleteandthejumperisjustsuspended(megahertz)and108MHzandtravelfromthecord,itisstretched1710m.Whatatthesamespeed.WhatistherangeofisthespringconstantofthebungeecordifFMwavelengths?thejumperhasamassof68kg?v3.0108m/sFmg(68kg)(9.80m/s2)f8.8107Hzkxx1710m540mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.4m0.57N/mv3.0108m/s8Hz93.Thetimeneededforawaterwavetochangef1.0810fromtheequilibriumleveltothecrestis2.8m0.18s.Rangeis2.8mto3.4m.a.Whatfractionofawavelengthisthis?191.Youarefloatingjustoffshoreatthebeach.wavelength4Eventhoughthewavesaresteadilyb.Whatistheperiodofthewave?movingintowardthebeach,youdon’tmoveanyclosertothebeach.T(4)(0.18s)0.72sa.Whattypeofwaveareyouexperiencingc.Whatisthefrequencyofthewave?asyoufloatinthewater?11f1.4HztransversewavesT0.72sb.Explainwhytheenergyinthewave94.Whena225-gmassishungfromaspring,doesnotmoveyouclosertoshore.thespringstretches9.4cm.ThespringandThedisplacementisperpendicularmassthenarepulled8.0cmfromthisnewtothedirectionofthewave—inequilibriumpositionandreleased.Findthethiscase,upanddown.springconstantofthespringandthemaxi-mumspeedofthemass.322SolutionsManualPhysics:PrinciplesandProblems

326Chapter14continuedFmgNew:kxxT2lg(0.225kg)(9.80m/s2)4223N/m0.094m(1.0101s)2(9.80m/s2)Maximumvelocityoccurswhenthemass42passesthroughtheequilibriumpoint,25mwherealltheenergyiskineticenergy.ThearmonthenewstructureisUsingtheconservationofenergy:11mlonger.PEspKEmasse.Iftheparkownerswantedtodoublethe1212periodoftheride,whatpercentagekxmv22increasewouldneedtobemadetothe2lengthofthependulum?2kxvmBecauseofthesquarerelationship,kx2(23N/m)(0.080m)2therewouldneedtobea4timesvm0.225kgincreaseinthelengthofthependu-lum,ora300%increase.0.81m/s96.ClocksThespeedatwhichagrandfather95.AmusementRideYounoticethatyourclockrunsiscontrolledbyaswingingfavoriteamusement-parkrideseemsbigger.pendulum.Therideconsistsofacarriagethatisa.Ifyoufindthattheclocklosestimeattachedtoastructuresoitswingslikeaeachday,whatadjustmentwouldyoupendulum.Yourememberthatthecarriageneedtomaketothependulumsoitwillusedtoswingfromonepositiontoanotherkeepbettertime?andbackagaineighttimesinexactly1min.Nowitonlyswingssixtimesin1min.GiveTheclockmustbemadetorunyouranswerstothefollowingquestionstofaster.Theperiodofthependulumtwosignificantdigits.canbeshortened,thusincreasingthespeedoftheclock,byshorten-a.Whatwastheoriginalperiodoftheride?ingthelengthofthependulum.11T7.5sb.Ifthependulumcurrentlyis15.0cm,f8swingsbyhowmuchwouldyouneedto60.0schangethelengthtomaketheperiodb.Whatisthenewperiodoftheride?lessenby0.0400s?111sT1.010f6swingsl2l1T2260.0sggc.Whatisthenewfrequency?Tl2l1112ggf0.10HzT1.0101sT11d.Howmuchlongeristhearmsupporting2gl2gl1thecarriageonthelargerride?T11Original:2l2l1ggCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.T2lg42Tg2l2l1(7.5s)2(9.80m/s2)4214mPhysics:PrinciplesandProblemsSolutionsManual323

327Chapter14continuedTgl22l1Tg2l22l1(0.0400s)9.80m/s220.150m20.135mThelengthwouldneedtoshortenbyl1l20.150m0.135m0.015m97.BridgeSwingingInthesummerovertheNewRiverinWestVirginia,severalteensswingfrombridgeswithropes,thendropintotheriverafterafewswingsbackandforth.a.IfPamisusinga10.0-mlengthofrope,howlongwillittakehertoreachthepeakofherswingattheotherendofthebridge?1swingtopeakT2l10.0m3.17sg9.80m/s2b.IfMikehasamassthatis20kgmorethanPam,howwouldyouexpecttheperiodofhisswingtodifferfromPam’s?Thereshouldbenodifference.Tisnotaffectedbymass.c.AtwhatpointintheswingisKEatamaximum?Atthebottomoftheswing,KEisatamaximum.d.AtwhatpointintheswingisPEatamaximum?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Atthetopoftheswing,PEisatamaximum.e.AtwhatpointintheswingisKEataminimum?Atthetopoftheswing,KEisataminimum.f.AtwhatpointintheswingisPEataminimum?Atthebottomoftheswing,PEisataminimum.98.Youhaveamechanicalfishscalethatismadewithaspringthatcompresseswhenweightisaddedtoahookattachedbelowthescale.Unfortunately,thecalibrationshavecompletelywornoffofthescale.However,youhaveoneknownmassof500.0gthatdisplacesthespring2.0cm.a.Whatisthespringconstantforthespring?Fmgkxmgkx(0.5000kg)(9.80m/s2)0.020m2.4102N/mb.Ifafishdisplacesthespring4.5cm,whatisthemassofthefish?Fmgkx324SolutionsManualPhysics:PrinciplesandProblems

328Chapter14continuedkxFmTgNowv,so(2.4102N/m)(4.5102m)F2Tv9.80m/s2(1.50102m/s)2(2.83103kg/m)1.1kg63.7N99.CarSpringsWhenyouadda45-kgloadtothetrunkofanewsmallcar,thetworearThinkingCriticallyspringscompressanadditional1.0cm.page400a.Whatisthespringconstantforeachof101.AnalyzeandConcludeA20-Nforceisthesprings?requiredtostretchaspringby0.5m.Fmg(45kg)(9.80m/s2)440Na.Whatisthespringconstant?forceperspring220NF20NFkx,sok40N/mx0.5mFFkx,sokxb.Howmuchenergyisstoredinthespring?220Nk22,000N/m0.010m12PEsp2kxb.Howmuchadditionalpotentialenergyisstoredineachofthecarsprings1(40N/m)(0.5m)25J2afterloadingthetrunk?c.Whyisn’ttheworkdonetostretchthe12PEkxspringequaltotheforcetimesthe2distance,or10J?12(22,000N/m)(0.010m)2Theforceisnotconstantasthe1.1Jspringisstretched.Theaverageforce,10N,timesthedistanceLevel3doesgivethecorrectwork.100.Thevelocityofawaveonastringdepends102.MakeandUseGraphsSeveralweightsonhowtightlythestringisstretched,andweresuspendedfromaspring,andtheonthemassperunitlengthofthestring.IfresultingextensionsofthespringwereFTisthetensioninthestring,andisthemeasured.Table14-1showsthemass/unitlength,thenthevelocity,v,cancollecteddata.bedeterminedbythefollowingequation.FvTTable14-1WeightsonaSpringApieceofstring5.30-mlonghasamassForce,F(N)Extension,x(m)of15.0g.Whatmustthetensionin2.50.12thestringbetomakethewavelengthofa5.00.26125-Hzwave120.0cm?7.50.35vf(1.200m)(125Hz)10.00.501.50102m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12.50.60mand15.00.71L1.50102kg5.30m2.83103kg/mPhysics:PrinciplesandProblemsSolutionsManual325

329Chapter14continueda.Makeagraphoftheforceappliedtothespringversusthespringlength.Plottheforceonthey-axis.20.015.010.0Force(N)5.00.00.200.400.600.80Stretch(m)b.Determinethespringconstantfromthegraph.Thespringconstantistheslope.F15.0N2.5Nkslope21N/mx0.71m0.12mc.Usingthegraph,findtheelasticpotentialenergystoredinthespringwhenitisstretchedto0.50m.Thepotentialenergyistheareaunderthegraph.1PEbhsparea21(0.50m)(10.0N)22.5J103.ApplyConceptsGravelroadsoftendevelopregularlyspacedridgesthatareperpendiculartotheroad,asshowninFigure14-25.Thiseffect,calledwashboarding,occursbecausemostcarstravelataboutthesamespeedandCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.thespringsthatconnectthewheelstothecarsoscillateataboutthesamefrequency.Iftheridgesonaroadare1.5mapartandcarstravelonitatabout5m/s,whatisthefrequencyofthesprings’oscillation?■Figure14-25vfv5m/sf3Hz1.5m326SolutionsManualPhysics:PrinciplesandProblems

330Chapter14continuedWritinginPhysicsc.Whatwastheaverageaccelerationoftheautomobile?page400104.ResearchChristiaanHuygens’workonavtwavesandthecontroversybetweenhimandNewtonoverthenatureoflight.112m/s9.8sCompareandcontrasttheirexplanationsofsuchphenomenaasreflectionandrefrac-11m/s2tion.Whosemodelwouldyouchooseasthebestexplanation?Explainwhy.106.Howmuchwaterwouldasteamenginehavetoevaporatein1stoproduce1kWHuygensproposedthewavetheoryofpower?AssumethattheengineisoflightandNewtonproposedthe20percentefficient.(Chapter12)particletheoryoflight.ThelawofreflectioncanbeexplainedusingW1000J/stboththeories.Huygen’sprincipleandNewton’sparticletheoryareopposed,Iftheengineisonly20percenthowever,intheirexplanationoftheefficientitmustusefivetimesmorelawofrefraction.heattoproducethe1000J/s.QmHv5000J/sCumulativeReviewttpage400m5000J/sTherefore,105.A1400-kgdragracerautomobilecantHvcompleteaone-quartermile(402m)5000J/scoursein9.8s.Thefinalspeedofthe2.26106J/kgautomobileis250mi/h(112m/s).2103kg/s(Chapter11)a.Whatisthekineticenergyoftheautomobile?ChallengeProblemKE1mv2page3802Acarofmassmrestsatthetopofahillof1(1400kg)(112m/s)2heighthbeforerollingwithoutfrictionintoa2crashbarrierlocatedatthebottomofthehill.8.8106JThecrashbarriercontainsaspringwithaspringb.Whatistheminimumamountofworkconstant,k,whichisdesignedtobringthecartothatwasdonebyitsengine?Whycan’trestwithminimumdamage.youcalculatethetotalamountofwork1.Determine,intermsofm,h,k,andg,thedone?maximumdistance,x,thatthespringwillTheminimumamountofworkbecompressedwhenthecarhitsit.mustequalKE,or8.8106J.TheConservationofenergyimpliesthattheenginehadtodomoreworkthangravitationalpotentialenergyofthecarwasdissipatedinworkdoneatthetopofthehillwillbeequaltotheagainstfriction.elasticpotentialenergyinthespringwhenithasbroughtthecartorest.TheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.equationsfortheseenergiescanbesetequalandsolvedforx.12PEkxgPEsp,somgh22mghxkPhysics:PrinciplesandProblemsSolutionsManual327

331Chapter14continued2.Ifthecarrollsdownahillthatistwiceashigh,howmuchfartherwillthespringbecompressed?Theheightisdoubledandxispropor-tionaltothesquarerootoftheheight,soxwillincreaseby2.3.Whatwillhappenafterthecarhasbeenbroughttorest?Inthecaseofanidealspring,thespringwillpropelthecarbacktothetopofthehill.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.328SolutionsManualPhysics:PrinciplesandProblems

332CHAPTER15SoundvPracticeProblemss24.6m/s115.1PropertiesandDetectionfd524Hz1(24.6m/s)ofSound343m/spages403–410489Hzpage4051.Findthewavelengthinairat20°Cofan7.Youareinanautotravelingat25.0m/s18-Hzsoundwave,whichisoneofthetowardapole-mountedwarningsiren.Iflowestfrequenciesthatisdetectablebythethesiren’sfrequencyis365Hz,whathumanear.frequencydoyouhear?Use343m/sasthespeedofsound.v343m/s19mf18Hzv343m/s,fs365Hz,vs0,2.Whatisthewavelengthofan18-Hzsoundvd25.0m/swaveinseawaterat25°C?vvfdv1533m/sdfsvv85msf18Hz343m/s25.0m/s3.Findthefrequencyofasoundwavemoving(365Hz)343m/sthroughironat25°Cwithawavelengthof392Hz1.25m.fv5130m/s4.10103Hz8.Youareinanautotravelingat55mph1.25m(24.6m/s).Asecondautoismovingtoward4.Ifyoushoutacrossacanyonandheartheyouatthesamespeed.Itshornissoundingecho0.80slater,howwideisthecanyon?at475Hz.Whatfrequencydoyouhear?dUse343m/sasthespeedofsound.vtv343m/s,fs475Hz,vs24.6m/s,sodvt(343m/s)(0.40s)140mvd24.6m/svv5.A2280-Hzsoundwavehasawavelengthoffddfsvv0.655minanunknownmedium.Identifysthemedium.343m/s24.6m/s(475Hz)v343m/s24.6m/sf548Hzsovf(0.655m)(2280Hz)9.Asubmarineismovingtowardanother1490m/ssubmarineat9.20m/s.Itemitsa3.50-MHzThisspeedcorrespondstowateratultrasound.Whatfrequencywouldthe25°C.secondsub,atrest,detect?ThespeedofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.soundinwateris1482m/s.v1482m/s,fs3.50MHz,page4096.RepeatExampleProblem1,butwiththecarvs9.20m/s,vd0m/smovingawayfromyou.Whatfrequencyvvfdwouldyouhear?dfsvvsPhysics:PrinciplesandProblemsSolutionsManual329

333Chapter15continued1482m/s13.SoundPropertiesWhatphysicalcharac-(3.50MHz)1482m/s9.20m/steristicofasoundwaveshouldbechanged3.52MHztochangethepitchofthesound?Tochangetheloudness?10.AsoundsourceplaysmiddleC(262Hz).frequency;amplitudeHowfastwouldthesourcehavetogotoraisethepitchtoCsharp(271Hz)?Use14.DecibelScaleHowmuchgreateristhe343m/sasthespeedofsound.soundpressurelevelofatypicalrockband’sv343m/s,fmusic(110dB)thananormalconversations262Hz,fd271Hz,(50dB)?vd0m/s,vsisunknownThesoundpressurelevelincreasesbyvvfdafactorof10forevery20-dBincreasedfsvvsinsoundlevel.Therefore,60dBcorre-Solvethisequationforvspondstoa1000-foldincreaseinSPL.s.fvs(vv15.EarlyDetectionInthenineteenthcentury,svfd)dpeopleputtheirearstoarailroadtrackto262Hzgetanearlywarningofanapproaching343m/s271Hz(343m/s0m/s)train.Whydidthiswork?11.4m/sThevelocityofsoundisgreaterinsolidsthaningases.Therefore,soundtravelsfasterinsteelrailsthaninair,SectionReviewandtherailshelpfocusthesoundsoitdoesnotdieoutasquicklyasinair.15.1PropertiesandDetectionofSound16.BatsAbatemitsshortpulsesofhigh-pages403–410frequencysoundanddetectstheechoes.page410a.InwhatwaywouldtheechoesfromCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11.GraphTheeardrummovesbackandforthlargeandsmallinsectscompareiftheyinresponsetothepressurevariationsofawerethesamedistancefromthebat?soundwave.Sketchagraphofthedisplace-Theywoulddifferinintensity.Largermentoftheeardrumversustimefortwoinsectswouldreflectmoreofthecyclesofa1.0-kHztoneandfortwocyclessoundenergybacktothebat.ofa2.0-kHztone.b.InwhatwaywouldtheechofromaninsectflyingtowardthebatdifferfromthatofaninsectflyingawayfromtheTime(ms)bat?0.51.01.52.0AninsectflyingtowardthebatDisplacementwouldreturnanechoofhigherfre-1.0kHzquency(Dopplershift).Aninsectfly-2.0kHzingawayfromthebatwouldreturnanechooflowerfrequency.12.EffectofMediumListtwosoundcharac-teristicsthatareaffectedbythemedium17.CriticalThinkingCanatrooperusingathroughwhichthesoundpassesandtworadardetectoratthesideoftheroadcharacteristicsthatarenotaffected.determinethespeedofacarattheinstantaffected:speedandwavelength;thecarpassesthetrooper?Explain.unaffected:periodandfrequencyNo.Thecarmustbeapproachingor330SolutionsManualPhysics:PrinciplesandProblems

334Chapter15continuedrecedingfromthedetectorforthev343m/sf64.7Hz15.30mDopplereffecttobeobserved.1Transversemotionproducesnob.FindthenexttworesonantfrequenciesDopplereffect.forthebugle.vv343m/sf129Hz2L2.65mPracticeProblems2v3v(3)(343m/s)f194Hz15.2ThePhysicsofMusic32L(2)(2.65m)3pages411–419page416SectionReview18.A440-Hztuningforkisheldaboveaclosed15.2ThePhysicsofMusicpipe.Findthespacingbetweenthereso-nanceswhentheairtemperatureis20°C.pages411–419vpage419Resonancespacingsousing2f22.OriginsofSoundWhatisthevibratingtheresonancespacingisobjectthatproducessoundsineachofthev343m/s0.39mfollowing?22f(2)(440Hz)a.ahumanvoice19.A440-Hztuningforkisusedwithavocalcordsresonatingcolumntodeterminetheb.aclarinetvelocityofsoundinheliumgas.Ifthespacingsbetweenresonancesare110cm,areedwhatisthevelocityofsoundinheliumgas?c.atubaResonancespacing1.1mtheplayer’slips2so2.2md.aviolinvf(2.2m)(440Hz)970m/sastring20.Thefrequencyofatuningforkisunknown.23.ResonanceinAirColumnsWhyistheAstudentusesanaircolumnat27°Candtubefromwhichatubaismademuchfindsresonancesspacedby20.2cm.Whatlongerthanthatofacornet?isthefrequencyofthetuningfork?UsetheThelongerthetube,thelowerthespeedcalculatedinExampleProblem2forresonantfrequencyitwillproduce.thespeedofsoundinairat27°C.v347m/sat27°C24.ResonanceinOpenTubesHowmustthelengthofanopentubecomparetotheResonancespacinggives0.202m,2wavelengthofthesoundtoproducetheor0.404mstrongestresonance?v347m/sf859HzThelengthofthetubeshouldbe0.404mone-halfthewavelength.21.Abuglecanbethoughtofasanopenpipe.Ifabuglewerestraightenedout,itwouldbe25.ResonanceonStringsAviolinsoundsa2.65-mlong.noteofFsharp,withapitchof370Hz.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Ifthespeedofsoundis343m/s,findWhatarethefrequenciesofthenextthreethelowestfrequencythatisresonantforharmonicsproducedwiththisnote?abugle(ignoringendcorrections).Astring’sharmonicsarewholenumbermultiplesofthefundamental,sothefre-12L(2)(2.65m)5.30mquenciesare:ThelowestfrequencyisPhysics:PrinciplesandProblemsSolutionsManual331

335Chapter15continuedfgreatlywhenitispressedagainstother22f1(2)(370Hz)740Hzfobjectsbecausetheyresonatelikea33f1(3)(370Hz)1110Hzsoundingboard.Theysounddifferent1100Hzbecausetheyresonatewithdifferentf44f1(4)(370Hz)1480Hzharmonics;therefore,theyhavedifferent1500Hztimbres.26.ResonanceinClosedPipesOneclosedorganpipehasalengthof2.40m.ChapterAssessmenta.WhatisthefrequencyofthenoteConceptMappingplayedbythispipe?page4244L(4)(2.40m)9.60m30.Completetheconceptmapbelowusingthevfollowingterms:amplitude,perception,pitch,fspeed.v343m/sf35.7Hz9.60mSoundb.Whenasecondpipeisplayedatthesametime,a1.40-Hzbeatnoteisheard.Byhowmuchisthesecondpipetoopropertiesperceptionlong?f35.7Hz1.40Hz34.3Hzfrequencypitchloudnessv343m/s10.0mf34.3Hzspeed4Lamplitude10.0mL2.50m44ThedifferenceinlengthsisMasteringConceptsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.50m2.40m0.10mpage42427.TimbreWhydovariousinstrumentssound31.Whatarethephysicalcharacteristicsofdifferentevenwhentheyplaythesamenote?soundwaves?(15.1)EachinstrumentproducesitsownsetSoundwavescanbedescribedbyoffundamentalandharmonicfrequen-frequency,wavelength,amplitude,cies,sotheyhavedifferenttimbres.andspeed.28.BeatsAtuningforkproducesthreebeats32.Whentimingthe100-mrun,officialsatthepersecondwithasecond,392-Hztuningfinishlineareinstructedtostarttheirstop-fork.Whatisthefrequencyofthefirstwatchesatthesightofsmokefromthetuningfork?starter’spistolandnotatthesoundofitsfiring.Explain.WhatwouldhappentotheItiseither389Hzor395Hz.Youcan’ttimesfortherunnersifthetimingstartedtellwhichwithoutmoreinformation.whensoundwasheard?(15.1)29.CriticalThinkingStrikeatuningforkwithLighttravelsat3.00108m/s,whilearubberhammerandholditatarm’ssoundtravelsat343m/s.Officialslength.Thenpressitshandleagainstadesk,wouldseethesmokebeforetheywouldadoor,afilingcabinet,andotherobjects.hearthepistolfire.ThetimeswouldbeWhatdoyouhear?Why?lessthanactualifsoundwereused.Thetuningfork’ssoundisamplified332SolutionsManualPhysics:PrinciplesandProblems

336Chapter15continued33.Nametwotypesofperceptionofsoundand39.MusicalInstrumentsWhydon’tmostthephysicalcharacteristicsofsoundwavesmusicalinstrumentssoundliketuningthatcorrespondtothem.(15.1)forks?(15.2)pitch—frequency,loudness—amplitudeTuningforksproducesimple,single-frequencywaves.Musicalinstruments34.DoestheDopplershiftoccurforonlysomeproducecomplexwavescontainingtypesofwavesorforalltypesofwaves?manydifferentfrequencies.Thisgives(15.1)themtheirtimbres.alltypesofwaves40.MusicalInstrumentsWhatpropertydis-35.Soundwaveswithfrequencieshigherthantinguishesnotesplayedonbothatrumpetcanbeheardbyhumans,calledultrasound,andaclarinetiftheyhavethesamepitchcanbetransmittedthroughthehumanbody.andloudness?(15.2)Howcouldultrasoundbeusedtomeasurethesoundqualityortimbrethespeedofbloodflowinginveinsorarter-ies?Explainhowthewaveschangetomake41.TrombonesExplainhowtheslideofathismeasurementpossible.(15.1)trombone,showninFigure15-21,changesDoctorscanmeasuretheDopplershiftthepitchofthesoundintermsofatrom-fromsoundreflectedbythemovingbonebeingaresonancetube.(15.2)bloodcells.Becausethebloodismov-ing,soundgetsDopplershifted,thecompressionseithergetpileduporspacedapart.Thisaltersthefrequencyofthewave.36.Whatisnecessaryfortheproductionandtransmissionofsound?(15.2)avibratingobjectandamaterialmedium■Figure15-21Theslideofatrombonevariespitchby37.SingingHowcanacertainnotesungbyanchangingthelengthoftheresonatingoperasingercauseacrystalglasstoshatter?columnofvibratingair.(15.2)ThefrequencyofthenoteisthesameApplyingConceptsasthenaturalresonanceofthecrystal,pages424–425causingitsmoleculestoincreasetheir42.EstimationToestimatethedistanceinamplitudeofvibrationasenergyfromkilometersbetweenyouandalightningthesoundisaccepted.flash,countthesecondsbetweentheflashandthethunderanddivideby3.Explain38.MarchingInthemilitary,asmarchinghowthisruleworks.Deviseasimilarrulesoldiersapproachabridge,thecommandformiles.“routestep”isgiven.ThesoldiersthenwalkThespeedofsound343m/sout-of-stepwitheachotherastheycrossthe0.343km/s(1/2.92)km/s;or,soundbridge.Explain.(15.2)travelsapproximately1kmin3s.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Whilemarchinginstep,acertainTherefore,dividethenumberofsec-frequencyisestablishedthatcouldondsbythree.Formiles,soundtravelsresonatethebridgeintodestructiveapproximately1milein5s.Therefore,oscillation.Nosinglefrequencyismain-dividethenumberofsecondsbyfive.tainedunder“routestep.”Physics:PrinciplesandProblemsSolutionsManual333

337Chapter15continued43.Thespeedofsoundincreasesbyabout47.Ifthepitchofsoundisincreased,whatare0.6m/sforeachdegreeCelsiuswhenthethechangesinthefollowing?airtemperaturerises.Foragivensound,asa.thefrequencythetemperatureincreases,whathappenstoFrequencywillincrease.thefollowing?b.thewavelengtha.thefrequencyWavelengthwilldecrease.Thereisnochangeinfrequency.c.thewavevelocityb.thewavelengthWavevelocitywillremainthesame.Thewavelengthincreases.d.theamplitudeofthewave44.MoviesInascience-fictionmovie,asatelliteAmplitudewillremainthesame.blowsup.Thecrewofanearbyshipimmedi-atelyhearsandseestheexplosion.Ifyouhad48.Thespeedofsoundincreaseswithtempera-beenhiredasanadvisor,whattwophysicsture.Wouldthepitchofaclosedpipeerrorswouldyouhavenoticedandcorrected?increaseordecreasewhenthetemperatureoftheairrises?AssumethatthelengthofFirst,ifyouhadheardasound,youthepipedoesnotchange.wouldhavehearditafteryousawtheexplosion.Soundwavestravelmuch4landvfsov4fl.Ifvmoreslowlythanelectromagneticincreasesandlremainsunchanged,fwaves.Second,inspacethedensityofincreasesandpitchincreases.matterissosmallthatthesoundwavesdonotpropagate.Consequently,no49.MarchingBandsTwoflutistsaretuningup.soundshouldhavebeenheard.Iftheconductorhearsthebeatfrequencyincreasing,arethetwoflutefrequenciesget-45.TheRedshiftAstronomershaveobservedtingclosertogetherorfartherapart?thatthelightcomingfromdistantgalaxiesThefrequenciesaregettingfartherappearsredderthanlightcomingfromapart.nearergalaxies.WiththehelpofFigureCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.15-22,whichshowsthevisiblespectrum,50.MusicalInstrumentsAcoveredorganexplainwhyastronomersconcludethatdis-pipeplaysacertainnote.IfthecoveristantgalaxiesaremovingawayfromEarth.removedtomakeitanopenpipe,isthepitchincreasedordecreased?Thepitchisincreased;thefrequencyistwiceashighforanopenpipeasforaclosedpipe.4107m5107m6107m7107m■Figure15-2351.StringedInstrumentsOnaharp,RedlighthasalongerwavelengthandFigure15-23a,longstringsproducelowtherefore,alowerfrequencythanothernotesandshortstringsproducehighnotes.colors.TheDopplershiftoftheirlighttoOnaguitar,Figure15-23b,thestringsarelowerfrequenciesindicatesthatdistantallthesamelength.Howcantheyproducegalaxiesaremovingawayfromus.notesofdifferentpitches?46.Doesasoundof40dBhaveafactorof100(102)timesgreaterpressurevariationthanthethresholdofhearing,orafactorof40timesgreater?A40-dBsoundhassoundpressures100timesgreater.334SolutionsManualPhysics:PrinciplesandProblems

338Chapter15continued■Figure15-23■Figure15-24ThestringshavedifferenttensionsandThetotaldistancethesoundmustmassesperunitlength.Thinner,tightertravelis6.00m.stringsproducehighernotesthandodvthicker,looserstrings.td6.00msot0.0175sMasteringProblemsv343m/s15.1PropertiesandDetectionofSound57.Soundwithafrequencyof261.6Hztravelspages425–426throughwaterat25°C.Findthesound’sLevel1wavelengthinwater.Donotconfusesound52.Youhearthesoundofthefiringofadistantwavesmovingthroughwaterwithsurfacecannon5.0safterseeingtheflash.Howfarwavesmovingthroughwater.areyoufromthecannon?v1493m/s5.707mdvt(343m/s)(5.0s)1.7kmf261.6Hz58.Ifthewavelengthofa4.40102-Hzsound53.Ifyoushoutacrossacanyonandhearaninfreshwateris3.30m,whatisthespeedofecho3.0slater,howwideisthecanyon?soundinfreshwater?dvt(343m/s)(3.0s)isthetotaldis-2Hz)vf(3.30m)(4.4010tancetraveled.Thedistancetothewall11.45103m/sis(343m/s)(3.0s)5.1102m259.Soundwithafrequencyof442Hztravels54.Asoundwavehasafrequencyof4700Hzthroughanironbeam.Findthewavelengthandtravelsalongasteelrod.Ifthedistanceofthesoundiniron.betweencompressions,orregionsofhighv5130m/spressure,is1.1m,whatisthespeedofthe11.6mf442Hzwave?60.AircraftAdam,anairportemployee,isvf(1.1m)(4700Hz)5200m/sworkingnearajetplanetakingoff.He55.BatsThesoundemittedbybatshasaexperiencesasoundlevelof150dB.wavelengthof3.5mm.Whatisthesound’sa.IfAdamwearsearprotectorsthatreducefrequencyinair?thesoundleveltothatofatypicalv343m/s4Hzrockconcert,whatdecreaseindBisf9.8100.0035mprovided?56.PhotographyAsshowninFigure15-24,Atypicalrockconcertis110dB,sosomecamerasdeterminethedistancetothe40dBreductionisneeded.subjectbysendingoutasoundwaveandb.IfAdamthenhearssomethingthatmeasuringthetimeneededfortheechotosoundslikeabarelyaudiblewhisper,returntothecamera.Howlongwoulditwhatwillapersonnotwearingtheeartakethesoundwavetoreturntosuchaprotectorshear?cameraifthesubjectwere3.00maway?Abarelyaudiblewhisperis10dB,sotheactuallevelwouldbe50dB,orthatofanaverageclassroom.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.61.RockMusicArockbandplaysatan80-dBsoundlevel.Howmanytimesgreateristhesoundpressurefromanotherrockbandplay-ingateachofthefollowingsoundlevels?3.00mPhysics:PrinciplesandProblemsSolutionsManual335

339Chapter15continueda.100dBLevel2Each20dBincreasespressurebya65.MedicalImagingUltrasoundwithafactorof10,so10timesgreaterfrequencyof4.25MHzcanbeusedtopressure.produceimagesofthehumanbody.Ifthespeedofsoundinthebodyisthesameasb.120dBinsaltwater,1.50km/s,whatisthelength(10)(10)100timesgreaterpressureofa4.25-MHzpressurewaveinthebody?v1.50103m/s4m62.Acoiled-springtoyisshakenatafrequency3.5310f4.25106Hzof4.0Hzsuchthatstandingwavesareobservedwithawavelengthof0.50m.What0.353mmisthespeedofpropagationofthewave?66.SonarAshipsurveyingtheoceanbottomvf(0.50m)(4.0s1)2.0m/ssendssonarwavesstraightdownintotheseawaterfromthesurface.Asillustratedin63.AbaseballfanonawarmsummerdayFigure15-26,thefirstreflection,offofthe(30°C)sitsinthebleachers152mawaymudattheseafloor,isreceived1.74safterfromhomeplate.itwassent.Thesecondreflection,fromthea.Whatisthespeedofsoundinairatbedrockbeneaththemud,returnsafter30°C?2.36s.TheseawaterisatatemperatureofThespeedincreases0.6m/sper°C,25°C,andthespeedofsoundinmudissotheincreasefrom20°Cto30°Cis1875m/s.6m/s.Thus,thespeedis3436349m/s.b.Howlongafterseeingtheballhitthebatdoesthefanhearthecrackofthebat?d152mt1.74st2.36st0.436sv349mSeawaterCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.64.Onadaywhenthetemperatureis15°C,apersonstandssomedistance,d,asshowninFigure15-25,fromacliffandclapshisMudhands.Theechoreturnsin2.0s.Howfarawayisthecliff?Bedrock■Figure15-26a.Howdeepisthewater?Thespeedofsoundintheseawateris1533m/sandthetimeforaone-waytripis0.87s,sodwvtw(1533m/s)(0.87s)1300mb.Howthickisthemud?dTheround-triptimeinthemudis■Figure15-252.36s1.74s0.62sAt15°C,thespeedofsoundis3m/sslowerthanat20°C.Thus,thespeedofTheone-waytimeinthemudis0.31s,soundis340m/s.sodmvtm(1875m/s)(0.31s)v340m/sand2t2.0s580mdvt(340m/s)(1.0s)3.4102m336SolutionsManualPhysics:PrinciplesandProblems

340Chapter15continued67.Determinethevariationinsoundpressure(305Hz)(343m/s0)ofaconversationbeingheldatasound343m/s(31.0m/s)levelof60dB.2.80102HzThepressurevariationat0dBisb.atrainmovingawayfromthefirsttrain2105Pa.Forevery20-dBincrease,theataspeedof21.0m/spressurevariationincreasesbyafactorvvfdof10.Therefore,60dBhasapressuredfsvvsvariationamplitudeof2102Pa.(305Hz)(343m/s21.0m/s)343m/s(31.0m/s)68.Afiretruckismovingat35m/s,andacarinfrontofthetruckismovinginthesame2.63102Hzdirectionat15m/s.Ifa327-Hzsirenblaresfromthetruck,whatfrequencyisheardby15.2ThePhysicsofMusicthedriverofthecar?pages426–427vs35m/s,v343m/s,vd15m/s,Level1fs327Hz71.Averticaltubewithatapatthebaseisvvdfilledwithwater,andatuningforkvibratesfdfsvvsoveritsmouth.Asthewaterlevelislowered34315inthetube,resonanceisheardwhenthe(327Hz)350Hz34335waterlevelhasdropped17cm,andagainafter49cmofdistanceexistsfromthewaterLevel3tothetopofthetube.Whatisthefrequency69.Atrainmovingtowardasounddetectorofthetuningfork?at31.0m/sblowsa305-Hzwhistle.What49cm17cm32cmor0.32mfrequencyisdetectedoneachofthe1existsbetweenpointsofresonancefollowing?2a.astationarytrain10.32mvv2fddfsvvs0.64m(305Hz)(343m/s0)fv343m/s540Hz343m/s31.0m/s0.64m335Hz72.HumanHearingTheauditorycanallead-b.atrainmovingtowardthefirsttrainatingtotheeardrumisaclosedpipethatis21.0m/s3.0cmlong.Findtheapproximatevalue(ignoringendcorrection)ofthelowestvvfddfsvvresonancefrequency.s(305Hz)(343m/s(21.0m/s))L4343m/s31.0m/sv356Hzfv343m/s70.Thetraininthepreviousproblemismov-f2.9kHz4L(4)(0.030m)ingawayfromthedetector.WhatfrequencyCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.isnowdetectedoneachofthefollowing?73.Ifyouholda1.2-maluminumrodinthecenterandhitoneendwithahammer,ita.astationarytrainwilloscillatelikeanopenpipe.Antinodesvvfdofpressurecorrespondtonodesofmolecu-dfsvvslarmotion,sothereisapressureantinodeinthecenterofthebar.ThespeedofsoundPhysics:PrinciplesandProblemsSolutionsManual337

341Chapter15continuedinaluminumis5150m/s.Whatwouldbe78.MusicalInstrumentsThelowestnoteonthebar’slowestfrequencyofoscillation?anorganis16.4Hz.1a.WhatistheshortestopenorganpipeTherodlengthis,so2.4m2thatwillresonateatthisfrequency?fv5150m/s2.1kHzv343m/s2.4mf16.4Hz20.9mand74.Onetuningforkhasa445-Hzpitch.WhenL,soasecondforkisstruck,beatnotesoccur2withafrequencyof3Hz.Whatarethetwov343m/sL10.5m2f(2)(16.4Hz)possiblefrequenciesofthesecondfork?b.Whatisthepitchifthesameorganpipe445Hz3Hz442Hzisclosed?and445Hz3Hz448HzSinceaclosedpipeproducesafun-75.FlutesAfluteactsasanopenpipe.Ifadamentalwithawavelengthtwiceasflutesoundsanotewitha370-Hzpitch,longasthatofanopenpipeofthewhatarethefrequenciesofthesecond,samelength,thefrequencywouldbe1third,andfourthharmonicsofthispitch?(16.4Hz)8.20Hz.2f22f1(2)(370Hz)740Hz79.MusicalInstrumentsTwoinstrumentsarefplayingmusicalA(440.0Hz).Abeatnote33f1(3)(370Hz)1110Hzwithafrequencyof2.5Hzisheard.1100HzAssumingthatoneinstrumentisplayingf44f1(4)(370Hz)1480Hzthecorrectpitch,whatisthefrequencyof1500Hzthepitchplayedbythesecondinstrument?Itcouldbeeither440.02.5442.5Hz76.ClarinetsAclarinetsoundsthesamenote,or440.02.5437.5Hz.withapitchof370Hz,asinthepreviousproblem.Theclarinet,however,actsasa80.Aflexible,corrugated,plastictube,shownCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.closedpipe.WhatarethefrequenciesoftheinFigure15-27,is0.85mlong.Whenitislowestthreeharmonicsproducedbythisswungaround,itcreatesatonethatistheinstrument?lowestpitchforanopenpipeofthislength.3fWhatisthefrequency?1(3)(370Hz)1110Hz1100Hz0.85m5f1(5)(370Hz)1850Hz1800Hz7f1(7)(370Hz)2590Hz2600Hz■Figure15-2777.StringInstrumentsAguitarstringisL0.85m,so1.7m265.0cmlongandistunedtoproduceav343m/s2Hzlowestfrequencyof196Hz.f2.0101.7ma.Whatisthespeedofthewaveonthe81.Thetubefromthepreviousproblemisstring?swungfaster,producingahigherpitch.12L(2)(0.650m)1.30mWhatisthenewfrequency?vf(1.30m)(196Hz)255m/sf2Hz)4.0102Hz22f1(2)(2.010b.Whatarethenexttwohigherresonantfrequenciesforthisstring?Level2f22f1(2)(196Hz)392Hz82.Duringnormalconversation,theamplitudefofapressurewaveis0.020Pa.33f1(3)(196Hz)588Hz338SolutionsManualPhysics:PrinciplesandProblems

342Chapter15continueda.Iftheareaofaneardrumis0.52cm2,Thelengthofthesteelfingerclampedwhatistheforceontheeardrum?atoneendandfreetovibrateatthe1FPAotheriswavelength.Therefore,4(0.020N/m2)(0.52104m2)4L4(0.024m)0.096m,and1.0106Nvf(1760Hz)(0.096m)b.Themechanicaladvantageofthethree1.7102m/sbonesinthemiddleearis1.5.IftheforceinpartaistransmittedundiminishedtoMixedReviewthebones,whatforcedothebonesexertpages427–428ontheovalwindow,themembranetoLevel1whichthethirdboneisattached?85.Anopenorganpipeis1.65mlong.WhatFMArsoFfundamentalfrequencynotewillitproduceFr(MA)(Fe)eifitisplayedinheliumat0°C?F6N)1.5106Nr(1.5)(1.010Anopenpipehasalengthequaltoc.Theareaoftheovalwindowisone-halfitsfundamentalwavelength.0.026cm2.WhatisthepressureincreaseTherefore,3.30m.Thespeedoftransmittedtotheliquidinthecochlea?soundinheliumis972m/s.Therefore,v972m/sF1.5106Nf295HzP0.58Pa3.30mA0.026104m286.Ifyoudropastoneintoawellthatis83.MusicalInstrumentsOneopenorgan122.5mdeep,asillustratedinFigurepipehasalengthof836mm.Asecond15-29,howsoonafteryoudropthestoneopenpipeshouldhaveapitchthatisonewillyouhearithitthebottomofthewell?majorthirdhigher.Howlongshouldthesecondpipebe?vL,so2Land2fv343m/sf205Hz2L(2)(0.836m)Theratioofafrequencyonemajorthird122.5m5higheris5:4,so(205Hz)256Hz.4Thelengthofthesecondpipeisv343m/s2mmL6.70102f(2)(256Hz)84.AsshowninFigure15-28,amusicbox■Figure15-29containsasetofsteelfingersclampedatFirstfindthetimeittakesthestonetooneendandpluckedontheotherendbyfalldowntheshaftbyd1gt2,so2pinsonarotatingdrum.Whatisthespeedofawaveonafingerthatis2.4cmlongd122.5mandplaysanoteof1760Hz?t115.00sg(9.80m/s2)22ThetimeittakesthesoundtocomesCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Steelfingersbackupisfoundwithdvst,sod122.5mt0.357sv343m/ssThetotaltimeis5.00s0.357s5.36s.■Figure15-28Physics:PrinciplesandProblemsSolutionsManual339

343Chapter15continued87.Abirdonanewlydiscoveredplanetfliesfs22.4kHz,vd0towardasurprisedastronautataspeedofvv1533fd19.5m/swhilesingingatapitchof945Hz.dfsvv(22.4kHz)15334.15sTheastronauthearsatoneof985Hz.What22.3kHzisthespeedofsoundintheatmosphereofthisplanet?90.Whenawetfingerisrubbedaroundtherimfd985Hz,fs945Hz,vs19.5m/s,ofaglass,aloudtoneoffrequency2100Hzv?isproduced.Iftheglasshasadiameteroffdv16.2cmandthevibrationcontainsonefsvvsvswavelengtharounditsrim,whatisthe1vspeedofthewaveintheglass?vfSos1s,Thewavelengthisequaltothevfdcircumferenceoftheglassrim,dvs19.5m/sTherefore,thespeedisorvfs945Hz11vfdff985Hzd4.80102m/s(0.062m)(2100Hz)4.1102m/s88.InNorthAmerica,oneofthehottestout-91.HistoryofScienceIn1845,Dutchscien-doortemperatureseverrecordedis57°CtistChristophBuys-Ballotdevelopedatestandoneofthecoldestis62°C.WhatareoftheDopplereffect.Hehadatrumpetthespeedsofsoundatthosetwotempera-playersoundanAnoteat440Hzwhiletures?ridingonaflatcarpulledbyalocomotive.Atthesametime,astationarytrumpeterv(T)v(0°C)(0.6m/s)T,whereplayedthesamenote.Buys-Ballotheardv(0°C)331m/s.So,v(57°C)3.0beatspersecond.Howfastwasthetrain0.6m/smovingtowardhim?(331m/s)(57°C)°Cfd440Hz3.0Hz443HzCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.365m/s0.6m/svvdv(62°C)(331m/s)°C(62°C)fdfsvvs294m/sso(vvs)fd(vvd)fsand(vvLevel2vd)fssvf89.Aship’ssonarusesafrequencyof22.5kHz.dThespeedofsoundinseawateris1533m/s.(343m/s0)(440Hz)343m/sWhatisthefrequencyreceivedontheship443Hzthatwasreflectedfromawhaletravelingat2.3m/s4.15m/sawayfromtheship?Assumethattheshipisatrest.92.YoutrytorepeatBuys-Ballot’sexperimentPart1.Fromshiptowhale:fromthepreviousproblem.Youplantohaveatrumpetplayedinarapidlymovingvd4.15m/s,v1533m/s,car.Ratherthanlisteningforbeatnotes,fs22.5kHz,vs0however,youwanttohavethecarmovefastvvd15334.15enoughsothatthemovingtrumpetsoundsfdfsvv(22.5kHz)s1533onemajorthirdaboveastationarytrumpet.22.4kHza.Howfastwouldthecarhavetomove?Part2.Fromwhaletoship:5majorthirdratiov4s4.15m/s,v1533m/s,340SolutionsManualPhysics:PrinciplesandProblems

344Chapter15continuedvvd367HzfdfsvvsPart2.vd37.5m/s,v343m/s,so(vvs)fd(vvd)fsfs367Hz(vvd)fsvvd343(37.5)andvsvffdfsvv(367Hz)343ds4343m/s(343m/s0)5407Hz68.6m/sb.Shouldyoutrytheexperiment?Explain.ThinkingCritically3600s1mipage428v(68.6m/s)95.MakeandUseGraphsThewavelengthsof1h1609m153mph,thesoundwavesproducedbyasetoftun-ingforkswithgivenfrequenciesareshownsothecarwouldbemovinginTable15-2below.dangerouslyfast.Donottrytheexperiment.Table15-2TuningForksLevel3Frequency(Hz)Wavelength(m)93.GuitarStringsTheequationforthespeed1312.62FofawaveonastringisvT,whereFTis1472.33thetensioninthestringandisthemass1652.08perunitlengthofthestring.Aguitarstring1961.75hasamassof3.2gandis65cmlong.2201.56Whatmustbethetensioninthestring2471.39toproduceanotewhosefundamentalfrequencyis147Hz?a.Plotagraphofthewavelengthversus0.0032kg4.9103kg/mthefrequency(controlledvariable).0.65mWhattypeofrelationshipdoesthe2L2(0.65m)1.30mgraphshow?vf(147Hz)(1.30m)191m/sThegraphshowsaninverserela-F2(191m/s)2(4.9103kg/m)tionshipbetweenfrequencyandTvwavelength.180N94.Atrainspeedingtowardatunnelat2.6037.5m/ssoundsitshornat327Hz.Thesoundbouncesoffthetunnelmouth.What2.20isthefrequencyofthereflectedsoundheardonthetrain?Hint:Solvetheproblem(m)1.80intwoparts.First,assumethatthetunnelisastationaryobserverandfindthe1.40frequency.Then,assumethatthetunnelisastationarysourceandfindthefrequency1.00Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.measuredonthetrain.100140180240260Part1.vs37.5m/s,v343m/s,f(Hz)fs327Hzb.Plotagraphofthewavelengthversusfvvd(327Hz)343theinverseofthefrequency(1/f).Whatdfsvv34337.5skindofgraphisthis?DeterminethePhysics:PrinciplesandProblemsSolutionsManual341

345Chapter15continuedspeedofsoundfromthisgraph.frequencythanlightfromtherightside.ThegraphshowsadirectrelationshipWhatdothesemeasurementstellyouaboutbetweenperiod(1/f)andwavelength.theSun’smotion?ThespeedofsoundisrepresentedTheSunmustberotatingonitsaxisinbytheslope,~343m/s.thesamemannerasEarth.TheDopplershiftindicatesthattheleftsideofthe2.60Suniscomingtowardus,whiletherightsideismovingaway.2.2099.DesignanExperimentDesignanexperi-(m)mentthatcouldtesttheformulaforthe1.80speedofawaveonastring.Explainwhatmeasurementsyouwouldmake,howyou1.40wouldmakethem,andhowyouwouldusethemtotesttheformula.1.00Measurethemassandlengthofthe4.06.08.01stringtodetermine.Thenclampthe–(ms)fstringtoatable,hangoneendoverthe96.MakeGraphsSupposethatthefrequencytableedge,andstretchthestringbyofacarhornis300Hzwhenitisstation-hangingweightsonitsendtoobtainFT.ary.WhatwouldthegraphofthefrequencyCalculatethespeedofthewaveusingversustimelooklikeasthecarapproachedtheformula.Next,pluckthestringinitsandthenmovedpastyou?Completeamiddleanddeterminethefrequencybyroughsketch.matchingittoafrequencygenerator,usingbeatstotunethegenerator.ThegraphshouldshowafairlysteadyMultiplythefrequencybytwicethestringfrequencyabove300Hzasitapproach-length,whichisequaltothewavelength,esandafairlysteadyfrequencybelowtoobtainthespeedfromthewaveequa-300Hzasitmovesaway.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tion.Comparetheresults.Repeatfordif-97.AnalyzeandConcludeDescribehowyouferenttensionsandotherstringswithcoulduseastopwatchtoestimatethespeeddifferentmassesperunitlength.ofsoundifyouwerenearthegreenona200-Considerpossiblecausesoferror.mgolfholeasanothergroupofgolfershittheirteeshots.WouldyourestimateoftheWritinginPhysicsspeedofsoundbetoolargeortoosmall?page428Youcouldstartthewatchwhenyou100.Researchtheconstructionofamusicalsawthehitandstopthewatchwheninstrument,suchasaviolinorFrenchthesoundreachedyou.Thespeedhorn.Whatfactorsmustbeconsideredwouldbecalculatedbydividingthedis-besidesthelengthofthestringsortube?tance,200m,bythemeasuredtime.TheWhatisthedifferencebetweenaqualitymeasuredtimewouldbetoolargeinstrumentandacheaperone?Howarebecauseyoucouldanticipatethetheytestedfortonequality?impactbysight,butyoucouldnotAnswerswillvary.Areportonviolinanticipatethesound.Thecalculatedconstructionmightincludeinformationspeedwouldbetoosmall.aboutthebridgeasalinkbetweenthestringsandbodyandinformation98.ApplyConceptsAlightwavecomingfromabouttheroleofthebodyincausingapointontheleftedgeoftheSunisfoundairmoleculesaroundtheviolintobyastronomerstohaveaslightlyhighervibrate.Studentsalsomightdiscuss342SolutionsManualPhysics:PrinciplesandProblems

346Chapter15continuedthewaysinwhichthewoodsandfin-b.CalculatethemomentumandvelocityishesusedinmakingviolinsaffecttheofballBafterthecollision.qualityofthesoundproducedbytheHorizontal:mAvA1mBvB2instruments.SomBvB2(1.0kg)(3.0m/s)101.ResearchtheuseoftheDopplereffectin3.0kgm/sthestudyofastronomy.WhatisitsroleinVertical:0mthebigbangtheory?HowisitusedtoAvA2mBvB2detectplanetsaroundotherstars?TostudySomBvB2(1.0kg)(2.0m/s)themotionsofgalaxies?2.0kgm/sStudentsshoulddiscusstheworkofThevectorsumisEdwinHubble,theredshiftandthemexpandinguniverse,spectroscopy,andvthedetectionofwobblesinthemotion(3.0kgm/s)2(2.0kgm/s)2ofplanet-starsystems.2.0kgm/s3.6kgm/sandtan3.0kgm/sCumulativeReviewso34°page428Therefore,mBvB23.6kgm/sat102.BallA,rollingwestat3.0m/s,hasamassof1.0kg.BallBhasamassof2.0kgand34°northofwestisstationary.AftercollidingwithballB,3.6kgm/svballAmovessouthat2.0m/s.(Chapter9)B22.0kga.Sketchthesystem,showingtheveloci-1.8m/sat34°northofwesttiesandmomentabeforeandafterthecollision.103.Chriscarriesa10-Ncartonofmilkalongalevelhalltothekitchen,adistanceofWestwardandsouthwardare3.5m.HowmuchworkdoesChrisdo?positive.(Chapter10)BeforeBANowork,becausetheforceandthedisplacementareperpendicular.pB1=0vA1=3.0m/swest104.Amoviestuntpersonjumpsfromafive-pA1=3.0kgm/sweststorybuilding(22mhigh)ontoalargeAfterpillowatgroundlevel.Thepillowcush-Aionsherfallsothatshefeelsadecelerationofnomorethan3.0m/s2.Ifsheweighs480N,howmuchenergydoesthepillowhavetoabsorb?Howmuchforcedoesthepillowexertonher?(Chapter11)TheenergytobeabsorbedequalstheAmechanicalenergythatshehad,whichequalsherinitialpotentialenergy.Umgh(480N)(22m)11kJ.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.TheforceonherisvB2=?Fg480Np=?Fma(a)2(3.0m/s2)B2g9.80m/svA2=2.0m/ssouthpA2=2.0kgm/ssouth150NPhysics:PrinciplesandProblemsSolutionsManual343

347Chapter15continuedChallengeProblempage4171.Determinethetension,FT,inaviolinstringofmassmandlengthLthatwillplaythefundamentalnoteatthesamefrequencyasaclosedpipealsooflengthL.Expressyouranswerintermsofm,L,andthespeedofsoundinFair,v.TheequationforthespeedofawaveonastringisuT.whereFTisthetensionstringandisthemassperunitlengthofthestring.Thewavelengthofthefundamentalinaclosedpipeisequalto4L,sovthefrequencyisf.Thewavelengthofthefundamentalonastringis4Luequalto2L,sothefrequencyofthestringisf,whereuisthespeed2LFofthewaveonthestring,uT.Themassperunitlengthofthestringm/L.Squaringthefrequenciesandsettingthemequalgivesv2u2FTFTLFT16L24L24L24L2m4Lmmv2Finally,rearrangingforthetensiongivesFT4L2.Whatisthetensioninastringofmass1.0gand40.0cmlongthatplaysthesamenoteasaclosedpipeofthesamelength?Forastringofmass1.0gandlength0.400m,thetensionismv2(0.0010kg)(343m/s)2F74NT4L4(0.400m)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.344SolutionsManualPhysics:PrinciplesandProblems

348CHAPTER16FundamentalsofLight4.A64-cdpointsourceoflightis3.0mabovePracticeProblemsthesurfaceofadesk.Whatistheillumina-16.1Illuminationtiononthedesk’ssurfaceinlux?pages431–438P4(64cd)256lmpage436P256lmsoE7.1lx1.Alampismovedfrom30cmto90cm4d24(3.0m)2abovethepagesofabook.Comparethe5.Apublicschoollawrequiresaminimumilluminationonthebookbeforeandafterilluminanceof160lxatthesurfaceofeachthelampismoved.student’sdesk.Anarchitect’sspecificationsPE4d2d2callforclassroomlightstobelocated2.0mafterafterbeforeEPd2abovethedesks.Whatistheminimumbeforeafterluminousfluxthatthelightsmustproduce?4d2beforeP(30cm)21E2;therefore,after4d(90cm)29P4Ed2thelampismovedtheilluminationisone-ninthoftheoriginalillumination.4(160lm/m2)(2.0m)22.Whatistheilluminationonasurfacethatis8.0103lm3.0mbelowa150-Wincandescentlampthatemitsaluminousfluxof2275lm?6.Ascreenisplacedbetweentwolampssothattheyilluminatethescreenequally,asP2275lm1lxE4d24(3.0m)22.010showninFigure16-7.Thefirstlampemitsaluminousfluxof1445lmandis2.5m3.Drawagraphoftheilluminanceproducedfromthescreen.Whatisthedistanceofbya150-Wincandescentlampbetweenthesecondlampfromthescreenifthe0.50mand5.0m.luminousfluxis2375lm?Illuminanceofa150-WbulbScreenP2275,d0.50,0.75,…,5.0PE(d)4d2P2375lm2.5m720P1445lm■Figure16-7(Nottoscale)(lx)EP2275lmE1E2PPSo12Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.7.2d2d2120.505.0Pr(m)ord22d1P12375(2.5m)14453.2mPhysics:PrinciplesandProblemsSolutionsManual345

349Chapter16continuedd2SectionReviewf1(1.0m)2216.1Illumination1dpages431–438f2mpage4380.71m7.UseofMaterialLightPropertiesWhymightyouchooseawindowshadethatis11.DistanceofLightTravelHowfardoestranslucent?Opaque?lighttravelinthetimeittakessoundtoYouwoulduseatranslucentwindowtravel1cminairat20°C?shadetokeeppeoplefromlookinginorSoundvelocityis343m/s,sosoundout,whilestillallowingdaylightin.Youtakes3105stotravel1cm.Inthatwoulduseanopaquewindowshadetotime,lighttravels9km.keepthedaylightout.vsound343m/s8.IlluminanceDoesonelightbulbprovidedtsoundvmoreilluminancethantwoidenticallight-soundbulbsattwicethedistance?Explain.1102mOnelightbulbprovidesanilluminance343m/sthatisfourtimeslargerthantwoofthe3105ssamelightbulbattwicethedistance,Pv8m/sbecauseE.light3.0010d2dlightvlighttsound9.LuminousIntensityTwolampsilluminate(3.00108m/s)(3105s)ascreenequally—lampAat5.0m,lampBat3.0m.IflampAisrated75cd,whatis9103m9kmlampBrated?12.DistanceofLightTravelThedistancetoIE2theMooncanbefoundwiththehelpofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.dmirrorsleftontheMoonbyastronauts.ASincetheilluminationisequal,pulseoflightissenttotheMoonandE1E2returnstoEarthin2.562s.UsingtheIImeasuredvalueofthespeedoflighttotheSo12d2d2sameprecision,calculatethedistancefrom12EarthtotheMoon.I2orI1d22d2dct11(75cd)(3.0m)2(299,800,000m/s)2(2.562s)27cd(5.0m)23.840108m10.DistanceofaLightSourceSupposethat13.CriticalThinkingUsethecorrecttimealightbulbilluminatingyourdeskprovidestakenforlighttocrossEarth’sorbit,onlyhalftheilluminancethatitshould.If16.5min,andthediameterofEarth’sorbit,itiscurrently1.0maway,howfarshouldit2.981011m,tocalculatethespeedoflightbetoprovidethecorrectilluminance?usingRoemer’smethod.DoesthismethodIlluminationdependson1/d2,appeartobeaccurate?Whyorwhynot?Ed21soifd3.01011Efd22vit(16min)(60s/min)3.1108m/s346SolutionsManualPhysics:PrinciplesandProblems

350Chapter16continuedPracticeProblems(vobs)c16.2TheWaveNaturevofLightobs1cpages439–4477m)6.55106(4.861013.00108m/spage447m/s14.Whatisthefrequencyofoxygen’sspectrallineifitswavelengthis513nm?4.97107mcUsefandsolveforf.17.Anastronomerislookingatthespectrumofagalaxyandfindsthatithasanoxygencfspectrallineof525nm,whilethelaboratoryvalueismeasuredat513nm.Calculatehow3.00108m/s5.13107mfastthegalaxywouldbemovingrelativetoEarth.Explainwhetherthegalaxyismoving5.851014HztowardorawayfromEarthandhowyouknow.15.Ahydrogenatominagalaxymovingwith6m/sawayfromAssumethattherelativespeedalongaspeedof6.5510theaxisismuchlessthanthespeedofEarthemitslightwithafrequencyof14Hz.Whatfrequencyoflightlight.Thus,youcanusetheDoppler6.1610shiftequation.fromthathydrogenatomwouldbeobservedbyanastronomeronEarth?(vobs)cTherelativespeedalongtheaxisismuchTheobserved(apparent)wavelengthlessthanthespeedoflight.Thus,youappearstobelongerthantheknowncanusetheobservedlightfrequency(actual)wavelengthoftheoxygenequation.Becausetheastronomerandspectralline.Thismeansthatthethegalaxyaremovingawayfromeachastronomerandthegalaxyaremovingother,usethenegativeformoftheawayfromeachother.Sousetheposi-observedlightfrequencyequation.tiveformoftheDopplershiftequation.vfobsf1cv(obs)c14Hz)6.55106m/s(6.161013.00108m/sSolvefortheunknownvariable.6.031014Hz(obs)vc16.Ahydrogenatominagalaxymovingwith8m/s)(525nm513nm)16m/sawayfrom(3.0010513nmaspeedof6.5510Earthemitslightwithawavelengthof7.02106m/s4.86107m.WhatwavelengthwouldbeobservedonEarthfromthathydrogenatom?Therelativespeedalongtheaxisismuchlessthanthespeedoflight.Thus,youcanusetheobservedDopplershiftCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.equation.Becausetheastronomerandthegalaxyaremovingawayfromeachother,usethepositiveformoftheDopplershiftequation.Physics:PrinciplesandProblemsSolutionsManual347

351Chapter16continued22.PolarizationDescribeasimpleexperimentSectionReviewthatyoucoulddotodeterminewhether16.2TheWaveNaturesunglassesinastorearepolarizing.ofLightSeeiftheglassesreduceglarefromthepages439–447reflectivesurfaces,suchaswindowsorroadways.Polarizationoflightallowspage447photographerstophotographobjects18.AdditionofLightColorsWhatcolorofwhileeliminatingglare.lightmustbecombinedwithbluelighttoobtainwhitelight?23.CriticalThinkingAstronomershavedeter-yellow(amixtureoftheothertwopri-minedthatAndromeda,aneighboringmaries,redandgreen)galaxytoourowngalaxy,theMilkyWay,ismovingtowardtheMilkyWay.Explainhow19.CombinationofPigmentsWhatprimarytheydeterminedthis.CanyouthinkofapigmentcolorsmustbemixedtoproducepossiblereasonwhyAndromedaismovingred?Explainhowredresultsusingcolortowardourgalaxy?subtractionforpigmentcolors.ThespectrallinesoftheemissionsofYellowandmagentapigmentsareusedknownatomsareblue-shiftedinthetoproducered.Yellowpigmentsub-lightweseecomingfromAndromeda.tractsblueandmagentapigmentsub-Andromedawouldbemovingtowardustractsgreen,neithersubtractsredsoduetogravitationalattraction.Thisthemixturewouldreflectred.gravitationalattractioncouldbeduetothemassoftheMilkyWayorother20.LightandPigmentInteractionWhatobjectslocatedneartheMilkyWay.colorwillayellowbananaappeartobewhenilluminatedbyeachofthefollowing?a.whitelightChapterAssessmentyellowConceptMappingCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.greenandredlightpage452yellow24.Completethefollowingconceptmapusingc.bluelightthefollowingterms:wave,c,Dopplereffect,blackpolarization.21.WavePropertiesofLightThespeedofLightModelsredlightisslowerinairandwaterthaninavacuum.Thefrequency,however,doesnotchangewhenredlightenterswater.Doesthewavelengthchange?Ifso,how?raywaveYes,becausevfandv/f,whenvdecreases,sodoes.straightDopplerdiffractionceffectpathpolarization0c/f348SolutionsManualPhysics:PrinciplesandProblems

352Chapter16continuedMasteringConcepts31.WhatdidGalileoassumeaboutthespeedoflight?(16.1)page45225.Sounddoesnottravelthroughavacuum.Thespeedoflightisveryfast,butfinite.Howdoweknowthatlightdoes?(16.1)32.WhyisthediffractionofsoundwavesmoreLightcomesthroughthevacuumoffamiliarineverydayexperiencethanisthespacefromtheSun.diffractionoflightwaves?(16.2)26.DistinguishbetweenaluminoussourceandDiffractionismostpronouncedaroundanilluminatedsource.(16.1)obstaclesapproximatelythesamesizeasthewavelengthofthewave.WeareAluminousbodyemitslight.Anillumi-moreaccustomedtoobstaclesofthenatedbodyisoneonwhichlightfallssizethatdiffractthemuchlargerwave-andisreflected.lengthsofsound.27.Lookcarefullyatanordinary,frosted,33.Whatcoloroflighthastheshortestincandescentbulb.Isitaluminousoranwavelength?(16.2)illuminatedsource?(16.1)Violetlighthastheshortestwave-Itismainlyilluminated.Thefilamentlength.isluminous;thefrostedglassisillumi-nated.Youbarelycanseethehot34.Whatistherangeofthewavelengthsoffilamentthroughthefrostedglass.light,fromshortesttolongest?(16.2)28.Explainhowyoucanseeordinary,nonlu-400nmto700nmminousclassroomobjects.(16.1)35.Ofwhatcolorsdoeswhitelightconsist?Ordinarynonluminousobjectsareillu-(16.2)minatedbyreflectedlight,allowingWhitelightisacombinationofallthethemtobeseen.colors,oratleasttheprimarycolors.29.Distinguishamongtransparent,translucent,36.Whydoesanobjectappeartobeblack?andopaqueobjects.(16.1)(16.2)AtransparentobjectisamaterialAnobjectappearstobeblackbecausethroughwhichlightcanpasswithoutlittle,ifany,lightisbeingreflecteddistortion.Atranslucentobjectallowsfromit.lighttopassbutdistortsthelighttothepointwhereimagesarenotdiscernable.37.Canlongitudinalwavesbepolarized?AnopaqueobjectdoesnotallowlightExplain.(16.2)topassthrough.No.Theyhavenoverticalorhorizontal30.Towhatistheilluminationofasurfacebyacomponents.lightsourcedirectlyproportional?Towhatisitinverselyproportional?(16.1)38.Ifadistantgalaxyweretoemitaspectrallineinthegreenregionofthelightspec-Theilluminationonasurfaceisdirectlytrum,wouldtheobservedwavelengthonproportionaltotheintensityoftheEarthshifttowardredlightortowardbluesourceandinverselyproportionaltotheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.light?Explain.(16.2)squareofthedistanceofthesurfacefromthesource.Becausethegalaxyisdistant,itismostlikelymovingawayfromEarth.Thewavelengthactuallywouldshiftawayfromthewavelengthofgreenlighttowardalongerredwavelength.IfitPhysics:PrinciplesandProblemsSolutionsManual349

353Chapter16continuedshiftedtowardthebluewavelength,theobservingorphotographingobjects.wavelengthwouldbeshorter,notlonger.Thiswouldindicatethegalaxyis43.EyeSensitivityTheeyeismostsensitivetogettingclosertoEarth,andnogalaxyyellow-greenlight.ItssensitivitytoredandoutsidetheLocalGrouphasbeendis-bluelightislessthan10percentasgreat.coveredmovingtowardus.Basedonthisknowledge,whatcolorwouldyourecommendthatfiretrucksandambu-39.Whathappenstothewavelengthoflightaslancesbepainted?Why?thefrequencyincreases?(16.2)Firetrucksshouldbepaintedyellow-Asthefrequencyincreases,thewave-green,550nm,becauselesslighthastolengthdecreases.bereflectedtotheeyeforthefiretrucktobeseen.ApplyingConcepts44.StreetlightColorSomeveryefficientpage452streetlightscontainsodiumvaporunder40.Apointsourceoflightis2.0mfromscreenhighpressure.TheyproducelightthatisAand4.0mfromscreenB,asshowninmainlyyellowwithsomered.ShouldaFigure16-21.HowdoestheilluminanceatcommunitythathastheselightsbuydarkscreenBcomparewiththeilluminanceatbluepolicecars?Whyorwhynot?screenA?BluepigmentofapolicecarwillabsorbScreenAScreenBtheredandyellowlight.Darkbluepolicecarswouldnotbeveryvisible.IfSourceacommunitywantsitspolicecarstobevisible,theyshouldbuyyellowcars.2m4mRefertoFigure16-22forproblems45and46.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.■Figure16-21IlluminationEl/r2;therefore,theillu-minationatscreenBwillbeone-fourthofthatatscreenAbecauseitistwiceasfarfromthesource.■Figure16-2241.ReadingLampYouhaveasmallreadinglampthatis35cmfromthepagesofa45.Whathappenstotheilluminanceatabook.Youdecidetodoublethedistance.bookasthelampismovedfartherawaya.Istheilluminanceatthebookthesame?fromthebook?No.Illuminancedecreases,asdescribedbyb.Ifnot,howmuchmoreorlessisit?theinverse-squarelaw.Distanceisdoubled,sotheillumina-46.Whathappenstotheluminousintensityoftionofthepageisone-fourthasthelampasitismovedfartherawayfromgreat.thebook?42.WhyaretheinsidesofbinocularsandNochange;distancedoesnotaffectthecameraspaintedblack?luminousintensityofthelamp.Theinsidesarepaintedblackbecauseblackdoesnotreflectanylight,andthusthereisnointerferencewhile350SolutionsManualPhysics:PrinciplesandProblems

354Chapter16continued47.PolarizedPicturesPhotographersoften52.TrafficViolationSupposethatyouareaputpolarizingfiltersovercameralensestotrafficofficerandyoustopadriverformakecloudsintheskymorevisible.Thegoingthrougharedlight.Furthersupposecloudsremainwhite,whiletheskylooksthatthedriverdrawsapictureforyoudarker.Explainthisbasedonyourknowl-(Figure16-23)andexplainsthatthelightedgeofpolarizedlight.lookedgreenbecauseoftheDopplereffectLightscatteredfromtheatmosphereiswhenhewentthroughit.Explaintohimpolarized,butlightscatteredfromtheusingtheDopplershiftequationjusthowcloudsisnot.Byrotatingthefilter,thefasthewouldhavehadtobegoingforthephotographercanreducetheamountofredlight(645nm),toappeargreen(polarizedlightreachingthefilm.545nm).Hint:Forthepurposeofthisprob-lem,assumethattheDopplershiftequationis48.Anappleisredbecauseitreflectsredlightvalidatthisspeed.andabsorbsblueandgreenlight.a.Whydoesredcellophanelookredinreflectedlight?Cellophanereflectsredlightandabsorbsblueandgreenlight.b.Whydoesredcellophanemakeawhitelightbulblookredwhenyouholdthe■Figure16-23cellophanebetweenyoureyeandthelightbulb?645nm545nm(3.00108m/s)Cellophanetransmitsredlight.645nm4.65107m/sc.Whathappenstotheblueandgreenlight?Thatisover100millionmph.IfhedoesBlueandgreenlightareabsorbed.notgetaticketforrunningaredlight,hewillgetaticketforspeeding.49.YouputapieceofredcellophaneoveroneflashlightandapieceofgreencellophaneMasteringProblemsoveranother.Youshinethelightbeamson16.1Illuminationawhitewall.Whatcolorwillyouseewherepage453thetwoflashlightbeamsoverlap?Level1yellow53.Findtheillumination4.0mbelowa405-lmlamp.50.Younowputboththeredandgreencello-P405lmphanepiecesoveroneoftheflashlightsinE24(4.0m)22.0lx4dProblem49.Ifyoushinetheflashlightbeamonawhitewall,whatcolorwillyou54.Lighttakes1.28stotravelfromtheMoontosee?Explain.Earth.Whatisthedistancebetweenthem?Black;almostnolightwouldgetdvt(3.00108m/s)(1.28s)throughbecausethelighttransmitted3.84108mthroughthefirstfilterwouldbeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.absorbedbythesecond.Level255.Athree-waybulbuses50,100,or150W51.Ifyouhaveyellow,cyan,andmagentapig-ofelectricpowertodeliver665,1620,orments,howcanyoumakeabluepigment?2285lminitsthreesettings.ThebulbisExplain.placed80cmaboveasheetofpaper.IfanMixcyanandmagenta.Physics:PrinciplesandProblemsSolutionsManual351

355Chapter16continuedilluminationofatleast175lxisneededonPPSo12thepaper,whatistheminimumsettingthatd2d212shouldbeused?P2orP1d2EP2d24d21P4Ed24(175lx)(0.80m)2(1750lm)(1.08m)2(1.25m)21.4103lm1.31103lmThus,the100-W(1620-lm)settingisneeded.58.Supposethatyouwantedtomeasurethespeedoflightbyputtingamirroronadis-56.Earth’sSpeedOleRoemerfoundthatthetantmountain,settingoffacameraflash,averageincreaseddelayinthedisappear-andmeasuringthetimeittakestheflashtoanceofIofromoneorbitaroundJupitertoreflectoffthemirrorandreturntoyou,asthenextis13s.showninFigure16-24.Withoutinstru-a.Howfardoeslighttravelin13s?ments,apersoncandetectatimeintervalof3.9109mabout0.10s.Howmanykilometersawaywouldthemirrorhavetobe?Comparethisb.EachorbitofIotakes42.5h.Earthtrav-distancewiththatofsomeknowndistances.elsthedistancecalculatedinpartain42.5h.FindthespeedofEarthinkm/s.dMirrorYouvtd3.9109m1km5s1000m1.531025km/sc.Checktomakesurethatyouranswerforpartbisreasonable.CalculateEarth’sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.speedinorbitusingtheorbitalradius,1.5108km,andtheperiod,1.0yr.■Figure16-24d2(1.5108km)1dvdvtt365d24h1h8m/s)(0.1s)1km(3.00101000m3600s3.0101km/s,sofairlyaccurate3104kmThemirrorwouldbehalfthisdistance,Level3or15,000kmaway.Earthis40,000km57.Astudentwantstocomparetheluminousincircumference,sothisisthree-fluxofalightbulbwiththatofa1750-lmeighthsofthewayaroundEarth.lamp.Thelightbulbandthelampillumi-nateasheetofpaperequally.The1750-lmlampis1.25mawayfromthesheetof16.2TheWaveNatureofLightpaper;thelightbulbis1.08maway.Whatpages453–454isthelightbulb’sluminousflux?Level1P59.Convert700nm,thewavelengthofredE4d2light,tometers.Sincetheilluminationisequal,1109m7m(700nm)7101nmE1E2352SolutionsManualPhysics:PrinciplesandProblems

356Chapter16continued60.GalacticMotionHowfastisagalaxymov-Level2ingrelativetoEarthifahydrogenspectral62.PolarizingSunglassesInwhichdirectionlineof486nmisred-shiftedto491nm?shouldthetransmissionaxisofpolarizingAssumethattherelativespeedalongsunglassesbeorientedtocuttheglarefromtheaxisismuchlessthanthespeedofthesurfaceofaroad:verticallyorhorizon-light.Thus,youcanusetheDopplertally?Explain.shiftequation.Thetransmissionaxisshouldbevorientedvertically,sincethelight(app)creflectingofftheroadwillbepartiallyThelightisred-shifted,sotheastron-polarizedinthehorizontaldirection.omerandthegalaxyaremovingawayAverticaltransmissionaxiswillfilterfromeachother.Sousethepositivehorizontalwaves.formoftheapparentlightwavelength63.GalacticMotionAhydrogenspectrallineequation.thatisknowntobe434nmisred-shiftedv(app)cby6.50percentinlightcomingfromadis-tantgalaxy.HowfastisthegalaxymovingSolvefortheunknownvariable.awayfromEarth?(app)Assumethattherelativespeedalongvctheaxisismuchlessthanthespeedof8m/s)491nm486nmlight.Thus,youcanusetheDoppler(3.0010486nmshiftequation.3.09106m/sv(app)cTheoriginalassumptionwasvalid.Thelightisred-shifted,sotheastron-61.Supposethatyouarefacingdueeastatsun-omerandthegalaxyaremovingawayrise.Sunlightisreflectedoffthesurfaceofafromeachother.Sousethepositivelake,asshowninFigure16-25.Istheformoftheapparentlightwavelengthreflectedlightpolarized?Ifso,inwhatequation.direction?v(app)cSolvefortheunknownvariable.(app)vc(3.00108m/s)(1.065)(434nm)434nm434nm1.95107m/sTheoriginalassumptionwasvalid.■Figure16-25ThereflectedlightispartiallypolarizedLevel3inthedirectionparalleltothesurfaceof64.Foranyspectralline,whatwouldbeanthelakeandperpendiculartothepathCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.unrealisticvalueoftheapparentwavelengthoftravelofthelightfromthelaketoforagalaxymovingawayfromEarth?Why?youreyes.Anunrealisticvaluewouldmakethegalaxyseemtobemovingawayfromusataspeedclosetoorgreaterthanthespeedoflight,orvc.IfthiswerethePhysics:PrinciplesandProblemsSolutionsManual353

357Chapter16continuedcase,useofthelow-speedDopplershiftilluminatedandequationwouldgiveawavelengthE1E2cdifferenceof(app)c.WhenI1I2Sosolved,thiswouldgiveanapparentd2d212wavelengthofapp2.ItwouldbeI60.0cdtwiceaslargeastheactualwavelength.ord2(6.0m)2d1I10.0cd1Soanyapparentwavelengthclosetoorgreaterthantwicetheactualwave-15mlengthwouldbeunrealistic.68.ThunderandLightningExplainwhyitMixedReviewtakes5stohearthunderwhenlightningis1.6kmaway.page454Thetimeforlighttotravel1.6kmisaLevel1smallfractionofasecond(5.3s).The65.StreetlightIlluminationAstreetlightsoundtravelsabout340m/s,whichiscontainstwoidenticalbulbsthatare3.3maboutone-fifthofthe1.6kmeveryabovetheground.Ifthecommunitywantssecond,andtakesabout4.7stotraveltosaveelectricalenergybyremovingone1.6km.bulb,howfarfromthegroundshouldthestreetlightbepositionedtohavethesameLevel3illuminationonthegroundunderthelamp?69.SolarRotationBecausetheSunrotatesonPEitsaxis,oneedgeoftheSunmovestoward4d2Earthandtheothermovesaway.TheSunIfPisreducedbyafactorof2,sorotatesapproximatelyonceevery25days,mustd2.andthediameteroftheSunis1.4109m.Thus,disreducedbyafactorof2,HydrogenontheSunemitslightoffre-becomingquency6.161014Hzfromthetwosidesof(3.3m)theSun.WhatchangesinwavelengthareCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.3mobserved?2Speedofrotationisequaltocircumfer-66.Anoctaveinmusicisadoublingoffre-encetimesperiodofrotation.quency.Comparethenumberofoctaves(1.4109m)vthatcorrespondtothehumanhearingrot(25days)(24h/day)(3600s/h)rangetothenumberofoctavesinthe2.04103m/shumanvisionrange.cHumanshearoverarangeofaboutfnineortenoctaves(20Hzto10,240or3.00108m/s20,480Hz);however,humanvisionis6.161014Hzlessthanone“octave.”4.87107mLevel2v67.A10.0-cdpoint-sourcelampanda60.0-cdcpoint-sourcelampcastequalintensitiesonvrotawall.Ifthe10.0-cdlampis6.0mfromthecwall,howfarfromthewallisthe60.0-cd(2.04103m/s)7m)(4.8710lamp?(3.00108m/s)EIandsincetheintensitiesonthe3.31012md2wallareequal,thewallisequally354SolutionsManualPhysics:PrinciplesandProblems

358Chapter16continuedThinkingCritically74.LookupinformationontheSIunitcandela,cd,andexplaininyourownpage454wordsthestandardthatisusedtoset70.ResearchWhydidGalileo’smethodforthevalueof1cd.measuringthespeedoflightnotwork?Answerswillvary.Beginwiththeele-Itwasnotpreciseenough.Hewasnotmentthorium.Heatittothemeltingabletomeasurethesmalltimeintervalspointofplatinum.Atthistemperature,involvedinaterrestrialmeasurement.thethoriumwillglow.Surroundthetho-71.MakeandUseGraphsA110-cdlightriumwithanopaquematerialthatcansourceis1.0mfromascreen.Determinetakethehightemperature.Leaveantheilluminationonthescreenoriginallyopeningthatisone-sixtiethofasquareandforeverymeterofincreasingdistancecentimeterinsize.Thecandelaisupto7.0m.Graphthedata.definedastheamountofsteadyflowoflightenergythatisemittedbythe110thoriumthroughtheopeningunder100theseconditions.908070CumulativeReview60P1.4103lm(lm)page454E504075.A2.0-kgobjectisattachedtoa1.5-mlong30stringandswunginaverticalcircleata20constantspeedof12m/s.(Chapter7)10a.Whatisthetensioninthestringwhen0.01.02.03.04.05.06.07.0theobjectisatthebottomofitspath?r(m)mv2(2.0kg)(12m/s)2a.Whatistheshapeofthegraph?Fnetr1.5mhyperbola1.9101Nb.Whatistherelationshipbetweenillumi-nanceanddistanceshownbythegraph?F2)gmg(2.0kg)(9.80m/sinversesquare2.0102N72.AnalyzeandConcludeIfyouweretoFnetTFgdriveatsunsetinacityfilledwithbuildingsTFnetFgthathaveglass-coveredwalls,thesettingSunreflectedoffthebuilding’swallsmight1.9102N0.20102Ntemporarilyblindyou.Wouldpolarizing2.1102Nglassessolvethisproblem?b.WhatisthetensioninthestringwhenYes.Lightreflectedoffglassispartiallytheobjectisatthetopofitspath?polarized,sopolarizingsunglasseswillFreducemuchoftheglare,ifthesun-netTFgglassesarealignedcorrectly.TFnetFg1.9102N0.20102NWritinginPhysics2NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page4541.71073.Writeanessaydescribingthehistoryofhumanunderstandingofthespeedoflight.Includesignificantindividualsandthecon-tributionthateachindividualmade.Answerswillvary.Physics:PrinciplesandProblemsSolutionsManual355

359Chapter16continued76.Aspaceprobewithamassof7.600103kg1.Youobservethatsomelightpassesistravelingthroughspaceat125m/s.throughfilter2,thoughnolightpassedMissioncontroldecidesthatacoursecor-throughfilter2previoustoinsertingtherectionof30.0°isneededandinstructstheanalyzerfilter.Whydoesthishappen?probetofirerocketsperpendiculartoitsTheanalyzerfilterallowssomelighttopresentdirectionofmotion.Ifthegaspassthrough,sinceitspolarizingaxisexpelledbytherocketshasaspeedofisnotperpendiculartothepolarizing3.200km/s,whatmassofgasshouldbeaxisofthefirstfilter.Thelastpolarizingreleased?(Chapter9)filternowcanpasslightfromtheana-mlyzerfilter,sincethepolarizingaxisofgvgtan30.0°theanalyzerfilterisnotperpendicularmpvp1tothepolarizingaxisofthelastpolariz-mpvp1(tan30.0°)ingfilter.mgvg2.Theanalyzerfilterisplacedatanangleof(7.600103kg)(125m/s)(tan30.0°)relativetothepolarizingaxesoffilter1.3.200103m/sDeriveanequationfortheintensityoflight171kgcomingoutoffilter2comparedtotheintensityoflightcomingoutoffilter1.77.Whena60.0-cm-longguitarstringisIpluckedinthemiddle,itplaysanoteoffre-1isthelightintensityoutofthefirstfil-ter,Iquency440Hz.Whatisthespeedoftheanalyzeristhelightintensityoutoftheanalyzerfilter,andIwavesonthestring?(Chapter14)2isthelightintensityoutofthelastfilter.2L2(0.600m)1.20mI2vf(1.20m)(440Hz)530m/sanalyzerI1cosI2(90°)78.Whatisthewavelengthofasoundwave2Ianalyzercoswithafrequencyof17,000HzinwateratI2()cos2(90°)2I1cos25°C?(Chapter15)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.v1493m/sFilter1Polarized0.0878m8.8cmf17,000HzlightAnalyzerChallengeProblempage444YouplaceananalyzerfilterbetweenthetwoFilter290°cross-polarizedfilters,suchthatitspolarizingaxisisnotparalleltoeitherofthetwofilters,asshowninthefiguretotheright.Polarizingaxes356SolutionsManualPhysics:PrinciplesandProblems

360CHAPTER17ReflectionandMirrorsPracticeProblemsr1i130°i290°r117.1ReflectionfromPlane90°30°60°Mirrorspages457–463page460SectionReview1.Explainwhythereflectionoflightoff17.1ReflectionfromPlanegroundglasschangesfromdiffusetospecularifyouspillwateronit.Mirrorspages457–463Waterfillsintheroughareasandmakesthesurfacesmoother.page4636.ReflectionAlightraystrikesaflat,smooth,2.Iftheangleofincidenceofarayoflightisreflectingsurfaceatanangleof80°tothe42°,whatiseachofthefollowing?normal.Whatistheanglethatthereflecteda.theangleofreflectionraymakeswiththesurfaceofthemirror?ri42°rib.theangletheincidentraymakeswith80°themirrorr,mirror90°r90°80°i,mirror90°i90°42°48°c.theanglebetweentheincidentrayand10°thereflectedray7.LawofReflectionExplainhowthelawofir2i84°reflectionappliestodiffusereflection.3.IfalightrayreflectsoffaplanemirroratanThelawofreflectionappliestoindivid-angleof35°tothenormal,whatwastheualraysoflight.Roughsurfacesmakeangleofincidenceoftheray?thelightraysreflectinmanydifferentdirections.ir35°8.ReflectingSurfacesCategorizeeachofthe4.Lightfromalaserstrikesaplanemirroratfollowingasaspecularoradiffusereflect-anangleof38°tothenormal.Ifthelaseringsurface:paper,polishedmetal,windowismovedsothattheangleofincidenceglass,roughmetal,plasticmilkjug,smoothincreasesby13°,whatisthenewangleofwatersurface,andgroundglass.reflection?Specular:windowglass,smoothwater,ii,initial13°polishedmetal.Diffuse:paper,rough38°13°51°metal,groundglass,plasticmilkjug.ri51°9.ImagePropertiesA50-cm-talldogstandsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5.Twoplanemirrorsarepositionedatright3mfromaplanemirrorandlooksatitsanglestooneanother.Arayoflightstrikesimage.Whatistheimageposition,height,onemirroratanangleof30°totheandtype?normal.Itthenreflectstowardtheseconddmirror.Whatistheangleofreflectionofido3mthelightrayoffthesecondmirror?Physics:PrinciplesandProblemsSolutionsManual357

361Chapter17continuedh13.Anobjectis36.0cminfrontofaconcaveiho50cmmirrorwitha16.0-cmfocallength.Determinetheimageposition.Theimageisvirtual.111fdd10.ImageDiagramAcarisfollowinganotheroicardownastraightroad.Thefirstcarhasadforearwindowtilted45°.DrawaraydiagramdidfoshowingthepositionoftheSunthatwould(36.0cm)(16.0cm)causesunlighttoreflectintotheeyesofthe36.0cm16.0cmdriverofthesecondcar.28.8cmTheSun14.A3.0-cm-tallobjectis20.0cmfromaRearwindow16.0-cm-radiusconcavemirror.Determine45°offirstcartheimagepositionandimageheight.Driverof45°111secondcarfddoi45°RoadgradedfodidfTheSun’spositiondirectlyoverheadowouldlikelyreflectlightintothedriver’s16.0cm(20.0cm)eyes,accordingtothelawofreflection.213.3cm16.0cm20.0cm211.CriticalThinkingExplainhowdiffusereflectionoflightoffanobjectenablesyouhidimtoseeanobjectfromanyangle.hodoTheincominglightreflectsoffthesur-diho(13.3cm)(3.0cm)hfaceoftheobjectinalldirections.Thisid20.0cmoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.enablesyoutoviewtheobjectfromany2.0cmlocation.15.Aconcavemirrorhasa7.0-cmfocallength.A2.4-cm-tallobjectis16.0cmfromthePracticeProblemsmirror.Determinetheimageheight.17.2CurvedMirrors111fddpages464–473oipage469dofd12.Usearaydiagram,drawntoscale,tosolveidfoExampleProblem2.(16.0cm)(7.0cm)16.0cm7.0cmO1Ray112.4cmRay2hdiimHorizontalscale:CFhdoo1block1.0cmI1Verticalscale:diho3blocks1.0cmhido(12.4cm)(2.4cm)16.0cm1.9cm358SolutionsManualPhysics:PrinciplesandProblems

362Chapter17continued16.Anobjectisnearaconcavemirrorofdofd10.0-cmfocallength.Theimageis3.0cmidoftall,inverted,and16.0cmfromthemirror.(60.0cm)(13.0cm)Whataretheobjectpositionandobject60.0cm(13.0cm)height?10.7cm111fdodihidimhddfoodiodfi(10.7cm)m(16.0cm)(10.0cm)60.0cm16.0cm10.0cm0.17826.7cmhimho(0.178)(6.0cm)hdii1.1cmmhdoo19.Aconvexmirrorisneededtoproduceandhoihodimagethatisthree-fourthsthesizeofaniobjectandlocated24cmbehindthemir-(26.7cm)(3.0cm)ror.Whatfocallengthshouldbespecified?16.0cm1115.0cmfdidoddodiifandmpage472dodido17.Anobjectislocated20.0cminfrontofadconvexmirrorwitha15.0-cmfocalisodomlength.Findtheimagepositionusingbothascalediagramandthemirrorequation.di24cmandm0.75,so(24cm)O1Ray1do0.75I1Ray2F32cmHorizontalscale:di8.6cm(32cm)(24cm)f1block1.0cm32cm(24cm)96cm111ddfoi20.A7.6-cm-diameterballislocated22.0cmdoffromaconvexmirrorwitharadiusofcur-sodidfvatureof60.0cm.Whataretheball’simageopositionanddiameter?(20.0cm)(15.0cm)20.0cm(15.0cm)111fdd8.57cmoidfod18.AconvexmirrorhasafocallengthofidfoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13.0cm.Alightbulbwithadiameterof(22.0cm)(30.0cm)6.0cmisplaced60.0cmfromthemirror.22.0cm(30.0cm)Whatisthelightbulb’simagepositionanddiameter?12.7cm111ddfoiPhysics:PrinciplesandProblemsSolutionsManual359

363Chapter17continuedhdii111mhdfddoooidihodofhididfdoo(12.7cm)(7.6cm)(20.0cm)(9.0cm)22.0cm20.0cm9.0cm4.4cm16.4cmdi21.A1.8-m-tallgirlstands2.4mfromastore’smdosecuritymirror.Herimageappearstobe16.4cm0.36mtall.Whatisthefocallengthofthe20.0cmmirror?0.82hdiimhdoo24.ObjectPositionTheplacementofandhobjectinfrontofaconcavemirrorwithaoidihfocallengthof12.0cmformsarealimageothatis22.3cmfromthemirror.Whatisthe(2.4m)(0.36m)objectposition?1.8m1110.48mfddoi111fdodidifdodfidfido(22.3cm)(12.0cm)ddio22.3cm12.0cm(0.48m)(2.4m)26.0cm0.48m2.4mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.60m25.ImagePositionandHeightA3.0-cm-tallobjectisplaced22.0cminfrontofaconcavemirrorhavingafocallengthofSectionReview12.0cm.Findtheimagepositionandheightbydrawingaraydiagramtoscale.17.2CurvedMirrorsVerifyyouranswerusingthemirrorandpages464–473magnificationequations.page47322.ImagePropertiesIfyouknowthefocalhi3.6cmO1Ray1lengthofaconcavemirror,whereshoulddi26.4cmFyouplaceanobjectsothatitsimageisCRay2uprightandlargercomparedtotheobject?Willthisproducearealorvirtualimage?I1Horizontalscale:1block1.0cmYoushouldplacetheobjectbetweenVerticalscale:themirrorandthefocalpoint.The1block1.0cmimagewillbevirtual.11123.MagnificationAnobjectisplaced20.0cmfddoiinfrontofaconcavemirrorwithafocaldflengthof9.0cm.Whatisthemagnificationodidfoftheimage?o360SolutionsManualPhysics:PrinciplesandProblems

364Chapter17continued(22.0cm)(12.0cm)hidim22.0cm12.0cmhdoo26.4cmdhoidhdihiiomhodo(16.4cm)(2.8cm)dh6.0cmiohid7.7cmo111(26.4cm)(3.0cm)fdodi22.0cmdd3.6cmfoiddoi26.RayDiagramA4.0-cm-tallobjectis(7.7cm)(16.4cm)located14.0cmfromaconvexmirrorwith7.7cm16.4cmafocallengthof12.0cm.Drawascale14.5cmraydiagramshowingtheimagepositionr2fandheight.Verifyyouranswerusingthemirrorandmagnificationequations.(2)(14.5cm)29cmO1Ray1I28.FocalLengthAconvexmirrorisusedto1Ray2produceanimagethatistwo-thirdstheFsizeofanobjectandlocated12cmbehindHorizontalscale:1block1.0cmhi1.8cmthemirror.WhatisthefocallengthoftheVerticalscale:di6.5cmmirror?3blocks2.0cmhdiimhdoo111fdodididomdfodidf(12cm)o2(14.0cm)(12.0cm)314.0cm(12.0cm)18cm6.46cm111fddhdoiiimhodododifddoidhiohid(12cm)(18cm)o12cm18cm(6.46cm)(4.0cm)36cm14.0cmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.8cm29.CriticalThinkingWouldsphericalaberra-tionbelessforamirrorwhoseheight,27.RadiusofCurvatureA6.0-cm-tallobjectcomparedtoitsradiusofcurvature,isisplaced16.4cmfromaconvexmirror.Ifsmallorlarge?Explain.theimageoftheobjectis2.8cmtall,whatItwouldbelessforamirrorwhoseistheradiusofcurvatureofthemirror?heightisrelativelysmallcomparedtoPhysics:PrinciplesandProblemsSolutionsManual361

365Chapter17continueditsradiusofcurvature;diverginglight34.Describethepropertiesofaplanemirror.raysfromanobjectthatstrikethemir-(17.1)roraremoreparaxialsotheyconvergeAplanemirrorisaflat,smoothmorecloselytocreateanimagethatissurfacefromwhichlightisreflectedbynotblurred.specularreflection.Theimagescreatedbyplanemirrorsarevirtual,upright,andasfarbehindthemirrorastheChapterAssessmentobjectisinfrontofit.ConceptMapping35.Astudentbelievesthatverysensitivepage478photographicfilmcandetectavirtual30.Completethefollowingconceptmapusingimage.Thestudentputsphotographicfilmthefollowingterms:convex,upright,inverted,atthelocationofavirtualimage.Doesthisreal,virtual.attemptsucceed?Explain.(17.1)No,theraysdonotconvergeatavirtualMirrorsimage.Noimageformsandthestudentwouldnotgetapicture.Somevirtualimagesarebehindthemirror.planeconcaveconvex36.Howcanyouprovetosomeonethatanimageisarealimage?(17.1)virtualrealvirtualvirtualPlaceasheetofplainpaperorphoto-graphicfilmattheimagelocationanduprightinverteduprightuprightyoushouldbeabletofindtheimage.37.AnobjectproducesavirtualimageinaMasteringConceptsconcavemirror.Whereistheobjectlocated?page478(17.2)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.31.HowdoesspecularreflectiondifferfromObjectmustbelocatedbetweenFanddiffusereflection?(17.1)themirror.Whenparallellightisreflectedfromasmoothsurface,theraysarereflected38.Whatisthedefectthatallconcavesphericalparalleltoeachother.Theresultisanmirrorshaveandwhatcausesit?(17.2)imageoftheoriginoftherays.WhenRaysparalleltotheaxisthatstrikethelightisreflectedfromaroughsurface,itedgesofaconcavesphericalmirrorareisreflectedinmanydifferentdirections.notreflectedthroughthefocalpoint.Theraysarediffusedorscattered.NoThiseffectiscalledsphericalaberration.imageofthesourceresults.39.Whatistheequationrelatingthefocalpoint,32.Whatismeantbythephrase“normaltotheobjectposition,andimageposition?(17.2)surface”?(17.1)111anylinethatisperpendiculartothefddiosurfaceatanypoint40.Whatistherelationshipbetweenthe33.Whereistheimageproducedbyaplanecenterofcurvatureandthefocallengthmirrorlocated?(17.1)ofaconcavemirror?(17.2)TheimageisonalinethatisC2fperpendiculartothemirrorandthesamedistancebehindthemirrorastheobjectisinfrontofthemirror.362SolutionsManualPhysics:PrinciplesandProblems

366Chapter17continued41.Ifyouknowtheimagepositionandobject46.Locateanddescribethephysicalpropertiespositionrelativetoacurvedmirror,howcanoftheimageproducedbyaconcavemirroryoudeterminethemirror’smagnification?whentheobjectislocatedatthecenterof(17.2)curvature.ThemagnificationisequaltotheTheimagewillbeatC,thecenterofnegativeoftheimagedistancedividedcurvature,inverted,real,andthesamebytheobjectdistance.sizeastheobject.42.Whyareconvexmirrorsusedasrearview47.Anobjectislocatedbeyondthecenterofmirrors?(17.2)curvatureofasphericalconcavemirror.ConvexmirrorsareusedasrearviewLocateanddescribethephysicalpropertiesmirrorsbecausetheyallowforawideoftheimage.rangeofview,allowingthedrivertoseeTheimagewillbebetweenCandF,andamuchlargerareathanisaffordedbywillbeinverted,real,andsmallerthanordinarymirrors.theobject.43.Whyisitimpossibleforaconvexmirrorto48.TelescopeYouhavetoorderalargeformarealimage?(17.2)concavemirrorforatelescopethatThelightraysalwaysdiverge.produceshigh-qualityimages.ShouldyouorderasphericalmirrororaparabolicApplyingConceptsmirror?Explain.pages478–479Youshouldorderaparabolicmirrorto44.WetRoadAdryroadismoreofaeliminatesphericalaberrations.diffusereflectorthanawetroad.Based49.DescribethepropertiesoftheimageseeninonFigure17-16,explainwhyawetthesingleconvexmirrorinFigure17-17.roadappearsblackertoadriverthanadryroaddoes.WetasphaltDryasphalt■Figure17-16Lesslightisreflectedbacktothecarfromawetroad.45.BookPagesWhyisitdesirablethatthe■Figure17-17pagesofabookberoughratherthanTheimageinasingleconvexmirrorisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.smoothandglossy?alwaysvirtual,erect,smallerthantheThesmootherandglossierthepagesobject,andlocatedclosertothemirrorare,thelesserthediffusereflectionofthantheobject.lightandthegreatertheglarefromthepages.Physics:PrinciplesandProblemsSolutionsManual363

367Chapter17continued50.Listallthepossiblearrangementsinwhich55.Arayoflightincidentuponamirrormakesyoucoulduseasphericalmirror,eitheranangleof36°withthemirror.Whatistheconcaveorconvex,toformarealimage.anglebetweentheincidentrayandtheYoucanuseonlyaconcavemirrorwithreflectedray?theobjectbeyondthefocalpoint.Acon-i90°36°vexmirrorwillnotformarealimage.54°51.Listallpossiblearrangementsinwhichyouricoulduseasphericalmirror,eitherconcave54°orconvex,toformanimagethatissmallerircomparedtotheobject.54°54°Youmayuseaconcavemirrorwiththe108°objectbeyondthecenterofcurvatureoraconvexmirrorwiththeobjectLevel2anywhere.56.PictureinaMirrorPennywishestotakeapictureofherimageinaplanemirror,as52.RearviewMirrorsTheoutsiderearviewshowninFigure17-18.Ifthecameraismirrorsofcarsoftencarrythewarning1.2minfrontofthemirror,atwhatdis-“Objectsinthemirrorarecloserthantheytanceshouldthecameralensbefocused?appear.”Whatkindofmirrorsaretheseandwhatadvantagedotheyhave?Convexmirror;itprovidesawiderfieldofview.MasteringProblems17.1ReflectionfromPlaneMirrorspage479Level1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.53.Arayoflightstrikesamirroratanangleof38°tothenormal.Whatistheanglethatthereflectedanglemakeswiththenormal?ri38°54.Arayoflightstrikesamirroratanangleof53°tothenormal.■Figure17-18a.Whatistheangleofreflection?Theimageis1.2mbehindthemirror,sorithecameralensshouldbesetto2.4m.53°b.Whatistheanglebetweentheincident57.Twoadjacentplanemirrorsformarightrayandthereflectedray?angle,asshowninFigure17-19.Alightrayisincidentupononeofthemirrorsatanirangleof30°tothenormal.53°53°106°364SolutionsManualPhysics:PrinciplesandProblems

368Chapter17continuedthefeethitsthemirrorhalfwaybetweentheeyesandthefeet.Thedistancebetweenthepointthetworayshitthemirrorishalfthetotalheight.Level330°59.Twoplanemirrorsareconnectedattheirsidessothattheyforma45°anglebetweenthem.Alightraystrikesonemirroratan■Figure17-19angleof30°tothenormalandthenreflectsa.Whatistheangleatwhichthelightrayoffthesecondmirror.Calculatetheangleofisreflectedfromtheothermirror?reflectionofthelightrayoffthesecondmirror.Reflectionfromthefirstmirror:Reflectionfromthefirstmirrorisr1i130°Reflectionfromthesecondmirror:r,1i,130°.Theangletherayformswiththemirroristhus90°30°i290°r160°.Becausethetwomirrorsforma45°90°30°angle,theangletherayreflectingoff60°thefirstmirrorformswiththesecondmirroris180°60°45°75°.Ther2i2angletherayformswiththesecond60°mirroristhusi,290°75°15°.b.AretroreflectorisadevicethatreflectsTheangleofreflectionfromthesecondincominglightraysbackinadirectionmirrorisoppositetothatoftheincidentrays.r,2i,215°.Drawadiagramshowingtheangleof60.Arayoflightstrikesamirroratanangleofincidenceonthefirstmirrorforwhich60°tothenormal.Themirroristhenrotatedthemirrorsystemactsasaretroreflector.18°clockwise,asshowninFigure17-20.Whatistheanglethatthereflectedraymakeswiththemirror?Incidentlight45°18°60°58.DrawaraydiagramofaplanemirrortoNormalshowthatifyouwanttoseeyourselffromyourfeettothetopofyourhead,themirrormustbeatleasthalfyourheight.EyeMirrorImageMirrorlevelCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Feet■Figure17-20ii,old18°Therayfromthetopoftheheadhits60°18°themirrorhalfwaybetweentheeyes42°andthetopofthehead.TherayfromPhysics:PrinciplesandProblemsSolutionsManual365

369Chapter17continuedri42°r,mirror90°r90°42°48°CF17.2CurvedMirrorspage480Level161.Aconcavemirrorhasafocallengthof■Figure17-2110.0cm.Whatisitsradiusofcurvature?real;inverted;largerr2f2(10.0cm)20.0cmLevel262.Anobjectlocated18cmfromaconvex65.StarImageLightfromastariscollectedbymirrorproducesavirtualimage9cmfromaconcavemirror.Howfarfromthemirrorthemirror.Whatisthemagnificationoftheistheimageofthestariftheradiusofimage?curvatureis150cm?diStarsarefarenoughawaythatthemdolightcomingintothemirrorcanbe(9cm)consideredtobeparallelandparallel18cmlightwillconvergeatthefocalpoint.0.5Sincer2f,r150cm63.FunHouseAboyisstandingnearaf2275cmconvexmirrorinafunhouseatafair.Henoticesthathisimageappearstobe66.FindtheimagepositionandheightfortheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.60mtall.IfthemagnificationoftheobjectshowninFigure17-22.1mirroris,whatistheboy’sheight?3himhohi3.8cmhFom0.60m16cm131cm31.8m64.Describetheimageproducedbytheobject■Figure17-22inFigure17-21asrealorvirtual,invertedorupright,andsmallerorlargerthanthe111fddobject.oidfodidfo(31cm)(16cm)31cm16cm366SolutionsManualPhysics:PrinciplesandProblems

370Chapter17continued33cmwhatisthemagnificationoftheimage?hdfr(40mm)20mmiim22hdoo111dihododifhidodf(16mm)(20mm)od80mm(33cm)(3.8cm)idf16mm20mmo31cmdi(80mm)4.1cmm5d16mmo67.RearviewMirrorHowfardoestheimage70.A3.0-cm-tallobjectis22.4cmfromaofacarappearbehindaconvexmirror,concavemirror.Ifthemirrorhasaradiusofwithafocallengthof6.0m,whenthecarcurvatureof34.0cm,whataretheimageis10.0mfromthemirror?positionandheight?111rfdodif2dof34.0cmdidf2o(10.0m)(6.0m)17.0cm10.0m(6.0m)111fdd3.8moidfo68.Anobjectis30.0cmfromaconcavemirrordidfoof15.0cmfocallength.Theobjectis(22.4cm)(17.0cm)1.8cmtall.Usethemirrorequationto22.4cm17.0cmfindtheimageposition.Whatistheimageheight?70.5cm111hdiiddfmoihdoodfodhdioidfhiodo(30.0cm)(15.0cm)(70.5cm)(3.0cm)30.0cm15.0cm22.4cm30.0cm9.4cmhdiimhdLevel3oo71.Jeweler’sMirrorAjewelerinspectsadhiohwatchwithadiameterof3.0cmbyplacingidoit8.0cminfrontofaconcavemirrorof(30.0cm)(1.8cm)12.0-cmfocallength.(30.0cm)a.WherewilltheimageofthewatchCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.8cmappear?11169.DentalMirrorAdentistusesasmallmir-ddfoirorwitharadiusof40mmtolocateacavityinapatient’stooth.Ifthemirrorisconcaveandisheld16mmfromthetooth,Physics:PrinciplesandProblemsSolutionsManual367

371Chapter17continueddof(8.0cm)(12.0cm)111didf8.0cm12.0cmdodifo24cmfdodb.Whatwillbethediameteroftheimage?idfohidi(10.0cm)(150cm)9.4cmhd150cm(10.0cm)oodiho(24cm)(3.0cm)di(9.4cm)hm0.063ido8.0cmdo150cm9.0cmhimho(0.063)(12cm)0.75cm72.SunlightfallsonaconcavemirrorandMixedReviewformsanimagethatis3.0cmfromthepages480–481mirror.Anobjectthatis24mmtallisLevel1placed12.0cmfromthemirror.74.Alightraystrikesaplanemirroratananglea.Sketchtheraydiagramtoshowtheof28°tothenormal.Ifthelightsourceislocationoftheimage.movedsothattheangleofincidenceincreasesby34°,whatisthenewangleofO1Ray1reflection?Ray2ii,initial34°CF28°34°Horizontalscale:I11block1.0cm62°Verticalscale:1block4mmri62°b.Usethemirrorequationtocalculatetheimageposition.75.CopyFigure17-23onasheetofpaper.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Drawraysonthediagramtodeterminethe111heightandlocationoftheimage.ddfoifd(3.0cm)(12.0cm)doid12.0cm3.0cmof4.0cmc.Howtallistheimage?3.0cmFdi4.0cmm0.338.0cm4.0cmd12.0cmohimho(0.33)(24mm)8.0mm73.Shinyspheresthatareplacedonpedestals■Figure17-23onalawnareconvexmirrors.Onesuchspherehasadiameterof40.0cm.A12-cm-tallrobinsitsinatreethatis1.5mfromthesphere.Whereistheimageoftherobinandhowtallistheimage?r20.0cm,f10.0cm368SolutionsManualPhysics:PrinciplesandProblems

372Chapter17continuedb.Whatistheimageheight?O1Ihi1.0cmhi1d2.7cmmihoHorizontalscale:Fdhio1block1.0cmhidVerticalscale:o2blocks1.0cm(22.9cm)(2.4cm)30.0cmTheimageheightis1.0cm,andits1.8cmlocationis2.7cmfromthemirror.78.WhatistheradiusofcurvatureofaconcaveLevel2mirrorthatmagnifiesanobjectbyafactor76.Anobjectislocated4.4cminfrontofaof3.2whentheobjectisplaced20.0cmconcavemirrorwitha24.0-cmradius.fromthemirror?Locatetheimageusingthemirrorequation.hirmfho224.0cmdimdo2(3.2)(20.0cm)12.0cm64cm111fdodi111fddoidfodidofdodifddoi(4.4cm)(12.0cm)4.4cm12.0cm(20.0cm)(64cm)20.0cm(64cm)6.9cm29cm77.Aconcavemirrorhasaradiusofcurvaturer2fof26.0cm.Anobjectthatis2.4cmtallis(2)(29cm)placed30.0cmfromthemirror.58cma.Whereistheimageposition?r79.Aconvexmirrorisneededtoproduceanf2imageone-halfthesizeofanobjectand26.0cmlocated36cmbehindthemirror.What2focallengthshouldthemirrorhave?13.0cmhdiim111hdoofddoidhiodofdohdiidfo(36cm)hoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(30.0cm)(13.0cm)ho30.0cm13.0cm222.9cm72cm111fddoiPhysics:PrinciplesandProblemsSolutionsManual369

373Chapter17continueddodia.Whatkindofmirrorwoulddothisjob?fddoiAnenlarged,uprightimageresults(72cm)(36cm)onlyfromaconcavemirror,withtheobjectinsidethefocallength.72cm(36cm)72cmb.Whatisitsradiusofcurvature?di80.SurveillanceMirrorAconveniencestoremdousesasurveillancemirrortomonitorthedstore’saisles.Eachmirrorhasaradiusofimdo(7.5)(14.0mm)curvatureof3.8m.105mma.Whatistheimagepositionofacustomer111whostands6.5minfrontofthemirror?dodifAmirrorthatisusedforsurveillancedodi(14.0mm)(105mm)fisaconvexmirror.Sothefocaldd14.0mm(105mm)iolengthisthenegativeofhalfthe16mmradiusofcurvature.r2f(2)(16mm)rf232mm3.8m82.TheobjectinFigure17-24movesfrom2position1toposition2.Copythediagram1.9montoasheetofpaper.Drawraysshowing111howtheimagechanges.fddoidfodidfo(6.5m)(1.9m)16.5m(1.9m)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C2F1.5m1.0mb.Whatistheimageheightofacustomer1.5mwhois1.7mtall?2.0mhd2.5miimhdoo■Figure17-24dhiohidoO1Ray1O2Ray1(1.5m)(1.7m)Ray2Ray26.5m1C2FI10.38mI2Horizontalscale:Level31block10cm81.InspectionMirrorAproduction-lineinspectorwantsamirrorthatproducesan83.Aballispositioned22cminfrontofaspheri-imagethatisuprightwithamagnificationcalmirrorandformsavirtualimage.Iftheof7.5whenitislocated14.0mmfromasphericalmirrorisreplacedwithaplanemir-machinepart.ror,theimageappears12cmclosertothemir-ror.Whatkindofsphericalmirrorwasused?370SolutionsManualPhysics:PrinciplesandProblems

374Chapter17continuedTheobjectpositionforbothmirrorsisORay1122cm.So,theimagepositionfortheRay2Horizontalscale:planemirroris22cm.1block1.0mBecausethesphericalmirrorformsaCFVerticalscale:hi2.4m2blocks1.0mvirtualimage,theimageislocateddi14mbehindthemirror.Thus,theimageposi-I1tionforthesphericalmirrorisnegative.Theimageis2.4mtall,anditis14mdidi,plane12cmfromthemirror.22cm12cm86.A4.0-cm-tallobjectisplaced12.0cmfroma34cmconvexmirror.Iftheimageoftheobjectis1112.0cmtall,andtheimageislocatedatfddoi6.0cm,whatisthefocallengthofthemir-ddror?Drawaraydiagramtoanswertheques-foition.Usethemirrorequationandthemag-ddoinificationequationtoverifyyouranswer.(22cm)(34cm)22cm(34cm)O1Ray1I162cmRay2Thefocallengthispositive,sotheHorizontalscale:Fsphericalmirrorisaconcavemirror.1block1.0cmf12cmVerticalscale:84.A1.6-m-tallgirlstands3.2mfromaconvex3blocks2.0cmmirror.Whatisthefocallengthofthemirrorifherimageappearstobe0.28mtall?111fddhdoiiimhodododifddhdoiiodiho(12.0cm)(6.0cm)12.0cm(6.0cm)(0.28m)(3.2m)1.6m12cm0.56mThinkingCritically111pages481–482fddoi87.ApplyConceptsTheballinFigure17-25dodislowlyrollstowardtheconcavemirroronthefddoiright.Describehowthesizeoftheball’simage(3.2m)(0.56m)changesasitrollsalong.3.2m(0.56m)0.68mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.85.MagicTrickAmagicianusesaconcaveCFmirrorwithafocallengthof8.0mtomakea3.0-m-tallhiddenobject,located18.0mfromthemirror,appearasarealimagethatisseenbyhisaudience.Drawascaleraydiagramtofindtheheightandlocationoftheimage.■Figure17-25Physics:PrinciplesandProblemsSolutionsManual371

375Chapter17continuedBeyondC,theimageissmallerthanthemirror,whatisthefocallengthofthecon-ball.Astheballrollstowardthemirror,cavemirror?theimagesizeincreases.Theimageisdi,initialdo,initialthesamesizeastheballwhentheballisatC.Theimagesizecontinuesto6.0cmincreaseuntilthereisnoimagewhendtheballisatF.PastF,thesizeoftheidi,initial(8.0cm)imagedecreasesuntilitequalsthe6.0cm(8.0cm)ball’ssizewhentheballtouchesthe14.0cmmirror.11188.AnalyzeandConcludeTheobjectinfdodiFigure17-26islocated22cmfromaddfoiconcavemirror.Whatisthefocallengthddoiofthemirror?(6.0cm)(14.0cm)f6.0cm(14.0cm)f1.0101cm91.AnalyzeandConcludeThelayoutofthetwo-mirrorsystemshowninFigure17-11isthatofaGregoriantelescope.Forthis22cmquestion,thelargerconcavemirrorhasaradiusofcurvatureof1.0m,andthesmallermirrorislocated0.75maway.Whyisthesecondarymirrorconcave?Thesmallermirrorisconcavetoproduce■Figure17-26arealimageattheeyepiecethatisrCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fupright.Thelightraysareinvertedbythe2firstconcavemirrorandtheninverteddoagainbythesecondaryconcavemirror.222cm92.AnalyzeandConcludeAnopticalarrange-2mentusedinsometelescopesisthe11cmCassegrainfocus,showninFigure17-27.Thistelescopeusesaconvexsecondary89.UseEquationsShowthatastheradiusofmirrorthatispositionedbetweenthecurvatureofaconcavemirrorincreasestoprimarymirrorandthefocalpointofinfinity,themirrorequationreducestothetheprimarymirror.relationshipbetweentheobjectpositionConvexConcaveandtheimagepositionforaplanemirror.secondarymirrorprimarymirrorAsf→,1/f→0.Themirrorequationthenbecomes1/do1/di,ordodi.F90.AnalyzeandConcludeAnobjectislocated6.0cmfromaplanemirror.Iftheplanemir-rorisreplacedwithaconcavemirror,theTelescopetubeEyepieceresultingimageis8.0cmfartherbehindthe■Figure17-27mirror.Assumingthattheobjectislocateda.Asingleconvexmirrorproducesonlybetweenthefocalpointandtheconcavevirtualimages.Explainhowtheconvex372SolutionsManualPhysics:PrinciplesandProblems

376Chapter17continuedmirrorinthistelescopefunctionswithinb.theprecisionopticalpolishingofalu-thesystemofmirrorstoproducerealminumtosuchadegreeofsmoothnessimages.thatnoglassisneededintheprocessofTheconvexmirrorisplacedtointer-makingamirrorcepttheraysfromaconcavemirrorAnswerswillvary.Studentanswersbeforetheyconverge.Theconvexmightincludeinformationaboutmirrorplacesthepointofconver-deformationofamirrorfromitsowngenceintheoppositedirectionbackweightassizeincreasesandhowatowardtheconcavemirror,andmirrormadeofaluminumcouldlengthensthetotaldistancethelightimpactthisproblem.travelsbeforeconverging.ThiseffectivelyincreasesthefocallengthCumulativeReviewcomparedtousingtheconcavemir-page482rorbyitself,thusincreasingthe95.Achildrunsdowntheschoolhallwayandtotalmagnification.thenslidesonthenewlywaxedfloor.Heb.Aretheimagesproducedbythewasrunningat4.7m/sbeforehestartedCassegrainfocusuprightorinverted?slidingandheslid6.2mbeforestopping.HowdoesthisrelatetothenumberofWhatwasthecoefficientoffrictionofthetimesthatthelightcrosses?waxedfloor?(Chapter11)Inverted;eachtimethelightraysTheworkdonebythewaxedfloorcrosstheimageinverts.equalsthechild’sinitialkineticenergy.KE1mv2WFd2kmgdWritinginPhysicsThemassofthechildcancelsout,givingpage482v293.Researchamethodusedforgrinding,polish-k2gding,andtestingmirrorsusedinreflecting2(4.7m/s)telescopes.Youmayreporteitheronmeth-2)(6.2m)(2)(9.80m/sodsusedbyamateurastronomerswhomake0.18theirowntelescopeoptics,oronamethodusedbyaprojectatanationallaboratory.96.A1.0gpieceofcopperfallsfromaheightofPrepareaone-pagereportdescribingthe4mfromanairplanetotheground.1.010method,andpresentittotheclass.BecauseofairresistanceitreachesthegroundAnswerswillvarydependingonthemovingatavelocityof70.0m/s.Assumingmirrorsandmethodschosenbythestu-thathalfoftheenergylostbythepiecewasdis-dents.Amateurmethodsusuallyinvolvetributedasthermalenergytothecopper,howrubbingtwo“blanks”againsteachothermuchdiditheatduringthefall?(Chapter12)withvaryinggritsbetweenthem.PotentialenergyofthepieceMethodsusedatnationallabsvary.Emgh94.Mirrorsreflectlightbecauseoftheirmetal-(0.0010kg)(9.80m/s2)(1.0104m)liccoating.Researchandwriteasummary9.8Jofoneofthefollowing:Finalenergya.thedifferenttypesofcoatingsusedandCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12theadvantagesanddisadvantagesofeachEf2mvAnswerswillvary.Studentanswers1(0.0010kg)(70.0m/s)2shouldincludeinformationabout2shininessaswellastarnish2.4Jresistance.Physics:PrinciplesandProblemsSolutionsManual373

377Chapter17continuedHeataddedtothepiece99.OrganpipesAnorganbuildermustdesign1apipeorganthatwillfitintoasmallspace.Q(EE2f)(Chapter15)1a.Shouldhedesigntheinstrumenttohave(9.8J2.4J)2openpipesorclosedpipes?Explain.3.7JTheresonantfrequencyofanopenQpipeistwicethatofaclosedpipeTmcofthesamelength.Therefore,the3.7Jpipesofaclosed-pipeorganneed(0.0010kg)(385J/kg°C)beonlyhalfaslongasopenpipes9.5°Ctoproducethesamerangeoffundamentalfrequencies.97.Itispossibletoliftapersonwhoissittingonab.Willanorganconstructedwithopenpillowmadefromalargesealedplasticgarbagepipessoundthesameasonecon-bagbyblowingairintothebagthroughasodastructedwithclosedpipes?Explain.straw.Supposethatthecross-sectionalareaof2andNo.Whilethetwoorganswillhavethepersonsittingonthebagis0.25mthesamefundamentaltones,theperson’sweightis600N.Thesodastrawhasacross-sectionalareaof2105m2.Withclosedpipesproduceonlytheoddharmonics,sotheywillhavediffer-whatpressuremustyoublowintothestrawtoenttimbresthanopenpipes.liftthepersonthatissittingonthesealedgarbagebag?(Chapter13)100.FiltersareaddedtoflashlightssothatoneApplyPascal’sprinciple.shinesredlightandtheothershinesgreenAlight.Thebeamsarecrossed.Explainin1F2F1Atermsofwaveswhythelightfromboth2flashlightsisyellowwherethebeams2105m2(600N)20.048Ncross,butrevertbacktotheiroriginal0.25mcolorsbeyondtheintersectionpoint.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.F20.048N(Chapter16)P2.4kPaA22105m2Wavescaninterfere,add,andthenor2kPatoonesignificantdigitpassthroughunaffected.Chapter14notaverylargepressureatallshowedtheamplitudeofwavesadding.Inthiscase,thewavesretain98.Whatwouldbetheperiodofa2.0-m-longtheircolorinformationastheycrosspendulumontheMoon’ssurface?Thethrougheachother.Moon’smassis7.341022kg,anditsradiusis1.74106m.WhatistheperiodofthispendulumonEarth?(Chapter14)ChallengeProblemGmgmpage470md2mAnobjectofheighthoislocatedatdorelativeto(6.671011aconcavemirrorwithfocallengthf.Nm2/kg2)(7.341022kg)1.Drawandlabelaraydiagramshowingthefocallengthandlocationoftheobjectifthe1.62m/s2imageislocatedtwiceasfarfromthel2.0mmirrorastheobject.ProveyouranswerT27.0sMoon2g1.62m/s2mathematically.Calculatethefocallengthl2.0masafunctionofobjectpositionforthisT22.8sEarth2g9.80m/s2placement.374SolutionsManualPhysics:PrinciplesandProblems

378Chapter17continued111ORay11fddoidi2dORay2fddiCFodf2dOif3f(2f)2ff1112ffddoihdiiddmhdfoiooddoidhhiodo(2do)idod2doo(2f)ho2do2f3ho2.Drawandlabelaraydiagramshowingthelocationoftheobjectiftheimageislocated3.Whereshouldtheobjectbelocatedsothattwiceasfarfromthemirrorasthefocalnoimageisformed?point.Proveyouranswermathematically.TheobjectshouldbeplacedatthefocalCalculatetheimageheightasafunctionofpoint.theobjectheightforthisplacement.O1Ray1Ray2di2fCFhihOI1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual375

379

380CHAPTER18RefractionandLenses5.AblockofunknownmaterialissubmergedPracticeProblemsinwater.Lightinthewaterisincidenton18.1RefractionofLighttheblockatanangleofincidenceof31°.pages485–492Theangleofrefractionofthelightintheblockis27°.Whatistheindexofrefractionpage487ofthematerialoftheblock?1.Alaserbeaminairisincidentuponethanolatanangleofincidenceof37.0°.Whatisn1sin1n2sin2theangleofrefraction?nn1sin1n2sin1sin1n2sin221n1sin1(1.33)(sin31°)2sinnsin27°21.51(1.00)(sin37.0°)sin1.3626.3°SectionReview2.Lightinairisincidentuponapieceof18.1RefractionofLightcrownglassatanangleofincidenceofpages485–49245.0°.Whatistheangleofrefraction?page492n1sin1n2sin26.IndexofRefractionYounoticethatwhen1n1sin1alightrayentersacertainliquidfrom2sinn2water,itisbenttowardthenormal,butwhenitentersthesameliquidfromcrown1(1.00)(sin45.0°)sin1.52glass,itisbentawayfromthenormal.Whatcanyouconcludeabouttheliquid’sindex27.7°ofrefraction?3.Lightpassesfromairintowaterat30.0°tonwaternliquidncrownglass,therefore,thenormal.Findtheangleofrefraction.nliquidmustbebetween1.33and1.52.n1sin1n2sin27.IndexofRefractionArayoflighthasann11sin1angleofincidenceof30.0°onablock2sinn2ofunknownmaterialandanangleof1(1.00)(sin30.0°)refractionof20.0°.Whatistheindexofsin1.33refractionofthematerial?22.1°n1sin1n2sin2n4.Lightisincidentuponadiamondfacetatn1sin12sin45.0°.Whatistheangleofrefraction?2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.n(1.00)(sin30.0°)1sin1n2sin2sin20.0°n11sin11.462sinn21(1.00)(sin45.0°)sin17.0°2.42Physics:PrinciplesandProblemsSolutionsManual377

381Chapter18continued8.SpeedofLightCouldanindexof14.CriticalThinkingInwhatdirectioncanrefractioneverbelessthan1?Whatwouldyouseearainbowonarainylateafter-thisimplyaboutthespeedoflightinthatnoon?Explain.medium?Intheeast,becausetheSunsetsintheNo,itwouldmeanthespeedoflightinwestandsunlightmustshinefromthemediumisfasterthanitisinabehindyouinorderforyoutoseeavacuum.rainbow.9.SpeedofLightWhatisthespeedoflightinchloroform(n1.51)?PracticeProblemscn18.2ConvexandConcavevcLensesvchloroformnchloroformpages493–4993.00108m/spage4961.5115.A2.25-cm-tallobjectis8.5cmtotheleftof1.99108m/saconvexlensof5.5-cmfocallength.Findtheimagepositionandheight.10.TotalInternalReflectionIfyouwereto111usequartzandcrownglasstomakeanfdodiopticalfiber,whichwouldyouusefortheddofcladdinglayer?Why?idofcrownglassbecauseithasalower(8.5cm)(5.5cm)indexofrefractionandwouldproduce8.5cm5.5cmtotalinternalreflection15.6cm,or16cm11.AngleofRefractionAbeamoflighthidimpassesfromwaterintopolyethylenewithhodoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.n1.50.Ifi57.5°,whatistheangleofdihorefractioninthepolyethylene?hidon1sin1n2sin2(15.6cm)(2.25cm)n8.5cm11sin12sinn4.1cm21(1.33)(sin57.5°)sin16.Anobjectnearaconvexlensproducesa1.501.8-cm-tallrealimagethatis10.4cmfrom48.4°thelensandinverted.Ifthefocallengthofthelensis6.8cm,whataretheobject12.CriticalAngleIsthereacriticalangleforpositionandheight?lighttravelingfromglasstowater?From111watertoglass?fddoiYes,becausenglassnwater.No.ddifod13.DispersionWhycanyouseetheimageofiftheSunjustabovethehorizonwhenthe(10.4cm)(6.8cm)Sunitselfhasalreadyset?10.4cm6.8cmbecauseofbendingoflightraysinthe2.0101cmatmosphere;refraction378SolutionsManualPhysics:PrinciplesandProblems

382Chapter18continuedmhididofdidhodoofd(25cm)(5.0cm)hohiod25cm5.0cmi6.2cm(19.6cm)(1.8cm)10.4cmhdmiihd3.4cmoodhiho17.Anobjectisplacedtotheleftofaconvexidolenswitha25-mmfocallengthsothatitsimageisthesamesizeastheobject.What(6.2cm)(2.0cm)25cmaretheimageandobjectpositions?0.50cm(invertedimage)111fddiowithdpage497odibecause20.Anewspaperisheld6.0cmfromaconvexdmiandm1lensof20.0-cmfocallength.Findtheimagedopositionofthenewsprintimage.Therefore,11112fdidofdidSodofdidi2fof2(25mm)(6.0cm)(20.0cm)6.0cm20.0cm5.0101mm8.6cmd1mmodi5.01021.Amagnifyingglasshasafocallengthof18.Useascaleraydiagramtofindtheimage12.0cm.Acoin,2.0cmindiameter,ispositionofanobjectthatis30cmtotheplaced3.4cmfromthelens.Locatetheleftofaconvexlenswitha10-cmfocalimageofthecoin.Whatisthediameteroflength.theimage?di15cm111fddioO1Ray1ddofRay2di15cmidofFF(3.4cm)(12.0cm)Horizontalscale:3.4cm12.0cm1block1cmI14.7cm19.Calculatetheimagepositionandheightofhodi(2.0cm)(4.7cm)ha2.0-cm-tallobjectlocated25cmfromaid3.4cmoconvexlenswithafocallengthof5.0cm.2.8cmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Whatistheorientationoftheimage?111fddoiPhysics:PrinciplesandProblemsSolutionsManual379

383Chapter18continued22.AconvexlenswithafocallengthofThelocationshouldbeabout15cmon22.0cmisusedtoviewa15.0-cm-longthesamesideofthelens(15cm)andpencillocated10.0cmaway.Findthetheimageshouldbeuprightandaboutheightandorientationoftheimage.1.5cmtall.111fddoiSectionReviewddofidof18.2ConvexandConcave(10.0cm)(22.0cm)Lenses10.0cm22.0cmpages493–49918.3cmpage499hidi25.MagnificationMagnifyingglassesnormal-mhodolyareusedtoproduceimagesthatarelargerthantherelatedobjects,buttheyalsocandhihoidproduceimagesthataresmallerthantheorelatedobjects.Explain.(18.3cm)(15.0cm)Iftheobjectislocatedfartherthan10.0cmtwicethefocallengthfromthelens,the27.5cm(uprightimage)sizeoftheimageissmallerthanthesizeoftheobject.23.Astampcollectorwantstomagnifyastampby4.0whenthestampis3.5cmfromthe26.ImagePositionandHeightA3.0-cm-talllens.Whatfocallengthisneededfortheobjectislocated2.0cmfromaconvexlenslens?havingafocallengthof6.0cm.Drawaraydidiagramtodeterminethelocationandsizemdooftheimage.Usethethinlensequationandthemagnificationequationtoverifydimdo(4.0)(3.5cm)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.youranswer.14cm111I1fddiod(3.5cm)(14cm)O1fodid3.5cm(14cm)FFodiHorizontalscale:4.7cm2blocks1.0cmhi4.5cmVerticalscale:24.Amagnifierwithafocallengthof30cmdi3.0cm1block1.0cmisusedtoviewa1-cm-tallobject.Useray111tracingtodeterminethelocationandfddoisizeoftheimagewhenthemagnifierisdpositioned10cmfromtheobject.dofidof(2.0cm)(6.0cm)I12.0cm6.0cmFFO13.0cmhi1.5cmHorizontalscale:hdiidi15cm1block2cmmhdVerticalscale:oo3blocks1cm380SolutionsManualPhysics:PrinciplesandProblems

384Chapter18continueddhihoidWatero(3.0cm)(3.0cm)2.0cm4.5cmLightraysAir27.TypesofLensesThecrosssectionsoffourdifferentthinlensesareshowninFigure18-16.Water■Figure18-17Thelightrayswilldiverge.Water■Figure18-16a.Whichoftheselenses,ifany,areconvex,orconverging,lenses?Lightraysn1.3n1.0n1.3Lensesaandcareconverging.b.Whichoftheselenses,ifany,areconcave,ordiverging,lenses?Lensesbanddarediverging.Water28.ChromaticAberrationAllsimplelenseshavechromaticaberration.Explain,then,SectionReviewwhyyoudonotseethiseffectwhenyoulookthroughamicroscope.18.3ApplicationsofLensesAllprecisionopticalinstrumentsusepages500–503acombinationoflenses,calledanpage503achromaticlens,tominimizechromatic31.RefractionExplainwhythecorneaistheaberration.primaryfocusingelementintheeye.Thedifferenceinindexofrefraction29.ChromaticAberrationYoushinewhitebetweentheairandthecorneaislightthroughaconvexlensontoascreengreaterthananyotherdifferencethatandadjustthedistanceofthescreenfromlightraysencounterwhentravelingthelenstofocustheredlight.Whichdirec-towardtheretina.tionshouldyoumovethescreentofocusthebluelight?32.LensTypesWhichtypeoflens,convexorclosertothelensconcave,shouldanearsightedpersonuse?Whichtypeshouldafarsightedpersonuse?30.CriticalThinkingAnairlensconstructedAnearsightedpersonshoulduseacon-oftwowatchglassesisplacedinatankofcavelens.AfarsightedpersonshouldCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.water.CopyFigure18-17anddrawtheuseaconvexlens.effectofthislensonparallellightraysincidentonthelens.Physics:PrinciplesandProblemsSolutionsManual381

385Chapter18continued33.FocalLengthSupposeyourcameraisChapterAssessmentfocusedonapersonwhois2maway.YounowwanttofocusitonatreethatisfartherConceptMappingaway.Shouldyoumovethelensclosertopage508thefilmorfartheraway?37.CompletethefollowingconceptmapusingCloser;realimagesarealwaysfartherthefollowingterms:inverted,larger,smaller,fromthelensthanthefocalpoint.Thevirtual.fartherawaytheobjectis,theclosertheimageistothefocalpoint.Lenses34.ImageWhyistheimagethatyouobserveconvexconcaveinarefractingtelescopeinverted?Afterthelightrayspassthroughtheobjectivelens,theycross,forminganrealvirtualvirtualimagethatisinverted.Theeyepiecemaintainsthisorientationwhenitusesinverteduprightuprightthisimageasitsobject.35.PrismsWhatarethreebenefitsofhavingsmallerunchangedlargerlargersmallersizeprismsinbinoculars?Theprismsextendthelight’spathlengthMasteringConceptstomakethebinocularsmorecompact,invertlightrayssothattheviewerseespage508anuprightimage,andincreasesepara-38.Howdoestheangleofincidencecomparetionbetweenobjectivelensestoimprovewiththeangleofrefractionwhenalightraythethree-dimensionalview.passesfromairintoglassatanonzeroangle?(18.1)36.CriticalThinkingWhenyouusetheTheangleofincidenceislargerthanCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.highestmagnificationonamicroscope,theangleofrefraction,becauseairhastheimageismuchdarkerthanitisatlowerasmallerindexofrefraction.magnifications.Whataresomepossiblereasonsforthedarkerimage?Whatcould39.Howdoestheangleofincidencecompareyoudotoobtainabrighterimage?withtheangleofrefractionwhenalightrayYouareusingthelightthatstrikesonlyleavesglassandentersairatanonzeroasmallareaoftheobject.Abrighterangle?(18.1)lampcouldbeused.Theangleofincidenceissmallerthantheangleofrefraction,becauseglasshasalargerindexofrefraction.40.Regardingrefraction,whatisthecriticalangle?(18.1)Thetermcriticalanglereferstotheinci-dentanglethatcausestherefractedraytolierightalongtheboundaryofthesubstancewhenarayispassingfromaregionofhigherindexofrefractiontoaregionoflowerindexofrefraction.Iftheincidentangleexceedsthecriticalangle,totalinternalreflectionwilloccur.382SolutionsManualPhysics:PrinciplesandProblems

386Chapter18continued41.AlthoughthelightcomingfromtheSunisAnotherlensisincludedintheopticsrefractedwhilepassingthroughEarth’ssystemoftheprojectortoinverttheatmosphere,thelightisnotseparatedintoimageagain.Asaresult,theimageisitsspectrum.Whatdoesthisindicateuprightcomparedtotheoriginalobject.aboutthespeedsofdifferentcolorsoflighttravelingthroughair?(18.1)47.Describewhyprecisionopticalinstrumentsuseachromaticlenses.(18.2)Thespeedsofthedifferentcolorsoflighttravelingthroughairarethesame.Alllenseshavechromaticaberration,whichmeansdifferentwavelengthsof42.ExplainwhytheMoonlooksredduringalightarebentatslightlydifferentangleslunareclipse.(18.1)neartheiredges.AnachromaticlensisDuringalunareclipse,EarthblockstheacombinationoftwoormorelensesSun’sraysfromtheMoon.However,withdifferentindicesofrefractionthatsunlightrefractingoffEarth’satmos-reducethiseffect.phereisdirectedinwardtowardthe48.Describehowtheeyefocuseslight.(18.3)Moon.Becausebluewavelengthsoflightaredispersedmore,redwave-LightenteringtheeyeisprimarilylengthsoflightreflectofftheMoonfocusedbythecornea.FinefocusingtowardEarth.occurswhenmuscleschangetheshapeofthelens,allowingtheeyetofocuson43.Howdotheshapesofconvexandconcaveeithernearorfarobjects.lensesdiffer?(18.2)49.WhatistheconditioninwhichthefocalConvexlensesarethickeratthecenterlengthoftheeyeistooshorttofocuslightthanattheedges.Concavelensesareontheretina?(18.3)thinnerinthemiddlethanattheedges.nearsightedness44.Locateanddescribethephysicalpropertiesoftheimageproducedbyaconvexlens50.Whattypeofimageisproducedbythewhenanobjectisplacedsomedistanceobjectivelensinarefractingtelescope?beyond2F.(18.2)(18.3)Itisarealimagethatislocatedrealimage,invertedbetweenFand2F,andthatisinvertedandsmallercomparedtotheobject.51.Theprismsinbinocularsincreasethedis-tancebetweentheobjectivelenses.Whyis45.Whatfactor,otherthanthecurvatureofthethisuseful?(18.3)surfacesofalens,determinesthelocationItimprovesthethree-dimensionalview.ofthefocalpointofthelens?(18.2)Theindexofrefractionofthematerial52.Whatisthepurposeofacamera’sreflexfromwhichthelensismadealsodeter-mirror?(18.3)minesthefocus.Thereflexmirrordivertstheimageontoaprismsothatitcanbeviewedbefore46.Toprojectanimagefromamovieprojectortakingaphotograph.Whentheshutterontoascreen,thefilmisplacedbetweenreleasebuttonispressed,thereflexFand2Fofaconverginglens.Thisarrange-mirrormovesoutofthewaysothattheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mentproducesanimagethatisinverted.lensfocusestheimageontothefilmorWhydoesthefilmedsceneappeartobeotherphotodetector.uprightwhenthefilmisviewed?(18.2)Physics:PrinciplesandProblemsSolutionsManual383

387Chapter18continuedApplyingConcepts11.00c,glass/airsin1.52pages508—50941.1°53.Whichsubstance,AorB,inFigure18-24hasalargerindexofrefraction?Explain.Airandglasshavethesmallercriticalangleof41.1°.Thecriticalangleforairandwateris48.8°.58.CrackedWindshieldIfyoucrackthewindshieldofyourcar,youwillseeasilverylinealongthecrack.Theglasshasseparatedatthecrack,andthereisairinthecrack.Thesilverylineindicatesthatlightisreflect-ingoffthecrack.Drawaraydiagramtoexplainwhythisoccurs.Whatphenomenondoesthisillustrate?AB■Figure18-24FromoutsideTheangleinsubstanceAissmaller,soCrackithasthelargerindexofrefraction.n1.054.Alightraystrikestheboundarybetweentwotransparentmedia.Whatistheangleofincidenceforwhichthereisnorefraction?1c1cAnangleofincidenceof0°allowsthennglassglasslighttogothroughunchanged.Oriftheangleofincidenceisgreaterthanthecriticalanglethereistotalinternalreflection.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.55.Howdoesthespeedoflightchangeastheindexofrefractionincreases?Tothedriver’seyesAstheindexofrefractionofamaterialThisillustrateslightreflectedatanglesincreases,thespeedoflightinthatlargerthanthecriticalangle,ortotalmaterialdecreases.internalreflection.56.Howdoesthesizeofthecriticalangle59.LegendaryMirageAccordingtolegend,changeastheindexofrefractionincreases?ErictheRedsailedfromIcelandanddiscov-ThecriticalangledecreasesastheeredGreenlandafterhehadseentheislandindexofrefractionincreases.inamirage.Describehowthemiragemighthaveoccurred.57.Whichpairofmedia,airandwaterorairEventhoughGreenlandisbelowtheandglass,hasthesmallercriticalangle?horizon,itisvisibleasamiragedueton12therefractionoflight.csinn111.00c,water/airsin1.3348.8°384SolutionsManualPhysics:PrinciplesandProblems

388Chapter18continuedThemagnificationismuchlessinwaterMiragethaninair.Thedifferenceintheindicesofrefractionforwaterandglassismuchlessthanthedifferenceforairandglass.Cold,deW64.Whyistherechromaticaberrationforlightnsearthatgoesthroughalensbutnotforlightamirathatreflectsfromamirror?irChromaticaberrationforlensesisduetoIcelandthedispersionoflight(differentwave-Greenlandlengthsoflighthavedifferentspeedsin60.Aprismbendsvioletlightmorethanitthelensandrefractwithslightlydiffer-bendsredlight.Explain.entangles).Mirrorsreflect,andreflec-tionisindependentofwavelength.Violetlighttravelsslowerinaprismthanredlightdoes.65.Whensubjectedtobrightsunlight,thepupilsofyoureyesaresmallerthanwhentheyare61.RainbowsWhywouldyouneverseearain-subjectedtodimmerlight.Explainwhyyourbowinthesouthernskyifyouwereintheeyescanfocusbetterinbrightlight.northernhemisphere?InwhichdirectionshouldyoulooktoseerainbowsifyouareEyescanfocusbetterinbrightlightinthesouthernhemisphere?becauseraysthatarerefractedintolargeranglesarecutoffbytheiris.YoucanseearainbowonlywhentheTherefore,allraysconvergeataSun’srayscomefrombehindyouatannarrowangle,sothereislesssphericalanglenotgreaterthan42°withthehori-aberration.zon.Whenyouarefacingsouthinthenorthernhemisphere,theSunisnever66.BinocularsTheobjectivelensesinbinocu-behindyouatanangleof42°orless.larsformrealimagesthatareuprightcomparedtotheirobjects.Wherearethe62.SupposethatFigure18-14isredrawnwithimageslocatedrelativetotheeyepiecealensofthesamefocallengthbutalargerlenses?diameter.Explainwhythelocationoftheimagedoesnotchange.WouldtheimageEachsideofthebinocularsislikeabeaffectedinanyway?refractingtelescope.Therefore,theobjectivelensimagemustbebetweenThelocationoftheimagedependsontheeyepiecelensanditsfocalpointtothefocallengthofthelensandthedis-magnifytheimage.tanceoftheobjectfromthelens.Therefore,thelocationoftheimagedoesn’tchange.MasteringProblems18.1RefractionofLight63.Aswimmerusesamagnifyingglasstopages509—510observeasmallobjectonthebottomofaLevel1swimmingpool.Shediscoversthatthemag-67.Arayoflighttravelsfromairintoaliquid,nifyingglassdoesnotmagnifytheobjectCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.asshowninFigure18-25.Therayisinci-verywell.Explainwhythemagnifyingglassdentupontheliquidatanangleof30.0°.isnotfunctioningasitwouldinair.Theangleofrefractionis22.0°.Physics:PrinciplesandProblemsSolutionsManual385

389Chapter18continuedb.Atwhatangledoesthebeamenterthewater?30°ngsingnwsinwAirn1gsingwsinnLiquidw1(1.50)(sin25.4°)sin1.3322°28.9°■Figure18-2570.RefertoTable18-1.Usetheindexofrefrac-a.UsingSnell’slaw,calculatetheindexoftionofdiamondtocalculatethespeedofrefractionoftheliquid.lightindiamond.n1sin1n2sin2cnnv1sin1n2sinc2vdiamondndiamond(1.00)(sin30.0°)8m/ssin22.0°3.00102.421.331.24108m/sb.Comparethecalculatedindexofrefrac-tiontothoseinTable18-1.Whatmight71.RefertoTable18-1.Findthecriticalangletheliquidbe?foradiamondinair.watern12c,diamond/airsinn168.Lighttravelsfromflintglassintoethanol.Theangleofrefractionintheethanolissin11.002.4225.0°.WhatistheangleofincidenceintheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.glass?24.4°n1sin1n2sin2nLevel212sin21sinn72.AquariumTankAthicksheetofplastic,1n1.500,isusedasthesideofanaquari-1(1.36)(sin25.0°)sinumtank.Lightreflectedfromafishinthe1.62waterhasanangleofincidenceof35.0°.20.8°Atwhatangledoesthelightentertheair?n69.Abeamoflightstrikestheflat,glasssideof1sin1n2sin2awater-filledaquariumatanangleof40.0°nwatersinwaternplasticsinplastictothenormal.Forglass,n1.50.na.Atwhatangledoesthebeamenterthe1watersinwaterplasticsinnglass?plasticn1(1.33)(sin35.0°)AsinAngsingsin1.500n1AsinA30.57°gsinngn1(1.00)(sin40.0°)plasticsinplasticnairsinairsin1.5025.4°386SolutionsManualPhysics:PrinciplesandProblems

390Chapter18continuedn74.Adiamond’sindexofrefractionforred1plasticsinplasticairsinnlight,656nm,is2.410,whilethatforblueairlight,434nm,is2.450.Supposethatwhite1(1.500)(sin30.57°)sinlightisincidentonthediamondat30.0°.1.00Findtheanglesofrefractionforredand49.7°bluelight.n73.Swimming-PoolLightsAlightsourceisAsinAndsindlocated2.0mbelowthesurfaceofaswim-n1AsinAmingpooland1.5mfromoneedgeofthedsinndpool,asshowninFigure18-26.ThepoolForredlightisfilledtothetopwithwater.1(1.00)(sin30.0°)dsin2.41012.0°Forbluelight2m1(1.00)(sin30.0°)dsin2.4501.5m11.8°75.Theindexofrefractionofcrownglassis1.53■Figure18-26forvioletlight,anditis1.51forredlight.a.Atwhatangledoesthelightreachinga.Whatisthespeedofvioletlightintheedgeofthepoolleavethewater?crownglass?11.5mc3.00108m/sitan2.0mvn1.5337°1.96108m/sThenfindtheangleinair.b.Whatisthespeedofredlightincrownnglass?AsinAnwsinwc3.00108m/s1nwsinwvn1.51AsinnA1.99108m/s1(1.33)(sin37°)sin1.0076.Thecriticalangleforaspecialglassinairis53°41.0°.Whatisthecriticalangleiftheglassb.Doesthiscausethelightviewedfromisimmersedinwater?thisangletoappeardeeperorshallowernsinAthanitactuallyis?c,airngsideoppositetan53°nA1.00sideadjacentngsinsin41.0°1.524c,airsideoppositesideadjacentntan53°sinwc,watern1.5mgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tan53°n1w1.1m,shallowerc,watersinng11.33sin1.52460.8°Physics:PrinciplesandProblemsSolutionsManual387

391Chapter18continuedLevel3c.Determine.277.Arayoflightinatankofwaterhasannangleofincidenceof55.0°.Whatistheairsin2nglasssin1angleofrefractioninair?n1glasssin1n2sinn1sin1n2sin2air1n1sin1sin1(1.5)(sin32°)2sinn1.00253°1(1.33)(sin55.0°)sin1.0079.Thespeedoflightinaclearplasticissin1(1.09)1.90108m/s.ArayoflightstrikestheThevaluesinplasticatanangleof22.0°.Atwhatangleis21.09isnotdefined.Therefore,totalinternalreflectiontherayrefracted?coccurs.nairsinairnpsinpandnpv,sopc78.TherayoflightshowninFigure18-27isnairsinairvsinppincidentupona60°-60°-60°glassprism,n1.5.vpnairsinairsinpcAirAirv1pnairsinairpsinc60°145°1(1.90108m/s)(1.00)(sin22.0°)sin8m/s3.0010PQ213.7°21R80.Alightrayentersablockofcrownglass,nglass1.5asillustratedinFigure18-28.Usearay60°Glass60°Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.diagramtotracethepathoftherayuntilitleavestheglass.■Figure18-27a.UsingSnell’slawofrefraction,determinetheangle,2,tothenearestdegree.45°nairsin1nglasssin2n1airsin12sinnglass11.00sin45°■Figure18-28sin1.5nAsinAngsing28°n1AsinAb.Usingelementarygeometry,determinegsinngthevalueof.11(1.00)(sin45°)sinP90°28°62°1.52Q180°62°60°58°28°Findthecriticalangleforcrownglass.190°58°32°nsinAcng388SolutionsManualPhysics:PrinciplesandProblems

392Chapter18continuedn83.Ifanobjectis10.0cmfromaconverging1Acsinnlensthathasafocallengthof5.00cm,howgfarfromthelenswilltheimagebe?11.00sin1.52111fdd41.1°oidWhenthelightrayintheglassstrikesdofidthesurfaceata62°angle,totalinternalofreflectionoccurs.(10.0cm)(5.00cm)10.0cm5.00cm10.0cm62°62°28°28°45°45°Level284.Aconvexlensisneededtoproduceanimagethatis0.75timesthesizeoftheobjectandlocated24cmfromthelensontheotherside.Whatfocallengthshouldbespecified?18.2ConvexandConcaveLenseshdmiipage510hdooLevel1d81.Thefocallengthofaconvexlensis17cm.diomAcandleisplaced34cminfrontofthelens.Makearaydiagramtolocatethe(24cm)0.75image.32cmdi34cm111O1Ray1fdodiRay2di34cmfdodidodiFF(32cm)(24cm)Horizontalscale:32cm24cmI11block2cm14cm85.Anobjectislocated14.0cmfromaconvex82.Aconverginglenshasafocallengthoflensthathasafocallengthof6.0cm.The25.5cm.Ifitisplaced72.5cmfromanobjectis2.4cmhigh.object,atwhatdistancefromthelenswilla.Drawaraydiagramtodeterminetheloca-theimagebe?tion,size,andorientationoftheimage.111fddoiRay1O1dofhi1.8cmdRay2iddi10.5cmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofFF(72.5cm)(25.5cm)72.5cm25.5cmI1Horizontalscale:39.3cm1block1.0cmVerticalscale:Theimageis39.3cmfromthelens.1block0.4cmPhysics:PrinciplesandProblemsSolutionsManual389

393Chapter18continuedb.Solvetheproblemmathematically.b.Iftheoriginallensisreplacedwithalens111havingtwicethefocallength,whatarefddtheimageposition,size,andorientation?oidoffnew2fdidof2(6.00cm)(14.0cm)(6.0cm)12.0cm14.0cm6.0cm10.5cm111fddoihdmiihododdofnewi,newdofnewdhihoid(15.0cm)(12.0cm)o15.0cm12.0cm(10.5cm)(2.4cm)60.0cm14.0cmhd1.8cm,sotheimageismiihdinvertedoodhi,newho86.A3.0-cm-tallobjectisplaced22cminfronti,newdoofaconverginglens.Arealimageisformed(60.0cm)(3.0cm)11cmfromthelens.Whatisthesizeofthe15cmimage?12cmhdmiihdTheimageisinvertedcomparedtoootheobject.dhihoido88.AdiverginglenshasafocallengthofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(11cm)(3.0cm)15.0cm.Anobjectplacednearitformsa22cm2.0-cm-highimageatadistanceof5.0cm1.5cmfromthelens.Theimageis1.5cmtall.a.Whataretheobjectpositionandobjectheight?111Level3fddoi87.A3.0-cm-tallobjectisplaced15.0cmindfrontofaconverginglens.Arealimageisdifodformed10.0cmfromthelens.ifa.Whatisthefocallengthofthelens?(5.0cm)(15.0cm)5.0cm(15.0cm)1117.5cmfddoihddiifodimhddooodi(15.0cm)(10.0cm)dohih15.0cm10.0cmodi6.00cm(7.5cm)(2.0cm)5.0cm3.0cm390SolutionsManualPhysics:PrinciplesandProblems

394Chapter18continuedb.Thediverginglensisnowreplacedbyab.A1000.0-mmlensisfocusedonanconverginglenswiththesamefocalobject125maway.Whatistheimagelength.Whataretheimageposition,position?height,andorientation?Isitavirtual111imageorarealimage?fddiofnewfdofSodid(15.0cm)of(125m)(1.0000m)15.0cm125m1.00m1111.01m1.01103mmfddnewoi,newd90.EyeglassesToclearlyreadabook25cmdofnewi,newdaway,afarsightedgirlneedstheimagetobeofnew45cmfromhereyes.Whatfocallengthis(7.5cm)(15cm)neededforthelensesinhereyeglasses?7.5cm15cm11115cmfddiohididmodihdSofoododidhi,newho(25cm)(45cm)i,newdo25cm(45cm)(15cm)(3.0cm)56cm7.5cm6.0cmLevel2Thisisavirtualimagethatisupright91.CopyMachineTheconvexlensofacopycomparedtotheobject.machinehasafocallengthof25.0cm.Alettertobecopiedisplaced40.0cmfromthelens.18.3ApplicationsofLensesa.Howfarfromthelensisthecopypages510–511paper?Level111189.CameraLensesCameralensesarefddiodescribedintermsoftheirfocallength.dA50.0-mmlenshasafocallengthofdofid50.0mm.ofa.Acamerawitha50.0-mmlensis(40.0cm)(25.0cm)40.0cm25.0cmfocusedonanobject3.0maway.Whatistheimageposition?66.7cm111b.Howmuchlargerwillthecopybe?fddiohdiiSoddofhodoiCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.dofd(66.7cm)(hhihoo)(3.0103mm)(50.0mm)ido40.0cm3.0103mm50.0mm1.67ho51mmThecopyisenlargedandinverted.Physics:PrinciplesandProblemsSolutionsManual391

395Chapter18continued92.CameraAcameralenswithafocallength94.TelescopeTheopticalsystemofatoyof35mmisusedtophotographadistantrefractingtelescopeconsistsofaconvergingobject.Howfarfromthelensistherealobjectivelenswithafocallengthofimageoftheobject?Explain.20.0cm,located25.0cmfromaconverging35mm;foradistantobject,deyepiecelenswithafocallengthofocanbeconsideredat,thus1/d4.05cm.Thetelescopeisusedtoviewaoiszero.Accordingtothethinlensequation,10.0-cm-highobject,located425cmfromdtheobjectivelens.if.a.Whataretheimageposition,height,Level3andorientationasformedbytheobjec-93.MicroscopeAslideofanonioncellistivelens?Isthisarealorvirtualimage?placed12mmfromtheobjectivelensof111amicroscope.Thefocallengthofthefdodiobjectivelensis10.0mm.ddofa.Howfarfromthelensistheimageidofformed?(425cm)(20.0cm)111425cm20.0cmfddio21.0cmdSodofhididmiofhdoo(12mm)(10.0mm)d12mm10.0mmhihoid1mmo6.010(21.0cm)(10.0cm)b.Whatisthemagnificationofthisimage?425cmdi6.0101mmm0.494cmod12mm5.0oThisisarealimagethatisinvertedCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.Therealimageformedislocatedcomparedtotheobject.10.0mmbeneaththeeyepiecelens.Ifthefocallengthoftheeyepieceisb.Theobjectivelensimagebecomesthe20.0mm,wheredoesthefinalimageobjectfortheeyepiecelens.Whatareappear?theimageposition,height,andorienta-tionthatapersonseeswhenlooking111fddintothetelescope?Isthisarealoriovirtualimage?ddofdido,new25.0cmdiof(10.0mm)(20.0mm)25.0cm21.0cm10.0mm20.0mm4.0cm20.0mm,or20.0mmbeneaththeeyepiece111fddnewo,newi,newd.Whatisthefinalmagnificationofthisdcompoundsystem?do,newfnewi,newddo,newfnewi(20.0mm)medo10.0mm2.00(4.0cm)(4.05cm)4.0cm4.05cmmtotalmome(5.0)(2.00)2cm3.2101.0101392SolutionsManualPhysics:PrinciplesandProblems

396Chapter18continuedh97.A3.0-cm-tallobjectisplaced20cminfronto,newhiofaconverginglens.Arealimageisformed0.494cm10cmfromthelens.Whatisthefocalhdlengthofthelens?miihodo111dfdodihi,newho,newi,newddo,newodif(3.2102cm)(0.494cm)dodi4.0cm(20cm)(10cm)1cm20cm10cm4.0107cmThisisavirtualimagethatisinvert-edcomparedtotheobject.c.WhatisthemagnificationoftheLevel298.Derivensintelescope?1/sin2fromthegeneralformofSnell’slawofrefraction,hmi,newn1sin1n2sin2.Stateanyassumptionshoandrestrictions.4.0101cmTheangleofincidencemustbeinair.If10.0cmweletsubstance1beair,thenn14.01.000.Letn2n.Therefore,nMixedReview1sin1n2sin2pages511—512sin1nsin2Level1sin1n95.Ablockofglasshasacriticalangleof45.0°.sin2Whatisitsindexofrefraction?99.AstronomyHowmanymoreminutesnsin2wouldittakelightfromtheSuntoreachcn1Earthifthespacebetweenthemwerefilledn2withwaterratherthanavacuum?TheSunn1sinc,n21.00forairis1.5108kmfromEarth.1.00Timethroughvacuumn1sin45.0°d(1.5108km)(1000m/1km)t1.41c3.00108m/s5.0102s96.Findthespeedoflightinantimonytrioxideifithasanindexofrefractionof2.35.Speedthroughwatercc3.00108m/snvvn1.33c2.26108m/svnTimethroughwater3.00108m/s8km)(1000m/1km)d(1.510Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2.35t8m/sv2.26101.28108m/s660st660s500s160s(160s)(1min/60s)2.7minPhysics:PrinciplesandProblemsSolutionsManual393

397Chapter18continued100.Whatisthefocallengthofthelensesinb.Atwhatdepthdoesthebottomoftheyoureyeswhenyoureadabookthatistankappeartobeifyoulookintothe35.0cmfromthem?Thedistancefromwater?Dividethisapparentdepthintoeachlenstotheretinais0.19mm.thetruedepthandcompareittothe111indexofrefraction.fdodiUsingrighttrianglegeometry,(actu-dodialdepth)(tan1)(apparentfdodidepth)(tan2)(350mm)(0.19mm)tan5.0°apparentdepth(12cm)350mm0.19mmtan6.7°0.19mm8.9cmTherefractedraysappeartointer-sect8.9cmbelowthesurface;thisLevel3istheapparentdepth.101.ApparentDepthSunlightreflectsdiffu-apparentdepth8.9cmsivelyoffthebottomofanaquarium.0.74truedepth12cmFigure18-29showstwoofthemanylightnair1.0raysthatwouldreflectdiffusivelyfromaAlso,0.75n1.33pointoffthebottomofthetankandtravelwatertothesurface.ThelightraysrefractintotheTherefore,airasshown.Thereddashedlineextendingapparentdepthnairbackfromtherefractedlightrayisasighttruedepthnwaterlinethatintersectswiththeverticalrayat102.Itisimpossibletoseethroughadjacentsidesthelocationwhereanobserverwouldseeofasquareblockofglasswithanindexoftheimageofthebottomofthetank.refractionof1.5.Thesideadjacenttothesidethatanobserverislookingthroughacts2asamirror.Figure18-30showsthelimitingcasefortheadjacentsidetonotactlikeaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mirror.Useyourknowledgeofgeometryandcriticalanglestoshowthatthisrayconfigu-rationisnotachievablewhennglass1.5.12cmnglass1.55.0°190°■Figure18-29a.ComputethedirectionthattheAir2refractedraywilltravelabovethesurfaceofthewater.n11sin1n2sin2n11sin12sinnAir2290°1(1.33)(sin5.0°)sin1.0■Figure18-306.7°Thelightrayenterstheglassatanangle1andisrefractedtoanangle2.394SolutionsManualPhysics:PrinciplesandProblems

398Chapter18continuedn1AsinAThinkingCritically2sinngpage5121(1.00)(sin90°)104.RecognizeSpatialRelationshipsWhitesin1.5lighttravelingthroughair(n1.0003)42°entersaslabofglass,incidentatexactly45°.Fordenseflintglass,n1.7708forTherefore,148°.bluelight(435.8nm)andn1.7273Butthecriticalangleforglassisforredlight(643.8nm).Whatisthenangulardispersionoftheredandblue1Acsinnlight?gFindtheanglesofrefractionforred11.00sinandbluelight,andfindthedifference1.5inthoseanglesindegrees.42°UseSnell’slaw,nBecause1sin1n2sin2.1c,thelightreflectsbackintotheglassandonecannotThus,1n1sin1seeoutofanadjacentside.2sinn2Forredlight103.BankTellerWindowA25-mm-thicksheetofplastic,n1.5,isusedinabankteller’s1(1.0003)(sin45.000°)2sin1.7273window.Arayoflightstrikesthesheetatanangleof45°.Therayleavesthesheetat24.173°45°,butatadifferentlocation.UsearayForbluelightdiagramtofindthedistancebetweenthe1(1.0003)(sin45.000°)raythatleavesandtheonethatwouldhave2sin1.7708leftiftheplasticwerenotthere.23.543°8mmDifference24.173°23.543°0.630°105.CompareandContrastFindthecriticalangleforice(n1.31).Inaverycold25mmworld,wouldfiber-opticcablesmadeofmiceorthosemadeofglassdoabetterjobm8ofkeepinglightinsidethecable?Explain.nsinaircnice45°1nair11.00csinnsin1.3149.8°iceIncomparison,thecriticalanglefor28°glass,n1.54,is40.5°.Thelargercrit-icalanglemeansthatfewerrayswould45°betotallyinternallyreflectedinaniceCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.corethaninaglasscore.Thus,theywouldnotbeabletotransmitasmuchlight.Fiberopticcablesmadeofglasswouldworkbetter.Physics:PrinciplesandProblemsSolutionsManual395

399Chapter18continued106.RecognizeCauseandEffectYourlabWritinginPhysicspartnerusedaconvexlenstoproduceanpage512imagewithdi25cmandhi4.0cm.109.Theprocessofaccommodation,wherebyYouareexaminingaconcavelenswithamusclessurroundingthelensintheeyefocallengthof15cm.Youplacethecontractorrelaxtoenabletheeyetoconcavelensbetweentheconvexlensandfocusoncloseordistantobjects,variesfortheoriginalimage,10cmfromtheimage.differentspecies.InvestigatethiseffectToyoursurprise,youseearealimageonfordifferentanimals.Prepareareportthewallthatislargerthantheobject.Youfortheclassshowinghowthisfinefocus-aretoldthattheimagefromtheconvexingisaccomplishedfordifferenteyelensisnowtheobjectfortheconcavelens,mechanisms.andbecauseitisontheoppositesideofAnswerswillvarydependingonthetheconcavelens,itisavirtualobject.Useanimalsselectedbythestudents.thesehintstofindthenewimagepositionandimageheightandtopredictwhether110.Investigatethelenssystemusedinanopticaltheconcavelenschangedtheorientationinstrumentsuchasanoverheadprojectororoftheoriginalimage.aparticularcameraortelescope.PrepareaThenewdo10cm.Thus,graphicsdisplayfortheclassexplaininghowfdo(15cm)(10cm)theinstrumentformsimages.did10cm(15cm)Answerswillvary.Studentsmayfindof30cmthatitisnecessarytosimplifytheirchosensystemforexplanationdi30cmm3purposes.d10cmohimho(3)(4.0cm)10cmCumulativeReviewTheimageorientationisnotchanged.page512111.Ifyoudropa2.0kgbagofleadshotfromaheightof1.5m,youcouldassumeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.107.DefineOperationallyNameanddescribetheeffectthatcausestherainbow-coloredthathalfofthepotentialenergywillbefringecommonlyseenattheedgesofaconvertedintothermalenergyinthelead.spotofwhitelightfromaslideorTheotherhalfwouldgotothermalenergyoverheadprojector.inthefloor.Howmanytimeswouldyouhavetodropthebagtoheatitby10°C?Thelightthatpassesthroughalens(Chapter12)neartheedgesofthelensisslightlydispersed,sincetheedgesofalensPEmghresembleaprismandrefractdifferent(2.0kg)(9.80m/s2)(1.5m)wavelengthsoflightatslightlydifferent29.4Jangles.TheresultisthatwhitelightisToheatthebagdispersedintoitsspectrum.Theeffectiscalledchromaticaberration.QmCT(2.0kg)(130J/kg°C)(10°C)108.ThinkCriticallyAlensisusedtoproject2600Jtheimageofanobjectontoascreen.Q2600JSupposethatyoucovertherighthalfofN11thelens.Whatwillhappentotheimage?PE(29.4J)22Itwillgetdimmer,becausefewerlight180timesrayswillconverge,butyouwillseeacompleteimage.396SolutionsManualPhysics:PrinciplesandProblems

400Chapter18continued112.Ablacksmithputsanironhooportireonb.HowwouldtheilluminanceonyourtheouterrimofawoodencarriagewheelhandfromtheSunatthenewpositionbyheatingthehoopsothatitexpandstoacomparetowhatitwaswhenyouwerediametergreaterthanthewoodenwheel.standingonitssurface?(Forsimplicity,Whenthehoopcools,itcontractstoholdassumethattheSunisapointsourcetheriminplace.Ifablacksmithhasaatbothpositions.)woodenwheelwitha1.0000-mdiameter116Itis110andwantstoputarimwitha0.9950-m100021,000,000diameteronthewheel,whatisthethevalueitwasoriginally.minimumtemperaturechangetheironc.Comparetheeffectofdistanceuponmustexperience?(6/°C)iron1210thegravitationalforceandilluminance.(Chapter13)TheybothfollowtheinversesquareLironLTwhereListhediameterlawofdistance.oftheironhoop.WewantLgreaterthan0.0050m.115.Beautician’sMirrorThenoseofaTherefore,customerwhoistryingsomefacepowderLis3.00-cmhighandislocated6.00cminTironLfrontofaconcavemirrorhavinga14.0-cmfocallength.Findtheimagepositionand0.0050(12106/°C)(0.9950m)heightofthecustomer’snosebymeansofthefollowing.(Chapter17)420°Ca.araydiagramdrawntoscaleActuallyhewouldheatitmuchhottertogiveroomtofitoverthewheeleasily.O1I1113.Acarsoundsitshornasitapproachesapedestrianinacrosswalk.WhatdoesthepedestrianhearasthecarbrakestoallowHorizontalscale:hi5.25cmhimtocrossthestreet?(Chapter15)1block1.0cmdi10.5cmVerticalscale:Thepitchofthehornheardbythe1block1.0cmpedestrianwilldecreaseasthecarslowsdown.b.themirrorandmagnificationequations111114.SupposethatyoucouldstandonthefdodisurfaceoftheSunandweighyourself.AlsosupposethatyoucouldmeasurethedofdidilluminanceonyourhandfromtheSun’sofvisiblespectrumproducedatthatposition.(6.00cm)(14.0cm)Next,imagineyourselftravelingtoa6.00cm14.0cmposition1000timesfartherawayfromthe10.5cmcenteroftheSunasyouwerewhenhdstandingonitssurface.(Chapter16)miihdooa.HowwouldtheforceofgravityonyoufromtheSunatthenewpositioncom-dihoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.hiparetowhatitwasatthesurface?doItis111106(10.5cm)(3.00cm)(1000)21,000,0006.00cmthevalueitwasoriginally.5.25cmPhysics:PrinciplesandProblemsSolutionsManual397

401Chapter18continued4.IfyouwanttheangleofrefractionfortheChallengeProblemlightrayinwatertobethesameasitispage501forair,whatshouldthenewangleofinci-Aslightenterstheeye,itfirstencountersthedencebe?air/corneainterface.Considerarayoflightthatn1sin1n2sin2strikestheinterfacebetweentheairandaper-nson’scorneaatanangleof30.0°tothenormal.12sin21sinnTheindexofrefractionofthecorneaisapproxi-1mately1.4.(1.4)(sin21°)sin11.UseSnell’slawtocalculatetheangleof1.33refraction.22°n1sin1n2sin2nAirCornea11sin12sinn2(1.0)(sin30.0°)sin11.430.0°21°22.Whatwouldtheangleofrefractionbeifthepersonwasswimmingunderwater?n1sin1n2sin2n11sin12sinn2(1.33)(sin30.0°)sin11.428°Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.Istherefractiongreaterinairorinwater?Doesthismeanthatobjectsunderwaterseemcloserormoredistantthantheywouldinair?Refractionisgreaterinairbecausetheangletothenormalissmaller.Objectsseemcloserinwater.398SolutionsManualPhysics:PrinciplesandProblems

402CHAPTERInterferenceand19Diffraction4.Yellow-orangelightwithawavelengthofPracticeProblems596nmpassesthroughtwoslitsthatare19.1Interferenceseparatedby2.25105mandmakesanpages515–523interferencepatternonascreen.Ifthedis-tancefromthecentrallinetothefirst-orderpage519yellowbandis2.00102m,howfaristhe1.Violetlightfallsontwoslitsseparatedby1.90105m.Afirst-orderbrightbandscreenfromtheslits?appears13.2mmfromthecentralbrightxdLbandonascreen0.600mfromtheslits.Whatis?Lxdxd25L(2.0010m)(2.2510m)959610m(13.2103m)(1.90105m)0.600m0.755m418nmpage5222.Yellow-orangelightfromasodiumlampof5.InthesituationinExampleProblem2,wavelength596nmisaimedattwoslitsthatareseparatedby1.90105m.Whatwhatwouldbethethinnestfilmthatwouldcreateareflectedred(635nm)band?isthedistancefromthecentralbandtothefirst-orderyellowbandifthescreenis12tm2n0.600mfromtheslits?oilLForthethinnestfilm,m0.xd1t(13.2103m)(1.90105m)4noil0.600m635nm1.88102m18.8mm(4)(1.45)109nm3.Inadouble-slitexperiment,physicsstu-dentsusealaserwith632.8nm.A6.Aglasslenshasanonreflectivecoatingstudentplacesthescreen1.000mfromtheplacedonit.Ifafilmofmagnesiumslitsandfindsthefirst-orderbrightbandfluoride,n1.38,isplacedontheglass,65.5mmfromthecentralline.Whatisthen1.52,howthickshouldthelayerbetoslitseparation?keepyellow-greenlightfrombeingreflected?xdBecausenfilmnair,thereisaphaseLinversiononthefirstreflection.LdBecausenCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.xglassnfilm,thereisaphase(632.8109m)(1.000m)inversiononthesecondreflection.65.5103mFordestructiveinterferencetokeepyellow-greenfrombeingreflected:9.66106m9.66m12tm2nfilmPhysics:PrinciplesandProblemsSolutionsManual399

403Chapter19continuedForthethinnestfilm,m0.Forthethinnestfilm,m1.1tt4nfilm2nfilm555nm521nm(4)(1.38)(2)(1.33)101nm196nm7.Asiliconsolarcellhasanonreflective9.Whatisthethinnestsoapfilm(n1.33)coatingplacedonit.Ifafilmofsodiumforwhichlightofwavelength521nmwillmonoxide,n1.45,isplacedonthesili-constructivelyinterferewithitself?con,n3.5,howthickshouldthelayerbeForconstructiveinterferencetokeepyellow-greenlight(555nm)1frombeingreflected?2tm2nfilmBecausenfilmnair,thereisaphaseForthethinnestfilm,m0.inversiononthefirstreflection.1Becausentsiliconnfilm,thereisaphase4nfilminversiononthesecondreflection.521nmFordestructiveinterferencetokeep(4)(1.33)yellow-greenfrombeingreflected:97.9nm12tm2nfilmForthethinnestfilm,m0.SectionReview1t19.1Interference4nfilmpages515–523555nmpage523(4)(1.45)10.FilmThicknessLucienisblowingbubblesCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.95.7nmandholdsthebubblewandupsothatasoapfilmissuspendedverticallyintheair.Whatis8.Youcanobservethin-filminterferencebythesecondthinnestwidthofthesoapfilmatdippingabubblewandintosomebubblewhichhecouldexpecttoseeabrightstripeifsolutionandholdingthewandintheair.thelightilluminatingthefilmhasawave-Whatisthethicknessofthethinnestsoaplengthof575nm?Assumethesoapsolutionfilmatwhichyouwouldseeablackstripehasanindexofrefractionof1.33.ifthelightilluminatingthefilmhasawavelengthof521nm?Usen1.33.Thereisonephaseinversion,soconstructiveinterferencewillbewhenBecausenfilmnair,thereisaphase1changeonthefirstreflection.Because2tm2nfilmnairnfilm,thereisnophasechangeonForthesecondthinnestthickness,m1.thesecondreflection.Fordestructiveinterferencetogeta3t4nblackstripefilmm(3)(575nm)2tn(4)(1.33)film324nm400SolutionsManualPhysics:PrinciplesandProblems

404Chapter19continued11.BrightandDarkPatternsTwoverynar-reflection.Forconstructiveinterfer-rowslitsarecutclosetoeachotherinalargeencetoreflectyellow-greenlight:pieceofcardboard.Theyareilluminatedby1monochromaticredlight.Asheetofwhite2tm2nfilmpaperisplacedfarfromtheslits,andapat-Forthethinnestfilm,m0.ternofbrightanddarkbandsisseenonthepaper.Describehowawavebehaveswhenit1colort4nencountersaslit,andexplainwhysomefilmregionsarebrightwhileothersaredark.555nmWhenawaveencountersaslit,the(4)(1.83)wavebends.Lightisdiffractedbythe75.8nmslits.Lightfromoneslitinterfereswithb.Unfortunately,afilmthisthincannotlightfromtheother.Ifinterferenceisbemanufactured.Whatisthenext-constructive,thereisabrightband;ifthinnestfilmthatwillproducethedestructive,theregionisdark.sameeffect?Forthenextthinnestfilm,m1.12.InterferencePatternsSketchthepatterndescribedinproblem11.3colort4nfilmRepeatsRepeatsandfadesandfades(3)(555nm)(4)(1.83)227nmBlackRedBlackRedBlackRedBlack15.CriticalThinkingTheequationforwave-13.InterferencePatternsSketchwhathap-lengthfromadouble-slitexperimentusespenstothepatterninproblem11whenthethesimplificationthatissmallsothatredlightisreplacedbybluelight.sintan.UptowhatangleisthisaRepeatsRepeatsgoodapproximationwhenyourdatahasandfadesandfadestwosignificantfigures?Wouldthemaxi-mumangleforavalidapproximationRedRedRedincreaseordecreaseasyouincreasetheBlackBlackBlackBlackprecisionofyouranglemeasurement?sintantotwosignificantdigitsupBlueBlackBlueBlackBlueBlackBlueBlackBlueto9.9°.Anincreaseintheprecisionofthemeasurementreducesthisangleto2.99°.Thelightbandsbecomemorecloselyspaced.PracticeProblems14.FilmThicknessAplasticreflectingfilm19.2Diffraction(n1.83)isplacedonanautoglasspages524–531window(n1.52).page526a.Whatisthethinnestfilmthatwill16.Monochromaticgreenlightofwavelengthreflectyellow-greenlight?546nmfallsonasingleslitwithawidthCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Becausenfilmnair,thereisaphaseof0.095mm.Theslitislocated75cmfromchangeonthefirstreflection.ascreen.HowwidewillthecentralbrightBecausenglassnfilm,thereisnotbandbe?aphasechangeonthesecondPhysics:PrinciplesandProblemsSolutionsManual401

405Chapter19continuedx20.Whitelightfallsonasingleslitthatisminw0.050mmwide.Ascreenisplaced1.00mLaway.Astudentfirstputsablue-violetfilterLxminw(441nm)overtheslit,thenaredfilter(5.46107m)(0.75m)(622nm).Thestudentmeasuresthe9.5105mwidthofthecentralbrightband.a.Whichfilterproducedthewiderband?4.3mmRed,becausecentralpeakwidthis17.Yellowlightwithawavelengthof589nmproportionaltowavelength.passesthroughaslitofwidth0.110mmandb.Calculatethewidthofthecentralbrightmakesapatternonascreen.Ifthewidthofbandforeachofthetwofilters.thecentralbrightbandis2.60102m,2Lhowfarisitfromtheslitstothescreen?2x1w2x2LForblue,1w2(4.41107m)(1.00m)(2x1)w2x15.0105mL218mm(2.60102m)(0.110103m)Forred,(2)(589109m)2(6.22107m)(1.00m)2x2.43m15.0105m25mm18.LightfromaHe-Nelaser(632.8nm)fallsonaslitofunknownwidth.Apatternisformedonascreen1.15maway,onpage529whichthecentralbrightbandis15mm21.Whitelightshinesthroughagratingontoawide.Howwideistheslit?screen.Describethepatternthatisproduced.2x2LAfullspectrumofcolorisseen.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1wBecauseofthevarietyofwavelengths,2Ldarkfringesofonewavelengthwouldw2x1befilledbybrightfringesofanother(2)(632.8109m)(1.15m)color.15103m22.Ifbluelightofwavelength434nmshines9.7105monadiffractiongratingandthespacing97moftheresultinglinesonascreenthatis1.05mawayis0.55m,whatisthespacing19.Yellowlightfallsonasingleslit0.0295mmbetweentheslitsinthegrating?wide.Onascreenthatis60.0cmaway,thedsincentralbrightbandis24.0mmwide.Whatisthewavelengthofthelight?dwheretan1xsinL2L2x1wsintan1x(2x)wL12L434109(24.0103m)(0.0295103m)sintan10.55m(2)(60.0102m)1.05m5.90102nm9.410–7m402SolutionsManualPhysics:PrinciplesandProblems

406Chapter19continued23.Adiffractiongratingwithslitsseparatedby8.60107misilluminatedbyvioletlightSectionReviewwithawavelengthof421nm.Ifthescreenis19.2Diffraction80.0cmfromthegrating,whatisthesepa-pages524–531rationofthelinesinthediffractionpattern?page531dsin26.DistanceBetweenFirst-OrderDarkBandsMonochromaticgreenlightofwave-sindlength546nmfallsonasingleslitofwidthx0.080mm.Theslitislocated68.0cmfromtanLascreen.WhatistheseparationofthefirstxLtandarkbandsoneachsideofthecentralbrightband?1Ltansind2L2xminw1421109m(0.800m)tansin8.60107m(2)(546109m)(68.0102m)0.080103m0.449m9.3mm24.BluelightshinesontheDVDinExampleProblem3.Ifthedotsproducedona27.DiffractionPatternsManynarrowslitsarewallthatis0.65mawayareseparatedclosetoeachotherandequallyspacedinaby58.0cm,whatisthewavelengthoflargepieceofcardboard.Theyareillumi-thelight?natedbymonochromaticredlight.Asheetofwhitepaperisplacedfarfromtheslits,1xdsindsintanandapatternofbrightanddarkbandsisLvisibleonthepaper.Sketchthepatternthat(7.41107m)sintan10.58mwouldbeseenonthescreen.0.65m490nmPatternPatternrepeatsrepeats25.Lightofwavelength632nmpassesthroughadiffractiongratingandcreatesapatternonascreenthatis0.55maway.Ifthefirstbrightbandis5.6cmfromthecentralbrightband,howmanyslitspercentimeterdoesthegratinghave?Brightredlinesdsin1Thereisoneslitperdistanced,soBandspacingisexactlythesameasindgivesslitspercentimeter.thepatternproducedbythetwoslits,butnowlightbandsaremuchthinnerdsinsintan1xandseparatedbywiderdarkbands.L632109m28.LineSpacingYoushinearedlaserlightthroughonediffractiongratingandformasintan10.056mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.55mpatternofreddotsonascreen.Thenyou6.210–6m6.210–4cmsubstituteaseconddiffractiongratingforthefirstone,formingadifferentpattern.1slit3slits/cmThedotsproducedbythefirstgratingare1.6106.210–4cmspreadoutmorethanthoseproducedbythesecond.WhichgratinghasmorelinesPhysics:PrinciplesandProblemsSolutionsManual403

407Chapter19continuedpermillimeter?Completethefollowingconceptmapusing,L,anddtoindicatehowyoucouldvaryxd,sothegreaterthedotspacings,themtoproducetheindicatedchangeintheLx,thenarrowertheslitspacing,d,andspacingbetweenadjacentbrightbands,x.thusmorelinespermillimeter.xd29.RayleighCriterionThebrighteststarintheLwinterskyinthenorthernhemisphereisSirius.Inreality,Siriusisasystemoftwotoincreasex,todecreasex,starsthatorbiteachother.IftheHubbleSpaceTelescope(diameter2.4m)ispointedattheSiriussystem,whichis8.44light-yearsfromincreasedecreaseincreasedecreaseEarth,whatistheminimumseparationtherewouldneedtobebetweenthestarsinorderLddLforthetelescopetobeabletoresolvethem?Assumethattheaveragelightcomingfromthestarshasawavelengthof550nm.MasteringConcepts1.22Lobjpage536xobjD32.Whyisitimportantthatmonochromaticlight1.22(330109m)(7.991016m)wasusedtomaketheinterferencepatternin2.4mYoung’sinterferenceexperiment?(19.1)2.21010mWhenmonochromaticlightisused,yougetasharpinterferencepattern;ifyou30.CriticalThinkingYouareshownausewhitelight,yougetsetsofcoloredspectrometer,butdonotknowwhetherbands.itproducesitsspectrumwithaprismoragrating.Bylookingatawhite-light33.Explainwhythepositionofthecentralspectrum,howcouldyoutell?brightbandofadouble-slitinterferenceCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.patterncannotbeusedtodeterminetheDetermineifthevioletortheredendofwavelengthofthelightwaves.(19.1)thespectrummakesthelargestanglewiththedirectionofthebeamofinci-Allwavelengthsproducethelineinthedentwhitelight.Aprismbendsthevio-sameplace.letendofthespectrumthemost,whereasagratingdiffractsredwave-34.Describehowyoucoulduselightofalengthsthemost.knownwave-lengthtofindthedistancebetweentwoslits.(19.1)Letthelightfallonthedoubleslit,andChapterAssessmentlettheinterferencepatternfallonasheetofpaper.MeasurethespacingsConceptMappingbetweenthebrightbands,x,andusepage536Ltheequationd.31.MonochromaticlightofwavelengthxilluminatestwoslitsinaYoung’sdouble-slit35.Describeinyourownwordswhathappensinexperimentsetupthatareseparatedbyathin-filminterferencewhenadarkbandisdistance,d.Apatternisprojectedontoaproducedbylightshiningonasoapfilmsus-screenadistance,L,awayfromtheslits.pendedinair.Makesureyouincludeinyourexplanationhowthewavelengthofthelightandthicknessofthefilmarerelated.(19.1)404SolutionsManualPhysics:PrinciplesandProblems

408Chapter19continuedWhenthelightstrikesthefrontofthe39.Foragivendiffractiongrating,whichcolorfilm,somereflectsoffthissurfaceandofvisiblelightproducesabrightlineclosestsomepassesthroughthefilmandtothecentralbrightband?(19.2)reflectsoffthebacksurfaceofthefilm.violetlight,thecolorwiththesmallestWhenlightreflectsoffamediumwithawavelengthhigherindexofrefraction,itundergoesaphaseshiftofone-halfwavelength;ApplyingConceptsthishappenstothelightthatinitiallypage536reflects.Inorderforadarkbandtobe40.Foreachofthefollowingexamples,indicateproduced,thetwolightraysmustbewhetherthecolorisproducedbythin-filmone-halfwavelengthoutofphase.Iftheinterference,refraction,orthepresenceofthicknessofthefilmissuchthattheraypigments.reflectingoffthebacksurfacegoesa.soapbubblesthroughawholenumberofcycleswhilepassingthroughthefilm,thelightraysinterferencearrivingatyoureyewillbeoutofphaseb.rosepetalsanddestructivelyinterfere.Rememberpigmentsthatthewavelengthisalteredbythec.oilfilmsindexofrefractionofthefilm,sothatthethicknessofthefilmmustequalainterferencemultipleofhalfawavelengthofthed.arainbowlight,dividedbythefilm’sindexofrefractionrefraction.41.Howcanyoutellwhetherapatternispro-36.Whitelightshinesthroughadiffractionducedbyasingleslitoradoubleslit?grating.AretheresultingredlinesspacedAdouble-slitinterferencepatterncon-morecloselyorfartherapartthanthesistsofequallyspacedlinesofalmostresultingvioletlines?Why?(19.2)equalbrightness.Asingle-slitdiffrac-Thespacingisdirectlyproportionaltotionpatternhasabright,broadcentralthewavelength,andbecauseredlightbandanddimmersidebands.hasalongerwavelengththanviolet,theredlineswillbespacedfartherapart42.Describethechangesinasingle-slitthanthevioletlines.diffractionpatternasthewidthoftheslitisdecreased.37.WhydodiffractiongratingshavelargeThebandsgetwideranddimmer.numbersofslits?Whyaretheseslitssoclosetogether?(19.2)43.ScienceFairAtasciencefair,oneexhibi-Thelargenumberofgroovesindiffrac-tionisaverylargesoapfilmthathasafairlytiongratingsincreasestheintensityofconsistentwidth.Itisilluminatedbyalightthediffractionpatterns.Thegroovesarewithawavelengthof432nm,andnearlytheclosetogether,producingsharperentiresurfaceappearstobealovelyshadeofimagesoflight.purple.Whatwouldyouseeinthefollowingsituations?38.Whywouldatelescopewithasmalldiam-a.thefilmthicknesswasdoubledCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.eternotbeabletoresolvetheimagesofcompletedestructiveinterferencetwocloselyspacedstars?(19.2)b.thefilmthicknesswasincreasedbyhalfSmallapertureshavelargediffractionawavelengthoftheilluminatinglightpatternsthatlimitresolution.completeconstructiveinterferencePhysics:PrinciplesandProblemsSolutionsManual405

409Chapter19continuedc.thefilmthicknesswasdecreasedbyonequarterofawavelengthoftheilluminatinglight19.0mcompletedestructiveinterference44.Whatarethedifferencesinthecharacteris-ticsofthediffractionpatternsformedby80.0cmdiffractiongratingscontaining104lines/cmDoubleslitScreenand105lines/cm?■Figure19-17(Nottoscale)Thelinesinthediffractionpatternarexdnarrowerforthe105lines/cmgrating.L(19.0106m)(1.90102m)45.LaserPointerChallengeYouhavetwo80.0102mlaserpointers,aredoneandagreenone.451nmYourfriendsMarkandCarlosdisagreeaboutwhichhasthelongerwavelength.Mark48.OilSlickAfterashortspringshower,Tominsiststhatredlighthasalongerwavelength,andAnntaketheirdogforawalkandwhileCarlosissurethatgreenhasthelongernoticeathinfilmofoil(n1.45)onawavelength.Youhaveadiffractiongratingpuddleofwater,producingdifferentcolors.handy.DescribewhatdemonstrationyouWhatistheminimumthicknessofaplacewoulddowiththisequipmentandhowyouwheretheoilcreatesconstructiveinterfer-wouldexplaintheresultstoCarlosandMarkenceforlightwithawavelengthof545nm?tosettletheirdisagreement.Thereisonephaseinversion,soShineeachlaserpointerthroughthegrat-constructiveinterferencewillbewheningontoanearbywall.Thecolorwiththe1longerwavelengthwillproducedotswith2tm2nfilmagreaterspacingonthewallbecausetheFortheminimumthickness,m0.spacingisdirectlyproportionaltothewavelength.(Markiscorrect;redlighthast1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4nalongerwavelengththangreenlight.)film545nm46.OpticalMicroscopeWhyisbluelightused(4)(1.45)forilluminationinanopticalmicroscope?94.0nmLessdiffractionresultsfromtheshortwavelengthofbluelight.Level249.Lightofwavelength542nmfallsonaMasteringProblemsdoubleslit.First-orderbrightbandsappear19.1Interference4.00cmfromthecentralbrightband.Thepages536—537screenis1.20mfromtheslits.Howfarapartaretheslits?Level147.Lightfallsonapairofslits19.0mapartxdand80.0cmfromascreen,asshowninLFigure19-17.Thefirst-orderbrightbandisLdx1.90cmfromthecentralbrightband.Whatisthewavelengthofthelight?(5.42107m)(1.20m)4.00102m16.3m50.InsulationFilmWinterisapproachingandAlejandroishelpingtocoverthewindows406SolutionsManualPhysics:PrinciplesandProblems

410Chapter19continuedinhishomewiththinsheetsofclearplasticfirst-orderbrightband,fromleasttomost(n1.81)tokeepthedraftsout.Aftertheseparation.Specificallyindicateanyties.plasticistapeduparoundthewindowssuchxdthatthereisairbetweentheplasticandtheLglasspanes,theplasticisheatedwithahairLxdryertoshrink-wrapthewindow.Thethick-dnessoftheplasticisalteredduringthisBecauseisthesameforeachsetup,process.Alejandronoticesaplaceonthexcalculatetocomparethesetups.plasticwherethereisabluestripeofcolor.Herealizesthatthisiscreatedbythin-filmxLinterference.Whatarethreepossiblethick-dnessesoftheportionoftheplasticwhereSetupA:thebluestripeisproducedifthewavelength0.60mofthelightis4.40102nm?1.50104mThereisonephaseinversion,so34.010constructiveinterferencewillbewhenSetupB:12tm0.80m2nfilm4m1.7510Threepossiblethicknessesoccurat4.6103m0,1,and2.SetupC:1tform00.80m4nfilm1.50104m4.40102nm(4)(1.81)5.3103x60.8nmCxBxA3tform119.2Diffraction4nfilmpage537(3)(4.40102nm)Level1(4)(1.81)52.Monochromaticlightpassesthroughasin-182nmgleslitwithawidthof0.010cmandfalls5onascreen100cmaway,asshownintform24nfilmFigure19-18.Ifthewidthofthecentral(5)(4.40102nm)bandis1.20cm,whatisthewavelengthof(4)(1.81)thelight?304nmLevel351.Samirshinesaredlaserpointerthrough0.010cmthreedifferentdouble-slitsetups.InsetupA,theslitsareseparatedby0.150mmandthescreenis0.60mawayfromtheslits.In100cmSingleslitScreenCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.setupB,theslitsareseparatedby0.175mmandthescreenis0.80maway.SetupChas■Figure19-18(Nottoscale)theslitsseparatedby0.150mmandthe2Lscreenadistanceof0.80maway.Rankthe2x1wthreesetupsaccordingtotheseparationxwbetweenthecentralbrightbandandtheLPhysics:PrinciplesandProblemsSolutionsManual407

411Chapter19continued(0.60cm)(0.010cm)xobj1.22100cmLDobj600nm1.22Lobjx3linesobjD53.Agooddiffractiongratinghas2.510percm.Whatisthedistancebetweentwo(1.22)(5.1107m)(1.0105m)linesinthegrating?2.4m12.6102md2.5103lines/cm2.6cm4.0104cmLevel3Level256.Monochromaticlightwithawavelengthof54.Lightwithawavelengthof4.5105cm425nmpassesthroughasingleslitandpassesthroughasingleslitandfallsonafallsonascreen75cmaway.Ifthecentralscreen100cmaway.Iftheslitis0.015cmbrightbandis0.60cmwide,whatisthewide,whatisthedistancefromthecenterwidthoftheslit?ofthepatterntothefirstdarkband?2L2x1w2L2x1w2LLw2xx2x1isthewidthofthebrightband,so11togetthedistancefromthecenterto1thefirstdarkband,divideby2.x12(2x1)0.30cmL(4.25105cm)(75cm)x1w0.30cm(4.5105cm)(100m)1.1102cm0.015cm0.3cm57.KaleidoscopeJenniferisplayingwithakaleidoscopefromwhichthemirrorshaveCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.55.HubbleSpaceTelescopeSupposetheHubblebeenremoved.TheeyeholeattheendisSpaceTelescope,2.4mindiameter,isinorbit7.0mmindiameter.Ifshecanjustdistin-1.0105maboveEarthandisturnedtoguishtwobluish-purplespecksontheotherviewEarth,asshowninFigure19-19.Ifendofthekaleidoscopeseparatedby40m,youignoretheeffectoftheatmosphere,whatisthelengthofthekaleidoscope?howlargeanobjectcanthetelescopeUse650nmandassumethattheresolve?Use5.1107m.resolutionisdiffractionlimitedthroughtheeyehole.1.22LobjxobjDxobjDLobj1.22(40106m)(7.0103m)(1.22)(650109m)0.4m■Figure19-19408SolutionsManualPhysics:PrinciplesandProblems

412Chapter19continued58.SpectroscopeAspectroscopeusesagrat-632.8109mingwith12,000lines/cm.Findtheanglesat121103msintanwhichredlight,632nm,andbluelight,4.0m421nm,havefirst-orderbrightlines.1.2104m1.2102cmd18.33105cm1112,000lines/cmd1.2102cm83ridges/cmdsinb.Mariechecksherresultsbynotingthattheridgesrepresentasongthatlastssind4.01minutesandtakesup16mmonForredlight,therecord.Howmanyridgesshould16.32105cmtherebeinacentimeter?sin8.33105cmNumberofridgesis49.3°(4.01min)(33.3rev/min)134ridgesForbluelight,134ridges84ridges/cm14.21105cm1.6cmsin8.33105cmLevel230.3°60.Ananti-reflectivecoating,n1.2,isappliedtoalens.IfthethicknessoftheMixedReviewcoatingis125nm,whatis(are)thepage538color(s)oflightforwhichcompleteLevel1destructiveinterferencewilloccur?Hint:159.RecordMarieusesanold33rpmrecordAssumethelensismadeoutofglass.3asadiffractiongrating.Sheshinesalaser,Becausenfilmnair,thereisaphase632.8nm,ontherecord,asshownininversiononthefirstreflection.Figure19-20.Onascreen4.0mfromtheBecausenrecord,aseriesofreddots21mmapartarelens1.52nfilm,thereisaphaseinversiononthesecondreflection.visible.Fordestructiveinterference:Laser12dm2nfilm21mm2dnfilm1m2(2)(125nm)(1.2)21mmRecord1m2Screen112nm)4.0mm(3.0102■Figure19-20(Nottoscale)Form0a.Howmanyridgesarethereinacen-112nm)(3.010timeteralongtheradiusoftherecord?2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.dsin6.0102nmThelightisreddish-orange.Forotherdsinsintan1xvaluesofm,thewavelengthisshorterLthanthatoflight.Physics:PrinciplesandProblemsSolutionsManual409

413Chapter19continuedLevel363.ApplyConceptsBluelightofwavelength61.CameraWhenacamerawitha50-mmpassesthroughasingleslitofwidthw.Alensissetatf/8,itsaperturehasanopeningdiffractionpatternappearsonascreen.If6.25mmindiameter.younowreplacethebluelightwithagreena.Forlightwith550nm,whatislightofwavelength1.5,towhatwidththeresolutionofthelens?Thefilmisshouldyouchangetheslittogettheoriginal50.0mmfromthelens.patternback?1.22LobjTheangleofdiffractiondependsonthexobjDratioofslitwidthtowavelength.Thus,(1.22)(5.5104mm)(50.0mm)youwouldincreasethewidthto1.5w.6.25mm64.AnalyzeandConcludeAtnight,thepupil5.4103mmofahumaneyehasanaperturediameterb.Theownerofacameraneedstodecideof8.0mm.Thediameterissmallerinwhichfilmtobuyforit.Theexpensivedaylight.Anautomobile’sheadlightsareone,calledfine-grainedfilm,hasseparatedby1.8m.200grains/mm.Thelesscostly,coarse-a.BaseduponRayleigh’scriterion,howfargrainedfilmhasonly50grains/mm.Ifawaycanthehumaneyedistinguishthetheownerwantsagraintobenosmallertwoheadlightsatnight?Hint:Assumeathanthewidthofthecentralbrightspotwavelengthof525nm.calculatedinparta,whichfilmshould1.22Lhepurchase?objxobjDCentralbrightbandwidthx3mmLobjD2x10.710obj1.221(8.0103m)(1.80m)The200grains/mmfilmhas200mm7m)(1.22)(5.2510betweengrains5103mm,so2.2104m22kmthisfilmwillwork.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1b.Canyouactuallyseeacar’sheadlightsThe50grains/mmhas50mmatthedistancecalculatedinparta?betweengrains20103mm,soDoesdiffractionlimityoureyes’sensingthisfilmwon’twork.ability?Hypothesizeastowhatmightbethelimitingfactors.ThinkingCriticallyNo;afewhundredmeters,notsever-page538alkilometers,isthelimit.Diffraction62.ApplyConceptsYellowlightfallsonadif-doesn’tlimitthesensingabilityoffractiongrating.Onascreenbehindtheyoureyes.Moreprobablefactorsaregrating,youseethreespots:oneatzerotherefractiveeffectsoftheatmos-degrees,wherethereisnodiffraction,andphere,likethosethatcausestarstooneeachat30°and30°.Younowaddtwinkle,orthelimitationsofthereti-abluelightofequalintensitythatisinthenaandtheopticareaofthebraintosamedirectionastheyellowlight.Whatseparatetwodimsources.patternofspotswillyounowseeonthescreen?WritinginPhysicsAgreenspotat0°,yellowspotsat30°page538and30°,andtwobluespotsslightly65.ResearchanddescribeThomasYoung’scon-closerin.tributionstophysics.Evaluatetheimpactofhisresearchonthescientificthoughtaboutthenatureoflight.410SolutionsManualPhysics:PrinciplesandProblems

414Chapter19continuedStudentanswerswillvary.Answersa.WhatisthewavelengthofthesoundinshouldincludeYoung’stwo-slitexperi-stillair?mentthatallowedhimtoprecisely1.50mmeasurethewavelengthoflight.b.Ifthespeedofsoundis330m/s,whatisthefrequencyofthesource?66.Researchandinterprettheroleofdiffrac-tioninmedicineandastronomy.Describevfatleasttwoapplicationsineachfield.v330m/sf220HzStudentanswerswillvary.Answers1.50mcouldincludediffractionintelescopesc.Whatisthespeedoftheairplane?andmicroscopes,aswellasTheplanemovesforward0.50mspectroscopy.forevery1.50mthatthesoundwavetravels,sotheplane’sspeedisone-CumulativeReviewthirdthespeedofsound,or110m/s.page53867.Howmuchworkmustbedonetopusha70.Aconcavemirrorhasa48.0-cmradius.A0.5-m3blockofwoodtothebottomofa2.0-cm-tallobjectisplaced12.0cmfrom4-m-deepswimmingpool?Thedensityofthemirror.Calculatetheimagepositionwoodis500kg/m3.(Chapter13)andimageheight.(Chapter17)Theblockwouldfloat,buttosubmergeitrfwouldrequireanextraforcedownward.2WFd48.0cm2FFbuoyancyFg24.0cmFgVg111(500kg/m3)(0.5m3)(9.80m/s2)fdodid2450NdofidFofbuoyancyVg(1000kg/m3)(0.5m3)(12.0cm)(24.0cm)12.0cm24.0cm(9.80m/s2)24.0cm4900NhdWork(4900N2450N)(4m)miihdoo10kJdhhioid68.Whatarethewavelengthsofmicrowavesinoanoveniftheirfrequencyis2.4GHz?(24.0cm)(2.0cm)(Chapter14)12.0cmcf4.0cmc3.00108m/s9Hzf2.41071.Thefocallengthofaconvexlensis21.0cm.0.12mA2.00-cm-tallcandleislocated7.50cmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fromthelens.Usethethin-lensequation69.Soundwavecreststhatareemittedbyantocalculatetheimagepositionandimageairplaneare1.00mapartinfrontoftheheight.(Chapter18)plane,and2.00mapartbehindtheplane.111(Chapter15)fddoiPhysics:PrinciplesandProblemsSolutionsManual411

415Chapter19continueddxdofUse(1)minw,(2)vidLsubstancef,of(7.50cm)(21.0cm)cand(3)nsubstancev.7.50cm21.0cmCombine(2)and(3).11.7cmfnvacuumvacuumhidisubstancef(4),msubstancesubstancehdoobecausethefrequencyremainsconstantdhihoasthelightcrossesaboundary.idoRewrite(1)intermsofasubstancein(11.7cm)(2.00cm)thespacebetweentheslitsandthe7.50cmscreen.3.11cmxminwsubstanceL(5)Combine(4)and(5)andsolveforx.ChallengeProblemvacuumnsubstancexwpage526minLYouhaveseveralunknownsubstancesandwishtouseasingle-slitdiffractionapparatustodeter-xvacuumLminnminewhateachoneis.Youdecidetoplaceasubstancewsampleofanunknownsubstanceintheregion2.Ifthesourceyouusedhadawavelengthofbetweentheslitandthescreenandusethedata634nm,theslitwidthwas0.10mm,thethatyouobtaintodeterminetheidentityofeachdistancefromtheslittothescreenwassubstancebycalculatingitsindexofrefraction.1.15m,andyouimmersedtheapparatusinwater(nsubstance1.33),thenwhatwouldyouexpectthewidthofthecenterUnknownbandtobe?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.substancexvacuumLnsubstancedx1(634109m)(1.15m)(1.33)(0.10103m)5.510–3mLIncominglight1.Comeupwithageneralformulafortheindexofrefractionofanunknownsub-stanceintermsofthewavelengthofthelight,vacuum,thewidthoftheslit,w,thedistancefromtheslittothescreen,L,andthedistancebetweenthecentralbrightbandandthefirstdarkband,x1.412SolutionsManualPhysics:PrinciplesandProblems

416CHAPTER20StaticElectricityThewoolbecomespositivelychargedSectionReviewbecauseitgivesupelectronstothe20.1ElectricChargerubberrod.pages541–5455.ConservationofChargeAnapplecon-page545tainstrillionsofchargedparticles.Why1.ChargedObjectsAfteracombisrubbeddon’ttwoapplesrepeleachotherwhenonawoolsweater,itisabletopickuptheyarebroughttogether?smallpiecesofpaper.WhydoesthecombEachapplecontainsequalnumbersoflosethatabilityafterafewminutes?positiveandnegativecharges,sotheyThecomblosesitschargetoitssur-appearneutraltoeachother.roundingsandbecomesneutralonceagain.6.ChargingaConductorSupposeyouhangalongmetalrodfromsilkthreadssothat2.TypesofChargeIntheexperimentstherodisisolated.Youthentouchadescribedearlierinthissection,howcouldchargedglassrodtooneendofthemetalyoufindoutwhichstripoftape,BorT,isrod.Describethechargesonthemetalrod.positivelycharged?TheglassrodattractselectronsoffBringapositivelychargedglassrodthemetalrod,sothemetalbecomesnearthetwostripsoftape.Theonethatpositivelycharged.Thechargeisdis-isrepelledbytherodispositive.tributeduniformlyalongtherod.3.TypesofChargeApithballisasmall7.ChargingbyFrictionYoucanchargeaspheremadeofalightmaterial,suchasplas-rubberrodnegativelybyrubbingitwithticfoam,oftencoatedwithalayerofgraphitewool.Whathappenswhenyourubacop-oraluminumpaint.Howcouldyoudeter-perrodwithwool?minewhetherapithballthatissuspendedBecausethecopperisaconductor,itfromaninsulatingthreadisneutral,isremainsneutralaslongasitisincon-chargedpositively,orischargednegatively?tactwithyourhand.Bringanobjectofknowncharge,suchasanegativelychargedhardrubber8.CriticalThinkingItoncewasproposedrod,nearthepithball.Ifthepithballisthatelectricchargeisatypeoffluidthatrepelled,ithasthesamechargeastheflowsfromobjectswithanexcessoftherod.Ifitisattracted,itmayhavethefluidtoobjectswithadeficit.Whyistheoppositechargeorbeneutral.Tofindcurrenttwo-chargemodelbetterthantheoutwhich,bringapositivelychargedsingle-fluidmodel?glassrodnearthepithball.IftheyThetwo-chargemodelcanbetterrepel,thepithballispositive;iftheyexplainthephenomenaofattractionattract,thepithballmustbeneutral.andrepulsion.ItalsoexplainshowCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.objectscanbecomechargedwhenthey4.ChargeSeparationArubberrodcanbearerubbedtogether.Thesingle-fluidchargednegativelywhenitisrubbedwithmodelindicatedthatthechargeshouldwool.Whathappenstothechargeofthebeequalizedonobjectsthatareincon-wool?Why?tactwitheachother.Physics:PrinciplesandProblemsSolutionsManual413

417Chapter20continuedPracticeProblems20.2ElectricForcepages546–553page5529.Anegativechargeof
2.010
4Candapositivechargeof8.010
4Careseparatedby0.30m.Whatistheforcebetweenthetwocharges?KqAqB(9.0109Nm2/C2)(2.0104C)(8.0104C)F

d2(0.30m)2AB
1.6104N10.Anegativechargeof
6.010
6Cexertsanattractiveforceof65Nonasecondchargethatis0.050maway.Whatisthemagnitudeofthesecondcharge?KqAqBF
d2ABFd22AB(65N)(0.050m)qB

Kq(9.0109Nm2/C2)(6.0106C)A
3.0106C11.ThechargeonBinExampleProblem1isreplacedbyachargeof3.00
C.DiagramthenewsituationandfindthenetforceonA.Magnitudesofallforcesremainthesame.Thedirectionchangesto42°abovethexaxis,or138°.12.SphereAislocatedattheoriginandhasachargeof2.010
6C.SphereBislocatedat0.60monthex-axisandhasachargeof
3.610
6C.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.SphereCislocatedat0.80monthex-axisandhasachargeof4.010
6C.DeterminethenetforceonsphereA.qAqB9Nm2/C2)(2.0106C)(3.6106C)
0.18NFBonA
Kd2
(9.010(0.60m)2ABdirection:towardtherightqAqC9Nm2/C2)(2.0106C)(4.0106C)
0.1125
NFConA
Kd2
(9.010(0.80m)2ACdirection:towardtheleftFnet
FBonAFConA
(0.18N)(0.1125N)
0.068Ntowardtheright13.DeterminethenetforceonsphereBinthepreviousproblem.qAqBFAonB
Kd2ABqAqBFConB
Kd2ABFnet
FConBFAonBqqBqCAqB
KKd2d2BCAB414SolutionsManualPhysics:PrinciplesandProblems

418Chapter20continued
(9.0109Nm2/C2)b.anegativerod.(3.6106C)(4.0106C)Bringthenegativerodnear,butnot(0.20m)2touchingtheelectroscope.Touch(9.0109Nm2/C2)(ground)theelectroscopewithyourfinger,allowingelectronstobe(2.0106C)(3.6106C)repelledoffoftheelectroscopeinto(0.60m)2yourfinger.Removeyourfingerand
3.1Ntowardtherightthenremovetherod.18.AttractionofNeutralObjectsWhattwoSectionReviewpropertiesexplainwhyaneutralobjectisattractedtobothpositivelyandnegatively20.2ElectricForcechargedobjects?pages546–553Chargeseparation,causedbythepage553attractionofoppositechargesandthe14.ForceandChargeHowareelectricforcerepulsionoflikecharges,movestheandchargerelated?Describetheforcewhenoppositechargesintheneutralbodythechargesarelikechargesandtheforceclosertothechargedobjectandthewhenthechargesareoppositecharges.likechargesfartheraway.TheinverserelationbetweenforceanddistanceElectricforceisdirectlyrelatedtoeachmeansthatthenearer,oppositecharge.Itisrepulsivebetweenlikechargeswillattractmorethanthemorechargesandattractivebetweenoppo-distant,likechargeswillrepel.Thesitecharges.overalleffectisattraction.15.ForceandDistanceHowareelectricforce19.ChargingbyInductionInanelectroscopeanddistancerelated?Howwouldtheforcebeingchargedbyinduction,whathappenschangeifthedistancebetweentwochargeswhenthechargingrodismovedawaybeforeweretripled?thegroundisremovedfromtheknob?ElectricforceisinverselyrelatedtoChargethathadbeenpushedintothethesquareofthedistancebetweengroundbytherodwouldreturntothecharges.Ifthedistanceistripled,theelectroscopefromtheground,leavingforcewillbeone-ninthasgreat.theelectroscopeneutral.16.ElectroscopesWhenanelectroscopeis20.ElectricForcesTwochargedspheresareheldcharged,theleavesrisetoacertainangleadistance,r,apart.Onespherehasachargeofandremainatthatangle.Whydotheynot3
C,andtheotherspherehasachargeofrisefarther?9
C.Comparetheforceofthe3
CAstheleavesmovefartherapart,thesphereonthe9
Cspherewiththeforceofelectricforcebetweenthemdecreasesthe9
Csphereonthe3
Csphere.untilitisbalancedbythegravitationalTheforcesareequalinmagnitudeandforcepullingdownontheleaves.oppositeindirection.17.CharginganElectroscopeExplainhowto21.CriticalThinkingSupposethatyouareCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.chargeanelectroscopepositivelyusingtestingCoulomb’slawusingasmall,posi-a.apositiverod.tivelychargedplasticsphereandalarge,Touchthepositiverodtotheelectro-positivelychargedmetalsphere.Accordingscope.NegativechargeswillmovetoCoulomb’slaw,theforcedependsontotherod,leavingtheelectroscope1/r2,whereristhedistancebetweenthepositivelycharged.centersofthespheres.AsthespheresgetPhysics:PrinciplesandProblemsSolutionsManual415

419Chapter20continuedclosetogether,theforceissmallerthan26.LaundryWhydosockstakenfromaclothesexpectedfromCoulomb’slaw.Explain.dryersometimesclingtootherclothes?(20.2)SomechargesonthemetalspherewillTheyhavebeenchargedbycontactasberepelledtotheoppositesidefromtheyrubagainstotherclothes,andtheplasticsphere,makingtheeffectivethus,areattractedtoclothingthatisdistancebetweenthechargesgreaterneutralorhasanoppositecharge.thanthedistancebetweenthespheres’centers.27.CompactDiscsIfyouwipeacompactdiscwithacleancloth,whydoestheCDthenattractdust?(20.2)ChapterAssessmentRubbingtheCDchargesit.Neutralparticles,suchasdust,areattractedtoConceptMappingachargedobject.page55822.Completetheconceptmapbelowusingthe28.CoinsThecombinedchargeofallelectronsfollowingterms:conduction,distance,elemen-inanickelishundredsofthousandsoftarycharge.coulombs.Doesthisimplyanythingaboutthenetchargeonthecoin?Explain.(20.2)ElectricForceNo.Netchargeisthedifferencebetweenpositiveandnegativecharges.ThecoinCoulomb’schargingstillcanhaveanetchargeofzero.law29.Howdoesthedistancebetweentwochargedistancechargesimpacttheforcebetweenthem?Ifthedistanceisdecreasedwhilethechargesremainthesame,whathappenselementarycoulombconductioninductionchargetotheforce?(20.2)ElectricforceisinverselyproportionalCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.MasteringConceptstothedistancesquared.Asdistancedecreasesandchargesremainthepage558same,theforceincreasesasthesquare23.Ifyoucombyourhaironadryday,theofthedistance.combcanbecomepositivelycharged.Canyourhairremainneutral?Explain.(20.1)30.Explainhowtochargeaconductornega-No.Byconservationofcharge,yourtivelyifyouhaveonlyapositivelychargedhairmustbecomenegativelycharged.rod.(20.2)Bringtheconductorcloseto,butnot24.Listsomeinsulatorsandconductors.(20.1)touching,therod.GroundtheconductorStudentanswerswillvarybutmayinthepresenceofthechargedrod;then,includedryair,wood,plastic,glass,removethegroundbeforeremovingthecloth,anddeionizedwaterasinsula-chargedrod.Theconductorwillhaveators;andmetals,tapwater,andyournetnegativecharge.bodyasconductors.ApplyingConcepts25.Whatpropertymakesmetalagoodconduc-page558torandrubberagoodinsulator?(20.1)31.HowdoesthechargeofanelectrondifferMetalscontainfreeelectrons;rubberfromthechargeofaproton?Howaretheyhasboundelectrons.similar?416SolutionsManualPhysics:PrinciplesandProblems

420Chapter20continuedThechargeoftheprotonisexactlytheLawofsamesizeastheelectron,buthastheUniversalGravitationCoulomb’sLawoppositesign.F
GmAmBqAqBr2F
K232.Usingachargedrodandanelectroscope,rhowcanyoufindwhetherornotanobjectmAmBqAqBisaconductor?

Useaknowninsulatortoholdoneendoftheobjectagainsttheelectroscope.rrTouchtheotherendwiththechargedrod.Iftheelectroscopeindicatesa■Figure20-13(Nottoscale)charge,theobjectisaconductor.Similar:inverse-squaredependenceondistance,forceproportionaltoproduct33.Achargedrodisbroughtnearapileoftinyoftwomassesortwocharges;differ-plasticspheres.Someofthespheresareent:onlyonesignofmass,sogravita-attractedtotherod,butassoonastheytionalforceisalwaysattractive;twotouchtherod,theyareflungoffindifferentsignsofcharge,soelectricforcecandirections.Explain.beeitherattractiveorrepulsive.Thenaturalspheresareinitiallyattract-edtothechargedrod,buttheyacquire37.Theconstant,K,inCoulomb’sequationisthesamechargeastherodwhentheymuchlargerthantheconstant,G,inthetouchit.Asaresult,theyarerepelleduniversalgravitationequation.Ofwhatsig-fromtherod.nificanceisthis?Theelectricforceismuchlargerthan34.LightningLightningusuallyoccurswhenathegravitationalforce.negativechargeinacloudistransportedtoEarth.IfEarthisneutral,whatprovidesthe38.ThetextdescribesCoulomb’smethodforattractiveforcethatpullstheelectronschargingtwospheres,AandB,sothatthetowardEarth?chargeonBwasexactlyhalfthechargeonThechargeinthecloudrepelselec-A.SuggestawaythatCoulombcouldhavetronsonEarth,causingachargesepa-placedachargeonsphereBthatwasexactlyrationbyinduction.ThesideofEarthone-thirdthechargeonsphereA.closesttothecloudispositive,result-AfterchangingspheresAandBequally,inginanattractiveforce.sphereBistouchedtotwootherequallysizedballsthataretouching35.Explainwhathappenstotheleavesofaeachother.ThechargeonBwillbepositivelychargedelectroscopewhenrodsdividedequallyamongallthreeballs,withthefollowingchargesarebroughtleavingone-thirdthetotalchargeonit.closeto,butnottouching,theelectroscope.a.positive39.CoulombmeasuredthedeflectionofsphereAwhenspheresAandBhadequalchargesTheleaveswillmovefartherapart.andwereadistance,r,apart.Hethenmadeb.negativethechargeonBone-thirdthechargeonA.Theleaveswilldropslightly.HowfarapartwouldthetwospheresthenCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.havehadtobeforAtohavehadthesame36.AsshowninFigure20-13,Coulomb’slawdeflectionthatithadbefore?andNewton’slawofuniversalgravitationTohavethesameforcewithone-thirdappeartobesimilar.Inwhatwaysarethethecharge,thedistancewouldhaveelectricandgravitationalforcessimilar?tobedecreasedsuchthatd2
1/3,orHowaretheydifferent?0.58timesasfarapart.Physics:PrinciplesandProblemsSolutionsManual417

421Chapter20continued40.Twochargedbodiesexertaforceof0.145Noneachother.Iftheyaremovedsothattheyareone-fourthasfarapart,whatforceisexerted?11F2andF,soF
16timestheoriginalforce.d12
441.Electricforcesbetweenchargesareenormousincomparisontogravitationalforces.Yet,wenormallydonotsenseelectricforcesbetweenusandoursurroundings,whilewedosensegravitationalinteractionswithEarth.Explain.Gravitationalforcesonlycanbeattractive.Electricforcescanbeeitherattractiveorrepulsive,andwecansenseonlytheirvectorsums,whicharegenerallysmall.ThegravitationalattractiontoEarthislargerandmorenoticeablebecauseofEarth’slargemass.MasteringProblems20.2ElectricForcepage559Level142.Twocharges,qAandqB,areseparatedbyadistance,r,andexertaforce,F,oneachother.AnalyzeCoulomb’slawandidentifywhatnewforcewouldexistunderthefollowingconditions.a.qAisdoubled2qA,thennewforce
2Fb.qAandqBarecutinhalf111112qAand2qB,thennewforce

2
2F
4Fc.ristripledCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.F13d,thennewforce

F(3)29d.riscutinhalf1F3d,thennewforce

F
4F2124
2e.qAistripledandrisdoubled(3)F33qAand2d,thennewforce

(2)24F43.LightningAstronglightningbolttransfersabout25CtoEarth.Howmanyelectronsaretransferred?1electron20electrons(25C)

1.6101.601019C44.AtomsTwoelectronsinanatomareseparatedby1.510
10m,thetypicalsizeofanatom.Whatistheelectricforcebetweenthem?KqAqB(9.0109Nm2/C2)(1.601019C)(1.601019C)F

d2(1.51010m)2
1.0108N,awayfromeachother418SolutionsManualPhysics:PrinciplesandProblems

422Chapter20continued45.Apositiveandanegativecharge,eachofmagnitude2.510
5C,areseparatedbyadistanceof15cm.Findtheforceoneachoftheparticles.KqAqB(9.0109Nm2/C2)(2.5105C)(2.5105C)F

2(1.5101m)2d
2.5102N,towardtheothercharge46.Aforceof2.4102Nexistsbetweenapositivechargeof8.010
5Candapositivechargeof3.010
5C.Whatdistanceseparatesthecharges?KqAqBF
d2KqAqB(9.0109Nm2/C2)(8.010
5C)(3.010
5C)d


F
2.4102N
0.30m47.Twoidenticalpositivechargesexertarepulsiveforceof6.410
9Nwhenseparatedbyadistanceof3.810
10m.Calculatethechargeofeach.KqAqBKq2F

d2d2Fd2(6.4109N)(3.81010m)2q



9Nm2/C2
3.21019CK9.010Level248.Apositivechargeof3.0
Cispulledonbytwo4.0
Cnegativecharges.AsshowninFigure20-14,one2.0
C3.0
Cnegativecharge,
2.0
C,is0.050mtothewest,andtheother,
4.0
C,is0.030mtotheeast.
Whattotalforceisexertedonthepositivecharge?0.050m0.030m(9.0109Nm2/C2)(3.0106C)(2.0106C)F1
(0.050m)2■Figure20-14
22Nwest(9.0109Nm2/C2)(3.0106C)(4.0106C)F2
(0.030m)2
120NeastF2N)(2.2101N)net
F2F1
(1.210
98N,east49.Figure20-15showstwopositivelychargedq3qspheres,onewiththreetimesthechargeofthe

other.Thespheresare16cmapart,andtheforceCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.betweenthemis0.28N.Whatarethecharges16cmonthetwospheres?■Figure20-15qAqBqA3qAF
K
Kd2d2Physics:PrinciplesandProblemsSolutionsManual419

423Chapter20continuedFd2(0.28N)(0.16m)2q7CA

3K

3(9.0109Nm2/C2)
5.210q6CB
3qA
1.51050.ChargeinaCoinHowmanycoulombsofchargeareontheelectronsinanickel?Usethefollowingmethodtofindtheanswer.a.Findthenumberofatomsinanickel.Anickelhasamassofabout5g.Anickelis75percentCuand25percentNi,soeachmoleofthecoin’satomswillhaveamassofabout62g.5gAcoinis
0.08mole.62gThus,ithas(0.08)(6.021023)
51022atomsb.Findthenumberofelectronsinthecoin.Onaverage,eachatomhas28.75electrons.(51022atoms)(28.75electrons/atom)
11024electronsc.Findthecoulombsontheelectrons.(1.61019coulombs/electron)(11024electrons)
2105coulombs51.Threeparticlesareplacedinaline.Theleftparticlehasachargeof
55
C,themiddleonehasachargeof45
C,andtherightonehasachargeof
78
C.Themiddleparticleis72cmfromeachoftheothers,asshowninFigure20-16.55
C45
C78
C
72cm72cm■Figure20-16Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Findthenetforceonthemiddleparticle.LetleftbethenegativedirectionKqmqlKqmqrFnet
Fl(Fr)
d2d2(9.0109Nm2/C2)(45106C)(55106C)
(0.72m)2(9.0109Nm2/C2)(45106C)(78106C)(0.72m)2
18N,rightb.Findthenetforceontherightparticle.KqKqlqrmqrFnet
Fl(Fm)
(2d)2d2(9.0109Nm2/C2)(55106C)(78106C)
(2(0.72m))2(9.0109Nm2/C2)(45106C)(78106C)(0.72m)2
42N,left420SolutionsManualPhysics:PrinciplesandProblems

424Chapter20continuedMixedReviewpage559Level152.Asmallmetalspherewithcharge1.210
5Cistouchedtoanidenticalneutralsphereandthenplaced0.15mfromthesecondsphere.Whatistheelectricforcebetweenthetwospheres?Thetwospheressharethechargeequally,soqAqB9Nm2C2)(6.0106C)(6.0106C)
14NF
K
(9.010d2(0.15m)253.AtomsWhatistheelectricforcebetweenanelectronandaprotonplaced5.310
11mapart,theapproximateradiusofahydrogenatom?qAqB9Nm2C2)(1.601019C)(1.601019C)
8.2108NF
K
(9.010d2(5.31011m)254.Asmallsphereofcharge2.4
Cexperiencesaforceof0.36Nwhenasecondsphereofunknownchargeisplaced5.5cmfromit.Whatisthechargeofthesecondsphere?qAqBF
Kd2Fd2(0.36N)(5.5102m)2q8CB

Kq(9.0109Nm2C2)(2.4106C)
5.010A55.Twoidenticallychargedspheresplaced12cmaparthaveanelectricforceof0.28Nbetweenthem.Whatisthechargeofeachsphere?qAqBF
K,whereqd2A
qBFd2(0.28N)(1.2101m)2q


K
(9.0109Nm2C2)
6.7107C56.InanexperimentusingCoulomb’sapparatus,aspherewithachargeof3.610
8Cis1.4cmfromasecondsphereofunknowncharge.Theforcebetweenthespheresis2.710
2N.Whatisthechargeofthesecondsphere?qAqBF
Kd2Fd2(2.7102N)(1.4102m)2qB

Kq
(9.0109Nm2C2)(3.6108C)A
1.6108C57.Theforcebetweenaprotonandanelectronis3.510
10N.Whatisthedistancebetweenthesetwoparticles?qAqBF
KCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.d2qAqBd
K
F219C)(1.61019C)
(9.0109Nm2/C2)(1.6010
8.11010m
3.51010NPhysics:PrinciplesandProblemsSolutionsManual421

425Chapter20continuedThinkingCriticallypage56058.ApplyConceptsCalculatetheratiooftheelectricforcetothegravitationalforcebetweentheelectronandtheprotoninahydrogenatom.qqepK2dFemmKqeqp

GepFgd2Gmemp(9.0109Nm2/C2)(1.601019C)239

2.310(6.671011Nm2/kg2)(9.111031kg)(1.671027kg)59.AnalyzeandConcludeSphereA,withachargeof64
C,ispositionedattheorigin.Asecondsphere,B,withachargeof
16
C,isplacedat11.00monthex-axis.a.Wheremustathirdsphere,C,ofcharge12
Cbeplacedsothereisnonetforceonit?Theattractiveandrepulsiveforcesmustcancel,soqqAqCBqCFAC
Kd2
Kd2
FBC,soACBCqqA
B,and16d2
64d2,ord2d2ACBCACBCd2
4d2,sodACBCAC
2dBCThethirdspheremustbeplacedat2.00monthex-axissoitistwiceasfarfromthefirstsphereasfromthesecondsphere.b.Ifthethirdspherehadachargeof6
C,whereshoulditbeplaced?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thethirdcharge,qc,cancelsfromtheequation,soitdoesn’tmatterwhatitsmagnitudeorsignis.c.Ifthethirdspherehadachargeof
12
C,whereshoulditbeplaced?Asinpartb,themagnitudeandsignofthethirdcharge,qc,donotmatter.60.ThreechargedspheresarelocatedatthepositionsshowninFigure20-17.FindthetotalforceonsphereB.y4.5
C8.2
CA
x4.0cmB3.0cm
6.0
CC■Figure20-17F1
FAonBKqAqB(9.0109Nm2/C2)(4.5106C)(8.2106C)

d2(0.040m)2422SolutionsManualPhysics:PrinciplesandProblems

426Chapter20continued
208N
208N,toleftThedistancebetweentheothertwochargesis(0.040m)2(0.030m)2
0.050m10.030m
1
tan
0.040m
37°belowthenegativex-axis,or217°fromthepositivex-axis.F2
FConBKqCqB(9.0109Nm2/C)(8.2106C)(6.0106C)

d2(0.050m)2
177N
177Nat217°fromthepositivex-axis(37°180°)ThecomponentsofF2are:F2x
F2cos

(177N)(cos217°)
142N
142NtotheleftF2y
F2sin

(177N)(sin217°)
106N
106NdownThecomponentsofthenet(resultant)forceare:Fnet,x
208N142N
350N
350N,toleftFnet,y
106N,downFnet
(350N)2(106N)2
366N
3.7102N1106N
2
tan
350N
17°belowthenegativex-axisF2Nat197°fromthepositivex-axisnet
3.71061.ThetwopithballsinFigure20-18eachhaveamassof1.0gandanequalcharge.Onepithballissuspendedbyaninsulatingthread.Theotherisbroughtto3.0cmfromthesuspendedball.Thesuspendedballisnowhangingwiththethreadforminganangleof30.0°withthevertical.TheballisinequilibriumwithFE,Fg,andFT.Calculateeachofthefollowing.30.0°FE3.0cm■Figure20-18a.FgonthesuspendedballCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.F3kg)(9.80m/s2)
9.8103Ng
mg
(1.010b.FEFEtan30.0°
FgPhysics:PrinciplesandProblemsSolutionsManual423

427Chapter20continuedFE
mgtan30.0°
(1.0103kg)(9.80m/s2)(tan30.0°)
5.7103Nc.thechargeontheballsKqqABF
d2Kq2F
d2Fd2(5.7103N)(3.0102m2)8Cq


2.410
K
(9.0109Nm2/C2)62.Twocharges,qAandqB,areatrestnearapositivetestcharge,qT,of7.2
C.Thefirstcharge,qA,isapositivechargeof3.6
Clocated2.5cmawayfromqTat35°;qBisanegativechargeof
6.6
Clocated6.8cmawayat125°.a.DeterminethemagnitudeofeachoftheforcesactingonqT.KqTqA(9.0109Nm2/C2)(7.2106C)(3.6106C)F
A
d2(0.025m)2
3.7102N,away(towardqT)KqTqA(9.0109Nm2/C2)(7.2106C)(6.6106C)F
B
d2(0.068m)2
92N,toward(awayfromqT)b.Sketchaforcediagram.qACopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.7102NqBFA125°FB35°92NqTc.GraphicallydeterminetheresultantforceactingonqT.21°35°F
3.8102NFA
3.7102NFB92N424SolutionsManualPhysics:PrinciplesandProblems

428Chapter20continuedWritinginPhysicsMeasurethelengthandperiodofthependulum,andusetheequationforthepage560periodofapendulumtosolveforg.63.HistoryofScienceResearchseveraldevicesthatwereusedintheseventeenth66.Asubmarinethatismoving12.0m/ssendsandeighteenthcenturiestostudystatic3Hzasonarpingoffrequency1.5010electricity.ExamplesthatyoumighttowardaseamountthatisdirectlyinfrontofconsiderincludetheLeydenjarandthethesubmarine.Itreceivestheecho1.800sWimshurstmachine.Discusshowtheylater.(Chapter15)wereconstructedandhowtheyworked.a.HowfaristhesubmarinefromtheStudentanswerswillvary,butshouldseamount?includeinformationsuchasthefollow-ing.TheLeydenjar,inventedinthed
vt
(1533m/s)(0.900s)
1380mmid-1740s,wastheearliestcapacitor.Itb.Whatisthefrequencyofthesonarwavewasusedthroughouttheeighteenthandthatstrikestheseamount?nineteenthcenturiestostorechargesforvvd3Hz)electricity-relatedexperimentsandfd
fs
vv
(1.5010sdemonstrations.TheWimshurstmachine1533m/s0.0m/swasadeviceusedinthenineteenthand
1533m/s12.0m/searlytwentiethcenturiestoproduceand
1510Hzdischargestaticcharges.Wimshurstmachines,whichwerereplacedbythec.WhatisthefrequencyoftheechoVandeGraaffgeneratorinthetwentiethreceivedbythesubmarine?century,usedLeydenjarstostorethechargespriortodischarge.vvdfd
fs
vv
(1510Hz)s64.InChapter13,youlearnedthatforcesexist1533m/s(12.0m/s)betweenwatermoleculesthatcausewaterto
1533m/s0.0m/sbedenserasaliquidbetween0°Cand4°C
1520Hzthanasasolidat0°C.Theseforcesareelec-trostaticinnature.Researchelectrostaticinter-67.SecurityMirrorAsecuritymirrorisusedmolecularforces,suchasvanderWaalsforcestoproduceanimagethatisthree-fourthsanddipole-dipoleforces,anddescribetheirthesizeofanobjectandislocated12.0cmeffectsonmatter.behindthemirror.WhatisthefocallengthAnswerswillvary,butstudentsshouldofthemirror?(Chapter17)describetheinteractionsbetweendipositiveandnegativechargesatthem
domolecularlevel.Studentsshoulddnotethatthestrengthoftheseforcesdio
maccountsfordifferencesinmeltingandboilingpointsandfortheunusual(12.0cm)
behaviorofwaterbetween0°Cand4°C.34CumulativeReview
16.0cmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page560111
65.Explainhowapendulumcanbeusedtofdodideterminetheaccelerationofgravity.dd(Chapter14)f
oiddoiPhysics:PrinciplesandProblemsSolutionsManual425

429Chapter20continued(16.0cm)(12.0cm)
ChallengeProblem16.0cm(12.0cm)page552
48.0cmAsshowninthefigurebelow,twospheresof68.A2.00-cm-tallobjectislocated20.0cmequalmass,m,andequalpositivecharge,q,awayfromadiverginglenswithafocalareadistance,r,apart.lengthof24.0cm.Whataretheimageposi-rtion,height,andorientation?Isthisarealoravirtualimage?(Chapter18)mass
mmass
m111charge
qcharge
q
fddoi■Figure20-11dfodi
df1.Deriveanexpressionforthecharge,q,thatomustbeoneachspheresothatthespheres(20.0cm)(24.0cm)
areinequilibrium;thatis,sothatthe20.0cm(24.0cm)attractiveandrepulsiveforcesbetween
10.9cmthemarebalanced.hidiTheattractiveforceisgravitation,andm

hodotherepulsiveforceiselectrostatic,sotheirexpressionsmaybesetequal.dhiohmqi
dAmBAqBoF
K
Fg
Gd2d2e(10.9cm)(2.00cm)
Themassesandchargesareequal,20.0cmandthedistancecancels,so
1.09cmGm2
Kq2,andThisisavirtualimagethatisuprightinorientation,relativetotheobject.Gq
m
Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.K69.SpectrometerAspectrometercontainsa(6.671011Nm2/kg2)gratingof11,500slits/cm.Findtheangle
m
(9.0109Nm2/C2)atwhichlightofwavelength527nmhasafirst-orderbrightband.(Chapter19)
(8.611011C/kg)mThenumberofcentimetersperslitis2.Ifthedistancebetweenthespheresisdou-theslitseparationdistance,d.bled,howwillthataffecttheexpressionfor1slit
11,500slits/cmthevalueofqthatyoudeterminedinthedpreviousproblem?Explain.d
8.70105cmThedistancedoesnotaffectthevalue

dsinofqbecausebothforcesareinverselyrelatedtothesquareofthedistance,1

sin
andthedistancecancelsoutofthedexpression.1527109m
sin
8.70103m3.Ifthemassofeachsphereis1.50kg,deter-
0.00347°minethechargeoneachsphereneededtomaintaintheequilibrium.q
(8.611011C/kg)(1.50kg)
1.291010C426SolutionsManualPhysics:PrinciplesandProblems

430CHAPTER21ElectricFieldsPracticeProblems3.2108CTheelectricforceisupward(opposite21.1CreatingandMeasuringthefield),sothechargeisnegative.ElectricFieldspages563–5685.Youareprobingtheelectricfieldofachargeofunknownmagnitudeandsign.Youfirstpage5656Cisinanmapthefieldwitha1.0106-Ctest1.Apositivetestchargeof5.010electricfieldthatexertsaforceof2.0104Ncharge,thenrepeatyourworkwitha2.0106-Ctestcharge.onit.Whatisthemagnitudeoftheelectricfieldatthelocationofthetestcharge?a.WouldyoumeasurethesameforcesatF2.0104N1N/CthesameplacewiththetwotestE4.010q5.0106Ccharges?Explain.No.Theforceonthe2.0-Ccharge2.Anegativechargeof2.0108Cexperi-wouldbetwicethatonthe1.0-Cencesaforceof0.060Ntotherightinancharge.electricfield.Whatarethefield’smagnitudeb.Wouldyoufindthesamefieldanddirectionatthatlocation?strengths?Explain.F0.060N6N/CEq8C3.010Yes.Youwoulddividetheforceby2.010thestrengthofthetestcharge,sodirectedtothelefttheresultswouldbethesame.3.Apositivechargeof3.0107Cislocatedinafieldof27N/Cdirectedtowardthepage566south.Whatistheforceactingonthecharge?6.WhatisthemagnitudeoftheelectricfieldFstrengthatapositionthatis1.2mfromaEqpointchargeof4.2106C?FEq(27N/C)(3.0107C)FqEK6Nqd28.1109Nm2/C2)(4.2106C)(9104.Apithballweighing2.1103Nisplaced(1.2m)2inadownwardelectricfieldof6.5104N/C.2.6104N/CWhatcharge(magnitudeandsign)mustbeplacedonthepithballsothattheelectric7.Whatisthemagnitudeoftheelectricfieldforceactingonitwillsuspenditagainstthestrengthatadistancetwiceasfarfromtheforceofgravity?pointchargeinproblem6?TheelectricforceandthegravitationalBecausethefieldstrengthvariesastheforcealgebraicallysumtozerobecausesquareofthedistancefromthepointtheballissuspended,i.e.notinmotion:charge,thenewfieldstrengthwillbeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FgFe0,soFeFgone-fourthoftheoldfieldstrength,orF6.5103N/C.Eeq8.WhatistheelectricfieldatapositionthatFeFg2.1103NqEE4N/Cis1.6meastofapointchargeof6.5107.2106C?Physics:PrinciplesandProblemsSolutionsManual427

431Chapter21continuedFqTodetectafieldatapoint,placeatestEKqd2chargeatthatpointanddetermineif9Nm2/C2)(7.2106C)thereisaforceonit.(9.010(1.6m)2Todeterminethemagnitudeofthefield,2.5104N/CdividethemagnitudeoftheforceonthetestchargebythemagnitudeofthetestThedirectionofthefieldiseast(awaycharge.fromthepositivepointcharge).Themagnitudeofthetestchargemust9.Theelectricfieldthatis0.25mfromabechosensothatitisverysmallcom-smallsphereis450N/Ctowardthesphere.paredtothemagnitudesofthechargesWhatisthechargeonthesphere?producingthefield.FqThenextthingyoushoulddoismea-EKqd2surethedirectionoftheforceonthe2testcharge.ThedirectionofthefieldisEdqthesameasthedirectionoftheforceifK2thetestchargeispositive;otherwise,it(450N/C)(0.25m)3.1109C(9.0109Nm2/C2)isintheoppositedirection.Thechargeisnegative,becausethe12.FieldStrengthandDirectionApositivefieldisdirectedtowardit.testchargeofmagnitude2.40108C6Cexperiencesaforceof1.50103Ntoward10.Howfarfromapointchargeof2.410theeast.Whatistheelectricfieldatthemustatestchargebeplacedtomeasureapositionofthetestcharge?fieldof360N/C?F1.50103NeastFKqEEqd2q2.40108C6.25104N/CeastKqdE13.FieldLinesInFigure21-4,canyoutellCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(9.0109Nm2/C2)(2.4106C)whichchargesarepositiveandwhichare360N/Cnegative?Whatwouldyouaddtocomplete7.7mthefieldlines?No.Thefieldlinesmusthavearrow-headsindicatingtheirdirections,fromSectionReviewpositivetonegativecharges.21.1CreatingandMeasuring14.FieldVersusForceHowdoestheelectricElectricFieldsfield,E,atthetestchargedifferfromthepages563–568force,F,onit?page568Thefieldisapropertyofthatregionof11.MeasuringElectricFieldsSupposeyouarespace,anddoesnotdependonthetestaskedtomeasuretheelectricfieldinspace.chargeusedtomeasureit.TheforceHowdoyoudetectthefieldatapoint?Howdependsonthemagnitudeandsignofdoyoudeterminethemagnitudeofthethetestcharge.field?Howdoyouchoosethemagnitudeofthetestcharge?Whatdoyoudonext?15.CriticalThinkingSupposethetopchargeinFigure21-2cisatestchargemeasuringthefieldresultingfromthetwonegativecharges.Isitsmallenoughtoproduceanaccuratemeasure?Explain.428SolutionsManualPhysics:PrinciplesandProblems

432Chapter21continuedNo.Thischargeislargeenoughtopage572distortthefieldproducedbytheother21.Whatworkisdonewhen3.0Cismovedchargeswithitsownfield.ComparethroughanelectricpotentialdifferencewithFigure21-4b.of1.5V?WqV(3.0C)(1.5V)4.5JPracticeProblems22.A12-Vcarbatterycanstore1.44106Cwhenitisfullycharged.Howmuchwork21.2ApplicationsofcanbedonebythisbatterybeforeitneedsElectricFieldsrecharging?pages569–579WqV(1.44106C)(12V)page5711.7107J16.Theelectricfieldintensitybetweentwolarge,charged,parallelmetalplatesis23.Anelectroninatelevisionpicturetube6000N/C.Theplatesare0.05mapart.passesthroughapotentialdifferenceofWhatistheelectricpotentialdifference18,000V.Howmuchworkisdoneonthebetweenthem?electronasitpassesthroughthatpotentialVEd(6000N/C)(0.05m)difference?300J/C3102VWqV(1.601019C)(1.8104V)2.91015J17.Avoltmeterreads400Vacrosstwocharged,parallelplatesthatare0.020mapart.What24.Ifthepotentialdifferenceinproblem18isistheelectricfieldbetweenthem?betweentwoparallelplatesthatare2.4cmVEdapart,whatisthemagnitudeoftheelectricEV400V2104N/Cfieldbetweenthem?d0.020mV5.00102V4N/CE2.110d0.024m18.Whatelectricpotentialdifferenceisappliedtotwometalplatesthatare0.200mapart25.Theelectricfieldinaparticle-acceleratoriftheelectricfieldbetweenthemis5N/C.Howmuchworkmachineis4.5102.50103N/C?isdonetomoveaproton25cmthroughVEd(2.50103N/C)(0.200m)thatfield?5.00102VWqVqEd(1.601019C)(4.5105N/C)(0.25m)19.Whenapotentialdifferenceof125Visappliedtotwoparallelplates,thefield1.81014Jbetweenthemis4.25103N/C.Howfarapartaretheplates?page574VEd26.AdropisfallinginaMillikanoil-dropappa-dV125V2.94102mratuswithnoelectricfield.WhatforcesareE4.25103N/Cactingontheoildrop,regardlessofitsaccel-eration?Ifthedropisfallingataconstant20.Apotentialdifferenceof275VisappliedtoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.velocity,describetheforcesactingonit.twoparallelplatesthatare0.35cmapart.Gravitationalforce(weight)downward,Whatistheelectricfieldbetweentheplates?frictionforceofairupward.ThetwoV275V4N/CforcesareequalinmagnitudeiftheE7.910d3.5103mdropfallsatconstantvelocity.Physics:PrinciplesandProblemsSolutionsManual429

433Chapter21continued27.Anoildropweighs1.91015N.Itissus-32.Thesametwocapacitorsasinproblem31pendedinanelectricfieldof6.0103N/C.areeachchargedto3.5104C.WhichhasWhatisthechargeonthedrop?Howmanythelargerelectricpotentialdifferenceacrossexcesselectronsdoesitcarry?it?Whatisit?FqgEqV,sothesmallercapacitorhasCFg1.91015Nthelargerpotentialdifference.qE6.0103N/C3.5104C2VV1.11019C3.5106F3.210q3.21019C33.A2.2-Fcapacitorfirstischargedsothat#electrons2qe1.601019Ctheelectricpotentialdifferenceis6.0V.Howmuchadditionalchargeisneededto28.Anoildropcarriesoneexcesselectronandincreasetheelectricpotentialdifferencetoweighs6.41015N.Whatelectricfield15.0V?strengthisrequiredtosuspendthedropsoqCVitismotionless?F6.41015NqC(V2V1)E4.0104N/C6F)(15.0V6.0V)q1.601019C(2.2102.0105C29.Apositivelychargedoildropweighing1.21014Nissuspendedbetweenparallel34.Whenachargeof2.5105Cisaddedtoaplatesseparatedby0.64cm.Thepotentialcapacitor,thepotentialdifferenceincreasesdifferencebetweentheplatesis240V.Whatfrom12.0Vto14.5V.Whatisthecapaci-isthechargeonthedrop?Howmanyelec-tanceofthecapacitor?tronsisthedropmissing?q2.5105CCEV240V3.8104N/CV2V114.5V12.0Vd6.4103m1.0105FFECopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.qF1.21014N19CqE4N/C3.210SectionReview3.810q3.41019C21.2Applicationsof#electrons2qe1.601019CElectricFieldspages569–579page578page57930.A27-Fcapacitorhasanelectricpotential35.PotentialDifferenceWhatisthediffer-differenceof45Vacrossit.Whatistheencebetweenelectricpotentialenergyandchargeonthecapacitor?electricpotentialdifference?qCV(27106F)(45V)Electricpotentialenergychangeswhen1.2103Cworkisdonetomoveachargeinanelectricfield.Itdependsontheamountofchargeinvolved.Electricpotential31.Botha3.3-Fanda6.8-Fcapacitoraredifferenceistheworkdoneperunitconnectedacrossa24-Velectricpotentialchargetomoveachargeinanelectricdifference.Whichcapacitorhasagreaterfield.Itisindependentoftheamountofcharge?Whatisit?chargethatismoved.qCV,sothelargercapacitorhasagreatercharge.q(6.8106F)(24V)1.6104C430SolutionsManualPhysics:PrinciplesandProblems

434Chapter21continued36.ElectricFieldandPotentialDifferenceThechargesonthemetaldomepro-Showthatavoltpermeteristhesameasaducenofieldinsidethedome.Thenewtonpercoulomb.chargesfromthebeltaretransferredV/mJ/CmNm/CmN/Cimmediatelytotheoutsideofthedome,wheretheyhavenoeffectonnew37.MillikanExperimentWhenthechargeonchargesarrivingatpointB.anoildropsuspendedinaMillikanappara-tusischanged,thedropbeginstofall.HowshouldthepotentialdifferenceontheChapterAssessmentplatesbechangedtobringthedropbackConceptMappingintobalance?page584Thepotentialdifferenceshouldbe42.Completetheconceptmapbelowusingtheincreased.followingterms:capacitance,fieldstrength,J/C,work.38.ChargeandPotentialDifferenceInproblem37,ifchangingthepotentialdif-ferencehasnoeffectonthefallingdrop,ElectricFieldwhatdoesthistellyouaboutthenewchargeonthedrop?forceworkThedropiselectricallyneutral(noelectronexcessordeficiency).testcharge39.CapacitanceHowmuchchargeisstoredfieldpotentialona0.47-Fcapacitorwhenapotentialstrengthdifferencedifferenceof12Visappliedtoit?qCV(4.7107F)(12V)capacitance5.6106CN/CJ/C40.ChargeSharingIfalarge,positivelycharged,conductingsphereistouchedbyasmall,negativelycharged,conductingsphere,C/Vwhatcanbesaidaboutthefollowing?a.thepotentialsofthetwospheresMasteringConceptsThesphereswillhaveequalpage584potentials.43.Whatarethetwopropertiesthatatestb.thechargesonthetwosphereschargemusthave?(21.1)ThelargespherewillhavemoreThetestchargemustbesmallinmag-chargethanthesmallsphere,butnituderelativetothemagnitudesofthetheywillbethesamesign.Thesignchargesproducingthefieldandbeofthechargewilldependonwhichpositive.spherebeganwithmorecharge.44.Howisthedirectionofanelectricfield41.CriticalThinkingReferringbacktodefined?(21.1)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Figure21-3a,explainhowchargecontinuesThedirectionofanelectricfieldisthetobuilduponthemetaldomeofaVandedirectionoftheforceonapositiveGraaffgenerator.Inparticular,whyisn’tchargeplacedinthefield.ThiswouldchargerepelledbackontothebeltatpointB?beawayfromapositiveobjectandtowardanegativeobject.Physics:PrinciplesandProblemsSolutionsManual431

435Chapter21continued45.Whatareelectricfieldlines?(21.1)48.InFigure21-15,wheredotheelectricfieldlinesofforcelinesleavethepositivechargeend?(21.1)46.Howisthestrengthofanelectricfieldindi-catedwithelectricfieldlines?(21.1)Theclosertogethertheelectricfieldlinesare,thestrongertheelectricfield.47.Drawsomeoftheelectricfieldlinesbetweeneachofthefollowing.(21.1)a.twolikechargesofequalmagnitude■ Figure21-15Theyendondistantnegativechargessomewherebeyondtheedgesofthediagram.b.twounlikechargesofequalmagnitude49.WhatSIunitisusedtomeasureelectricpotentialenergy?WhatSIunitisusedtomeasureelectricpotentialdifference?(21.2)electricpotentialenergy:joule;electricpotential:voltc.apositivechargeandanegativechargehavingtwicethemagnitudeoftheposi-50.Definevoltintermsofthechangeinpoten-tivechargetialenergyofachargemovinginanelectricfield.(21.2)Avoltisthechangeinelectricpotentialenergy,PE,resultingfrommovingaunittestcharge,q,adistance,d,ofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1minanelectricfield,E,of1N/C.VPE/qEd51.Whydoesachargedobjectloseitschargewhenitistouchedtotheground?(21.2)ThechargeissharedwiththesurfaceofEarth,whichisanextremelylargeobject.52.Achargedrubberrodthatisplacedonatablemaintainsitschargeforsometime.d.twooppositelychargedparallelplatesWhyisthechargedrodnotdischargedimmediately?(21.2)Thetableisaninsulator,oratleastaverypoorconductor.53.Ametalboxischarged.Comparethecon-centrationofchargeatthecornersoftheboxtothechargeconcentrationonthesidesofthebox.(21.2)Theconcentrationofchargeisgreateratthecorners.432SolutionsManualPhysics:PrinciplesandProblems

436Chapter21continued54.ComputersEDDelicatepartsinelectronicequipment,xCsuchasthosepicturedinABFigure21-16,arecontainedwithinametalyzboxinsidea■ Figure21-17plasticcase.■ Figure21-16Why?(21.2)SpherexwillfollowpathC.ItwillThemetalboxshieldsthepartsfromexperienceforcesshownbyDandB.externalelectricfields,whichdonotThevectorsumisC.existinsideahollowconductor.59.Whatistheunitofelectricpotentialdiffer-ApplyingConceptsenceintermsofm,kg,s,andC?pages584–585VJ/CNm/C(kgm/s2)(m/C)55.Whathappenstothestrengthofanelectrickgm2/s2Cfieldwhenthechargeonthetestchargeishalved?60.WhatdotheelectricfieldlineslooklikeNothing.Becausetheforceonthetestwhentheelectricfieldhasthesamechargealsowouldbehalved,theratiostrengthatallpointsinaregion?F/qandtheelectricfieldwouldremainTheyareparallel,equallyspacedlines.thesame.61.MillikanOil-DropExperimentWhen56.DoesitrequiremoreenergyorlessenergydoingaMillikanoil-dropexperiment,itistomoveaconstantpositivechargethroughbesttoworkwithdropsthathavesmallanincreasingelectricfield?charges.Therefore,whentheelectricfieldisturnedon,shouldyoutrytofinddropsthatEnergyisproportionaltotheforce,andaremovingrapidlyorslowly?Explain.theforceisproportionaltotheelectricfield.Therefore,itrequiresmoreenergy.Slowly.Thelargerthecharge,thestrongertheforce,andthus,thelarger57.Whatwillhappentotheelectricpotentialthe(terminal)velocity.energyofachargedparticleinanelectricfieldwhentheparticleisreleasedandfree62.Twooildropsareheldmotionlessinatomove?Millikanoil-dropexperiment.Theelectricpotentialenergyofthea.Canyoubesurethatthechargesaretheparticlewillbeconvertedintokineticsame?energyoftheparticle.No.Theirmassescouldbedifferent.b.Theratiosofwhichtwopropertiesof58.Figure21-17showsthreesphereswiththeoildropshavetobeequal?chargesofequalmagnitude,withtheirchargetomassratio,q/m(orm/q)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.signsasshown.Spheresyandzareheldinplace,butspherexisfreetomove.63.JoséandSuearestandingonaninsulatingInitially,spherexisequidistantfromplatformandholdinghandswhentheyarespheresyandz.Choosethepaththatgivenacharge,asinFigure21-18.Joséisspherexwillbegintofollow.AssumethatlargerthanSue.Whohasthelargeramountnootherforcesareactingonthespheres.ofcharge,ordotheybothhavethesameamount?Physics:PrinciplesandProblemsSolutionsManual433

437Chapter21continuedF0.30N4N/CE3.010q1.0105Cinthesamedirectionastheforce68.Atestchargeexperiencesaforceof0.30Nonitwhenitisplacedinanelectricfieldintensityof4.5105N/C.Whatisthemag-nitudeofthecharge?FEq■ Figure21-18F0.30N7Cq6.710E4.5105N/CJoséhasalargersurfacearea,sohewillhavealargeramountofcharge.69.Theelectricfieldintheatmosphereisabout150N/Cdownward.64.Whichhasalargercapacitance,analu-a.Whatisthedirectionoftheforceonaminumspherewitha1-cmdiameteroronenegativelychargedparticle?witha10-cmdiameter?upwardThe10-cmdiameterspherehasalargerb.Findtheelectricforceonanelectroncapacitancebecausethechargescanwithcharge1.61019C.befartherapart,reducingpotentialriseasitischarged.EFq65.HowcanyoustoredifferentamountsofFqE(1.61019C)(150N/C)chargeinacapacitor?2.41017NChangethevoltageacrossthecapacitor.F2.41017NdirectedupwardMasteringProblemsc.ComparetheforceinpartbwiththeforceofgravityonthesameelectronCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.21.1CreatingandMeasuringElectricFields(mass9.11031kg).pages585–58619C.Fmg(9.11031kg)(9.80m/s2)Thechargeofanelectronis1.6010Level18.91030N66.WhatchargeexistsonatestchargethatF8.91030N(downward),moreexperiencesaforceof1.4108Natathanonetrilliontimessmallerpointwheretheelectricfieldintensityis5.0104N/C?70.Carefullysketcheachofthefollowing.Fa.theelectricfieldproducedbya1.0-CEqchargeF1.4108N5Cq2.810E5.0104N/C67.Apositivechargeof1.0105C,showninFigure21-19,0.30N1.0Cexperiencesaforceof0.30Nwhenitislocatedatacertain1.0105Cpoint.Whatistheelectricfieldintensityatthatpoint?■ Figure21-19434SolutionsManualPhysics:PrinciplesandProblems

438Chapter21continuedb.theelectricfieldresultingfroma2.0-Cb.ChargeZnowhasasmallpositivecharge(Makethenumberoffieldlineschargeonit.Drawanarrowrepresent-proportionaltothechangeincharge.)ingtheforceonit.F2.0CFyFxZ71.Apositivetestchargeof6.0106Cis73.Inatelevisionpicturetube,electronsareplacedinanelectricfieldof50.0-N/Cacceleratedbyanelectricfieldhavingaintensity,asinFigure21-20.Whatisthevalueof1.00105NC.strengthoftheforceexertedonthetesta.Findtheforceonanelectron.charge?FEqq6.0106CFEq(1.601019C)(1.00105N/C)1.601014Nb.Ifthefieldisconstant,findtheE50.0N/Caccelerationoftheelectron■ Figure21-20(mass9.111031kg).FFmaEqF1.601014NFqE(6.0106C)(50.0N/C)am9.111031kg3.0104N1.761016m/s2Level274.Whatistheelectricfieldstrength20.0cm72.ChargesX,Y,andZallareequidistantfromfromapointchargeof8.0107C?eachother.Xhasa1.0-Ccharge,YhasaFKqqE,andF2.0-Ccharge,andZhasasmallnegativeqd2charge.KqsoEa.Drawanarrowrepresentingtheforceond2chargeZ.(9.0109Nm2/C2)(8.0107C)(0.200m)2ZFx1.8105N/CFyCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FXYPhysics:PrinciplesandProblemsSolutionsManual435

439Chapter21continued75.Thenucleusofaleadatomhasachargeof78.Anelectronismovedthroughanelectric82protons.potentialdifferenceof450V.Howmucha.Whatarethedirectionandmagnitudeworkisdoneontheelectron?oftheelectricfieldat1.01010mWVfromthenucleus?qQ(82protons)WqV(1.601019C/proton)(1.601019C)(450V)1.311017C7.21017JF1KqQKQEqq2279.A12-Vbatterydoes1200Jofworkddtransferringcharge.Howmuchcharge(9.0109Nm2/C2)(1.311017C)istransferred?(1.01010m)2W1.21013N/C,outwardVqb.WhatarethedirectionandmagnitudeW1200J2Cq1.010oftheforceexertedonanelectronV12Vlocatedatthisdistance?80.TheelectricfieldintensitybetweentwoFEq3N/C.Theplateschargedplatesis1.510(1.21013N/C)(1.601019C)are0.060mapart.Whatistheelectric1.9106N,towardthenucleuspotentialdifference,involts,betweentheplates?VEd21.2ApplicationsofElectricFieldspages586–587(1.5103N/C)(0.060m)Level19.0101V76.If120Jofworkisperformedtomove2.4Cofchargefromthepositiveplatetothenega-81.AvoltmeterindicatesthattheelectricCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tiveplateshowninFigure21-21,whatpotentialdifferencebetweentwoplatesispotentialdifferenceexistsbetweentheplates?70.0V.Theplatesare0.020mapart.Whatelectricfieldintensityexistsbetweenthem?VEd2.4CV70.0VE3500V/md0.020m3500N/C82.Acapacitorthatisconnectedtoa45.0-Vsourcecontains90.0Cofcharge.Whatis■ Figure21-21thecapacitor’scapacitance?W120J1VVq2.4C5.010q90.0106CC2.00µFV45.0V77.Howmuchworkisdonetotransfer0.15C83.Whatelectricpotentialdifferenceexistsofchargethroughanelectricpotentialdif-acrossa5.4-Fcapacitorthathasachargeferenceof9.0V?of8.1104C?WVqqCVWqV(0.15C)(9.0V)1.4Jq8.1104CVC5.4106F1.5102V436SolutionsManualPhysics:PrinciplesandProblems

440Chapter21continued84.TheoildropshowninFigure21-22isneg-87.PhotoflashTheenergystoredinacapaci-ativelychargedandweighs4.51015N.torwithcapacitanceC,andanelectricThedropissuspendedinanelectricfieldpotentialdifference,V,isrepresentedbyintensityof5.6103N/C.W1CV2.Oneapplicationofthisisin2theelectronicphotoflashofastrobelight,liketheoneinFigure21-24.Insuchaunit,acapacitorof10.0Fischargedto3.0102V.Findtheenergystored.8.51015N■ Figure21-22a.Whatisthechargeonthedrop?FEqF4.51015NqE5.6103N/C■ Figure21-248.01019CW1CV22b.Howmanyexcesselectronsdoesitcarry?1(10.0106F)(3.0102V)2(8.01019C)1electron21.601019C0.45J5electrons88.Supposeittook25stochargethecapacitor85.Whatisthechargeona15.0-pFcapacitorinproblem87.whenitisconnectedacrossa45.0-Vsource?a.FindtheaveragepowerrequiredtoqCVchargethecapacitorinthistime.qCV(15.01012F)(45.0V)PW0.45J1.8102Wt25s6.751010Cb.Whenthiscapacitorisdischargedthroughthestrobelamp,ittransfersallLevel2itsenergyin1.0104s.Findthe86.Aforceof0.065Nisrequiredtomoveapowerdeliveredtothelamp.chargeof37Cadistanceof25cminauniformelectricfield,asinFigure21-23.PW0.45J4.5103Wt1.0104sWhatisthesizeoftheelectricpotentialdifferencebetweenthetwopoints?c.Howissuchalargeamountofpowerpossible?37CPowerisinverselyproportionalto0.053Nthetime.Theshorterthetimeforagivenamountofenergytobe25cmexpended,thegreaterthepower.■ Figure21-23Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.WFdWFdandVqq(0.065N)(0.25m)37106C4.4102VPhysics:PrinciplesandProblemsSolutionsManual437

441Chapter21continued89.LasersLasersareusedtotrytoproducebetweenthemis300V,whatisthecapaci-controlledfusionreactions.Theselaserstanceofthesystem?requirebriefpulsesofenergythatarestored0.060C0.060Cinlargeroomsfilledwithcapacitors.Onesuchroomhasacapacitanceof61103FV300Vchargedtoapotentialdifferenceof10.0kV.12,findtheenergya.GiventhatWCV25cm2storedinthecapacitors.■ Figure21-25W1CV2q6.0108C10F2C210V300V1(61103F)(1.00104V)2293.Theplatesofa0.047Fcapacitorare3.1106J0.25cmapartandarechargedtoapoten-tialdifferenceof120V.Howmuchchargeb.Thecapacitorsaredischargedin10ns(1.0108s).Whatpowerisproduced?isstoredonthecapacitor?q6JCW3.11014WVP3.110t1.0108sqCVc.Ifthecapacitorsarechargedbyagener-(4.7108F)(120V)atorwithapowercapacityof1.0kW,howmanysecondswillberequiredto5.6106C5.6Cchargethecapacitors?6J94.WhatisthestrengthoftheelectricfieldW3.1103stP3W3.110betweentheplatesofthecapacitorin1.010Problem93above?MixedReviewVEdpage587VELevel1dCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.90.Howmuchworkdoesittaketomove120V4.8104V/m0.25Cbetweentwoparallelplatesthat2.5103mare0.40cmapartifthefieldbetweenthe95.Anelectronisplacedbetweentheplatesofplatesis6400N/C?thecapacitorinProblem93above,asinWqVqEdFigure21-26.Whatforceisexertedonthat(2.5107C)(6400N/C)(4.0103m)electron?6.4106JV120VC0.047F91.Howmuchchargeisstoredona0.22-Fparallelplatecapacitoriftheplatesare1.2cmapartandtheelectricfieldbetweenthemis2400N/C?qCVCEd7F)(2400N/C)(1.2102m)(2.2106.3C0.25cm■ Figure21-2692.Twoidenticalsmallspheres,25cmapart,carryequalbutoppositechargesof0.060C,asinFigure21-25.Ifthepotentialdifference438SolutionsManualPhysics:PrinciplesandProblems

442Chapter21continuedF1EWareabhq2FEq1(25V)(12.5C)2(4.8104V/m)(1.61019C)160J7.71015N101.TheworkfoundinProblem100aboveis96.HowmuchworkwouldittaketomoveannotequaltoqV.Whynot?additional0.010CbetweentheplatesatThepotentialdifferenceisnotcon-120VinProblem93?stantasthecapacitorischarged.WTherefore,theareaunderthegraphVqmustbeusedtofindwork,notjustWqVsimplemultiplication.(1.0108C)(120V)1.2106J102.GraphtheelectricfieldstrengthnearapositivepointchargeasafunctionofLevel2distancefromit.97.ThegraphinFigure21-27representsthechargestoredinacapacitorasthechargingpotentialincreases.Whatdoestheslopeofthelinerepresent?ChargeStoredEonCapacitor15C)d10(q103.Whereisthefieldofapointchargeequal5tozero?0Nowhere,orataninfinitedistance102030fromthepointcharge.V(V)■Figure21-27104.Whatistheelectricfieldstrengthatadis-capacitanceofthecapacitortanceofzerometersfromapointcharge?Istheresuchathingasatruepointcharge?98.WhatisthecapacitanceofthecapacitorInfinite.No.representedbyFigure21-27?Cslope0.50FThinkingCriticallypages587–58899.Whatdoestheareaunderthegraphlinein105.ApplyConceptsAlthoughalightningrodFigure21-27represent?isdesignedtocarrychargesafelytotheworkdonetochargethecapacitorground,itsprimarypurposeistopreventlightningfromstrikinginthefirstplace.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.100.HowmuchworkisrequiredtochargetheHowdoesitdothat?capacitorinproblem98toapotentialdif-Thesharppointontheendoftherodferenceof25V?leakschargeintotheatmospherebeforeithasthechancetobuildupenoughpotentialdifferencetocausealightningstrike.Physics:PrinciplesandProblemsSolutionsManual439

443Chapter21continued106.AnalyzeandConcludeInanearlysetofexperimentsin1911,Millikanobservedthatthefollowingmeasuredchargescouldappearonasingleoildrop.Whatvalueofelementarychargecanbededucedfromthesedata?a.6.5631019Cf.18.081019Cb.8.2041019Cg.19.711019Cc.11.501019Ch.22.891019Cd.13.131019Ci.26.131019Ce.16.481019C1.631019C.Subtractingadjacentvalues,ba,cb,dc,etc.yields1.6411019C,3.301019C,1.631019C,3.351019C,1.601019C,1.631019C,3.181019C,3.241019C.Therearetwonumbers,approximately1.6311019Cand3.21019C,thatarecommon.Averagingeachsimilargroupproducesonechargeof1.631019Candonechargeof3.271019C(whichistwotimes1.6411019C).Dividing1.631019Cintoeachpieceofdatayieldsnearlywhole-num-berquotients,indicatingitisthevalueofanelementarycharge.107.AnalyzeandConcludeTwosmallspheres,AandB,lieonthex-axis,asinFigure21-28.SphereAhasachargeof3.00106C.SphereBis0.800mtotherightofsphereAandhasachargeof5.00106C.Findthemagnitudeanddirectionoftheelectricfieldstrengthatapointabovethex-axisthatwould3.00106C5.00106CformtheapexofanequilateraltrianglewithABspheresAandB.0.800mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Drawthespheresandvectorsrepresenting■ Figure21-28thefieldsduetoeachchargeatthegivenpoint.EAEREB0.800mABNowdothemath:FAKqA(9.0109Nm2/C2)(3.00106C)4N/CEAqd2(0.800m)24.2210FBKqB(9.0109Nm2/C2)(5.00106C)4N/CEBqd2(0.800m)27.0310E4N/C)(cos60.0°)2.11104N/CAxEAcos60.0°(4.2210E4N/C)(sin60.0°)3.65104N/CAyEAsin60.0°(4.2210440SolutionsManualPhysics:PrinciplesandProblems

444Chapter21continuedE4N/C)(cos60.0°)3.52104N/CBxEBcos(60.0°)(7.0310E4N/C)(sin60.0°)6.09104N/CByEBsin(60.0°)(7.0310E4N/C)(3.52104N/C)5.63104N/CxEAxEBx(2.1110E4N/C)(6.09104N/C)2.44104N/CyEAyEBy(3.6510EREx2Ey26.14104N/CEytanExE1ytanEx12.44104N/Ctan4N/C5.631023.4°108.AnalyzeandConcludeInanink-jetprinter,dropsof1.5cminkaregivenacertainamountofchargebeforetheymovebetweentwolarge,parallelplates.ThepurposeoftheplatesistodeflectthechargessothattheyareqGutterstoppedbyagutteranddonotreachthepaper.ThisisshowninFigure21-29.Theplatesare1.5-cmlongmvandhaveanelectricfieldofE1.2106N/CbetweenE1.2106N/Cthem.Dropswithamassm0.10ng,andacharge■ Figure21-29q1.01016C,aremovinghorizontallyataspeed,v15m/s,paralleltotheplates.Whatistheverticaldisplacementofthedropswhentheyleavetheplates?Toanswerthisquestion,completethefollowingsteps.a.Whatistheverticalforceonthedrops?FEq(1.01016C)(1.2106N/C)1.21010Nb.Whatistheirverticalacceleration?F1.21010N3m/s2a1.210m1.01013kgc.Howlongaretheybetweentheplates?L1.5102m3st1.010v15m/sd.Howfararetheydisplaced?12yat2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13m/s2)(1.0103s)2(1.21026.0104m0.60mmPhysics:PrinciplesandProblemsSolutionsManual441

445Chapter21continued109.ApplyConceptsSupposetheMoonhadanetnegativechargeequaltoq,andEarthhadanetpositivechargeequalto10q.Whatvalueofqwouldyieldthesamemagnitudeofforcethatyounowattributetogravity?EquatetheexpressionsforgravitationalforceandCoulombicforcebetweenEarthandtheMoon:GmKq10Kq2FEmMEqMd2d2d2whereqisthenetnegativechargeoftheMoonandqE,thenetpositivechargeofEarth,is10q.Solvesymbolicallybeforesubstitutingnumbers.GmqEmM10K(6.671011Nm2/kg2)(6.001024kg)(7.311022kg)(10)(9.0109Nm2/C2)1.81013CWritinginPhysicspage588110.Choosethenameofanelectricunit,suchascoulomb,volt,orfarad,andresearchthelifeandworkofthescientistforwhomitwasnamed.Writeabriefessayonthispersonandincludeadiscussionoftheworkthatjustifiedthehonorofhavingaunitnamedforhim.Studentanswerswillvary.SomeexamplesofscientiststheycouldchooseareVolta,Coulomb,Ohm,andAmpère.CumulativeReviewCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page588111.Michelsonmeasuredthespeedoflightbysendingabeamoflighttoamirroronamountain35kmaway.(Chapter16)a.Howlongdoesittakelighttotravelthedistancetothemountainandback?(35km/trip)(2trips)(1000m/1km)/3.00108m/s2.3104sb.AssumethatMichelsonusedarotatingoctagonwithamirroroneachfaceoftheoctagon.Alsoassumethatthelightreflectsfromonemirror,travelstotheothermountain,reflectsoffofafixedmirroronthatmountain,andreturnstotherotatingmirrors.Iftherotatingmirrorhasadvancedsothatwhenthelightreturns,itreflectsoffofthenextmirrorintherotation,howfastisthemirrorrotating?2.3104s3s/revT(8mirrors/rev)1.8101mirror112rev/sf5.610T1.8103s/revNotethatifstudentscarryextradigitsfrompartatopreventround-ingerrors,theywillgetananswerof5.4102rev/s.442SolutionsManualPhysics:PrinciplesandProblems

446Chapter21continuedc.Ifeachmirrorhasamassof1.0101ghyouhmountainandrotatesinacirclewithanaveragedyoudmountainradiusof1.0101cm,whatisthehyoudmountainapproximatecentripetalforceneededhmountaindyoutoholdthemirrorwhileisitrotating?F2mf2r(2m)(50,000m)c450m42(0.010kg)(5.6102rev/s)22000m(0.10m)113.Aconverginglenshasafocallengthof1.2104N38.0cm.Ifitisplaced60.0cmfromanobject,atwhatdistancefromthelenswillNotethattheanswershouldbetheimagebe?(Chapter18)1.2104Nregardlessofwhether111thestudentsuse5.4102rev/sorfdd5.6104rev/sforf.oiddofid112.MountainSceneYoucanseeanimageofofadistantmountaininasmoothlakejust(60.0cm)(38.0cm)asyoucanseeamountainbikernextto60.0cm38.0cmthelakebecauselightfromeachstrikesthe104cmsurfaceofthelakeataboutthesameangleTheimageis104cmfromthelens.ofincidenceandisreflectedtoyoureyes.Ifthelakeisabout100mindiameter,the114.Aforce,F,ismeasuredbetweentworeflectionofthetopofthemountainischarges,Qandq,separatedbyadistance,aboutinthemiddleofthelake,themoun-r.Whatwouldthenewforcebeforeachoftainisabout50kmawayfromthelake,thefollowing?(Chapter20)andyouareabout2mtall,thenapproxi-a.ristripledmatelyhowhighabovethelakedoesthetopofthemountainreach?(Chapter17)F/9Sincetheangleofincidenceoftheb.Qistripledlightfromthetopofthemountainis3Fequaltoitsangleofreflectionfromthec.bothrandQaretripledlake,youandthereflectionofthetopF/3ofthemountainformatrianglethatissimilartoatriangleformedbythed.bothrandQaredoubledmountainandthetopofitsreflectionF/2inthelake.Yourheightmakesuponee.allthree,r,Q,andq,aretripledside,hyou2mandthetopoftheFmountainishalfwayacrossthelake,dyou50m.Themountainisadis-tancedmountain50,000mfromitsreflection.FindhmountainbyequatingtheratiosofthesidesofthetwoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.similartriangles.Physics:PrinciplesandProblemsSolutionsManual443

447Chapter21continuedChallengeProblempage579Theplatesofacapacitorattracteachotherbecausetheycarryoppositecharges.Acapacitorconsistingoftwoparallelplatesthatareseparatedbyadistance,d,hascapacitance,C.1.Deriveanexpressionfortheforcebetweenthetwoplateswhenthecapacitorhascharge,q.Combinethefollowingequations:dVqFEq,E,andVdCqq2VCFEqqqddCd2.Whatchargemustbestoredona22-Fcapacitortohaveaforceof2.0Nbetweentheplatesiftheyareseparatedby1.5mm?q2FCdsoqFCd(2.0N)(2.2105F)(1.5103m)2.6104CCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.444SolutionsManualPhysics:PrinciplesandProblems

448CHAPTER22CurrentElectricitypage598PracticeProblemsForallproblems,assumethatthebatteryvoltageand22.1CurrentandCircuitslampresistancesareconstant,nomatterwhatcurrentpages591–600ispresent.page5946.Anautomobilepanellampwitharesistance1.Thecurrentthroughalightbulbconnectedof33isplacedacrossa12-Vbattery.acrosstheterminalsofa125-VoutletisWhatisthecurrentthroughthecircuit?0.50A.AtwhatratedoesthebulbconvertV12VI0.36Aelectricenergytolight?(Assume100percentR33efficiency.)7.Amotorwithanoperatingresistanceof32PIV(0.50A)(125V)63J/s63Wisconnectedtoavoltagesource.Thecurrentinthecircuitis3.8A.Whatisthevoltageof2.Acarbatterycausesacurrentof2.0Athesource?throughalampandproduces12Vacrossit.Whatisthepowerusedbythelamp?VIR(3.8A)(32)1.2102VPIV(2.0A)(12V)24W8.Asensoruses2.0104Aofcurrentwhenitisoperatedbya3.0-Vbattery.Whatisthe3.Whatisthecurrentthrougha75-Wlight-resistanceofthesensorcircuit?bulbthatisconnectedtoa125-Voutlet?V3.0V4PIVRI2.0104A1.510P75WI0.60AV125V9.Alampdrawsacurrentof0.50Awhenitisconnectedtoa120-Vsource.4.Thecurrentthroughthestartermotorofaa.Whatistheresistanceofthelamp?caris210A.Ifthebatterymaintains12Vacrossthemotor,howmuchelectricenergyRV120V2.4102I0.50Aisdeliveredtothestarterin10.0s?b.WhatisthepowerconsumptionofthePIVandEPtlamp?Thus,EIVt(210A)(12V)(10.0s)1WPIV(0.50A)(120V)6.0102.5104J10.A75-Wlampisconnectedto125V.5.Aflashlightbulbisratedat0.90W.Ifthea.Whatisthecurrentthroughthelamp?lightbulbdrops3.0V,howmuchcurrentP75Wgoesthroughit?I0.60AV125VPIVb.Whatistheresistanceofthelamp?P0.90WI0.30ARV125V2.1102V3.0VI0.60ACopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual445

449Chapter22continued11.Aresistorisaddedtothelampintheprevi-Aousproblemtoreducethecurrenttohalfof85mAitsoriginalvalue.4.5V53a.Whatisthepotentialdifferenceacrossthelamp?IThenewvalueofthecurrentisV4.5V0.60AR530.30AI0.085A2VIR(0.30A)(2.1102)14.Addavoltmetertomeasurethepotential6.3101Vdifferenceacrosstheresistorsinproblems12and13andrepeattheproblems.b.HowmuchresistancewasaddedtotheBothcircuitswilltakethefollowingform.circuit?ThetotalresistanceofthecircuitAisnowRV125V2VtotalI0.30A4.210ITherefore,RresRtotalRIampBecausetheammeterresistanceis4.21022.1102assumedzero,thevoltmeterreadings2.1102willbe60.0VforPracticeProblem12and4.5VforPracticeProblem13.c.Howmuchpowerisnowdissipatedinthelamp?15.Drawacircuitusingabattery,alamp,aPIV(0.30A)(6.3101V)19Wpotentiometertoadjustthelamp’sbright-ness,andanon-offswitch.page600LampCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12.Drawacircuitdiagramtoincludea60.0-Vbattery,anammeter,andaresistanceof12.5inseries.Indicatetheammeterread-Batteryingandthedirectionofthecurrent.SwitchPotentiometerA4.80A16.Repeatthepreviousproblem,addingan60.0V12.5ammeterandavoltmeteracrossthelamp.IVV60.0VI4.80AR12.5A13.Drawaseries-circuitdiagramshowinga4.5-Vbattery,aresistor,andanammeterthatreads85mA.Determinetheresistanceandlabeltheresistor.Chooseadirectionfortheconventionalcurrentandindicatethepositiveterminalofthebattery.446SolutionsManualPhysics:PrinciplesandProblems

450Chapter22continuedP2/R2/1212WSectionReview1V1(12V)P2/R2/9.016W22.1CurrentandCircuits2V2(12V)PPpages591–6002P116W12W4.0Wpage6004.0Wincrease17.SchematicDrawaschematicdiagramofacircuitthatcontainsabatteryandalight-21.EnergyAcircuitconverts2.2103Jofbulb.Makesurethelightbulbwilllightinenergywhenitisoperatedfor3.0min.thiscircuit.Determinetheamountofenergyitwillconvertwhenitisoperatedfor1h.2.2103JE(60.0min)3.0min4.4104J22.CriticalThinkingWesaythatpoweris“dissipated”inaresistor.Todissipateistouse,towaste,ortosquander.Whatis“used”whenchargeflowsthrougharesistor?18.ResistanceJoestatesthatbecauseRV/I,Thepotentialenergyofthechargesifheincreasesthevoltage,theresistancedecreasesastheyflowthroughtheresis-willincrease.IsJoecorrect?Explain.tor.ThisdecreaseinpotentialenergyisNo,resistancedependsonthedevice.usedtoproduceheatintheresistor.WhenVincreases,sowillI.19.ResistanceYouwanttomeasuretheresis-PracticeProblemstanceofalongpieceofwire.Showhowyou22.2UsingElectricEnergywouldconstructacircuitwithabattery,avoltmeter,anammeter,andthewiretobepages601–605testedtomakethemeasurement.Specifypage603whatyouwouldmeasureandhowyou23.A15-electricheateroperatesona120-Vwouldcomputetheresistance.outlet.a.Whatisthecurrentthroughtheheater?AV120VI8.0AR15b.HowmuchenergyisusedbytheheaterVin30.0s?EI2Rt(8.0A)2(15)(30.0s)2.9104JMeasurethecurrentthroughthewirec.Howmuchthermalenergyisliberatedandthepotentialdifferenceacrossit.inthistime?Dividethepotentialdifferencebythe2.9104J,becauseallelectricenergycurrenttoobtainthewireresistance.isconvertedtothermalenergy.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.20.PowerAcircuithas12ofresistanceand24.A39-resistorisconnectedacrossa45-Visconnectedtoa12-Vbattery.Determinebattery.thechangeinpoweriftheresistancea.Whatisthecurrentinthecircuit?decreasesto9.0.V45VI1.2AR39Physics:PrinciplesandProblemsSolutionsManual447

451Chapter22continuedb.Howmuchenergyisusedbytheresistor27.A120-Vwaterheatertakes2.2htoheatain5.0min?givenvolumeofwatertoacertaintempera-2ture.Howlongwoulda240-VunitoperatingVEtwiththesamecurrenttaketoaccomplishR2thesametask?(45V)(5.0min)(60s/min)(39)tEIVtI(2V)21.6104JForagivenamountofenergy,doublingthevoltagewilldividethetimeby2.25.A100.0-Wlightbulbis22percentefficient.2.2hThismeansthat22percentoftheelectrict1.1h2energyisconvertedtolightenergy.a.Howmanyjoulesdoesthelightbulbpage605convertintolighteachminuteitisin28.Anelectricspaceheaterdraws15.0Afromaoperation?120-Vsource.Itisoperated,ontheaverage,EPtfor5.0heachday.(0.22)(100.0J/s)(1.0min)a.Howmuchpowerdoestheheateruse?(60s/min)PIV(15.0A)(120V)1.3103J1800W1.8kWb.Howmanyjoulesofthermalenergydoesb.HowmuchenergyinkWhdoesitcon-thelightbulbproduceeachminute?sumein30days?EPtEPt(1.8kW)(5.0h/day)(30days)(0.78)(100.0J/s)(1.0min)270kWh(60.0s/min)c.At$0.12perkWh,howmuchdoesit4.7103Jcosttooperatetheheaterfor30days?Cost($0.12/kWh)(270kWh)26.TheresistanceofanelectricstoveelementatCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.$32.40operatingtemperatureis11.a.If220Vareappliedacrossit,whatis29.Adigitalclockhasaresistanceof12,000thecurrentthroughthestoveelement?andispluggedintoa115-Voutlet.IV220V2.0101Aa.Howmuchcurrentdoesitdraw?R11IV115V9.6103Ab.HowmuchenergydoestheelementR12,000converttothermalenergyin30.0s?b.Howmuchpowerdoesituse?EI2Rt(2.0101A)2(11)(30.0PVI(115V)(9.6103A)1.1Ws)c.Iftheowneroftheclockpays$0.12per1.3105JkWh,howmuchdoesitcosttooperatec.Theelementisusedtoheatakettlecon-theclockfor30days?taining1.20kgofwater.AssumethatCost(1.1103kWh)($0.12/kWh)65percentoftheheatisabsorbedby(30days)(24h/day)thewater.Whatisthewater’sincreaseintemperatureduringthe30.0s?$0.10QmCTwithQ0.65E30.Anautomotivebatterycandeliver55Aat0.65E(0.65)(1.3105J)12Vfor1.0handrequires1.3timesasTmC(1.20kg)(4180J/kg°C)muchenergyforrechargeduetoitsless-17°Cthan-perfectefficiency.Howlongwillit448SolutionsManualPhysics:PrinciplesandProblems

452Chapter22continuedtaketochargethebatteryusingacurrentofPV2/R(0.5V2/R121)0.257.5A?AssumethatthechargingvoltageisP2V2/RV211thesameasthedischargingvoltage.E35.EfficiencyEvaluatetheimpactofresearchcharge(1.3)IVttoimprovepowertransmissionlineson(1.3)(55A)(12V)(1.0h)societyandtheenvironment.858WhResearchtoimprovepowertransmis-E858Wht9.5hsionlineswouldbenefitsocietyincostIV(7.5A)(12V)ofelectricity.Also,iflesspowerwas31.Reworkthepreviousproblembyassuminglostduringtransmission,lesscoalandthatthebatteryrequirestheapplicationofotherpower-producingresourceswould14Vwhenitisrecharging.havetobeused,whichwouldimprovethequalityofourenvironment.Echarge(1.3)IVt(1.3)(55A)(12V)(1.0h)36.VoltageWhywouldanelectricrangeandanelectrichot-waterheaterbeconnectedto858Wha240-Vcircuitratherthana120-Vcircuit?E858Wht8.2hForthesamepower,attwicethevolt-IV(7.5A)(14V)age,thecurrentwouldbehalved.TheI2RlossinthecircuitwiringwouldbeSectionReviewdramaticallyreducedbecauseitispro-portionaltothesquareofthecurrent.22.2UsingElectricEnergypages601–60537.CriticalThinkingWhendemandforelec-page605tricpowerishigh,powercompaniessome-32.EnergyAcarenginedrivesagenerator,timesreducethevoltage,therebyproducingwhichproducesandstoreselectricchargeina“brown-out.”Whatisbeingsaved?thecar’sbattery.Theheadlampsusetheelec-Power,notenergy;mostdeviceswilltricchargestoredinthecarbattery.Listthehavetorunlonger.formsofenergyinthesethreeoperations.Mechanicalenergyfromtheenginecon-vertedtoelectricenergyinthegenera-ChapterAssessmenttor;electricenergystoredaschemicalConceptMappingenergyinthebattery;chemicalenergypage610convertedtoelectricenergyinthebat-38.Completetheconceptmapusingthefol-teryanddistributedtotheheadlamps;lowingterms:watt,current,resistance.electricenergyconvertedtolightandthermalenergyinheadlamps.Electricity33.ResistanceAhairdryeroperatingfrom120Vhastwosettings,hotandwarm.Inwhichsettingistheresistancelikelytoberateofrateofoppositionflowconversiontoflowsmaller?Why?Hotdrawsmorepower,PIV,sotheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fixedvoltagecurrentislarger.BecausecurrentpowerresistanceIV/Rtheresistanceissmaller.34.PowerDeterminethepowerchangeinacircuitiftheappliedvoltageisdecreasedbyamperewattohmone-half.Physics:PrinciplesandProblemsSolutionsManual449

453Chapter22continuedMasteringConceptsb.Whichdeviceconvertschemicalenergytoelectricenergy?page61039.Definetheunitofelectriccurrentinterms1offundamentalMKSunits.(22.1)c.Whichdeviceturnsthecircuitonandoff?1A1C/1s2d.Whichdeviceprovidesawaytoadjust40.Howshouldavoltmeterbeconnectedinspeed?Figure22-12tomeasurethemotor’s3voltage?(22.1)44.Describetheenergyconversionsthatoccurineachofthefollowingdevices.(22.1)4a.anincandescentlightbulbelectricenergytoheatandlight3b.aclothesdryer1electricenergytoheatandkineticenergyc.adigitalclockradio2electricenergytolightandsound■Figure22-1245.WhichwireconductselectricitywithThepositivevoltmeterleadconnectstheleastresistance:onewithalargetotheleft-handmotorlead,andthecross-sectionaldiameteroronewithanegativevoltmeterleadconnectstosmallcross-sectionaldiameter?(22.1)theright-handmotorlead.Alarger-diameterwirehasasmallerresistancebecausetherearemore41.Howshouldanammeterbeconnectedinelectronstocarrythecharge.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Figure22-12tomeasurethemotor’scur-rent?(22.1)46.Asimplecircuitconsistsofaresistor,aBreakthecircuitbetweenthebatterybattery,andconnectingwires.(22.1)andthemotor.Thenconnecttheposi-a.Drawacircuitschematicofthissimpletiveammeterleadtothepositivesideofcircuit.thebreak(thesideconnectedtothepositivebatteryterminal)andthenega-tiveammeterleadtothenegativesideIInearestthemotor.42.WhatisthedirectionoftheconventionalmotorcurrentinFigure22-12?(22.1)b.Howmustanammeterbeconnectedinfromlefttorightthroughthemotoracircuitforthecurrenttobecorrectlyread?43.RefertoFigure22-12toanswerthefollow-Theammetermustbeconnectediningquestions.(22.1)series.a.Whichdeviceconvertselectricenergytomechanicalenergy?4450SolutionsManualPhysics:PrinciplesandProblems

454Chapter22continuedVApplyingConceptspages610–61151.BatteriesWhenabatteryisconnectedtoaIIAcompletecircuit,chargesflowinthecircuitalmostinstantaneously.Explain.Apotentialdifferenceisfeltovertheentirecircuitassoonasthebatteryisc.Howmustavoltmeterbeconnectedtoconnectedtothecircuit.Thepotentialaresistorforthepotentialdifferencedifferencecausesthechargestobeginacrossittoberead?toflow.Note:ThechargesflowslowlyThevoltmetermustbeconnectedincomparedtothechangeinpotentialparallel.difference.V52.Explainwhyacowexperiencesamildshockwhenittouchesanelectricfence.IIABytouchingthefenceandtheground,thecowencountersadifferenceinpotentialandconductscurrent,thusreceivingashock.47.Whydolightbulbsburnoutmorefrequentlyjustastheyareswitchedonratherthan53.PowerLinesWhycanbirdsperchonhigh-whiletheyareoperating?(22.2)voltagelineswithoutbeinginjured?ThelowresistanceofthecoldfilamentNopotentialdifferenceexistsalongtheallowsahighcurrentinitiallyandawires,sothereisnocurrentthroughgreaterchangeintemperature,subject-thebirds’bodies.ingthefilamenttogreaterstress.54.Describetwowaystoincreasethecurrentin48.Ifabatteryisshort-circuitedbyaheavycop-acircuit.perwirebeingconnectedfromoneterminalEitherincreasethevoltageordecreasetotheother,thetemperatureofthecoppertheresistance.wirerises.Whydoesthishappen?(22.2)Theshortcircuitproducesahighcur-55.LightbulbsTwolightbulbsworkonarent,whichcausesmoreelectronsto120-Vcircuit.Oneis50Wandtheotheriscollidewiththeatomsofthewire.This100W.Whichbulbhasahigherresistance?raisestheatoms’kineticenergiesandExplain.thetemperatureofthewire.50-WbulbV2V249.WhatelectricquantitiesmustbekeptsmallP,soRRPtotransmitelectricenergyeconomicallyTherefore,thelowerPiscausedbyaoverlongdistances?(22.2)higherR.theresistanceofthewireandthecurrentinthewire56.Ifthevoltageacrossacircuitiskeptcon-stantandtheresistanceisdoubled,whatCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.50.Definetheunitofpowerintermsoffunda-effectdoesthishaveonthecircuit’scurrent?mentalMKSunits.(22.2)Iftheresistanceisdoubled,thecurrentm2Jkg2kgm2ishalved.CJsWsCsss3Physics:PrinciplesandProblemsSolutionsManual451

455Chapter22continued57.WhatistheeffectonthecurrentinacircuitMotorifboththevoltageandtheresistanceare1.5Adoubled?Explain.Noeffect.VIR,soIV/R,andifthevoltageandtheresistancebotharedoubled,thecurrentwillnotchange.12V58.Ohm’sLawSuefindsadevicethatlookslikearesistor.Whensheconnectsittoa■Figure22-131.5-Vbattery,shemeasuresonly45106A,a.Howmuchpowerisdeliveredtothebutwhensheusesa3.0-Vbattery,shemea-motor?sures25103A.DoesthedeviceobeyPVI(12V)(1.5A)18WOhm’slaw?b.HowmuchenergyisconvertediftheNo.VIR,soRV/I.At1.5V,motorrunsfor15min?1.5V4R451063.310EPt(18W)(15min)(60s/min)3.0V1.6104JAt3.0V,R25103A120AdevicethatobeysOhm’slawhasa62.RefertoFigure22-14toanswerthefollow-resistancethatisindependentoftheingquestions.appliedvoltage.A59.IftheammeterinFigure22-4aonpage596weremovedtothebottomofthediagram,wouldtheammeterhavethesamereading?27V18VExplain.Yes,becausethecurrentisthesameeverywhereinthiscircuit.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.I60.Twowirescanbeplacedacrossthetermi-nalsofa6.0-Vbattery.Onehasahighresis-■Figure22-14tance,andtheotherhasalowresistance.a.Whatshouldtheammeterreadingbe?Whichwirewillproducethermalenergyat27Vafasterrate?Why?IV/R1.5A18thewirewiththesmallerresistanceb.Whatshouldthevoltmeterreadingbe?V227VPRc.HowmuchpowerisdeliveredtotheSmallerRproduceslargerpowerPresistor?dissipatedinthewire,whichproducesPVI(27V)(1.5A)41Wthermalenergyatafasterrate.d.HowmuchenergyisdeliveredtotheMasteringProblemsresistorperhour?22.1CurrentandCircuitsEPt(41W)(3600s)1.5105Jpages611–612Level161.Amotorisconnectedtoa12-Vbattery,asshowninFigure22-13.452SolutionsManualPhysics:PrinciplesandProblems

456Chapter22continued63.RefertoFigure22-15toanswerthefollow-d.Howmuchenergyisdeliveredtotheingquestions.resistorperhour?EPt(4.5W)(3600s)1.6104JA65.ToastersThecurrentthroughatoasterthatisconnectedtoa120-Vsourceis8.0A.27V9.0VWhatpowerisdissipatedbythetoaster?2WPIV(8.0A)(120V)9.61066.LightbulbsAcurrentof1.2AismeasuredIthroughalightbulbwhenitisconnectedacrossa120-Vsource.Whatpowerisdissi-■Figure22-15patedbythebulb?a.Whatshouldtheammeterreadingbe?PIV(1.2A)(120V)1.4102W27VIV/R3.0A9.067.Alampdraws0.50Afroma120-Vgenerator.b.Whatshouldthevoltmeterreadingbe?a.Howmuchpowerisdelivered?27VPIV(0.50A)(120V)6.0101Wc.Howmuchpowerisdeliveredtotheresistor?b.Howmuchenergyisconvertedin5.0min?PVI(27V)(3.0A)81WEd.HowmuchenergyisdeliveredtotheThedefinitionofpowerisP,sotresistorperhour?EPtEPt(81W)(3600s)2.9105J1W)5.0min60s(6.0101min64.RefertoFigure22-16toanswerthefollow-4J18,000J1.810ingquestions.68.A12-VautomobilebatteryisconnectedtoAanelectricstartermotor.Thecurrentthroughthemotoris210A.a.Howmanyjoulesofenergydoesthe9.0V18Vbatterydelivertothemotoreachsecond?PIV(210A)(12V)2500J/sor2.5103J/sIb.Whatpower,inwatts,doesthemotoruse?■Figure22-163WP2.510a.Whatshouldtheammeterreadingbe?9.0V69.DryersA4200-Wclothesdryerisconnect-IV/R0.50A18edtoa220-Vcircuit.Howmuchcurrentb.Whatshouldthevoltmeterreadingbe?doesthedryerdraw?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9.0VPIVc.HowmuchpowerisdeliveredtotheP4200WI19Aresistor?V220VPVI(9.0V)(0.50A)4.5WPhysics:PrinciplesandProblemsSolutionsManual453

457Chapter22continued70.FlashlightsAflashlightbulbisconnectedTable22-2acrossa3.0-Vpotentialdifference.Thecur-Voltage,VCurrent,IResistance,RV/Irentthroughthebulbis1.5A.(volts)(amps)(amps)a.Whatisthepowerratingofthebulb?2.000.0140______________PIV(1.5A)(3.0V)4.5W4.000.0270______________b.Howmuchelectricenergydoesthebulb6.000.0400______________convertin11min?8.000.0520______________EThedefinitionofpowerisP,so10.000.0630______________t2.000.0140______________EPt4.000.0280______________60s(4.5W)(11min)min6.000.0390______________8.000.0510______________3.0103J10.000.0620______________71.BatteriesAresistorof60.0hasacurrenta.Foreachmeasurement,calculatetheof0.40Athroughitwhenitisconnectedtoresistance.theterminalsofabattery.Whatisthevolt-R143,R148,R150,ageofthebattery?R154,R159,R143,VIR(0.40A)(60.0)24VR143,R154,R157,R16172.Whatvoltageisappliedtoa4.0-resistorifthecurrentis1.5A?b.GraphIversusV.VIR(1.5A)(4.0)6.0V0.0673.Whatvoltageisplacedacrossamotorwith0.04a15-operatingresistanceifthereis8.0Aofcurrent?0.02Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.VIR(8.0A)(15)1.2102V12.008.004.000.004.008.0012.0074.Avoltageof75Visplacedacrossa15-I(amps)resistor.Whatisthecurrentthroughtheresistor?0.04VIRV75V0.06I5.0AR15V(volts)Level2c.DoesthenichromewireobeyOhm’s75.Somestudentsconnectedalengthoflaw?Ifnot,forallthevoltages,specifythenichromewiretoavariablepowersupplyvoltagerangeforwhichOhm’slawholds.toproducebetween0.00Vand10.00Vacrossthewire.TheythenmeasuredtheOhm’slawisobeyedwhenthecurrentthroughthewireforseveralvolt-resistanceofadeviceisconstantages.Thestudentsrecordedthedatafortheandindependentofthepotentialvoltagesusedandthecurrentsmeasured,asdifference.TheresistanceoftheshowninTable22-2.nichromewireincreasessomewhatasthemagnitudeofthevoltageincreases,sothewiredoesnotquiteobeyOhm’slaw.454SolutionsManualPhysics:PrinciplesandProblems

458Chapter22continued76.Drawaseriescircuitdiagramtoincludea79.Thecurrentthroughalampconnected16-resistor,abattery,andanammeteracross120Vis0.40Awhenthelampison.thatreads1.75A.Indicatethepositiveter-a.Whatisthelamp’sresistancewhenitminalandthevoltageofthebattery,theison?positiveterminaloftheammeter,andtheVIRdirectionofconventionalcurrent.V120V2R3.010I0.40AAb.Whenthelampiscold,itsresistanceisI1.75A1asgreatasitiswhenthelampishot.V28V165Whatisthelamp’scoldresistance?I1(3.0102)6.01015VIR(1.75A)(16)28Vc.Whatisthecurrentthroughthelampas77.Alampdrawsa66-mAcurrentwhencon-itisturnedonifitisconnectedtoanectedtoa6.0-Vbattery.Whena9.0-Vpotentialdifferenceof120V?batteryisused,thelampdraws75mA.VIRa.DoesthelampobeyOhm’slaw?V120VI2.0AR6.0101No.Thevoltageisincreasedbya9.0factorof1.5,butthecurrentLevel36.075isincreasedbyafactorof1.180.ThegraphinFigure22-17showsthecur-66rentthroughadevicecalledasilicondiode.b.Howmuchpowerdoesthelampdissi-patewhenitisconnectedtothe6.0-VCurrentinaDiodebattery?PIV(66103A)(6.0V)0.40Wc.Howmuchpowerdoesitdissipateat0.029.0V?PIV(75103A)(9.0V)0.68WCurrent(A)0.0178.LightbulbsHowmuchenergydoesa60.0-Wlightbulbuseinhalfanhour?Ifthelight-00.20.40.60.8bulbconverts12percentofelectricenergytoVoltage(V)lightenergy,howmuchthermalenergydoes■Figure22-17itgenerateduringthehalfhour?a.Apotentialdifferenceof0.70VisEPplacedacrossthediode.Whatisthetresistanceofthediode?EPt(60.0W)(1800s)Fromthegraph,I22m/A,and1.08105JVIR,soIfthebulbis12percentefficient,V0.70VR3288percentoftheenergyislosttoheat,soI2.2102ACopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Q(0.88)(1.08105J)9.5104Jb.Whatisthediode’sresistancewhena0.60-Vpotentialdifferenceisused?V0.60V2R1.210I5.2103Ac.DoesthediodeobeyOhm’slaw?No.Resistancedependsonvoltage.Physics:PrinciplesandProblemsSolutionsManual455

459Chapter22continued81.Drawaschematicdiagramtoshowacircuitincludinga90-Vbattery,anammeter,andaresistanceof45connectedinseries.Whatistheammeterreading?Drawarrowsshow-ingthedirectionofconventionalcurrent.V40.0A2A90V45II■Figure22-18a.themaximumsafecurrentVIRPI2RV90VI2A1WR45P5.010I1AR40.0b.themaximumsafevoltage22.2UsingElectricEnergyPV2/Rpages612–613Level1VPR(5.0101W)(40.0)82.BatteriesA9.0-Vbatterycosts$3.00and45Vwilldeliver0.0250Afor26.0hbeforeitmustbereplaced.CalculatethecostperkWh.86.UtilitiesFigure22-19representsanelec-EIVt(0.0250A)(9.0V)(26.0h)tricfurnace.Calculatethemonthly(30-day)5.9Wh5.9103kWhheatingbillifelectricitycosts$0.10perkWhandthethermostatisonone-fourthofcost$3.00RateE5.9103kWhthetime.$510/kWhThermostatCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.83.Whatisthemaximumcurrentallowedina5.0-W,220-resistor?PI2R240.0V4.80P5.0WI0.15AR22084.A110-Velectricirondraws3.0Aofcurrent.Howmuchthermalenergyisdevelopedin■Figure22-19anhour?V2QEVIt(110V)(3.0A)(1.0h)(3600s/h)ER(t)1.2106J(240.0V)24.80(30d)(24h/d)(0.25)2160kWhLevel285.ForthecircuitshowninFigure22-18,theCost(2160kWh)($0.100/kWh)$216maximumsafepoweris5.0101W.Usethefiguretofindthefollowing:87.AppliancesAwindowairconditionerisesti-matedtohaveacostofoperationof$50per30days.Thisisbasedontheassumptionthattheairconditionerwillrunhalfofthetime456SolutionsManualPhysics:PrinciplesandProblems

460Chapter22continuedandthatelectricitycosts$0.090perkWh.90.Acurrentof1.2AismeasuredthroughaDeterminehowmuchcurrenttheaircondi-50.0-resistorfor5.0min.Howmuchheattionerwilltakefroma120-Voutlet.isgeneratedbytheresistor?Cost(E)(rate)QEI2RtECost$502(50.0)(5.0min)60s(1.2A)rate$0.090/kWhmin556kWh2.2104JEIVt91.A6.0-resistorisconnectedtoa15-VE(556kWh)(1000W/kW)battery.IVt(120V)(30d)(24h/d)(0.5)a.Whatisthecurrentinthecircuit?12.9AVIRV15V88.RadiosAtransistorradiooperatesbyI2.5AR6.0meansofa9.0-Vbatterythatsuppliesitb.Howmuchthermalenergyisproducedwitha50.0-mAcurrent.in10.0min?a.Ifthecostofthebatteryis$2.49andit2RtQEIlastsfor300.0h,whatisthecostperkWhtooperatetheradiointhismanner?(2.5A)2(6.0)(10.0min)60sminPIV(0.050A)(9.0V)0.45W4J2.3104.5104kW$2.49Level2Cost(4.5104kW)(300.0h)92.LightbulbsAnincandescentlightbulbwith$18.00/kWharesistanceof10.0whenitisnotlitandaresistanceof40.0whenitislithasb.Thesameradio,bymeansofaconverter,120Vplacedacrossit.ispluggedintoahouseholdcircuitbyahomeownerwhopays$0.12perkWh.a.WhatisthecurrentdrawwhenthebulbWhatdoesitnowcosttooperatetheislit?radiofor300.0h?V120VI3.0AR40.0Cost($0.12/kWh)(4.5104kW)(300h)b.Whatisthecurrentdrawattheinstantthebulbisturnedon?$0.02V120VI12AR10.0MixedReviewc.Whendoesthelightbulbusethemostpage613power?Level1theinstantitisturnedon89.Ifapersonhas$5,howlongcouldheorsheplaya200Wstereoifelectricitycosts93.A12-Velectricmotor’sspeediscontrolled$0.15perkWh?byapotentiometer.Atthemotor’sslowestCostsetting,ituses0.02A.Atitshighestsetting,EPtRatethemotoruses1.2A.WhatistherangeofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Costthepotentiometer?t(Rate)(P)Theslowestspeed’sresistanceis$5RV/I12V/0.02A600.The($0.15/kWh)(200W)1kWfastestspeed’sresistanceis1000WRV/I12V/1.2A1.0101.200hTherangeis1.0101to600.Physics:PrinciplesandProblemsSolutionsManual457

461Chapter22continuedLevel3QT94.Anelectricmotoroperatesapumpthatirri-mCgatesafarmer’scropbypumping1.0104L1.1106Jofwateraverticaldistanceof8.0mintoa(20.0kg)(4180J/kgC°)fieldeachhour.Themotorhasanoperating13°Cresistanceof22.0andisconnectedacrossd.At$0.08perkWh,howmuchdoesita110-Vsource.costtooperatetheheatingcoil30mina.Whatcurrentdoesthemotordraw?perdayfor30days?VIR1.1106J30minCost(30days)V110V5mindayI5.0AR22.01kWh$0.08b.Howefficientisthemotor?3.6106JkWhEWmgd$4.40(1104kg)(9.80m/s2)(8.0m)96.AppliancesAnelectricheaterisratedat8105J500W.Ea.HowmuchenergyisdeliveredtothemIVt(5.0A)(110V)(3600s)heaterinhalfanhour?2.0106JEPt(5102W)(1800s)EEfficiencyw100E9105Jm8105Jb.Theheaterisbeingusedtoheataroom2.0106J100containing50kgofair.Ifthespecificheatofairis1.10kJ/kg°C,and50per-40%centofthethermalenergyheatstheair95.Aheatingcoilhasaresistanceof4.0andintheroom,whatisthechangeinairoperateson120V.temperatureinhalfanhour?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.WhatisthecurrentinthecoilwhileitisQmCToperating?QTmCVIR(0.5)(9105J)IV120V3.0101A(50.0kg)(1100J/kgC°)R4.0b.Whatenergyissuppliedtothecoilin8°C5.0min?c.At$0.08perkWh,howmuchdoesit2Rtcosttoruntheheater6.0hperdayforEI30days?(3.0101A)2(4.0)(5.0min)60smin500J6.0h3600sCostsdayh1.1106J1kWH$0.08c.Ifthecoilisimmersedinaninsulated(30days)6JkWh3.610containerholding20.0kgofwater,$7whatwillbetheincreaseinthetemper-atureofthewater?Assume100percentThinkingCriticallyoftheheatisabsorbedbythewater.page614QmCT97.FormulateModelsHowmuchenergyisstoredinacapacitor?Theenergyneededtoincreasethepotentialdifferenceofacharge,q,458SolutionsManualPhysics:PrinciplesandProblems

462Chapter22continuedisrepresentedbyEqV.Butinacapacitor,Figure22-2bonpage593.LabeltheVq/C.Thus,aschargeisadded,thepoten-functionofeachblockaccordingtotialdifferenceincreases.Asmorechargeistotaljoulespersecond.added,however,ittakesmoreenergytoaddtheadditionalcharge.Considera1.0-FMicrowaveHeatUsefulInput→→→converterconverteroutput“supercap”usedasanenergystoragedevice1440J/s1440J/s1080J/s810J/sinapersonalcomputer.PlotagraphofVas→→thecapacitorischargedbyadding5.0Ctoit.Whatisthevoltageacrossthecapacitor?TheWastedWastedareaunderthecurveistheenergystoredin360J/s270J/sthecapacitor.Findtheenergyinjoules.Isitequaltothetotalchargetimesthefinalb.Deriveanequationfortherateoftem-potentialdifference?Explain.peratureincrease(T/s)fromtheinfor-mationpresentedinChapter12.Solve5.0fortherateoftemperaturerisegiventherateofenergyinput,themass,andthe4.0specificheatofasubstance.3.0T1QtmCt2.0c.Useyourequationtosolvefortherate1.0oftemperatureriseindegreesCelsiusPotentialdifference(V)persecondwhenusingthisoventoheat0.01.02.03.04.05.0250gofwateraboveroomtemperature.Charge(C)T1QtmCtq5.0CVoltageV5.0VC1.0F810J/s(0.25kg)(4180J/kg°C)EnergyEareaundercurve0.78°C/s12(5.0V)(5.0C)d.Reviewyourcalculationscarefullyfortheunitsusedanddiscusswhyyour13Janswerisinthecorrectform.No.Graphically,totalchangetimesfinalThekgunitcancelsandtheJunitpotentialdifferenceisexactlytwicethecancels,leaving°C/s.areaunderthecurve.Physically,ite.Discuss,ingeneralterms,differentwaysmeansthateachcoulombwouldrequireinwhichyoucouldincreasetheefficiencythesamemaximumamountofenergyofmicrowaveheating.toplaceitonthecapacitor.Actually,theTheefficiencyofconversionfromamountofenergyneededtoaddeachelectricenergytomicrowaveenergyischargeincreasesaschargeaccumu-75percent.Itmightbepossibletofindlatesonthecapacitor.awaytoconvertelectricenergytoradiationusingadifferentapproach98.ApplyConceptsAmicrowaveovenoper-thatwouldbemoreefficient.Theeffi-atesat120Vandrequires12Aofcurrent.ciencyofconversionfrommicrowaveCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Itselectricefficiency(convertingACtoradiationtothermalenergyinwaterismicrowaveradiation)is75percent,andits75percent.Itmightbepossibletouseconversionefficiencyfrommicrowaveradia-adifferentfrequencyofelectromag-tiontoheatingwaterisalso75percent.neticradiationtoimprovethisrating.a.DrawablockpowerdiagramsimilarOr,itmightbepossibletofindanewtotheenergydiagramshowninPhysics:PrinciplesandProblemsSolutionsManual459

463Chapter22continuedgeometryofradiatingobjectstobe100.ApplyConceptsThesizesof10-resis-heatedtoimprovetheefficiency.torsrangefromapinheadtoasoupcan.f.Discuss,inefficiencyterms,whyExplain.microwaveovensarenotusefulforheat-Thephysicalsizeofaresistorisdeter-ingeverything.minedbyitspowerrating.ResistorsTheconversionefficiencyfromratedat100Waremuchlargerthanmicrowaveenergytothermalenergythoseratedat1W.isgoodforwater.It’snotasgoodforothermaterials.Thecontainers101.MakeandUseGraphsThediodegraphanddishesdesignedforusewithshowninFigure22-17onpage612ismicrowaveovensconvertlittleofthemoreusefulthanasimilargraphforaenergy.resistorthatobeysOhm’slaw.Explain.g.Discuss,ingeneralterms,whyitisnotThevolt–amperegraphforaresistoragoodideatorunmicrowaveovensobeyingOhm’slawisastraightlinewhentheyareempty.andisseldomnecessary.Theemptyovenmeansthatthe102.MakeandUseGraphsBasedonwhatmicrowaveenergyhastobedissi-youhavelearnedinthischapter,identifypatedintheoven.Thiscanleadtoandpreparetwoparabolicgraphs.overheatingoftheovencompo-voltage–powerandcurrent–powernentsandtotheirfailure.PI2R99.AnalyzeandConcludeAsalesclerkinan40appliancestorestatesthatmicrowaveovensarethemostelectricallyefficient30meansofheatingobjects.a.Formulateanargumenttorefutethe20clerk’sclaim.Hint:Thinkaboutheatingaspecificobject.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Wattsdissipated10Inthecaseofheatingacupofwater,animmersionheaterusesonlyresistanceforenergyconver-05101520sionandisnearly100percenteffi-Voltageacrossa10-resistorcient.Amicrowaveovenusestwoenergyconversions(electricitytoV2microwaveradiationtoheat)andisPRtypicallyaround50percentefficient.40b.Formulateanargumenttosupporttheclerk’sclaim.Hint:Thinkaboutheating30aspecificobject.Inthecaseofheatingapotato,a20microwaveovenheatsmostlythepotatoandismoreefficientthananWattsdissipatedelectricovenorskillet,whichalso10heatstheair,cabinets,racks,etc.c.Formulateadiplomaticreplytothe012clerk.Amperesthrougha10-resistor“Itcanbetrue,butitdependsonthespecificapplication.”460SolutionsManualPhysics:PrinciplesandProblems

464Chapter22continuedWritinginPhysicsS(20kg)(3.34105J/kg)/(273K)page6142.4104J/K103.Therearethreekindsofequationsencoun-teredinscience:(1)definitions,(2)laws,106.Whenyougouptheelevatorofatalland(3)derivations.Examplesoftheseare:building,yourearsmightpopbecauseof(1)anampereisequaltoonecoulombtherapidchangeinpressure.Whatisthepersecond,(2)forceisequaltomasspressurechangecausedbyridinginanele-timesacceleration,(3)powerisequaltovatorupa30-storybuilding(150m)?Thevoltagesquareddividedbyresistance.densityofairisabout1.3kg/m3atseaWriteaone-pageexplanationofwherelevel.(Chapter13)“resistanceisequaltovoltagedividedbyPghcurrent”fits.Beforeyoubegintowrite,(1.3kg/m3)(9.80m/s2)(150m)firstresearchthethreecategoriesgivenabove.1.9kPaorabout2/100ofThestudent’sanswershouldincludethetotalairpressuretheidea(1)that,fordevicesobeyingOhm’slaw,thevoltagedropispropor-107.Whatisthewavelengthinairofa17-kHztionaltocurrentthroughthedevicesoundwave,whichisattheupperendofand(2)thattheformulaRV/I,thethefrequencyrangeofhumanhearing?definitionofresistance,isaderivation(Chapter15)fromOhm’slaw.vfv343m/s104.InChapter13,youlearnedthatmatter0.020m2.0cmf17,000Hzexpandswhenitisheated.Researchtherelationshipbetweenthermalexpansion108.Lightofwavelength478nmfallsonaandhigh-voltagetransmissionlines.doubleslit.First-orderbrightbandsappearAnswerswillvary,butstudentsshould3.00mmfromthecentralbrightband.determinethattransmissionlinescanThescreenis0.91mfromtheslits.Howbecomehotenoughtoexpandandsagfarapartaretheslits?(Chapter19)whentheyhavehighcurrents.SaggingxdlinescanbedangerousiftheytouchLobjectsbeneaththem,suchastreesorLdotherpowerlines.x(478109m)(0.91m)CumulativeReview(3.00103m)page6141.4104m105.Apersonburnsenergyattherateofabout8.4106Jperday.Howmuchdoesshe109.Achargeof3.0106Cis2.0mfromaincreasetheentropyoftheuniverseinsecondchargeof6.0105C.Whatisthatday?Howdoesthiscomparetothethemagnitudeoftheforcebetweenthem?entropyincreasecausedbymelting20kg(Chapter20)ofice?(Chapter12)qFKAqBSQ/TwhereTisthebody2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.dtemperatureof310K.(9.0109Nm2/C2)S(8.4106J)/(310K)(3.0106C)(6.0105C)2.7104J/K2(2.0m)Formeltingice0.41NPhysics:PrinciplesandProblemsSolutionsManual461

465Chapter22continuedChallengeProblemSwitch2page604Switch1Usethefiguretotherighttohelpyouanswerthe1.5Fquestionsbelow.15V1.Initially,thecapacitorisuncharged.Switch11200isclosed,andSwitch2remainsopen.Whatisthevoltageacrossthecapacitor?15V2.Switch1isnowopened,andSwitch2remainsopen.Whatisthevoltageacrossthecapacitor?Why?Itremains15Vbecausethereisnopathforthechargetoberemoved.3.Next,Switch2isclosed,whileSwitch1remainsopen.Whatisthevoltageacrossthecapacitorandthecurrentthroughtheresis-torimmediatelyafterSwitch2isclosed?15Vand13mA4.Astimegoeson,whathappenstothevolt-ageacrossthecapacitorandthecurrentthroughtheresistor?Thecapacitorvoltageremainsat15Vbecausethereisnopathtodischargethecapacitor;thecurrentremainsatCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13mAbecausethebatteryvoltageisconstantat15V.However,ifthebat-teryandcapacitorwererealcompo-nentsinsteadofidealcircuitcompo-nents,thecapacitorvoltageeventuallywouldbecomezeroduetoleakage,andthecurrenteventuallywouldbecomezeroduetobatterydepletion.462SolutionsManualPhysics:PrinciplesandProblems

466CHAPTER23SeriesandParallelCircuitsa.WhatistheequivalentresistanceofthePracticeProblemscircuit?23.1SimpleCircuitsV120V3R210pages617–626I0.06Apage619b.Whatistheresistanceofeachbulb?1.Three20-resistorsareconnectedinseriesR21032Racrossa120-Vgenerator.Whatistheequiv-bulb1010210alentresistanceofthecircuit?Whatisthe5.Calculatethevoltagedropsacrossthethreecurrentinthecircuit?resistorsinproblem2,andverifythattheirRR1R2R3sumequalsthevoltageofthebattery.202020V1IR1(3A)(10)30V60V2IR2(3A)(15)45VV120VV3IR3(3A)(5)15VI2AR60V1V2V330V45V15V2.A10-,15-,and5-resistorareconnect-90Vedinaseriescircuitwitha90-Vbattery.voltageofbatteryWhatistheequivalentresistanceofthecir-cuit?Whatisthecurrentinthecircuit?page622RR1R2R36.ThecircuitshowninExampleProblem1is1015530producingthesesymptoms:theammeterreads0A,VV90VAreads0V,andVBreads45V.I3AWhathashappened?R30R3.A9-Vbatteryisinacircuitwiththreeresis-Bhasfailed.Ithasinfiniteresistance,andthebatteryvoltageappearsacrosstorsconnectedinseries.it.a.Iftheresistanceofoneoftheresistorsincreases,howwilltheequivalentresis-7.SupposethecircuitshowninExampletancechange?Problem1hasthesevalues:RA255,Itwillincrease.RB292,andVA17.0V.Nootherb.Whatwillhappentothecurrent?informationisavailable.Va.Whatisthecurrentinthecircuit?I,soitwilldecrease.RV17.0VI66.7mAc.WilltherebeanychangeinthebatteryR255.0voltage?b.Whatisthebatteryvoltage?No.ItdoesnotdependontheFirst,findthetotalresistance,thenresistance.solveforvoltage.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.RRARB4.Astringofholidaylightshastenbulbswithequalresistancesconnectedinseries.When255292thestringoflightsisconnectedtoa120-V547outlet,thecurrentthroughthebulbsisVIR(66.7mA)(547)36.5V0.06A.Physics:PrinciplesandProblemsSolutionsManual463

467Chapter23continuedc.WhatarethetotalpowerdissipationVsourceVAVBVCandtheindividualpowerdissipations?VCVsource(VAVB)PIV(66.7mA)(36.5V)2.43W12.0V(1.21V3.33V)7.46VP2RAIA(66.7mA)2(255)page6231.13W11.A22-resistoranda33-resistorarecon-P2Rnectedinseriesandplacedacrossa120-VBIB(66.7mA)2(292)potentialdifference.a.Whatistheequivalentresistanceofthe1.30Wcircuit?d.DoesthesumoftheindividualpowerRRdissipationsinthecircuitequalthetotal1R2223355powerdissipationinthecircuit?Explain.b.Whatisthecurrentinthecircuit?Yes.Thelawofconservationofener-V120VI2.2Agystatesthatenergycannotbecre-R55atedordestroyed;therefore,theratec.Whatisthevoltagedropacrosseachatwhichenergyisconverted,orresistor?powerdissipated,willequalthesumV1IR1ofallparts.VRR18.Holidaylightsoftenareconnectedinseriesandusespeciallampsthatshortoutwhen120V(22)55thevoltageacrossalampincreasestothelinevoltage.Explainwhy.Alsoexplainwhy48Vtheselightsetsmightblowtheirfusesafter120VVmanybulbshavefailed.2IR255(33)72VIfnotfortheshortingmechanism,thed.WhatisthevoltagedropacrossthetwoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.entiresetwouldgooutwhenonelampresistorstogether?burnsout.AfterseverallampsfailandV48V72V1.20102Vthenshort,thetotalresistanceoftheremainingworkinglampsresultsinan12.Threeresistorsof3.3k,4.7k,and3.9increasedcurrentthatissufficienttokareconnectedinseriesacrossa12-Vblowthefuse.battery.a.Whatistheequivalentresistance?9.ThecircuitinExampleProblem1hasunequalresistors.ExplainwhytheresistorR3.3k4.7k3.9kwiththelowerresistancewilloperateata1.2101klowertemperature.b.Whatisthecurrentthroughtheresistors?TheresistorwiththelowerresistanceV12VIwilldissipatelesspower,andthuswillR12104becooler.1.0mA1.0103A10.Aseriescircuitismadeupofa12.0-Vbat-c.Whatisthevoltagedropacrosseachteryandthreeresistors.Thevoltageacrossresistor?oneresistoris1.21V,andthevoltageacrossVIRanotherresistoris3.33V.Whatisthevolt-V3A)3.3Vageacrossthethirdresistor?1(3.3k)(1.010V3A)4.7V2(4.7k)(1.010464SolutionsManualPhysics:PrinciplesandProblems

468Chapter23continuedV3A)3.9V16.A120.0-resistor,a60.0-resistor,anda3(3.9k)(1.01040.0-resistorareconnectedinparallelsoV3.3V,4.7V,and3.9Vandplacedacrossa12.0-Vbattery.d.Findthetotalvoltagedropacrossthea.Whatistheequivalentresistanceofthethreeresistors.parallelcircuit?V3.3V4.7V3.9V11.9V1111RR1R2R313.Astudentmakesavoltagedividerfroma45-Vbattery,a475-kresistor,anda235-k111120.060.040.0resistor.Theoutputismeasuredacrossthesmallerresistor.Whatisthevoltage?R20.0VRB(45V)(235k)b.WhatisthecurrentthroughtheentireVBRARB475k235k15Vcircuit?14.Selectaresistortobeusedaspartofavolt-IV12.0V0.600AR20.0agedivideralongwitha1.2-kresistor.Thedropacrossthe1.2-kresistoristobe2.2Vc.Whatisthecurrentthrougheachbranchwhenthesupplyis12V.ofthecircuit?VRV12.0VBIVB1R1120.00.100ARARBV12.0VVRIRB2R60.00.200AAVRB2BV12.0V(12.0V)(1.2k)I3R40.00.300A1.2k32.2V17.Supposethatoneofthe15.0-resistorsin5.3kproblem15isreplacedbya10.0-resistor.a.Doestheequivalentresistancechange?page626Ifso,how?15.Three15.0-resistorsareconnectedinpar-Yes,itgetssmaller.allelandplacedacrossa30.0-Vbattery.b.Doestheamountofcurrentthroughthea.Whatistheequivalentresistanceoftheentirecircuitchange?Ifso,inwhatway?parallelcircuit?Yes,itgetslarger.1111RR1R2R3c.Doestheamountofcurrentthroughtheother15.0-resistorschange?Ifso,111how?15.015.015.0No,itremainsthesame.Currents315.0areindependent.R5.0018.A150-branchinacircuitmustbereducedb.Whatisthecurrentthroughtheentireto93.Aresistorwillbeaddedtothiscircuit?branchofthecircuittomakethischange.V30.0VWhatvalueofresistanceshouldbeusedandI6.00AR5.00howmusttheresistorbeconnected?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.WhatisthecurrentthrougheachbranchAparallelresistorwillberequiredtoofthecircuit?reducetheresistance.V30.0VI2.00A111R115.0RRARBPhysics:PrinciplesandProblemsSolutionsManual465

469Chapter23continuedI11111TI1I2I3I4RARRB93150120mA250mA380mA2.1R2AA2.4102.4102inparallelwiththe150-0.12A0.25A0.38A2.1Aresistance2.9A19.A12-,2-Wresistorisconnectedinparal-22.TotalCurrentAseriescircuithasfourlelwitha6.0-,4-Wresistor.Whichwillresistors.Thecurrentthroughoneresistorisbecomehotterifthevoltageacrossthem810mA.Howmuchcurrentissuppliedbykeepsincreasing?thesource?Neither.Theybothreachmaximumdis-810mA.Currentisthesameevery-sipationatthesamevoltage.whereinaseriescircuit.V2P23.CircuitsAswitchisconnectedinseriesRwitha75-Wbulbtoasourceof120V.VPRa.WhatisthepotentialdifferenceacrossThevoltageisequalacrossparalleltheswitchwhenitisclosed(turnedon)?resistors,so:0V;VIRwithR0VP1R1P2R2b.Whatisthepotentialdifferenceacross(2W)(12)theswitchifanother75-Wbulbisaddedinseries?(4W)(6.0)0V;VIRwithR05Vmaximum24.CriticalThinkingThecircuitinFigure23-8hasfouridenticalresistors.SupposethataSectionReviewwireisaddedtoconnectpointsAandB.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.23.1SimpleCircuitsAnswerthefollowingquestions,andexplainyourreasoning.pages617–626page62620.CircuitTypesCompareandcontrasttheRRvoltagesandthecurrentsinseriesandpar-allelcircuits.ABThestudent’sanswershouldincludethefollowingideas:RR(1)Inaseriescircuit,thecurrentineachofthedevicesisthesame,andthesumofthedevicevoltagedropsequals■Figure23-8thesourcevoltage.a.Whatisthecurrentthroughthewire?(2)Inaparallelcircuit,thevoltagedrop0A;thepotentialsofpointsAandBacrosseachdeviceisthesameandthearethesame.sumofthecurrentsthrougheachloopb.Whathappenstothecurrentthroughequalsthesourcecurrent.eachresistor?nothing21.TotalCurrentAparallelcircuithasfourbranchcurrents:120mA,250mA,380mA,c.Whathappenstothecurrentdrawnand2.1A.Howmuchcurrentissuppliedfromthebattery?bythesource?nothing466SolutionsManualPhysics:PrinciplesandProblems

470Chapter23continuedd.Whathappenstothepotentialdiffer-SectionReviewenceacrosseachresistor?nothing23.2ApplicationsofCircuitspages627–631page631PracticeProblemsRefertoFigure23-13forquestions29–33,and35.Thebulbsinthecircuitareidentical.23.2ApplicationsofCircuitspages627–631Apage630BI3AI1I2A25.Aseries-parallelcircuithasthreeresistors:onedissipates2.0W,thesecond3.0W,andAthethird1.5W.Howmuchcurrentdoes12thecircuitrequirefroma12-Vbattery?CByconservationofenergy(andpower):V3PTP1P2P32.0W3.0W1.5W■Figure23-136.5WP29.BrightnessHowdothebulbbrightnessesTIVcompare?PT6.5WI0.54ABulbs2and3areequalinbrightnessV12Vbutdimmerthanbulb1.26.Thereare11lightsinseries,andtheyareinserieswithtwolightsinparallel.Ifthe1330.CurrentIfI3measures1.7AandI1mea-lightsareidentical,whichofthemwillburnsures1.1A,howmuchcurrentisflowinginbrightest?bulb2?The11lightsinserieswillburnI3I1I2brighter.TheparallellightseachwillIconducthalfofthecurrentoftheseries2I3I11.7A1.1A0.6Alightsandtheywillburnatone-fourth31.CircuitsinSeriesThewireatpointCistheintensityoftheserieslightssincebrokenandasmallresistorisinsertedinPI2R.serieswithbulbs2and3.Whathappenstothebrightnessesofthetwobulbs?Explain.27.Whatwillhappentothecircuitinproblem26ifoneoftheparallellightsburnsout?Bothdimequally.Thecurrentineachisreducedbythesameamount.Then,alloftheworkinglightsareinseries.The12workinglightswillburn32.BatteryVoltageAvoltmeterconnectedwithequalintensity.acrossbulb2measures3.8V,andavolt-meterconnectedacrossbulb3measures28.Whatwillhappentothecircuitinproblem4.2V.Whatisthebatteryvoltage?26ifoneoftheparallellightsshortsout?Thesebulbsareinseries,so:Then,theshortedlightwillreducetheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.voltageacrossitselfanditsparallelVTV1V23.8V4.2V8.0Vcompanionto0.The11serieslightswillburnwithequal,butincreased,intensityandthetwoparallellightswillgoout.Physics:PrinciplesandProblemsSolutionsManual467

471Chapter23continued33.CircuitsUsingtheinformationfromprob-MasteringConceptslem32,determineifbulbs2and3arepage636identical.37.WhyisitfrustratingwhenonebulbburnsNo.Identicalbulbsinserieswouldhaveoutonastringofholidaytreelightscon-identicalvoltagedropssincetheircur-nectedinseries?(23.1)rentsarethesame.Whenonebulbburnsout,thecircuitisopenandallthebulbsgoout.34.CircuitProtectionDescribethreecom-monsafetydevicesassociatedwithhouse-38.Whydoestheequivalentresistancedecreaseholdwiring.asmoreresistorsareaddedtoaparallelfuses,circuitbreakers,ground-faultcircuit?(23.1)circuitinterruptersEachnewresistorprovidesanaddition-alpathforthecurrent.35.CriticalThinkingIsthereawaytomakethethreebulbsinFigure23-13burnwith39.Severalresistorswithdifferentvaluesareequalintensitywithoutusinganyadditionalconnectedinparallel.Howdothevaluesofresistors?Explain.theindividualresistorscomparewiththeYes.Becauseintensityisproportionalequivalentresistance?(23.1)topower,itwouldbenecessarytouseTheequivalentresistancewillbelessabulbatlocation1thathasfourtimesthanthatofanyoftheresistors.theoperatingresistanceofeachofthoseatlocations2and3.40.WhyishouseholdwiringconstructedinV2/4R(V/2)2/Rparallelinsteadofinseries?(23.1)Appliancesinparallelcanberuninde-pendentlyofoneanother.ChapterAssessment41.WhyisthereadifferenceinequivalentConceptMappingresistancebetweenthree60-resistorscon-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page636nectedinseriesandthree60-resistors36.Completetheconceptmapusingthefol-connectedinparallel?(23.1)lowingterms:seriescircuit,RR1R2Inaseriescircuit,thecurrentisR3,constantcurrent,parallelcircuit,constantopposedbyeachresistanceinturn.Thepotential.totalresistanceisthesumoftheresis-tors.Inaparallelcircuit,eachresis-Resistorsincircuitstanceprovidesanadditionalpathforcurrent.Theresultisadecreaseintotalresistance.seriescircuitTypeofcircuitparallelcircuit42.Comparetheamountofcurrententeringajunctioninaparallelcircuitwiththatleav-ingthejunction.(AjunctionisapointconstantconstantwherethreeormoreconductorsarePrinciplecurrentpotentialjoined.)(23.1)Theamountofcurrententeringajunc-tionisequaltotheamountofcurrent111RR1R2R3...ResistanceRRR...12leaving.468SolutionsManualPhysics:PrinciplesandProblems

472Chapter23continued43.ExplainhowafusefunctionstoprotectanApplyingConceptselectriccircuit.(23.2)page636Thepurposeofafuseistopreventcon-48.Whathappenstothecurrentintheotherductorsfrombeingoverloadedwithcur-twolampsifonelampinathree-lamprent,causingfiresduetooverheating.Aseriescircuitburnsout?fuseissimplyashortlengthofwirethatIfoneofthelampfilamentsburnsout,willmeltfromtheheatingeffectifthethecurrentwillceaseandallthelampscurrentexceedsacertainmaximum.willgoout.44.Whatisashortcircuit?Whyisashortcir-49.Supposetheresistor,Rcuitdangerous?(23.2)A,inthevoltagedividerinFigure23-4ismadetobeavari-Ashortcircuitisacircuitthathasableresistor.Whathappenstothevoltageextremelylowresistance.Ashortcir-output,VB,ofthevoltagedividerifthecuitisdangerousbecauseanypotentialresistanceofthevariableresistorisdifferencewillproducealargecurrent.increased?TheheatingeffectofthecurrentcanVcauseafire.BVRB/(RARB).AsRAincreases,VBwilldecrease.45.Whyisanammeterdesignedtohaveavery50.CircuitAcontainsthree60-resistorsinlowresistance?(23.2)series.CircuitBcontainsthree60-resis-Anammetermusthavelowresistancetorsinparallel.Howdoesthecurrentinthebecauseitisplacedinseriesinthecir-second60-resistorofeachcircuitchangecuit.Ifitsresistancewerehigh,itwouldifaswitchcutsoffthecurrenttothefirstsignificantlychangethetotalresistance60-resistor?ofthecircuitandthusservetoreduceCircuitA:Therewillbenocurrentinthethecurrentinthecircuit,therebychang-resistor.ingthecurrentitismeanttomeasure.CircuitB:Thecurrentintheresistorwill46.Whyisavoltmeterdesignedtohaveaveryremainthesamehighresistance?(23.2)51.WhathappenstothecurrentintheotherAvoltmeterisplacedinparallelwiththetwolampsifonelampinathree-lamppar-portionofthecircuitwhosedifferenceallelcircuitburnsout?inpotentialistobemeasured.Avolt-Ifoneofthefilamentsburnsout,themetermusthaveveryhighresistanceresistanceandthepotentialdifferenceforthesamereasonthatanammeteracrosstheotherlampswillnotchange;haslowresistance.Ifthevoltmeterhadtherefore,theircurrentswillremainthelowresistance,itwouldlowertheresis-same.tanceoftheportionofthecircuititisacrossandincreasethecurrentinthe52.Anengineerneedsa10-resistorandacircuit.Thiswouldproduceahigher15-resistor,butthereareonly30-resis-voltagedropacrossthepartofthecir-torsinstock.Mustnewresistorsbepur-cuitwherethevoltmeterislocated,chased?Explain.changingthevoltageitismeasuring.No,the30-resistorscanbeusedinCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.47.Howdoesthewayinwhichanammeterisparallel.Three30-resistorsinparallelconnectedinacircuitdifferfromthewayinwillgivea10-resistance.Two30-whichavoltmeterisconnected?(23.2)resistorsinparallelwillgivea15-resistance.Anammeterisconnectedinseries;avoltmeterisconnectedinparallel.Physics:PrinciplesandProblemsSolutionsManual469

473Chapter23continued53.Ifyouhavea6-Vbatteryandmany1.5-Vf.Addingaresistortothecircuitincreasesbulbs,howcouldyouconnectthemsothatthetotalresistance.theylightbutdonothavemorethan1.5Vseriesacrosseachbulb?g.IfthecurrentthroughoneresistorintheConnectfourofthebulbsinseries.circuitgoestozero,thereisnocurrentThevoltagedropacrosseachwillbeintheentirecircuit.(6.0V)/41.5V.series54.Twolampshavedifferentresistances,oneh.Ifthecurrentthroughoneresistorinthelargerthantheother.circuitgoestozero,thecurrentthroughallotherresistorsremainsthesame.a.Ifthelampsareconnectedinparallel,whichisbrighter(dissipatesmoreparallelpower)?i.Thisformissuitableforhousewiring.Thelampwiththelowerresistance:parallelPIVandIV/R,soPV2/R.Becausethevoltagedropisthe56.HouseholdFusesWhyisitdangeroustosameacrossbothlamps,thesmallerreplacethe15-AfuseusedtoprotectaRmeanslargerP,andthuswillbehouseholdcircuitwithafusethatisratedatbrighter.30A?b.Whenthelampsareconnectedinseries,The30-Afuseallowsmorecurrenttowhichlampisbrighter?flowthroughthecircuit,generatingmoreheatinthewires,whichcanbeThelampwiththehigherresistance;PIVandVIR,soPI2R.dangerous.Becausethecurrentisthesameinbothlamps,thelargerRmeanslarg-MasteringProblemserP,andthuswillbebrighter.23.1SimpleCircuitspages637–63855.Foreachofthefollowing,writetheformofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Level1circuitthatapplies:seriesorparallel.57.Ammeter1inFigure23-14reads0.20A.a.Thecurrentisthesameeverywherethroughouttheentirecircuit.A122seriesA2b.Thetotalresistanceisequaltothesumoftheindividualresistances.Vseries15c.Thevoltagedropacrosseachresistorinthecircuitisthesame.A3parallel■Figure23-14d.Thevoltagedropinthecircuitispro-portionaltotheresistance.a.Whatshouldammeter2indicate?series0.20A,becausecurrentisconstantinaseriescircuit.e.Addingaresistortothecircuitdecreasesthetotalresistance.b.Whatshouldammeter3indicate?parallel0.20A,becausecurrentisconstantinaseriescircuit.470SolutionsManualPhysics:PrinciplesandProblems

474Chapter23continued58.Calculatetheequivalentresistanceofthese64.A22-lampanda4.5-lampareconnect-series-connectedresistors:680,1.1k,edinseriesandplacedacrossapotentialdif-and10k.ferenceof45VasshowninFigure23-15.R680110010,00012k2259.Calculatetheequivalentresistanceofthese45V4.5parallel-connectedresistors:680,1.1k,and10.2k.1111RR1RR■Figure23-1523a.Whatistheequivalentresistanceofthe1Rcircuit?1110.68k1.1k10.2k224.5260.40kb.Whatisthecurrentinthecircuit?V45V60.Aseriescircuithastwovoltagedrops:5.50VIR271.7Aand6.90V.Whatisthesupplyvoltage?c.WhatisthevoltagedropacrosseachV5.50V6.90V12.4Vlamp?VIR(1.7A)(22)37V61.Aparallelcircuithastwobranchcurrents:VIR(1.7A)(4.5)7.7V3.45Aand1.00A.Whatisthecurrentintheenergysource?d.Whatisthepowerdissipatedineachlamp?I3.45A1.00A4.45APIV(1.7A)(37V)63WPIV(1.7A)(7.7V)13WLevel262.Ammeter1inFigure23-14reads0.20A.65.RefertoFigure23-16toanswerthefollow-a.Whatisthetotalresistanceofthecircuit?ingquestions.RR1R2152237V1b.Whatisthebatteryvoltage?VIR(0.20A)(37)7.4VAc.Howmuchpowerisdeliveredtothe3522-resistor?PI2R(0.20A)2(22)0.88W10.0V15V2d.Howmuchpowerissuppliedbythebattery?PIV(0.20A)(7.4V)1.5W■Figure23-1663.Ammeter2inFigure23-14reads0.50A.a.Whatshouldtheammeterread?a.Findthevoltageacrossthe22-resistor.RR1R23515Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.VIR(0.50A)(22)11VIV/Rb.Findthevoltageacrossthe15-resistor.(10.0V)/(3515)VIR(0.50A)(15)7.5V0.20Ac.Whatisthebatteryvoltage?b.Whatshouldvoltmeter1read?VV1V2(11V)(7.5V)19VVIR(0.20A)(35)7.0VPhysics:PrinciplesandProblemsSolutionsManual471

475Chapter23continuedc.Whatshouldvoltmeter2read?67.ForFigure23-18,thebatterydevelops110V.VIR(0.20A)(15)3.0Vd.HowmuchenergyissuppliedbytheA2A3A4batteryperminute?VEPt20.050.010.0VIt(10.0V)(0.20A)(1min)(60s/min)A1120J■Figure23-1866.ForFigure23-17,thevoltmeterreads70.0V.a.Whichresistoristhehottest?10.0.SincePV2/RandVisVconstantinaparallelcircuit,thesmallestresistorwilldissipatetheAmostpower.35b.Whichresistoristhecoolest?V1550.0.SincePV2/RandVisconstantinaparallelcircuit,the50largestresistorwilldissipatetheleastpower.■Figure23-17c.Whatwillammeter1read?a.Whichresistoristhehottest?111150.SincePI2RandIiscon-RR1R2R3stantinaseriescircuit,thelargest1Rvalueofresistancewillproducethe111RRRmostpower.123b.Whichresistoristhecoolest?1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.15.SincePI2RandIiscon-11120.050.010.0stantinaseriescircuit,thesmallest5.88valueofresistancewillproducetheV1.1102Vleastpower.I19AR5.88c.Whatwilltheammeterread?d.Whatwillammeter2read?UseOhm’slaw:IV/RV1.1102VI5.5A(70.0V)/(35)R20.02.0Ae.Whatwillammeter3read?d.WhatisthepowersuppliedbytheV1.1102VI2.2Abattery?R50.0First,findthetotalresistance:f.Whatwillammeter4read?RR1R2R3V1.1102VI11AR10.035155068.ForFigure23-18,ammeter3reads0.40A.0.1ka.Whatisthebatteryvoltage?PI2RVIR(0.40A)(50.0)2.0101V(2.0A)2(0.1k)(1000/k)4102W472SolutionsManualPhysics:PrinciplesandProblems

476Chapter23continuedb.Whatwillammeter1read?b.Whatistheresistanceofasinglelight?Findtheequivalentresistance:Risthesumoftheresistancesof111118lamps,soeachresistanceisRR1R2R32.310213118R111c.Whatpowerisdissipatedbyeachlight?RRR12364W3.6W11811120.050.010.072.Oneofthelightsinproblem71burnsout.1Thelightshortsoutthebulbfilamentwhen0.17itburnsout.Thisdropstheresistanceofthelamptozero.5.881Va.WhatistheresistanceofthelightstringV2.010I3.4Anow?R5.88c.Whatwillammeter2read?Therearenow17lampsinseries1Vinsteadof18lamps.TheresistanceV2.010I1.0AR20.0is17(2.3102)2.210218d.Whatwillammeter4read?b.Findthepowerdissipatedbythestring.V2.0101VI2.0AV2(120V)2R10.0P265WR2.21069.Whatisthedirectionoftheconventionalcur-c.Didthepowerincreaseordecreaserentinthe50.0-resistorinFigure23-18?whenthebulbburnedout?downItincreased.70.Theloadacrossabatteryconsistsoftwo73.A16.0-anda20.0-resistorareconnect-resistors,withvaluesof15and47,edinparallel.Adifferenceinpotentialofconnectedinseries.40.0Visappliedtothecombination.a.Whatisthetotalresistanceoftheload?a.ComputetheequivalentresistanceofRR1R21547theparallelcircuit.62111RR1R2b.Whatisthevoltageofthebatteryifthecurrentinthecircuitis97mA?1R11VIR(97mA)(62)6.0VRR12171.HolidayLightsAstringof18identical11holidaytreelightsisconnectedinseriestoa16.020.0120-Vsource.Thestringdissipates64W.8.89a.Whatistheequivalentresistanceoftheb.Whatisthetotalcurrentinthecircuit?lightstring?V40.0V2I4.50ACopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.PVR8.89Reqc.Whatisthecurrentinthe16.0-V2(120V)22resistor?ReqP64W2.310V40.0VI1R116.02.50APhysics:PrinciplesandProblemsSolutionsManual473

477Chapter23continued74.Amyneeds5.0Vforanintegrated-circuitc.A12-hairdryerispluggedintotheexperiment.Sheusesa6.0-Vbatteryandsameoutlet.Findtheequivalentresis-tworesistorstomakeavoltagedivider.Onetanceofthetwoappliances.resistoris330.Shedecidestomakethe111otherresistorsmaller.WhatvalueshoulditRRARBhave?1RVR211V2R1R2RARBVR12R1VR21125212(6.0V)(330)330669.85.0Vd.Findthevoltagedropacrossthetelevi-75.Peteisdesigningavoltagedividerusingasionandthehairdryer.12-Vbatteryanda82-resistorasRB.VRA(120V)(9.8)WhatresistorshouldbeusedasRAiftheV1R9.82.596VARBoutputvoltageacrossRBistobe4.0V?VRVB23.2ApplicationsofCircuitsBRARBpages638–639VRBLevel1RARBV77.RefertoFigure23-19andassumethatallBVRtheresistorsare30.0.FindtheequivalentRBAVRBresistance.B(12V)(82)824.0VIA1.6102IBICCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Level3■Figure23-1976.TelevisionAtypicaltelevisiondissipates275Wwhenitispluggedintoa120-VTheparallelcombinationofthetwooutlet.30.0-resistorshasanequivalentresistanceof15.0.a.Findtheresistanceofthetelevision.2SoR30.015.045.0VVPIVandI,soP,orRR78.RefertoFigure23-19andassumethateachV2(120V)2R52resistordissipates120mW.FindthetotalP275Wdissipation.b.Thetelevisionand2.5-wiresconnect-P3(120mW)360mWingtheoutlettothefuseformaseriescir-cuitthatworkslikeavoltagedivider.Find79.RefertoFigure23-19andassumethatthevoltagedropacrossthetelevision.IA13mAandIB1.7mA.FindIC.VRVAIARCIAIBARB13mA1.7mA(120V)(52)522.511mA110V474SolutionsManualPhysics:PrinciplesandProblems

478Chapter23continued80.RefertoFigure23-19andassumethatHalfthetotalcurrentisineachIparallelbranchbecausethesumofB13mAandIC1.7mA.FindIA.ItheresistancesineachbranchareAIBICequal.13mA1.7mAPI2R(0.25A)2(30.0)1.9W15mAPI2R(0.25A)2(20.0)1.2WLevel2PI2R(0.25A)2(10.0)0.62W81.RefertoFigure23-20toanswerthefollow-PI2R(0.25A)2(40.0)2.5Wingquestions.The25.0-resistoristhehottest.25.0The10.0-resistoristhecoolest.30.010.082.Acircuitcontainssix60-Wlampswitha25Vresistanceof240-eachanda10.0-20.040.0heaterconnectedinparallel.Thevoltageacrossthecircuitis120V.Findthecurrentinthecircuitforthefollowingsituations.■Figure23-20a.Fourlampsareturnedon.a.Determinethetotalresistance.11111The30.0-and20.0-resistorsareRRRRR1234inseries.111R130.020.050.0240240240The10.0-and40.0-resistorsare1240inseries.4R210.040.050.0240R1andR2areinparallel.R2400.060k4111V120VRR1R2I2.0AR0.060k1b.Allofthelampsareturnedon.R1116RR12R240111240R0.040k50.050.0625.0andisinserieswiththeIV120V3.0AR0.040k25.0-resistorc.Sixlampsandtheheaterareoperating.RTotal25.025.050.0111b.Determinethecurrentthroughthe25-R0.040k10.0resistor.5Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.useOhm’slawandRTotal4.0101V25V4.0101I0.50AR8.0RTotal50.05c.Whichresistoristhehottest?Coolest?V120VI15AR8.0PI2R(0.50A)2(25.0)6.25WPhysics:PrinciplesandProblemsSolutionsManual475

479Chapter23continued83.Ifthecircuitinproblem82hasa12-Afuse,86.HomeCircuitAtypicalhomecircuitiswillthefusemeltifallthelampsandtheshowninFigure23-21.Thewirestotheheaterareon?kitchenlampeachhavearesistanceofYes.The15-Acurrentwillmeltthe12-A0.25.Thelamphasaresistanceoffuse.0.24k.Althoughthecircuitisparallel,theleadlinesareinserieswitheachof84.Duringalaboratoryexercise,youaresup-thecomponentsofthecircuit.pliedwithabatteryofpotentialdifference0.25KitchenV,twoheatingelementsoflowresistancelight240thatcanbeplacedinwater,anammeterofverysmallresistance,avoltmeterofextreme-SwitchPowerboxlyhighresistance,wiresofnegligibleresis-saw120Vtance,abeakerthatiswellinsulatedandhasnegligibleheatcapacity,and0.10kgofwaterWalloutletsat25°C.Bymeansofadiagramandstan-0.25dardsymbols,showhowthesecomponents■Figure23-21shouldbeconnectedtoheatthewaterasa.Computetheequivalentresistanceofrapidlyaspossible.thecircuitconsistingofjustthelampAandtheleadlinestoandfromthelamp.HeatingHeatingVelementelementVR0.250.250.24k#1#20.24kb.Findthecurrenttothelamp.V120V85.Ifthevoltmeterusedinproblem84holdsI0.50AR0.24ksteadyat45Vandtheammeterreadingc.Findthepowerdissipatedinthelamp.holdssteadyat5.0A,estimatethetimeinsecondsrequiredtocompletelyvaporizethePIV(0.50A)(120V)6.0101WCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.waterinthebeaker.Use4.2kJ/kg°Casthespecificheatofwaterand2.3106J/kgasMixedReviewtheheatofvaporizationofwater.page639QmCTLevel187.Aseriescircuithastwovoltagedrops:3.50(0.10kg)(4.2kJ/kg°C)(75°C)Vand4.90V.Whatisthesupplyvoltage?32kJ(energyneededtoraiseV3.50V4.90V8.40Vtemperatureofwaterto100°C)QmH6J/kg)88.Aparallelcircuithastwobranchcurrents:v(0.10kg)(2.3102kJ(energyneededto1.45Aand1.00A.Whatisthecurrentin2.310theenergysource?vaporizethewater)I1.45A1.00A2.45AQ2kJtotal32kJ2.3102kJ(totalenergyneed-89.Aseries-parallelcircuithasthreeresistors,2.610dissipating5.50W,6.90W,and1.05W,ed)Energyisprovidedattherespectively.Whatisthesupplypower?rateofP5.50W6.90W1.05W13.45WPIV(5.0A)(45V)0.23kJ/s.Thetimerequiredis2.6102kJ3st1.1100.23kJ/s476SolutionsManualPhysics:PrinciplesandProblems

480Chapter23continuedLevel2c.avoltmeterwitharesistanceof1010690.DeterminethemaximumsafepowerinThevoltmeterresistanceactsineachofthree150-,5-Wresistorsparallel:connectedinseries.111Allwilldissipatethesamepower.RBR1R2P(3)(5W)15W11471031010691.Determinethemaximumsafepowerin1eachofthree92-,5-Wresistorsconnected2.1105inparallel.REachresistorwilldevelopthesameB47kpower.VRBVBRP(3)(5W)15WARB(12V)(47k)92.Avoltagedividerconsistsoftwo47-k47k47kresistorsconnectedacrossa12-Vbattery.6.0VDeterminethemeasuredoutputfortheThemeterapproachestheidealfollowing.voltmeter.a.anidealvoltmeterVRVBBRLevel3ARB93.Determinethemaximumsafevoltagethat(12V)(47k)canbeappliedacrossthethreeseriesresistors47k47kinFigure23-22ifallthreeareratedat5.0W.6.0V92150220b.avoltmeterwitharesistanceof85kThevoltmeterresistanceactsin■Figure23-22parallel:Currentisconstantinaseriescircuit,111sothelargestresistorwilldeveloptheRRRB12mostpower.1147k85kPI2R1P5.0W5I0.151A3.310R220R1kThetotalresistanceisnowneeded.B3.010VRBRTotal92150220VBRARB462(12V)(3.0101k)UseOhm’slawtofindthevoltage.47k3.0101kVIR4.7V(0.151A)(462)7.0101VCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.94.Determinethemaximumsafetotalpowerforthecircuitinproblem93.1V)2PV2/R(7.01011W462Physics:PrinciplesandProblemsSolutionsManual477

481Chapter23continued95.Determinethemaximumsafevoltagethat97.ApplyConceptsThree-waylamps,ofthecanbeappliedacrossthreeparallelresistorstypeinFigure23-24,havingaratingofof92,150,and220,asshownin50W,100W,and150W,arecommon.Figure23-23,ifallthreeareratedat5.0W.Drawfourpartialschematicdiagramsthatshowthelampfilamentsandtheswitch92positionsforeachbrightnesslevel,aswellastheoffposition.(Youdonotneedtoshow150theenergysource.)Labeleachdiagram.220■Figure23-23The92-resistorwilldevelopthemostpowerbecauseitwillconductthemostcurrent.V2PRVPR(5.0W)(92)21VThinkingCritically■Figure23-24pages639–64096.ApplyMathematicsDeriveequationsforLowerRtheresistanceoftwoequal-valueresistorsinHigherRparallel,threeequal-valueresistorsinparallel,andNequal-valueresistorsinparallel.1112100WCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Req2RRR150W50WRReq221111398.ApplyConceptsDesignacircuitthatwillReq3RRRRlightonedozen12-Vbulbs,alltothecor-Rrect(same)intensity,froma48-Vbattery.Req33a.DesignArequiresthatshouldonebulbRRburnout,allotherbulbscontinuetoeqNNproducelight.48V478SolutionsManualPhysics:PrinciplesandProblems

482Chapter23continuedb.DesignBrequiresthatshouldonebulbburnout,thosebulbsthatcontinueworkingmustproducethecorrectintensity.48Vc.DesignCrequiresthatshouldonebulbburnout,oneotherbulbalsowillgoout.48Vd.DesignDrequiresthatshouldonebulbburnout,eithertwootherswillgooutornootherswillgoout.48V99.ApplyConceptsAbatteryconsistsofanidealsourceofpotentialdifferenceinserieswithasmallresistance.Theelectricenergyofthebatteryisproducedbychemicalreactionsthatoccurinthebattery.However,thesereactionsalsoresultinasmallresistancethat,unfortunately,cannotbecompletelyeliminated.Aflash-lightcontainstwobatteriesinseries,asshowninFigure23-25.Eachhasapoten-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tialdifferenceof1.50Vandaninternalresistanceof0.200.Thebulbhasaresistanceof22.0.Physics:PrinciplesandProblemsSolutionsManual479

483Chapter23continuedBatteryBatteryb.Theleadsarenowconnectedtoanunknownresistance.Whatresistancewouldproduceacurrentofhalf-scale,0.50mA?Quarter-scale,0.25mA?1.50V0.2001.50V0.200Three-quarters-scale,0.75mA?V6.0VR12kI0.50103A22.0andRTR1Re,soReRTR1■Figure23-2512k6.0ka.Whatisthecurrentthroughthebulb?6.0kThecircuithastwo1.50-VbatteriesinV6.0Vserieswiththreeresistors:0.200,R24kI0.25103A0.200,and22.0.TheequivalentandRresistanceis22.4.ThecurrentiseRTR1V24k6.0kIR18k2(1.50)VV6.0V(2(0.200)22.0)R8.0kI0.75103A0.134AandReRTR1b.Howmuchpowerdoesthebulb8.0k6.0kdissipate?2.0kThepowerdissipatedisc.Istheohmmeterscalelinear?Explain.PI2RNo.Zeroohmsisatfull-scale,6k(0.134A)2(22.0)isatmidscale,andinfinite(oropen-circuit)isatzero-scale.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.395Wc.HowmuchgreaterwouldthepowerbeifWritinginPhysicsthebatterieshadnointernalresistance?page640V2(3.00V)2PIV0.409W101.ResearchGustavKirchhoffandhislaws.R22.0Writeaone-pagesummaryofhowtheyP0.409W0.395W0.014WapplytothethreetypesofcircuitsPowerwouldbe0.014Wgreater.presentedinthischapter.Keyideasare:100.ApplyConceptsAnohmmeterismadebyconnectinga6.0-Vbatteryinserieswith(1)Kirchhoff’sVoltageLaw(KVL)isanadjustableresistorandanidealamme-conservationofenergyappliedtoter.Theammeterdeflectsfull-scalewithaelectriccircuits.currentof1.0mA.Thetwoleadsare(2)Kirchhoff’sCurrentLaw(KCL)istouchedtogetherandtheresistanceisconservationofchargeappliedtoadjustedsothat1.0mAflows.electriccircuits.a.Whatistheresistanceoftheadjustable(3)KVLstatesthatthealgebraicsumresistor?ofvoltagedropsaroundaclosedVIRloopiszero.Inaseriescircuitthereisoneclosedloop,andthesumofV6.0VRI36.0kvoltagedropsintheresistances1.010Aequalsthesourcevoltage.Ina480SolutionsManualPhysics:PrinciplesandProblems

484Chapter23continuedparallelcircuit,thereisaclosede.Whatisthefrequencydetectedbyanloopforeachbranch,andKVLobserverlocateddirectlybehindtheimpliesthatthevoltagedropinairplane?eachbranchisthesame.fv/(4)KCLstatesthatthealgebraicsum(340m/s)/(1.50m)ofcurrentsatanodeiszero.Ina230Hz,orseriescircuit,ateverypointthevvcurrentinequalscurrentout;there-fddfsvvfore,thecurrentisthesameevery-swhere.Inaparallelcircuit,thereis340m/s0(340Hz)340m/s(170m/s)acommonnodeateachendofthebranches.KCLimpliesthatthesum230Hzofthebranchcurrentsequalsthe103.Anobjectislocated12.6cmfromacon-sourcecurrent.vexmirrorwithafocallengthofCumulativeReview18.0cm.Whatisthelocationoftheobject’simage?(Chapter17)page640111102.AirplaneAnairplaneflyingthroughstillfdodiairproducessoundwaves.Thewavefrontsinfrontoftheplanearespaced0.50mdofdapartandthosebehindtheplaneareidofspaced1.50mapart.Thespeedofsound(12.6cm)(18.0cm)is340m/s.(Chapter15)12.6cm(18.0cm)a.Whatwouldbethewavelengthofthe7.41cmsoundwavesiftheairplanewerenotmoving?104.Thespeedoflightinaspecialpieceof1.00mglassis1.75108m/s.Whatisitsindexofrefraction?(Chapter18)b.Whatisthefrequencyofthesoundwavesproducedbytheairplane?cnvfv/(340m/s)/(1.00m)340Hz3.00108m/sc.Whatisthespeedoftheairplane?1.75108m/sTheairplanemovesforward0.50m1.71forevery1.00mthatthesoundwavesmove,soitsspeedisone-105.MonocleAnantireflectivecoatingwithanhalfthespeedofsound,170m/s.indexofrefractionof1.4isappliedd.Whatisthefrequencydetectedbyantoamonoclewithanindexofrefractionobserverlocateddirectlyinfrontoftheof1.52.Ifthethicknessofthecoatingisairplane?75nm,whatis/arethewavelength(s)oflightforwhichcompletedestructivefv/interferencewilloccur?(Chapter19)(340m/s)/(0.50m)Becausen680Hz,orfilmnair,thereisaphaseinversiononthefirstreflection.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.vvdBecausenfdfsvvmonoclenfilm,thereisasphaseinversiononthesecond340m/s0reflection.(340Hz)340m/s170m/sFordestructiveinterference:680Hz12tm2nfilmPhysics:PrinciplesandProblemsSolutionsManual481

485Chapter23continued2tnfilm1m2(2)(75nm)(1.4)1m211m(2.1102nm)2Form011(2.1102nm)24.2102nmForm131(2.1102nm)22.8102nmThisisanultravioletwavelength,soitisnotlight.Allothervaluesofmgivewavelengthsthatareshorterthanthelight.So4.2102nmistheonlywavelengthoflightforwhichdestructiveinterferenceoccurs.106.Twochargesof2.0105Cand8.0106Cexperienceaforcebetweenthemof9.0N.Howfarapartarethetwocharges?(Chapter20)qKqFKAqB,sodAqBd2F(9.0109Nm2/C2)(2.0105C)(8.0106C)0.40m9.0NCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.107.Afieldstrength,E,ismeasuredadistance,d,fromapointcharge,Q.WhatwouldhappentothemagnitudeofEinthefollowingsituations?(Chapter21)KQEd2a.distripledE9b.Qistripled3Ec.bothdandQaretripledE3d.thetestchargeqistripledE;bydefinition,fieldstrengthisforceperunittestcharge.e.allthree,d,Q,andq,aretripledE3482SolutionsManualPhysics:PrinciplesandProblems

486Chapter23continued108.Thecurrentflowina12-VcircuitdropsR3V(I1I2)R1R3from0.55Ato0.44A.CalculatethesubstituteVR3R2R3changeinresistance.(Chapter22)RRandV5V(I1I2)R1R51V/I12V/0.55A21.8R5R4R5R2V/I12V/0.44A27.3VRRR3VR5.RemovingR3fromtheleft2R1numeratorandR5fromtheright12V12Vnumeratorgivesthefollowing:0.44A0.55AV(IV(I1I2)R11I2)R15.5RR2141RR3511ChallengeProblemRR2141RRpage62835Whenthegalvanometer,adeviceusedtomea-R3R5sureverysmallcurrentsorvoltages,inthiscircuitR2R4measureszero,thecircuitissaidtobebalanced.3.Whichoftheresistorscanbereplacedwithavariableresistorandthenusedtobalancethecircuit?R1R2R4anyresistorbutR1VAB4.Whichoftheresistorscanbereplacedwithavariableresistorandthenusedasasensi-R3R5tivitycontrol?Whywouldthisbenecessary?Howwoulditbeusedinpractice?R1.Agalvanometerisasensitiveinstru-1.Yourlabpartnerstatesthattheonlywaytomentandcanbedamagedbytoomuchbalancethiscircuitistomakealltheresis-currentflow.IfR1isadjustable,itissettorsequal.Willthisbalancethecircuit?Isforahighvaluebeforethecircuitistheremorethanonewaytobalancethiscir-energized.Thislimitsthecurrentflowcuit?Explain.throughthegalvanometerifthecircuitYes;yes.Youalsocanbalancethecir-happenstobewayoutofbalance.AscuitbyadjustingtheresistancevaluesthebalancingresistorisadjustedandsothatRasthemeterreadingapproacheszero,7/R3R4/R5remainsequal.thesensitivitythenisincreasedby2.DeriveageneralequationforabalanceddecreasingR1.circuitusingthegivenlabels.Hint:Treatthecircuitasavoltagedivider.OnedefinitionofbalanceisthatVAB0.Ifthisisso,thenVR3VR5.ThesevoltagedropscanbedefinedusingOhm’slaw:Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.VR3I1R3andVR5I2R5V(I1I2)R1also,I1R2R3V(IandI1I2)R12R4R5Physics:PrinciplesandProblemsSolutionsManual483

487

488CHAPTER24MagneticFields4.WhydomagneticcompassessometimesPracticeProblemsgivefalsereadings?24.1Magnets:PermanentandbecauseEarth’smagneticfieldisdis-Temporarytortedbyobjectsmadeofiron,nickel,orpages643–651cobaltinthevicinityofthecompass,andbyoredepositsofthesesamemetalspage6471.Ifyouholdabarmagnetineachhandandbringyourhandsclosetogether,willthepage650forcebeattractiveorrepulsiveifthemag-5.Along,straight,current-carryingwirerunsnetsareheldinthefollowingways?fromnorthtosouth.a.thetwonorthpolesarebroughtclosea.Acompassneedleplacedabovethewiretogetherpointswithitsnorthpoletowardtheeast.Inwhatdirectionisthecurrentrepulsiveflowing?b.anorthpoleandasouthpolearefromsouthtonorthbroughttogetherb.Ifacompassisputunderneaththewire,attractiveinwhichdirectionwillthecompass2.Figure24-7showsfivediskmagnetsfloat-needlepoint?ingaboveeachother.Thenorthpoleofthewesttop-mostdiskfacesup.Whichpolesareonthetopsideofeachoftheothermagnets?6.Howdoesthestrengthofamagneticfield,1cmfromacurrent-carryingwire,comparewitheachofthefollowing?a.thestrengthofthefieldthatis2cmfromthewireBecausemagneticfieldstrengthvariesinverselywiththedistancefromthewire,themagneticfieldat1cmwillbetwiceasstrongasthemagneticfieldat2cm.b.thestrengthofthefieldthatis3cmfromthewireBecausemagneticfieldstrengthvariesinverselywiththedistance■Figure24-7fromthewire,themagneticfieldat1cmwillbethreetimesasstrongassouth,north,south,norththemagneticfieldat3cm.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.Amagnetattractsanail,which,inturn,attractsmanysmalltacks,asshowninFigure24-3onpage645.Ifthenorthpoleofthepermanentmagnetistheleftend,asshown,whichendofthenailisthesouthpole?thebottom(thepoint)Physics:PrinciplesandProblemsSolutionsManual485

489Chapter24continued7.AstudentmakesamagnetbywindingwireStudentanswersmayvary.Answersaroundanailandconnectingittoabattery,couldincludemagnetsonarefrigeratorasshowninFigure24-13.WhichendofandEarth’smagneticfield.Theeffectsthenail,thepointedendorthehead,willoftheseforcescanbedemonstratedbybethenorthpole?bringinganothermagnet,oramaterialthatcanbemagnetized,nearby.12.MagneticFieldsAcurrent-carryingwireispassedthroughacardonwhichironfilingsaresprinkled.Thefilingsshowthemagneticfieldaroundthewire.Asecondwireisclose■Figure24-13toandparalleltothefirstwire.Thereisanthepointedendidenticalcurrentinthesecondwire.Ifthetwocurrentsareinthesamedirection,how8.Youhaveaspoolofwire,aglassrod,anwillthefirstmagneticfieldbeaffected?ironrod,andanaluminumrod.WhichrodHowwillitbeaffectedifthetwocurrentsshouldyouusetomakeanelectromagnetareinoppositedirections?topickupsteelobjects?Explain.ItwouldbeapproximatelytwiceasUsetheironrod.Ironwouldbeattractedlarge;itwouldbeapproximatelyzero.toapermanentmagnetandtakeonprop-ertiesofamagnet,whereasaluminumor13.DirectionofaMagneticFieldDescribeglasswouldnot.Thiseffectwouldsup-theright-handruleusedtodeterminetheportthemagneticfieldinthewirecoilanddirectionofamagneticfieldaroundathusmakethestrongestelectromagnet.straight,current-carryingwire.Ifyougraspthewirewithyourrighthand,9.Theelectromagnetinproblem8workswell,withyourthumbpointinginthedirectionbutyoudecidethatyouwouldliketomakeitsoftheconventionalcurrent,yourfingersstrengthadjustablebyusingapotentiometercurlinthedirectionofthefield.asavariableresistor.Isthispossible?Explain.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Yes.Connectthepotentiometerin14.ElectromagnetsAglasssheetisplacedserieswiththepowersupplyandtheoveranactiveelectromagnet,andironfil-coil.Adjustingthepotentiometerforingssprinkledonthesheetcreateapatternmoreresistancewilldecreasethecur-onit.Ifthisexperimentisrepeatedwiththerentflowandthestrengthofthefield.polarityofthepowersupplyreversed,whatobservabledifferenceswillresult?Explain.None.ThefilingswouldshowthesameSectionReviewfieldpatternbutacompasswouldshow24.1Magnets:Permanentandthemagneticpolarityreversal.Temporary15.CriticalThinkingImagineatoycontain-pages643–651ingtwoparallel,horizontalmetalrods,onepage651abovetheother.Thetoprodisfreetomove10.MagneticFieldsIsamagneticfieldreal,orupanddown.isitjustameansofscientificmodeling?a.Thetoprodfloatsabovethelowerone.IfFieldlinesarenotreal.Thefieldisreal.thetoprod’sdirectionisreversed,however,itfallsdownontothelowerrod.Explain11.MagneticForcesIdentifysomemagneticwhytherodscouldbehaveinthisway.forcesaroundyou.Howcouldyoudemon-Themetalrodscouldbemagnetsstratetheeffectsofthoseforces?withtheiraxesparallel.Ifthetop486SolutionsManualPhysics:PrinciplesandProblems

490Chapter24continuedmagnetispositionedsothatits19.A40.0-cm-longcopperwirecarriesacur-northandsouthpolesareabovetherentof6.0Aandweighs0.35N.Acertainnorthandsouthpolesofthebottommagneticfieldisstrongenoughtobalancemagnet,itwillberepelledandfloattheforceofgravityonthewire.Whatistheabove.Ifthetopmagnetisturnedstrengthofthemagneticfield?endforend,itwillbeattractedtotheFBIL,whereFweightofthewirebottommagnet.F0.35NB0.15Tb.AssumethatthetoprodwaslostandIL(6.0A)(0.400m)replacedwithanotherone.Inthiscase,20.Howmuchcurrentwillberequiredtopro-thetoprodfallsontopofthebottomduceaforceof0.38Nona10.0cmlengthrodnomatterwhatitsorientationis.ofwireatrightanglestoa0.49-Tfield?WhattypeofreplacementrodmustFBILhavebeenused?F0.38NIfanordinaryironbarisusedonI7.8ABL(0.49T)(0.100m)top,itwillbeattractedtothebottommagnetinanyorientation.page65821.Inwhatdirectiondoesthethumbpointwhenusingthethirdright-handruleforanelectronPracticeProblemsmovingatrightanglestoamagneticfield?24.2ForcesCausedbyoppositetothedirectionoftheelectronMagneticFieldsmotionpages652–65922.Anelectronpassesthroughamagneticfieldpage654atrightanglestothefieldatavelocityof16.Whatisthenameoftheruleusedtopredict4.0106m/s.Thestrengthofthemagneticthedirectionofforceonacurrent-carryingfieldis0.50T.Whatisthemagnitudeofthewireatrightanglestoamagneticfield?forceactingontheelectron?Identifywhatmustbeknowntousethisrule.FBqvThirdright-handrule.Thedirectionof(0.50T)(1.601019C)(4.0106m/s)thecurrentandthedirectionofthefield3.21013Nmustbeknown.23.Astreamofdoublyionizedparticles(miss-17.Awirethatis0.50mlongandcarryingaingtwoelectrons,andthus,carryinganetcurrentof8.0Aisatrightanglestoa0.40-Tchargeoftwoelementarycharges)movesatmagneticfield.Howstrongistheforcethatavelocityof3.0104m/sperpendiculartoactsonthewire?amagneticfieldof9.0102T.WhatistheFBIL(0.40N/Am)(8.0A)(0.50m)magnitudeoftheforceactingoneachion?1.6NFBqv18.Awirethatis75cmlong,carryingacurrent(9.0102T)(2)(1.601019C)of6.0A,isatrightanglestoauniform(3.0104m/s)magneticfield.Themagnitudeoftheforce8.61016Nactingonthewireis0.60N.WhatistheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.strengthofthemagneticfield?24.TriplyionizedparticlesinabeamcarryanetpositivechargeofthreeelementaryFBILchargeunits.ThebeamentersamagneticF0.60NB0.13Tfieldof4.0102T.TheparticleshaveaIL(6.0A)(0.75m)speedof9.0106m/s.Whatisthemagni-tudeoftheforceactingoneachparticle?Physics:PrinciplesandProblemsSolutionsManual487

491Chapter24continuedFBqvloop,themagneticfieldoftheperma-(4.0102T)(3)(1.601019C)nentmagnetexertsaforceontheloop.(9.0106m/s)Theloopinagalvanometercannot13Nrotatemorethan180°.Theloopinan1.710electricmotorrotatesthroughmany360°turns.Themotor’ssplit-ringcom-25.Doublyionizedheliumatoms(alphaparti-mutatorallowsthecurrentinthelooptocles)aretravelingatrightanglestoamag-4m/s.Thereverseastheloopbecomesverticalinneticfieldataspeedof4.010fieldstrengthis5.0102T.Whatforcethemagneticfield,enablingthelooptospininthemagneticfield.Thegal-actsoneachparticle?vanometermeasuresunknowncurrents;FBqvtheelectricmotorhasmanyuses.(5.0102T)(2)(1.601019C)(4.0104m/s)29.MotorsWhentheplaneofthecoilina6.41016Nmotorisperpendiculartothemagneticfield,theforcesdonotexertatorqueonthecoil.DoesthismeanthatthecoildoesnotSectionReviewrotate?Explain.Notnecessarily;ifthecoilisalreadyin24.2ForcesCausedbyrotation,thenrotationalinertiawillcarryMagneticFieldsitpastthepointofzerotorque.Itisthepages652–659coil’saccelerationthatiszero,notthepage659velocity.26.MagneticForcesImaginethatacurrent-carryingwireisperpendiculartoEarth’s30.ResistanceAgalvanometerrequires180Amagneticfieldandrunseast-west.Ifthecur-forfull-scaledeflection.Whatisthetotalrentiseast,inwhichdirectionistheforceresistanceofthemeterandthemultiplieronthewire?resistorfora5.0-Vfull-scaledeflection?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.up,awayfromthesurfaceofEarthRV50V28kI180A27.DeflectionAbeamofelectronsina31.CriticalThinkingHowdoyouknowthatcathode-raytubeapproachesthedeflectingtheforcesonparallelcurrent-carryingwiresmagnets.Thenorthpoleisatthetopofthearearesultofmagneticattractionbetweentube;thesouthpoleisonthebottom.Ifyouwires,andnotaresultofelectrostatics?arelookingatthetubefromthedirectionofHint:Considerwhatthechargesarelikewhenthephosphorscreen,inwhichdirectionaretheforceisattractive.Thenconsiderwhatthetheelectronsdeflected?forcesarewhenthreewirescarrycurrentsinthetotheleftsideofthescreensamedirection.Ifthecurrentsareinthesamedirection,28.GalvanometersComparethediagramofatheforceisattractive.IfitwereduetogalvanometerinFigure24-18onpage655electrostaticforces,thelikechargeswiththeelectricmotorinFigure24-20onwouldmaketheforcerepulsive.Threepage656.Howisthegalvanometersimilarwireswouldallattracteachother,whichtoanelectricmotor?Howaretheydifferent?couldneverhappeniftheforceswereBoththegalvanometerandtheelectricduetoelectrostaticcharges.motorusealoopofwirepositionedbetweenthepolesofapermanentmag-net.Whenacurrentpassesthroughthe488SolutionsManualPhysics:PrinciplesandProblems

492Chapter24continuedChapterAssessmentConceptMappingpage664NS32.Completethefollowingconceptmapusingthefollowing:right-handrule,FqvB,andFILB.Forceresultingfromamagneticfield37.Drawthemagneticfieldbetweentwolikemagneticpolesandthenbetweentwounlikemagneticpoles.Showthedirectionsexertedonaofthefields.(24.1)current-carryingmovingchargewireNNhasahasahasamagnitudedirectionmagnitudeNSgivenbygivenbygivenby38.Ifyoubrokeamagnetintwo,wouldyouhaveisolatednorthandsouthpoles?FILBright-handFqvBExplain.(24.1)ruleNo,newpoleswouldformoneachofthebrokenends.MasteringConcepts39.Describehowtousethefirstright-handrulepage664todeterminethedirectionofamagneticfield33.Statetheruleformagneticattractionandaroundastraightcurrent-carryingwire.(24.1)repulsion.(24.1)Graspthewirewiththerighthand,Likepolesrepeloneanother;oppositekeepingthumbpointinginthedirectionpolesattract.oftheconventionalcurrentthroughthewire.Fingerswillencirclethewireand34.Describehowatemporarymagnetdifferspointinthedirectionofthefield.fromapermanentmagnet.(24.1)Atemporarymagnetislikeamagnet40.Ifacurrent-carryingwireisbentintoaonlywhileundertheinfluenceofloop,whyisthemagneticfieldinsidetheanothermagnet.Apermanentmagnetloopstrongerthanthemagneticfieldout-needsnooutsideinfluence.side?(24.1)Themagneticfieldlinesareconcentrat-35.Namethethreemostimportantcommonedinsidetheloop.magneticelements.(24.1)iron,cobalt,andnickelCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.41.Describehowtousethesecondright-handruletodeterminethepolarityofanelectro-36.Drawasmallbarmagnetandshowthemagnet.(24.1)magneticfieldlinesastheyappeararoundthemagnet.Usearrowstoshowthedirec-tionofthefieldlines.(24.1)Physics:PrinciplesandProblemsSolutionsManual489

493Chapter24continuedGraspthecoilwiththerighthand,keep-Useacompass.Thenorthpoleoftheingthefingersencirclingthecoilinthecompassneedleisattractedtothesouthdirectionoftheconventionalcurrentpoleofthemagnetandviceversa.flowthroughtheloops.Thethumboftherighthandwillpointtowardthe48.Apieceofmetalisattractedtoonepoleofanorthpoleoftheelectromagnet.largemagnet.Describehowyoucouldtellwhetherthemetalisatemporarymagnetor42.Eachelectroninapieceofironislikeatinyapermanentmagnet.magnet.Theiron,however,maynotbeaMoveittotheotherpole.Ifthesamemagnet.Explain.(24.1)endisattracted,itisatemporaryTheelectronsarenotallorientedandmagnet;ifthesameendisrepelledandmovinginthesamedirection;theirtheotherendisattracted,itisaperma-magneticfieldshaverandomdirections.nentmagnet.43.Whywilldroppingorheatingamagnet49.IsthemagneticforcethatEarthexertsonaweakenit?(24.1)compassneedlelessthan,equalto,orThedomainsarejostledoutofalignment.greaterthantheforcethatthecompassneedleexertsonEarth?Explain.44.Describehowtousethethirdright-handTheforcesareequalaccordingtoruletodeterminethedirectionofforceonaNewton’sthirdlaw.current-carryingwireplacedinamagneticfield.(24.2)50.CompassSupposeyouarelostinthePointthefingersofyourrighthandinwoodsbuthaveacompasswithyou.thedirectionofthemagneticfield.PointUnfortunately,theredpaintmarkingtheyourthumbinthedirectionofthecon-northpoleofthecompassneedlehaswornventionalcurrentinthewire.Thepalmoff.Youhaveaflashlightwithabatteryandofyourhandthenfacesinthedirectionalengthofwire.Howcouldyouidentifyoftheforceonthewire.thenorthpoleofthecompass?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Connectthewiretothebatterysothatthe45.Astrongcurrentsuddenlyisswitchedoninacurrentisawayfromyouinonesection.wire.Noforceactsonthewire,however.CanHoldthecompassdirectlyaboveandyouconcludethatthereisnomagneticfieldclosetothatsectionofthewire.Bytheatthelocationofthewire?Explain.(24.2)right-handrule,theendofthecompassNo,ifafieldisparalleltothewire,noneedlethatpointsrightisthenorthpole.forcewouldresult.51.Amagnetcanattractapieceofironthatis46.Whatkindofmeteriscreatedwhenashuntnotapermanentmagnet.Achargedrubberisaddedtoagalvanometer?(24.2)rodcanattractanunchargedinsulator.Describethedifferentmicroscopicprocessesanammeterproducingthesesimilarphenomena.ApplyingConceptsThemagnetcausesthedomainsintheirontopointinthesamedirection.Thepages664–665chargedrodseparatesthepositiveand47.Asmallbarmagnetishiddeninafixednegativechargesintheinsulator.positioninsideatennisball.Describeanexperimentthatyoucoulddotofindthe52.Acurrent-carryingwirerunsacrossalabora-locationofthenorthpoleandthesouthtorybench.Describeatleasttwowaysinpoleofthemagnet.whichyoucouldfindthedirectionofthecurrent.490SolutionsManualPhysics:PrinciplesandProblems

494Chapter24continuedUseacompasstofindthedirectionofFacingthefrontoftheroom,thethemagneticfield.Bringupastrongvelocityisforward,theforceisupward,magnetandfindthedirectionoftheforceandtherefore,usingthethirdright-handonthewire,thenusetheright-handrule.rule,Bistotheleft.53.Inwhichdirection,inrelationtoamagnetic58.Earth’smagneticfieldlinesareshowninfield,wouldyourunacurrent-carryingwireFigure24-23.Atwhatlocation,polesorsothattheforceonit,resultingfromtheequator,isthemagneticfieldstrengthfield,isminimized,orevenmadetobezero?greatest?Explain.Runthewireparalleltothemagneticfield.MagneticpoleNorthpole54.Twowirescarryequalcurrentsandrunpar-alleltoeachother.Sa.Ifthetwocurrentsareinoppositedirec-tions,wherewillthemagneticfieldNfromthetwowiresbelargerthanthefieldfromeitherwirealone?Themagneticfieldwillbelargerany-SouthpoleMagneticpolewherebetweenthetwowires.■Figure24-23b.Wherewillthemagneticfieldfrombothbeexactlytwiceaslargeasfromonewire?Earth’smagneticfieldstrengthisgreatestatthepoles.ThefieldlinesareThemagneticfieldwillbetwiceasclosertogetheratthepoles.largealongalinedirectlybetweenthewiresthatisequalindistanceMasteringProblemsfromeachwire.24.1Magnets:PermanentandTemporaryc.Ifthetwocurrentsareinthesamedirec-pages665–666tion,wherewillthemagneticfieldbeexactlyzero?Level159.AsthemagnetbelowinFigure24-24Themagneticfieldwillbezeroalongmovestowardthesuspendedmagnet,whatalinedirectlybetweenthewiresthatwillthemagnetsuspendedbythestringdo?isequalindistancefromeachwire.55.Howistherangeofavoltmeterchangedwhentheresistor’sresistanceisincreased?SNNSTherangeofthevoltmeterincreases.56.Amagneticfieldcanexertaforceonachargedparticle.Canthefieldchangethe■Figure24-24particle’skineticenergy?Explain.Movetotheleftorbegintoturn.LikeNo,theforceisalwaysperpendicularpolesrepel.tothevelocity.Noworkisdone.Theenergyisnotchanged.60.AsthemagnetinFigure24-25movesCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.towardthesuspendedmagnet,whatwillthe57.Abeamofprotonsismovingfromthemagnetthatissuspendedbythestringdo?backtothefrontofaroom.Itisdeflectedupwardbyamagneticfield.Whatisthedirectionofthefieldcausingthedeflection?Physics:PrinciplesandProblemsSolutionsManual491

495Chapter24continued64.Aconventionalcurrentflowsthroughawire,asshowninFigure24-28.CopythewiresegmentandsketchthemagneticfieldSNSNthatthecurrentgenerates.I■Figure24-25Movetotheright.Unlikepolesattract.61.RefertoFigure24-26toanswerthefollow-ingquestions.■Figure24-281I423■Figure24-26a.Wherearethepoles?65.Thecurrentiscomingstraightoutofthe4and2,bydefinitionpageinFigure24-29.Copythefigureandb.Whereisthenorthpole?sketchthemagneticfieldthatthecurrentgenerates.2,bydefinitionandfielddirectionc.Whereisthesouthpole?4,bydefinitionandfielddirection62.Figure24-27showstheresponseofacompassCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.intwodifferentpositionsnearamagnet.■Figure24-29Whereisthesouthpoleofthemagnetlocated?■Figure24-27Attherightend,unlikepolesattract.66.Figure24-30showstheendviewofanelec-tromagnetwithcurrentflowingthroughit.63.Awirethatis1.50mlongandcarryingacurrentof10.0Aisatrightanglestoauni-formmagneticfield.Theforceactingonthewireis0.60N.Whatisthestrengthofthemagneticfield?FILBF0.60NIB0.040N/AmIL(10.0A)(1.50m)0.040T■Figure24-30492SolutionsManualPhysics:PrinciplesandProblems

496Chapter24continueda.Whatisthedirectionofthemagnetic24.2ForcesCausedbyMagneticFieldsfieldinsidetheloops?pages666–667downintothepageLevel1b.Whatisthedirectionofthemagnetic68.ThearrangementshowninFigure24-31isfieldoutsidetheloops?usedtoconvertagalvanometertowhattypeofdevice?up(outofthepage)Level267.CeramicMagnetsTherepulsiveforceGbetweentwoceramicmagnetswasmeasuredandfoundtodependondistance,asgiveninTable24-1.■Figure24-31Table24-1Ammeter,muchofthecurrentflowsthroughtheresistorandallowstheSeparation,d(cm)Force,F(N)measurementofhighercurrents.1.03.931.20.4069.WhatistheresistorshowninFigure24-31called?1.40.13Shunt;bydefinitionshuntisanother1.60.057wordforparallel.1.80.0302.00.01870.ThearrangementshowninFigure24-32isusedtoconvertagalvanometertowhat2.20.011typeofdevice?2.40.00762.60.00532.80.0038G3.00.0028a.Plottheforceasafunctionofdistance.■Figure24-32Voltmeter;theaddedresistance4.0decreasesthecurrentforanygivenvoltage.3.071.WhatistheresistorshowninFigure24-32(N)called?F2.0Multiplier;bydefinitionsinceitmulti-pliesthevoltagerangeofthemeter1.072.Acurrent-carryingwireisplacedbetweenthepolesofamagnet,asshowninFigure24-33.0.01.01.41.82.22.63.0Whatisthedirectionoftheforceonthewire?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.d(cm)b.Doesthisforcefollowaninversesquarelaw?No.Physics:PrinciplesandProblemsSolutionsManual493

497Chapter24continued77.Awirethatis625mlongisperpendicularIStoa0.40-Tmagneticfield.A1.8-Nforceactsonthewire.Whatcurrentisinthewire?NFILBF1.8NIBL(0.40T)(625m)0.0072A■Figure24-337.2mAIS78.Theforceona0.80-mwirethatisperpen-diculartoEarth’smagneticfieldis0.12N.NWhatisthecurrentinthewire?UseF5.0105TforEarth’smagneticfield.FILBF0.12NIBL(5.0105T)(0.80m)73.Awirethatis0.50mlongandcarryingacurrentof8.0Aisatrightanglestoauni-3.0103Aformmagneticfield.Theforceonthewire3.0kAis0.40N.Whatisthestrengthofthemag-neticfield?79.Theforceactingonawirethatisatrightanglestoa0.80-Tmagneticfieldis3.6N.FILBThecurrentinthewireis7.5A.HowlongBF0.40N0.10Tisthewire?IL(8.0A)(0.50m)FILB74.Thecurrentthroughawirethatis0.80mF3.6NL0.60mlongis5.0A.ThewireisperpendiculartoaBI(0.80T)(7.5A)0.60-Tmagneticfield.Whatisthemagni-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tudeoftheforceonthewire?Level2FILB(5.0A)(0.80m)(0.60N/Am)80.Apowerlinecarriesa225-Acurrentfromeasttowest,paralleltothesurfaceofEarth.2.4Na.Whatisthemagnitudeoftheforce75.Awirethatis25cmlongisatrightanglesresultingfromEarth’smagneticfieldtoa0.30-Tuniformmagneticfield.Thecur-actingoneachmeterofthewire?Userentthroughthewireis6.0A.WhatistheB5T.Earth5.010magnitudeoftheforceonthewire?FILBFILB(6.0A)(0.25m)(0.30N/Am)FIB(225A)(5.0105T)0.45NL0.011N/m76.Awirethatis35cmlongisparalleltoa0.53-Tuniformmagneticfield.Thecurrentb.Whatisthedirectionoftheforce?throughthewireis4.5A.WhatforceactsTheforcewouldbedownward.onthewire?c.Inyourjudgment,wouldthisforcebeIfthewireisparalleltothefield,noimportantindesigningtowerstoholdcuttingistakingplace,sonoforceisthispowerline?Explain.produced.No;theforceismuchsmallerthantheweightofthewires.494SolutionsManualPhysics:PrinciplesandProblems

498Chapter24continued81.GalvanometerAgalvanometerdeflectsFBqvfull-scalefora50.0-Acurrent.(6.0102T)(1.61019C)a.Whatmustbethetotalresistanceofthe(2.5106m/s)seriesresistorandthegalvanometerto2.41014Nmakeavoltmeterwith10.0-Vfull-scaledeflection?84.SubatomicParticleAmuon(aparticleVIRwiththesamechargeasanelectron)istravelingat4.21107m/satrightanglesV10.0V5R2.0010I50.0106Atoamagneticfield.Themuonexperiences2kaforceof5.001012N.2.0010a.Howstrongisthemagneticfield?b.Ifthegalvanometerhasaresistanceof1.0k,whatshouldbetheresistanceofFqvBtheseries(multiplier)resistor?FBTotalresistance2.00102k,soqvtheseriesresistoris2.00102k5.001012N1.0k199k.(1.601019C)(4.21107m/s)0.742T82.Thegalvanometerinproblem81isusedtob.Whataccelerationdoesthemuonexpe-makeanammeterthatdeflectsfull-scaleforrienceifitsmassis1.881028kg?10mA.Fmaa.Whatisthepotentialdifferenceacrossthegalvanometer(1.0kresistance)whenaF5.001012Nam1.881028kgcurrentof50Apassesthroughit?VIR(50106A)(1.0103)2.661016m/s20.05V85.Asinglyionizedparticleexperiencesaforceb.Whatistheequivalentresistanceofpar-of4.11013Nwhenittravelsatrightallelresistorshavingthepotentialdiffer-anglesthrougha0.61-Tmagneticfield.encecalculatedinacircuitwithatotalWhatisthevelocityoftheparticle?currentof10mA?FqvBVIRF4.11013NV5102VvBq(0.61T)(1.601019C)R5I0.01A6m/s4.210c.Whatresistorshouldbeplacedparallelwiththegalvanometertomakethe86.Aroomcontainsastrong,uniformmagneticresistancecalculatedinpartb?field.Aloopoffinewireintheroomhas111currentflowingthroughit.AssumethatyousoRR1R2rotatetheloopuntilthereisnotendencyforittorotateasaresultofthefield.Whatisthe11111R1RR251.0103directionofthemagneticfieldrelativetotheplaneofthecoil?soR15ThemagneticfieldisperpendiculartoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.theplaneofthecoil.Theright-handrule83.Abeamofelectronsmovesatrightanglestoamagneticfieldof6.0102T.Theelectronswouldbeusedtofindthedirectionofhaveavelocityof2.5106m/s.Whatisthethefieldproducedbythecoil.Thefieldmagnitudeoftheforceoneachelectron?intheroomisinthesamedirection.Physics:PrinciplesandProblemsSolutionsManual495

499Chapter24continued87.Aforceof5.781016Nactsonan(24V)(0.075m)(1.9T)unknownparticletravelingata90°angle5.5throughamagneticfield.Ifthevelocityof0.62Ntheparticleis5.65104m/sandthefieldisc.Determinetheforceonthewire(direc-3.20102T,howmanyelementarytionandmagnitude)whentheswitchischargesdoestheparticlecarry?closedandthebatteryisreversed.FqvBDown,0.62N.ThedirectionoftheF5.781016Nforceisgivenbythethirdright-handqBv(3.20102T)(5.65104m/s)ruleandthemagnitudeoftheforce3.201019Cisthesameasinpartb.1charged.Determinetheforceonthewire(direc-N(3.201019C)tionandmagnitude)whentheswitchis1.601019Cclosedandthewireisreplacedwithadif-2chargesferentpiecehavingaresistanceof5.5.MixedReviewUp,0.31N.Thedirectionoftheforceisgivenbythethirdright-handrule.pages667–668Level2IVR88.AcopperwireofinsignificantresistanceisplacedinthecenterofanairgapFILBVLBRbetweentwomagneticpoles,asshowninFigure24-34.Thefieldisconfined(24V)(0.075m)(1.9T)5.55.5tothegapandhasastrengthof1.9T.0.31N5.5W7.5cmN89.Twogalvanometersareavailable.Onehas50.0-Afull-scalesensitivityandtheotherShas500.0-Afull-scalesensitivity.BothCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.havethesamecoilresistanceof855.Yourchallengeistoconvertthemtomeasurea24Vcurrentof100.0mA,full-scale.a.Determinetheshuntresistorforthe50.0-Ameter.■Figure24-34Findthevoltageacrossthemetera.Determinetheforceonthewire(direc-coilatfullscale.tionandmagnitude)whentheswitchisVIR(50.0A)(855)0.0428Vopen.Calculatetheshuntresistor.0N.Withnocurrent,thereisnomag-neticfieldproducedbythewireandRV0.0428VI100.0mA50.0Acopperisnotamagneticmaterial.0.428b.Determinetheforceonthewire(direc-tionandmagnitude)whentheswitchisb.Determinetheshuntresistorfortheclosed.500.0-Ameter.Up,0.62N.ThedirectionoftheforceFindthevoltageacrossthemeterisgivenbythethirdright-handrule.coilatfullscale.VVIR(500.0A)(855)0.428VIRVLBFILBR496SolutionsManualPhysics:PrinciplesandProblems

500Chapter24continuedCalculatetheshuntresistor.L(#ofturns)(circumference)ndV0.428VFBILRI100.0mA500.0ABVndF4.30Rc.Determinewhichofthetwoisbetterfor(0.15T)(15V)(250)()(0.025m)actualuse.Explain.8.0The50.0-Ameterisbetter.Its5.5Nmuchlowershuntresistancewilldolesstoalterthetotalresistanceof94.Awirecarrying15Aofcurrenthasalengththecircuitbeingmeasured.Anidealof25cminamagneticfieldof0.85T.Theammeterhasaresistanceof0.forceonacurrent-carryingwireinauni-formmagneticfieldcanbefoundusingthe90.SubatomicParticleAbetaparticle(high-equationFILBsin.Findtheforceonspeedelectron)istravelingatrightanglestothewirewhenitmakesthefollowinganglesa0.60-Tmagneticfield.Ithasaspeedofwiththemagneticfieldlinesof2.5107m/s.Whatsizeforceactsonthea.90°particle?FILBsinFBqv(15A)(0.25m)(0.85T)(sin90°)(0.60T)(1.61019C)(2.5107m/s)3.2N2.41012Nb.45°91.Themassofanelectronis9.111031kg.FILBsinWhatisthemagnitudeoftheaccelerationof(15A)(0.25m)(0.85T)(sin45°)thebetaparticledescribedinproblem90?2.3NFmac.0°F2.41012Nam9.111031kgsin0°02.61018m/s2soF0N92.Amagneticfieldof16Tactsinadirection95.Anelectronisacceleratedfromrestthroughduewest.Anelectronistravelingduesouthapotentialdifferenceof20,000V,whichat8.1105m/s.WhatarethemagnitudeexistsbetweenplatesP1andP2,showninandthedirectionoftheforceactingontheFigure24-35.Theelectronthenpasseselectron?throughasmallopeningintoamagneticfieldofuniformfieldstrength,B.Asindi-FBqvcated,themagneticfieldisdirectedinto(16T)(1.61019C)(8.1105m/s)thepage.2.11012N,upward(right-handP1P2rule—rememberingthatelectronflowisXXXXXXXoppositetocurrentflow)XXXXXXXXXXXXXX93.LoudspeakerThemagneticfieldinaloud-ElectronXXXXXXXspeakeris0.15T.Thewireconsistsof250XXXXXXXturnswoundona2.5-cm-diametercylindri-XXXXXXXCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.calform.Theresistanceofthewireis8.0.■Figure24-35Findtheforceexertedonthewirewhena.Statethedirectionoftheelectricfield15Visplacedacrossthewire.betweentheplatesaseitherP1toP2orVPI2toP1.RfromP2toP1Physics:PrinciplesandProblemsSolutionsManual497

501Chapter24continuedb.Intermsoftheinformationgiven,cal-fieldthatis0.5mfromsuchawireculatetheelectron’sspeedatplatePcomparetoEarth’smagneticfield?2.KEqV(1.61019C)I10A,d0.5m,so(20,000J/C)(2107Tm/A)IB3.21015Jd1(2107Tm/A)(10A)KEmv220.5m2KE(2)(3.21015J)4106Tvm9.111031kgEarth’sfieldis5105T,soEarth’s8107m/sfieldisabout12timesstrongerthanthatofthewire.c.Describethemotionoftheelectronthroughthemagneticfield.b.High-voltagepowertransmissionlinesoftencarry200Aatvoltagesashighasclockwise765kV.Estimatethemagneticfieldonthegroundundersuchaline,assumingThinkingCriticallythatitisabout20mhigh.Howdoespage668thisfieldcomparewithamagneticfield96.ApplyConceptsAcurrentissentthroughinyourhome?averticalspring,asshowninFigure24-36.I200A,d20m,soTheendofthespringisinacupfilledwithmercury.Whatwillhappen?Why?(2107Tm/A)IBd(2107Tm/A)(200A)Spring20m2106TThisishalfasstrongasthefieldinparta.MercuryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.Someconsumergroupshaverecommend-edthatpregnantwomennotuseelectric■Figure24-36blanketsincasethemagneticfieldscausehealthproblems.EstimatethedistanceWhenthecurrentpassesthroughthethatafetusmightbefromsuchawire,coil,themagneticfieldincreasesandclearlystatingyourassumptions.Ifsuchaforcescausethespringtocompress.blanketcarries1A,findthemagneticfieldThewirecomesoutofthemercury,atthelocationofthefetus.Comparethisthecircuitopens,themagneticfieldwithEarth’smagneticfield.decreases,andthespringdropsdown.Thespringwilloscillateupanddown.Assumeonlyonewirerunsover“thefetus,andusethecenterofthe97.ApplyConceptsThemagneticfieldpro-fetus(wherethevitalorgansare)asducedbyalong,current-carryingwireisareferencepoint.Atanearlystage,representedbyB(2107Tm/A)(I/d),thefetusmightbe5cmfromthewhereBisthefieldstrengthinteslas,Iisblanket.Atlaterstages,thecenterthecurrentinamps,anddisthedistanceofthefetusmaybe10cmaway.fromthewireinmeters.UsethisequationI1A,d0.05m,sotoestimatesomemagneticfieldsthatyou(2107Tm/A)Iencounterineverydaylife.Bda.Thewiringinyourhomeseldomcarries(2107Tm/A)(1A)morethan10A.Howdoesthemagnetic0.05m498SolutionsManualPhysics:PrinciplesandProblems

502Chapter24continued4106TStudentanswersmayvary.Super-Earth’smagneticfieldof5105Tisconductingmagnetscurrentlyareusedinmagneticresonanceimaging(MRI),aabout12timesstronger.medicaltechnology.Theyarebeingtest-98.AddVectorsInalmostallcasesdescribededforuseinmagneticallylevitatedhigh-inproblem97,asecondwirecarriesthespeedtrains,anditishopedthatsuper-samecurrentintheoppositedirection.Findconductingmagnetswillhelptomakethenetmagneticfieldthatisadistanceofnuclearfusionenergypractical.Adraw-0.10mfromeachwirecarrying10A.Thebackofsuperconductingmagnetsisthatwiresare0.01mapart.Makeascaledraw-theyrequireextremelylowtemperaturesingofthesituation.Calculatethemagni-(nearabsolutezero).Scientistsaretry-tudeofthefieldfromeachwireanduseaingtodevelopmaterialsthataresuper-right-handruletodrawvectorsshowingtheconductiveathighertemperatures.directionsofthefields.Finally,findthevec-torsumofthetwofields.Stateitsmagni-CumulativeReviewtudeanddirection.page668100.HowmuchworkisrequiredtomoveaB1chargeof6.40103Cthroughapoten-tialdifferenceof2500V?(Chapter21)BWqV(6.40103C)(2500V)16JBIinWireB0.10m101.Thecurrentflowina120-Vcircuitincreas-0.005m0.005mesfrom1.3Ato2.3A.Calculatethe0.10mWireAchangeinpower.(Chapter22)IoutBAPIVP1I1V,P2I2VPP2P1I2VI1VB1V(I2I1)FromeachwireI10A,d0.10m,so(120V)(2.3A1.3A)(2107Tm/A)(10A)5TB210120W0.10mFromthediagram,onlythecomponents102.Determinethetotalresistanceofthree,paralleltothelinefromthecenterof55-resistorsconnectedinparallelandthewirescontributetothenetfieldthenseries-connectedtotwo55-resistorsstrength.Thecomponentfromeachconnectedinseries.(Chapter23)wireisB1Bsin,wheresin0.005m0.05,soB5T)111110.10m1(210RparallelRRR55(0.05)1106T.But,eachwirecon-113tributesthesameamount,sothetotal555555fieldis2106T,about1/25Earth’sfield.Rparallel18WritingInPhysicsRequivRparallelRRCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page66818555599.Researchsuperconductingmagnetsandwrite128aone-pagesummaryofproposedfutureusesforsuchmagnets.Besuretodescribeanyhurdlesthatstandinthewayofthepracticalapplicationofthesemagnets.Physics:PrinciplesandProblemsSolutionsManual499

503Chapter24continuedSimplifyingandreplacing(depth)(width)ChallengeProblemwitharea,A,gives:page656nBIAThefigureshowstwoidenticalmotorswithaThetorqueproducedbythemotorcommonshaft.Forsimplicity,thecommutatorsarmature,inthepositionshown,isarenotshown.Eacharmaturecoilconsistsofequaltothenumberofturnstimesthe48turnsofwirewithrectangulardimensionsoffieldstrengthtimesthearmaturecur-17cmwideby35cmdeep.Thearmaturerenttimestheareaofthearmaturecoil.resistanceis12.Theredwiretravelstotheleft(alonghalfthewidth)andthenbacktotherear2.WithS1closedandS2open,determinetheofthemotor(alongthedepth).Themagnetictorqueontheshaftandtheforceonthefieldis0.21T.Thediameterofthepulleyisspringscale.7.2cm.AropefixedtothepulleyandthefloornBIApreventsthemotorshaftfromturning.120V(48)(0.21T)(0.35m)(0.17m)12S6.0NmNBecausetheshaftcannotturn,thesys-temisinequilibriumandtheforceonShaftSthespringscaleisfoundbyconsider-S2inghalfthepulleydiameter:N6.0Nm35VFspringscale0.036m170NPulleyS13.Withbothswitchesclosed,determinethetorqueontheshaftandtheforceontheSpringspringscale.120VscaleBothmotorsproducecounterclockwiseCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.torque:1.GivenFILB,deriveanequationforthe120Vtorqueonthearmaturefortheposition1(48)(0.21T)12(0.35m)(0.17m)shown.6.0NmTorqueisdefinedastheproductofthe35Vforceandtheleverarm.Inthecaseof2(48)(0.21T)12(0.35m)(0.17m)themotorarmaturepositionshown,the1.7Nmleverarmisequaltohalfofthewidthofthearmaturecoil(theshaftisthecenternet7.7Nmcounterclockwiseofrotationandisatthemidpointofthe7.7NmF210Nspringscale0.036mcoilwidth).Thelengthofthewireacteduponbythefieldisequaltothedepth4.Whathappenstotorqueifthearmatureisofthecoil.Thislengthiseffectivelyinadifferentposition?increasedbyn,thenumberofturnsinThetorqueisreducedwhenthereisthecoil.Finally,thetorqueisdoubledanyrotationfromthepositionshownbecauseasonesideispushedupbybecausetheleverarmisreduced.Withthemagneticfield,theothersideis90°rotation,theforceonthearmaturepusheddownaccordingtothethirdwillbeupanddown(canceling)withright-handrule.theeffectiveleverarmequaltozero.2nBI(depth)(width/2)Withtheshownpositionas0°:nBIAcos500SolutionsManualPhysics:PrinciplesandProblems

504CHAPTER25ElectromagneticInduction4.ApermanenthorseshoemagnetismountedPracticeProblemssothatthemagneticfieldlinesarevertical.If25.1ElectricCurrentfromastudentpassesastraightwirebetweentheChangingMagneticFieldspolesandpullsittowardherself,thecurrentflowthroughthewireisfromrighttoleft.pages671–678Whichisthenorthpoleofthemagnet?page675Usingtheright-handrule,thenorth1.Astraightwire,0.5mlong,ismovedpoleisatthebottom.straightupataspeedof20m/sthrougha0.4-Tmagneticfieldpointedinthehorizontaldirection.page678a.WhatEMFisinducedinthewire?5.AgeneratordevelopsamaximumvoltageEMFBLvof170V.(0.4T)(0.5m)(20m/s)a.Whatistheeffectivevoltage?4VVeff(0.707)Vmax(0.707)(170V)b.Thewireispartofacircuitoftotal1.2102Vresistanceof6.0.Whatisthecurrentb.A60-Wlightbulbisplacedacrossinthecircuit?thegeneratorwithanImaxof0.70A.EMF4VWhatistheeffectivecurrentthroughI0.7AR6.0thebulb?2.Astraightwire,25mlong,ismountedonIeff(0.707)Imax(0.707)(0.70A)anairplaneflyingat125m/s.Thewire0.49AmovesinaperpendiculardirectionthroughEarth’smagneticfield(B5.0105T).c.Whatistheresistanceofthelightbulbwhenitisworking?WhatEMFisinducedinthewire?VEMFBLvmax5T)(25m)(125m/s)Veff2Vmax170V(5.010RIeffImaxImax0.70A0.16V22.41023.Astraightwire,30.0mlong,movesat2.0m/sinaperpendiculardirection6.TheRMSvoltageofanAChouseholdoutletthrougha1.0-Tmagneticfield.is117V.Whatisthemaximumvoltagea.WhatEMFisinducedinthewire?acrossalampconnectedtotheoutlet?IfEMFBLvtheRMScurrentthroughthelampis5.5A,(1.0T)(30.0m)(2.0m/s)whatisthemaximumcurrentinthelamp?6.0101VVeff117VV165Vmax0.7070.707Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.ThetotalresistanceofthecircuitofIwhichthewireisapartis15.0.WhatIeff5.5A7.8Amax0.7070.707isthecurrent?EMF6.0101V7.AnACgeneratordeliversapeakvoltageofI4.0AR1.50425V.Physics:PrinciplesandProblemsSolutionsManual501

505Chapter25continueda.WhatistheVwhenthemagneticfieldismadestronger.effinacircuitplacedacrossthegenerator?WhatelseisaffectedbystrengtheningtheVmax425Vmagneticfield?V2Veff3.0110Themagnitudeoftheinducedvoltageis22directlyrelatedtothestrengthoftheb.Theresistanceis5.0102.Whatisthemagneticfield.Agreatervoltageiseffectivecurrent?inducedintheconductor(s)ifthefieldVeff3.01102Vstrengthisincreased.ThecurrentandIeffR5.01020.60Athepowerinthegeneratorcircuitalso8.Iftheaveragepowerdissipatedbyanelectricwereaffected.lightis75W,whatisthepeakpower?14.GeneratorExplainthefundamental1PPoperatingprincipleofanelectricgenerator.2maxMichaelFaradaydiscoveredthatavolt-P2Wmax(2)P(2)(75W)1.510ageisinducedinalengthofelectricwiremovinginamagneticfield.TheinducedvoltagemaybeincreasedbyusingaSectionReviewstrongermagneticfield,increasingthe25.1ElectricCurrentfromvelocityoftheconductor,orincreasingChangingMagneticFieldstheeffectivelengthoftheconductor.pages671–67815.CriticalThinkingAstudentasks,“Whypage678doesACdissipateanypower?Theenergy9.GeneratorCouldyoumakeageneratorbygoingintothelampwhenthecurrentismountingpermanentmagnetsonarotatingpositiveisremovedwhenthecurrentisshaftandkeepingthecoilstationary?negative.Thenetiszero.”ExplainwhythisExplain.reasoningiswrong.Yes,onlyrelativemotionbetweenthePoweristherateatwhichenergyisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.coilandthemagneticfieldisimportant.transferred.PoweristheproductofIandV.WhenIispositive,soisVand10.BikeGeneratorAbikegeneratorlightsthetherefore,Pispositive.WhenIisnega-headlamp.Whatisthesourceoftheenergytive,soisV;thus,Pispositiveagain.forthebulbwhentheridertravelsalongaEnergyisalwaystransferredinthelamp.flatroad?thestoredchemicalenergyofthebikeriderPracticeProblems11.MicrophoneConsiderthemicrophone25.2ChangingMagneticshowninFigure25-3.WhenthediaphragmFieldsInduceEMFispushedin,whatisthedirectionofthepages679–685currentinthecoil?page684clockwisefromtheleftForthefollowingproblems,effectivecurrentsandvoltagesareindicated.12.FrequencyWhatchangestothegenerator16.Astep-downtransformerhas7500turnsarerequiredtoincreasethefrequency?onitsprimarycoiland125turnsonitsincreasethenumberofmagneticpolesecondarycoil.Thevoltageacrossthepri-pairsmarycircuitis7.2kV.Whatvoltageisbeingappliedacrossthesecondarycircuit?Ifthe13.OutputVoltageExplainwhytheoutputcurrentinthesecondarycircuitis36A,voltageofanelectricgeneratorincreases502SolutionsManualPhysics:PrinciplesandProblems

506Chapter25continuedwhatisthecurrentintheprimarycircuit?19.MotorsIfyouunpluggedarunningvacu-umcleanerfromawalloutlet,youwouldVNssbemuchmorelikelytoseeasparkthanVNppyouwouldbeifyouunpluggedalightedVpNs(7.2103V)(125)lampfromthewall.Why?VsN7500pTheinductanceofthemotorcreatesa2Vback-EMFthatcausesthespark.The1.210bulbhasverylowself-inductance,soVpIpVsIsthereisnoback-EMF.VsIs(1.2102V)(36A)IpV7.2103V0.60A20.TransformersandCurrentExplainwhyaptransformermayonlybeoperatedonalter-17.Astep-uptransformerhas300turnsonnatingcurrent.itsprimarycoiland90,000turnsonitsInordertomagneticallylinkthesecondarycoil.TheEMFofthegeneratorprimaryandsecondarycoils,avaryingtowhichtheprimarycircuitisattachediscurrentmustflowintheprimarycoil.60.0V.WhatistheEMFinthesecondaryThischangingcurrentsetsupamag-circuit?Thecurrentinthesecondarycircuitisneticfieldthatbuilds,expandingout-0.50A.Whatcurrentisintheprimarycircuit?wards,andcollapsesasthecurrentVpNs(60.0V)(90,000)flowdirectionchanges.VsN300p21.TransformersFrequently,transformercoils1.80104VthathaveonlyafewturnsaremadeofveryVI4V)(0.50A)thick(low-resistance)wire,whilethosewithss(1.8010IpV60.0Vmanyturnsaremadeofthinwire.Why?pMorecurrentflowsthroughthecoilwith1.5102Afewerturns,sotheresistancemustbekeptlowtopreventvoltagedropsandI2Rpowerlossandheating.SectionReview22.Step-UpTransformersRefertothestep-up25.2ChangingMagnetictransformershowninFigure25-13.ExplainFieldsInduceEMFwhatwillhappentotheprimarycurrentifpages679–685thesecondarycoilisshort-circuited.page685Accordingtothetransformerequations,18.CoiledWireandMagnetsYouhangacoiltheratioofprimarytosecondarycur-ofwirewithitsendsjoinedsothatitcanrentisequaltotheratioofturnsandswingeasily.Ifyounowplungeamagnetdoesn’tchange.Thus,ifthesecondaryintothecoil,thecoilwillswing.Whichwaycurrentincreases,sodoestheprimary.willitswingrelativetothemagnetandwhy?23.CriticalThinkingWouldpermanentmag-Awayfromthemagnet.Thechangingnetsmakegoodtransformercores?Explain.magneticfieldinducesacurrentintheNo,theinducedvoltagedependsoncoil,producingamagneticfield.ThisachangingmagneticfieldthroughCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fieldopposesthefieldofthemagnet,thecore.Permanentmagnetsareandthus,theforcebetweencoiland“permanent”becausetheyaremadeofmagnetisrepulsive.materialsthatresistsuchchangesinmagneticfields.Physics:PrinciplesandProblemsSolutionsManual503

507Chapter25continued28.WhatisthepolarityofthevoltageinducedChapterAssessmentinthewirewhenitpassesthesouthpoleofConceptMappingthemagneticfield?(25.1)page690Aconductormovingbyasouthmag-24.Completethefollowingconceptmapusingneticpolewillhaveapositiveinducedthefollowingterms:generator,back-EMF,voltage.Lenz’slaw.29.Whatistheeffectofincreasingthenetcon-ductorlengthinanelectricgenerator?(25.1)generatormotorIncreasingtheconductorlengthresultsinanetincreaseininducedvoltage.BoutofpageEMFback-EMFMichaelFaradayLenz’slawWireMasteringConcepts■ Figure25-16page69025.Whatisthearmatureofanelectric30.HowwereOersted’sandFaraday’sresultsgenerator?(25.1)similar?Howweretheydifferent?(25.1)ThearmatureofanelectricgeneratorTheyaresimilarinthattheyeachshowaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.consistsofanumberofwireloopsrelationshipbetweenelectricityandmag-woundaroundanironcoreandplacednetism.Theyaredifferentinthatasteadyinastrongmagneticfield.Asitrotateselectriccurrentproducesamagneticinthemagneticfield,theloopscutfield,butachangeinmagneticfieldisthroughmagneticfieldlinesandanneededtoproduceanelectriccurrent.electriccurrentisinduced.31.Youhaveacoilofwireandabarmagnet.26.Whyisironusedinanarmature?(25.1)Describehowyoucouldusethemtogener-Ironisusedinanarmaturetoincreaseateanelectriccurrent.(25.1)thestrengthofthemagneticfield.Eithermovethemagnetintooroutofthecoil,ormovethecoilupanddownForproblems27–29,refertoFigure25-16.overtheendofthemagnet.27.Asingleconductormovesthroughamag-neticfieldandgeneratesavoltage.Inwhat32.WhatdoesEMFstandfor?Whyisthenamedirectionshouldthewirebemoved,relativeinaccurate?(25.1)tothemagneticfieldtogeneratethemini-Electromotiveforce;itisnotaforcebutmumvoltage?(25.1)anelectricpotential(energyperunitofTheminimumamountofvoltage(0V),ischarge).generatedwhentheconductorismovingparalleltothemagneticlinesofforce.33.Whatisthedifferencebetweenageneratorandamotor?(25.1)504SolutionsManualPhysics:PrinciplesandProblems

508Chapter25continuedInagenerator,mechanicalenergyturnsThisisLenz’slaw.Oncethemotorstartsanarmatureinamagneticfield.Theturning,itbehavesasageneratorandinducedvoltageproducescurrent,thuswillgeneratecurrentinoppositiontotheproducingelectricenergy.Inamotor,currentbeingputintothemotor.voltageisplacedacrossanarmaturecoilinamagneticfield.Thevoltageproduces39.Whyistherenosparkwhenyoucloseacurrentinthecoilandthearmatureswitchandputcurrentthroughaninductor,turns,producingmechanicalenergy.butthereisasparkwhenyouopentheswitch?(25.2)34.ListthemajorpartsofanACgenerator.(25.1)Thesparkisfromtheback-EMFthatAnACgeneratorconsistsofaperma-triestokeepthecurrentflowing.Thenentmagnet,anarmature,asetofback-EMFislargebecausethecurrentbrushes,andaslipring.hasdroppedquicklytozero.Whenclos-ingtheswitch,thecurrentincreaseisn’t35.WhyistheeffectivevalueofanACcurrentsofastbecauseoftheresistanceinthelessthanitsmaximumvalue?(25.1)wires.Inanalternating-currentgenerator,asthearmatureturns,thegenerated40.Whyistheself-inductanceofacoilamajorpowervariesbetweensomemaximumfactorwhenthecoilisinanACcircuitbutavalueandzero.TheaveragepowerisminorfactorwhenthecoilisinaDCcircuit?equaltoone-halfthemaximumpower.(25.2)TheeffectivecurrentistheconstantAnalternatingcurrentisalwayschang-valueofcurrentthatwouldcausetheinginthemagnitudeanddirection.averagepowertobedissipatedintheTherefore,self-inductionisaconstantload,R.factor.AdirectcurrenteventuallyP1P1I2RI2Rbecomessteady,andthus,afterashortavg2max2maxefftime,thereisnochangingmagneticfield.IImaxI41.Explainwhythewordchangeappearssoeffmax2ofteninthischapter.(25.2)36.HydroelectricityWatertrappedbehindaAsFaradaydiscovered,onlyachangingdamturnsturbinesthatrotategenerators.magneticfieldinducesEMF.Listalltheformsofenergythattakepartinthecyclethatincludesthestoredwaterand42.UponwhatdoestheratiooftheEMFinthetheelectricityproduced.(25.1)primarycircuitofatransformertotheEMFThereispotentialenergyinthestoredinthesecondarycircuitofthetransformerwater,kineticenergyinthefallingwaterdepend?(25.2)andturningturbine,andelectricenergyTheratioofnumberofturnsofwireininthegenerator.Inaddition,therearetheprimarycoiltothenumberofturnsfrictionallossesintheturbineandgen-ofwireinthesecondarycoildetermineseratorresultinginthermalenergy.theEMFratio.37.StateLenz’slaw.(25.2)ApplyingConceptsAninducedcurrentalwaysactsinsuchaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.pages690–692directionthatitsmagneticfieldopposes43.UseunitsubstitutiontoshowthattheunitsthechangebywhichthecurrentisofBLvarevolts.induced.TheunitofBLvis(T)(m)(m/s).38.Whatcausesback-EMFinanelectricmotor?TN/AmandAC/s.(25.2)So,BLv(Ns/Cm)(m)(m/s)Nm/CPhysics:PrinciplesandProblemsSolutionsManual505

509Chapter25continuedBecauseJNmandVJ/C,theunitofBLvisV(volts).44.Whenawireismovedthroughamagneticfield,doestheresistanceoftheclosedcircuitaffectcurrentonly,EMFonly,both,orneither?currentonly■ Figure25-1845.BikingAsLoganslowshisbike,whatItislargerseveraltenthsofasecondhappenstotheEMFproducedbyhisbike’saftertheconnectionismade.Theback-generator?UsethetermarmatureinyourEMFopposescurrentjustafterthecon-explanation.nectionismade.AsLoganslowshisbike,therotationofthearmatureinthemagneticfieldofthe49.Asegmentofawireloopismovinggeneratorslows,andtheEMFisreduced.downwardthroughthepolesofamagnet,asshowninFigure25-19.Whatisthe46.ThedirectionofACvoltagechangesdirectionoftheinducedcurrent?120timeseachsecond.DoesthismeanthatadeviceconnectedtoanACvoltagealter-Nnatelydeliversandacceptsenergy?SNo;thesignsofthecurrentandvoltagereverseatthesametime,and,therefore,vtheproductofthecurrentandthevolt-■ Figure25-19ageisalwayspositive.Thecurrentdirectionisout-of-pageto47.Awireismovedhorizontallybetweenthetheleftalongthepathofthewire.polesofamagnet,asshowninFigure25-17.Whatisthedirectionoftheinducedcurrent?50.AtransformerisconnectedtoabatteryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.throughaswitch.Thesecondarycircuitcon-tainsalightbulb,asshowninFigure25-20.SNWillthelampbelightedaslongastheswitchisclosed,onlyatthemomenttheswitchisclosed,oronlyatthemomenttheswitchisopened?Explain.■ Figure25-17Nocurrentisinducedbecausethedirectionofthevelocityisparalleltothemagneticfield.PrimarySecondary48.Youmakeanelectromagnetbywindingwirearoundalargenail,asshownin■ Figure25-20Figure25-18.IfyouconnectthemagnettoThebulbwilllightbecausethereisaabattery,isthecurrentlargerjustafteryoucurrentinthesecondarycircuit.Thismaketheconnectionorseveraltenthsofawillhappenwhenevertheprimarycur-secondaftertheconnectionismade?Or,isrentchanges,sothebulbwillglowitalwaysthesame?Explain.eitherwhentheswitchisclosedorwhenitisopened.506SolutionsManualPhysics:PrinciplesandProblems

510Chapter25continued51.Earth’sMagneticFieldThedirectionofa.WhatisthedirectionofthecurrentEarth’smagneticfieldinthenorthernhemi-inducedinthepipebythefallingmagnetsphereisdownwardandtothenorthasifthesouthpoleistowardthebottom?showninFigure25-21.Ifaneast-westwireInducedEMFisperpendiculartobothmovesfromnorthtosouth,inwhichdirec-thefieldandvelocity,sothecurrenttionisthecurrent?mustbecircumferential.FieldlinesNorthmagneticpoleNorthpolemoveintowardthesouthpoleandoutfromthenorthpole.Bytheright-handrule,currentisclockwisenearSthesouthpoleandcounterclockwisenearthenorthpole.b.TheinducedcurrentproducesamagneticNfield.Whatisthedirectionofthefield?Nearthesouthpole,thefieldinsideSouthpoleSouthmagneticpolethepipeisdown;nearthenorthpole,■ Figure25-21itisup.Thecurrentisfromwesttoeast.c.Howdoesthisfieldreducetheaccelerationofthefallingmagnet?52.YoumovealengthofcopperwiredownInducedfieldexertsanupwardforcethroughamagneticfield,B,asshowninonbothpoles.Figure25-19.a.Willtheinducedcurrentmovetothe54.GeneratorsWhyisageneratormorediffi-rightorleftinthewiresegmentintheculttorotatewhenitisconnectedtoacir-diagram?cuitandsupplyingcurrentthanitiswhenitTheright-handrulewillshowtheisstandingalone?currentmovingleft.Whenthearmatureofageneratorisb.Assoonasthewireismovedintherotated,aforcethatopposesthedirec-field,acurrentappearsinit.Thus,thetionofrotationisproducedasaresultwiresegmentisacurrent-carryingwireofinducedcurrent(Lenz’slaw).Whenlocatedinamagneticfield.Aforcemuststandingalone,however,nocurrentisactonthewire.Whatwillbethedirec-generatedandconsequentlynooppos-tionoftheforceactingonthewireasaingforceisproduced.resultoftheinducedcurrent?55.Explainwhytheinitialstart-upcurrentissoTheforcewillactinanupwardhighinanelectricmotor.Alsoexplainhowdirection.Lenz’slawappliesattheinstantt0.53.AphysicsinstructordropsamagnetIfthearmature(conductors)arenotthroughacopperpipe,asillustratedinrotating,nolinesofforcearebeingcut,Figure25-22.Themagnetfallsveryslowly,andnovoltageisinduced.Therefore,andthestudentsintheclassconcludethattheback-EMFiszero.Sincethereisnotheremustbesomeforceopposinggravity.currentinthearmature,nomagneticfieldisformedaroundthestationaryconductor.ItshouldbenotedthatthisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Nexplanationonlyholdstrueattheinstantofstartup,attimejustgreaterSthan0.Theinstantthearmaturebeginstorotate,itwillbecuttingthelinesofforceandwillhaveaninducedvoltage.■ Figure25-22Thisvoltage,theback-EMF,willhaveaPhysics:PrinciplesandProblemsSolutionsManual507

511Chapter25continuedpolaritysuchthatitproducesamagnetic59.Shawndropsamagnet,northpoledown,fieldopposingthefieldthatcreatedit.throughaverticalcopperpipe.Thisreducesthecurrentinthemotor.a.WhatisthedirectionoftheinducedTherefore,themotionofthemotorcurrentinthecopperpipeasthebot-increasesitsapparentresistance.tomofthemagnetpasses?clockwisearoundthepipe,as56.UsingFigure25-10inconjunctionwithviewedfromaboveLenz’slaw,explainwhyallpracticaltrans-formercoresincorporatealaminatedcore.b.Theinducedcurrentproducesamagnet-icfield.WhatisthedirectionoftheAlaminatedcoreisconstructedfrominducedmagneticfield?thinsheetsofsteel,separatedbyaverythincoatingofvarnish(insulation).downthepipe,atthelocationoftheEddycurrentsaregreatlyreducedsouthpoleofthemagnet(oroppo-becauseofthisinsulation.Currentinsitethemagnet’sfield)thecoreiscausedbythechangingmagneticfluxwithinthecore.TheeddyMasteringProblemscurrentsexistduetotheinducedvolt-25.1ElectricCurrentfromChangingagewithinthemagneticcore.MagneticFieldspages692–69357.ApracticaltransformerisconstructedwithaLevel1laminatedcorethatisnotasuperconductor.60.Awire,20.0-mlong,movesat4.0m/sper-Becausetheeddycurrentscannotbecom-pendicularlythroughamagneticfield.Anpletelyeliminated,thereisalwaysasmallEMFof40Visinducedinthewire.Whatiscoreloss.Thisresults,inpart,inanetlossthestrengthofthemagneticfield?ofpowerwithinthetransformer.Whatfun-EMFBLvdamentallawmakesitimpossibletobringthislosstozero?EMF40VBLv(20.0m)(4.0m/s)Lenz’slawCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.5T58.Explaintheprocessofmutualinductionwithinatransformer.61.AirplanesAnairplanetravelingat9.50102km/hpassesoveraregionwhereAnACcurrentappliedtotheprimarycoilEarth’smagneticfieldis4.5105Tandofatransformer,resultsinachangingisnearlyvertical.Whatvoltageisinducedcurrentflowthroughthecoilwinding.betweentheplane’swingtips,whichareThiscurrent,inturn,generatesamagnet-75mapart?icflux,alternatelybuildingandcollapsingasthedirectionofcurrentchanges.ThisEMFBLvstrongmagneticfieldradiatesoutwardin(4.5105T)(75m)alldirectionsfromtheprimarycoil.When(9.50102km/h)(1000m/km)themagneticfieldreachesthestationary(1h/3600s)secondarycoilontheothersideofthecore,avoltageisinducedwithinthatcoil.0.89VThevoltageorEMFinduceddepends62.Astraightwire,0.75-mlong,movesupwardupontherateatwhichthemagneticfieldthroughahorizontal0.30-Tmagneticfield,alternates(supplyfrequency),numberofasshowninFigure25-23,ataspeedofturnsonthecoil,andthestrengthofthe16m/s.magneticflux.Sincethemagneticfluxisresponsibleforinducingavoltageina.WhatEMFisinducedinthewire?thesecondarycoil,wesaythatitisEMFBLvmagneticallylinked.508SolutionsManualPhysics:PrinciplesandProblems

512Chapter25continued(0.30T)(0.75m)(16m/s)66.ElectricStoveAnelectricstoveisconnected3.6VtoanACsourcewithaneffectivevoltageof240V.b.Thewireispartofacircuitwithatotalresistanceof11W.Whatisthecurrent?a.Findthemaximumvoltageacrossoneofthestove’selementswhenitisoperating.EMFIRVEMF3.6Veff(0.707)VmaxI0.33AR11Veff240VV340Vmax0.7070.707Nb.TheresistanceoftheoperatingelementSis11.Whatistheeffectivecurrent?VeffIeffRvV■Figure25-23eff240VIeffR1122A63.Atwhatspeedwoulda0.20-mlengthofwirehavetomoveacrossa2.5-Tmagnetic67.YouwishtogenerateanEMFof4.5VbyfieldtoinduceanEMFof10V?movingawireat4.0m/sthrougha0.050-TEMFBLvmagneticfield.Howlongmustthewirebe,andwhatshouldbetheanglebetweentheEMF10VvfieldanddirectionofmotiontousetheBL(2.5T)(0.20m)shortestwire?20m/sEMFBLv64.AnACgeneratordevelopsamaximumEMFEMF4.5VLof565V.WhateffectiveEMFdoestheBv(0.050T)(4.0m/s)generatordelivertoanexternalcircuit?23mVmax565V4.00102VThisistheshortestlengthofwireassum-Veffingthatthewireandthedirectionof22motionareeachperpendiculartothefield.65.AnACgeneratordevelopsamaximumvoltageof150V.ItdeliversamaximumLevel2currentof30.0Atoanexternalcircuit.68.A40.0-cmwireismovedperpendicularlythroughamagneticfieldof0.32Twithaa.Whatistheeffectivevoltageofthevelocityof1.3m/s.Ifthiswireisconnectedgenerator?intoacircuitof10.0-resistance,whatisVeff(0.707)Vmax(0.707)(150V)thecurrent?110VEMFBLvb.Whateffectivecurrentdoesthegenerator(0.32T)(0.400m)(1.3m/s)delivertotheexternalcircuit?0.17VIeff(0.707)Imax(0.707)(30.0V)EMF0.17VI17mAR10.021.2Ac.Whatistheeffectivepowerdissipatedin69.YouconnectbothendsofacopperwirewithCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.thecircuit?atotalresistanceof0.10totheterminalsofIVagalvanometer.Thegalvanometerhasaresis-PmaxmaxeffIeffVefftanceof875.Youthenmovea10.0-cm22segmentofthewireupwardat1.0m/s1I1througha2.0102-Tmagneticfield.What2maxVmax2(150V)(30.0A)currentwillthegalvanometerindicate?2.3kWPhysics:PrinciplesandProblemsSolutionsManual509

513Chapter25continuedEMFBLvPE4.4108Jm2T)(0.100m)(1.0m/s)gh(9.80m/s2)(22m)(2.0103V2.0106kg2.010V2.0103V6A72.AconductorrotatinginamagneticfieldhasI2.310R875alengthof20cm.Ifthemagnetic-fluxden-2.3Asityis4.0T,determinetheinducedvoltagewhentheconductorismovingperpendicu-70.Thedirectionofa0.045-Tmagneticfieldislartothelineofforce.Assumethatthe60.0ºabovethehorizontal.Awire,2.5-mconductortravelsataconstantvelocityoflong,moveshorizontallyat2.4m/s.1m/s.a.WhatistheverticalcomponentoftheWhentheconductorismovingperpen-magneticfield?diculartothelineofforceTheverticalcomponentofmagneticEindBLvfieldis(4.0T)(0.20m)(1m/s)Bsin60.0°(0.045T)(sin60.0°)0.8V0.039Tb.WhatEMFisinducedinthewire?73.RefertoExampleProblem1andFigure25-24todeterminethefollowing.EMFBLv(0.039T)(2.5m)(2.4m/s)R1.0A0.23Vv3.6m/s71.DamsAgeneratoratadamcansupply375MW(375106W)ofelectricalpower.EMFAssumethattheturbineandgeneratorareL0.50m85percentefficient.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.FindtherateatwhichfallingwaterBoutofpagemustsupplyenergytotheturbine.AB7.0102TPouteff100■ Figure25-24Pina.inducedvoltageintheconductor100PinPouteffEMFindBLv(7.0102T)(0.50m)100375MW(3.6m/s)85440MWinput0.13Vb.Theenergyofthewatercomesfromab.current(I)changeinpotentialenergy,PEmgh.EMFind0.13VWhatisthechangeinPEneededeachI0.13AR1.0second?c.directionoffluxrotationaroundthe440MW440MJ/sconductor4.4108JeachsecondFluxrotatesclockwisearoundthec.Ifthewaterfalls22m,whatisthemassconductorwhenviewedfromabove.ofthewaterthatmustpassthroughthed.polarityofpointArelativetopointBturbineeachsecondtosupplythisPointAisnegativerelativetopointB.power?PEmgh510SolutionsManualPhysics:PrinciplesandProblems

514Chapter25continued25.2ChangingMagneticFieldsInduceEMFVpIp(120V)(30.0A)3.6kWpage693VsIs(1800V)(2.0A)3.6kWLevel174.Theprimarycoilofatransformerhas76.LaptopComputersThepowersupplyina150turns.Itisconnectedtoa120-Vsource.laptopcomputerrequiresaneffectivevolt-Calculatethenumberofturnsontheageof9.0Vfroma120-Vline.secondarycoilneededtosupplythefollow-a.Iftheprimarycoilhas475turns,howingvoltages.manydoesthesecondarycoilhave?a.625VVNssVsNsVpNpVNppV(9.0V)(475)sNpNVsV120Vs625VpNsVNp120V(150)p36turns781turns,whichroundsto780b.A125-mAcurrentisinthecomputer.Whatcurrentisintheprimarycircuit?b.35VVVpIpVsIss35VNsVNp120V(150)pVsIs(9.0V)(125mA)IpV7200V44turnspc.6.0V9.4mAVs6.0VNsVNp120V(150)Level2p77.HairDryersAhairdryermanufacturedfor7.5turnsuseintheUnitedStatesuses10Aat120V.ItisusedwithatransformerinEngland,75.Astep-uptransformerhas80turnsonitswherethelinevoltageis240V.primarycoiland1200turnsonitssec-a.Whatshouldbetheratiooftheturnsofondarycoil.Theprimarycircuitissuppliedthetransformer?withanalternatingcurrentat120V.VNpp240V2.0a.WhatvoltageisbeingappliedacrosstheVN120V1.0sssecondarycircuit?or2to1VNppb.WhatcurrentwillthehairdryernowVNssdraw?VpNs(120V)(1200)VV1.8kVpIpVsIssN80pVsIs(120V)(10A)b.ThecurrentinthesecondarycircuitisIpV240V5Ap2.0A.Whatcurrentisintheprimarycircuit?78.A150-WtransformerhasaninputvoltageVpIpVsIsof9.0Vandanoutputcurrentof5.0A.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.VsIs(1.8103V)(2.0A)a.Isthisastep-uporstep-downIpV120Vtransformer?pPoutVsIs3.0101APout150W1Vc.WhatarethepowerinputandoutputofV3.010sI5.0Athetransformer?sstep-uptransformerPhysics:PrinciplesandProblemsSolutionsManual511

515Chapter25continuedb.WhatistheratioofVoutputtoVinput?V1A)pIp(120V)(9.010PVsIs1.1104WP150W1VVsIs(3600V)(3.0A)1.1104WV3.010sI5.0As81.Withwhatspeedmusta0.20-m-longwireVoutput3.0101V1.0101cutacrossamagneticfieldforwhichBisV9.0V3.0input2.5TifitistohaveanEMFof10Vinducedor10to3init?EMFBLv79.Scottconnectsatransformertoa24-VEMF10Vsourceandmeasures8.0VatthesecondaryvBL(2.5T)(0.20m)circuit.Iftheprimaryandsecondarycircuitswerereversed,whatwouldthenewoutput20m/svoltagebe?82.Atwhatspeedmustawireconductor50-cmTheturnsratioisNVlongbemovedatrightanglestoamagneticss8.0V1.0fieldofinduction0.20TtoinduceanEMFNV24V3.0ppof1.0Vinit?3.0Reversed,itwouldbe.1.0EMFBLvThus,thevoltagewouldnowbefoundbyEMF1.0VNVvssBL(0.20T)(0.5m)NVppN1101m/ssVVsNp(3.0)(24V)72Vp83.Ahouselightingcircuitisratedat120-Veffectivevoltage.WhatisthepeakvoltageMixedReviewthatcanbeexpectedinthiscircuit?pages693–694Veff(0.707)VmaxLevel1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.V80.Astep-uptransformer’sprimarycoilhasVeff120V170Vmax0.7070.707500turns.Itssecondarycoilhas15,000turns.Theprimarycircuitisconnectedto84.ToasterAtoasterdraws2.5AofalternatinganACgeneratorhavinganEMFof120V.current.Whatisthepeakcurrentthrougha.CalculatetheEMFofthesecondarycircuit.thistoaster?VNppIeff(0.707)ImaxVNssIeff2.5AVpNs(120V)(15,000)Imax0.7070.7073.5AVsN500p85.Theinsulationofacapacitorwillbreakdown3.6103Viftheinstantaneousvoltageexceeds575V.b.FindthecurrentintheprimarycircuitifWhatisthelargesteffectivealternatingvolt-thecurrentinthesecondarycircuitis3.0A.agethatmaybeappliedtothecapacitor?VpIpVsIsVmax575V407VeffVsIs(3600V)(3.0A)1A22I9.010pV120Vp86.CircuitBreakerAmagneticcircuitbreakerc.Whatpowerisdrawnbytheprimarywillopenitscircuitiftheinstantaneouscircuit?Whatpowerissuppliedbythecurrentreaches21.25A.Whatisthelargestsecondarycircuit?effectivecurrentthecircuitwillcarry?512SolutionsManualPhysics:PrinciplesandProblems

516Chapter25continuedImax21.25ALevel2I15.03Aeff2291.Awire,0.40-mlong,cutsperpendicularlyacrossamagneticfieldforwhichBis2.0T87.Theelectricityreceivedatanelectricalatavelocityof8.0m/s.substationhasapotentialdifferenceof240,000V.Whatshouldtheratioofthea.WhatEMFisinducedinthewire?turnsofthestep-downtransformerbetoEMFBLvhaveanoutputof440V?(2.0T)(0.40m)(8.0m/s)NsVs440V16.4VNV240,000V545ppb.Ifthewireisinacircuitwitharesis-tanceof6.4W,whatisthesizeoftheprimary:secondary545:1currentinthewire?88.Analternating-currentelectricgeneratorEMFIRsuppliesa45-kWindustrialelectricheater.EMF6.4VI1.0AIfthesystemvoltageis660VR6.4rms,whatisthepeakcurrentsupplied?92.Acoilofwire,whichhasatotallengthof45kWI68A7.50m,ismovedperpendicularlytoEarth’srms660Vmagneticfieldat5.50m/s.Whatisthesize68ATherefore,Iofthecurrentinthewireifthetotalresis-peak0.70796Atanceofthewireis5.0102m?Assume89.Acertainstep-downtransformerhas100turnsEarth’smagneticfieldis5105T.ontheprimarycoiland10turnsonthesec-EMFBLvandVIR,butEMFV,ondarycoil.Ifa2.0-kWresistiveloadiscon-soIRBLv,andnectedtothetransformer,whatistheeffectiveBLv(5.0105T)(7.50m)(5.50m/s)primarycurrentthatflows?AssumethattheIR5.0102msecondaryvoltageis60.0Vpk.4.1102A41mAVs,peak60.0VV42.4Vs,eff2293.ThepeakvalueofthealternatingvoltageP2.0103Wappliedtoa144-resistoris1.00102V.I47As,effV42.4VWhatpowermusttheresistorbeabletos,effhandle?Ns10VIp,effNIs,eff10047A4.7APIVandVIR,soItherefore,pR90.Atransformerratedat100kVAhasanVV2(1.00102V)2PmaxRVR144efficiencyof98percent.a.Iftheconnectedloadconsumes98kW69.4Wofpower,whatistheinputpowertotheTheaveragepowerisPmax/2sothetransformer?resistormustdissipate34.7W.Pout98kW94.TelevisionTheCRTinatelevisionusesa98kW2kWPin0.981.010step-uptransformertochange120Vto48,000V.Thesecondarysideofthetrans-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Whatisthemaximumprimarycurrentformerhas20,000turnsandanoutputofwiththetransformerconsumingits1.0mA.ratedreactivepower?AssumethatVa.HowmanyturnsdoestheprimarysideP600V.have?100kVAI200A600VPhysics:PrinciplesandProblemsSolutionsManual513

517Chapter25continuedVsNsPrimarycurrent:VNppPp757WI6.05AN(20,000)(120V)pVp125VsVpNpV48000Vs97.AnalyzeandConcludeAtransformerthat50turnssupplieseighthomeshasanefficiencyof95percent.Alleighthomeshaveoperatingb.Whatistheinputcurrent?electricovensthateachdraw35AfromVpIpVsIs240-Vlines.HowmuchpowerissuppliedVsIs(48,000V)(1.0103A)totheovensintheeighthomes?HowIpV120Vmuchpowerisdissipatedasheatintheptransformer?0.40ASecondarypower:PThinkingCriticallys(#ofhomes)VsIs(8)(240V)(35A)67kWpage69495.ApplyConceptsSupposethatan“anti-67kWissuppliedtotheovensintheLenz’slaw”existedthatmeantaforcewaseighthomes.exertedtoincreasethechangeinamagneticPrimarypower:field.Thus,whenmoreenergywasdemand-(100)P(100)(67W)Psed,theforceneededtoturnthegeneratorpe9571kWwouldbereduced.WhatconservationlawThedifferencebetweenthesetwoisthewouldbeviolatedbythisnew“law”?Explain.powerdissipatedasheat,4kW.Itwouldviolatethelawofconversationofenergy.MoreenergywouldcomeoutWritinginPhysicsthanwentin.Ageneratorwouldcreatepage694energy,notjustchangeitfromoneform98.Commontools,suchasanelectricdrill,aretoanother.typicallyconstructedusingauniversalCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.motor.Usingyourlocallibrary,andother96.AnalyzeRealtransformersarenotsources,explainhowthistypeofmotor100percentefficient.WriteanexpressionformayoperateoneitherACorDCcurrent.transformerefficiencyinpercentusingpower.Astep-downtransformerthathasanAseriesDCMotorusesbothanarma-efficiencyof92.5percentisusedtoobtaintureandseriescoil.Whenoperated28.0Vfroma125-Vhouseholdvoltage.Theonalternatingcurrent,thepolarityoncurrentinthesecondarycircuitis25.0A.bothfieldschangessimultaneously.Whatisthecurrentintheprimarycircuit?Therefore,thepolarityofthemagneticfieldremainsunchanged,andhencetheEfficiencydirectionofrotationisconstant.Pse100PpCumulativeReviewSecondarypower:page694P99.LightisemittedbyadistantstaratasVsIs(28.0V)(25.0A)frequencyof4.561014Hz.Ifthestaris7.00102WmovingtowardEarthataspeedof2750Primarypower:km/s,whatfrequencylightwillbedetected2W)byobserversonEarth?(Chapter16)(100)P(100)(7.0010Pspe92.5757Wfvobsf1c514SolutionsManualPhysics:PrinciplesandProblems

518Chapter25continuedBecausetheyaremovingtowardeachPV2/Rother5.0WvVPR(22)7.4Vfobsf12c14Hz)2.75106m/s103.Determinethetotalresistanceofthree,(4.561013.00108m/s85-resistorsconnectedinparalleland4.60104Hzthenseries-connectedtotwo85-resistorsconnectedinparallel,asshownin100.AdistantgalaxyemitslightatafrequencyFigure25-25.(Chapter23)of7.291014Hz.ObserversonEarthreceivethelightatafrequencyof6.141014Hz.Howfastisthegalaxymov-858585ing,andinwhatdirection?(Chapter16)ThegalaxyismovingawayfromEarthbecausetheobservedfrequencyislowerthantheemittedfrequency.To85findthespeed:vfobsf1c85■ Figure25-25Becausetheobservedlighthasalowerfrequency,thegalaxymustbe1111movingawayfromEarth.So,usetheR3inparallel858585negativeformoftheequationaboveR3inparallel28.3vfobsf1c111R8585f2inparallelobsv1fcR2inparallel42.5vfobsRRc1f3inparallelR2inparallel28.342.5fvc1obsf716.141014Hz6m/s(3.00108m/s)114Hz104.Anelectronwithavelocityof2.1107.2910isatrightanglestoa0.81-Tmagneticfield.4.73107m/sWhatistheforceontheelectronproducedbythemagneticfield?Whatistheelec-101.Howmuchchargeisona22-Fcapacitortron’sacceleration?Themassofanelec-with48Vappliedtoit?(Chapter21)tronis9.111031kg.(Chapter24)qFBqvCV(0.81T)(1.601019C)(2.1106m/s)qCV2.71013N(22106F)(48V)FmaCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.1103CF2.71013Nam9.111031kg102.Findthevoltageacrossa22-,5.0-W3.01017m/s2resistoroperatingathalfofitsrating.(Chapter22)Physics:PrinciplesandProblemsSolutionsManual515

519Chapter25continued3.Whatisthesecondarycurrentoftrans-ChallengeProblemformerT1?page68513V)Adistributiontransformer(TVs15(3.0101)hasitsprimarycoilconnectedtoa3.0-kVACsource.Thesecondarycoil6.0102Visconnectedtotheprimarycoilofasecondtrans-P210.3103Wformer(T2)bycopperconductors.Finally,thesec-Is1V6.0102V17As1ondarycoiloftransformerT2connectstoaloadthatuses10.0kWofpower.TransformerT1hasa4.HowmuchcurrentistheACsourcesupply-turnratioof5:1,andT2hasaloadvoltageof120V.ingtoT1?Thetransformerefficienciesare100.0percentand1197.0percent,respectively.Ip15Is15(17A)3.4A1.Calculatetheloadcurrent.PL10.0kWILVL120V83A2.HowmuchpowerisbeingdissipatedbytransformerT2?PL10.0kWP20.9700.97010.3kWP2ispowerinputtotransformerT2.Ofthe10.3kW,0.3kWisdissipatedbyT2;theother10.0kWisdissipatedintheload.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.516SolutionsManualPhysics:PrinciplesandProblems

520CHAPTER26ElectromagnetismPracticeProblems26.1InteractionsofElectricandMagneticFieldsandMatterpages697–704page700Assumethatallchargedparticlesmoveperpendiculartoauniformmagneticfield.1.Aprotonmovesataspeedof7.5103m/sasitpassesthroughamagneticfieldof0.60T.Findtheradiusofthecircularpath.Notethatthechargecarriedbytheprotonisequaltothatoftheelectron,butispositive.mv2BqvrmvrBq(1.671027kg)(7.5103m/s)(0.60T)(1.601019C)1.3104m2.Electronsmovethroughamagneticfieldof6.0102Tbalancedbyanelectricfieldof3.0103N/C.Whatisthespeedoftheelectrons?BqvEqE3.0103N/CvB6.0102T5.0104m/s3.Calculatetheradiusofthecircularpaththattheelectronsinproblem2followintheabsenceoftheelectricfield.mv2BqvrmvrBq(9.111031kg)(5.0104m/s)4.7106m(6.0102T)(1.601019C)4.Protonspassingwithoutdeflectionthroughamagneticfieldof0.60Tarebalancedbyanelectricfieldof4.5103N/C.Whatisthespeedofthemovingprotons?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BqvEqE4.5103N/CvB0.60T7.5103m/sPhysics:PrinciplesandProblemsSolutionsManual517

521Chapter26continuedpage7035.Abeamofsinglyionized(1)oxygenatomsissentthroughamassspectrometer.ThevaluesareB7.2102T,q1.601019C,r0.085m,andV110V.Findthemassofanoxygenatom.B2r2q(7.2102T)2(0.085m)2(1.601019C)m2.71026kg2V(2)(110V)6.Amassspectrometeranalyzesandgivesdataforabeamofdoublyionized(2)argonatoms.Thevaluesareq2(1.601019C),B5.0102T,r0.106m,andV66.0V.Findthemassofanargonatom.B2r2q(5.0102T)2(0.106m)2(2)(1.601019C)m6.81026kg2V(2)(66.0V)7.Astreamofsinglyionized(1)lithiumatomsisnotdeflectedasitpassesthroughamagneticfieldof1.5103Tthatisperpendiculartoanelectricfieldof6.0102N/C.Whatisthespeedofthelithiumatomsastheypassthroughthetwofields?BqvEqE6.0102N/CvB1.5103T4.0105m/s8.InExampleProblem2,themassofaneonisotopeisdetermined.Anotherneonisotopeisfoundtohaveamassof22protonmasses.Howfarapartonthephotographicfilmwouldthesetwoisotopesland?Usethecharge-to-massratiotofindtheratiooftheradiiofthetwoisotopes.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.q2VmB2r22Vm122rBqm12Vm2222Thus,rand,Bqr2Vmm2012020BqTheradiusoftheisotopewithamassof22protonmasses,then,ism22r22r20m2022mrp2020mp22r202022(0.053m)200.056mThedifferenceintheradiiisr22r200.056m0.053m0.003m3mm518SolutionsManualPhysics:PrinciplesandProblems

522Chapter26continuedSectionReview26.1InteractionsofElectricandMagneticFieldsandMatterpages697–704page7049.Cathode-RayTubeDescribehowacathode-raytubeformsanelectronbeam.Electronsareemittedbythecathode,acceleratedbyapotentialdifference,andpassedthroughslitstoformabeam.10.MagneticFieldTheradiusofthecircularpathofanioninamassspectrometerisgivenbyr(1/B)2Vm/q.Usethisequationtoexplainhowamassspectrometerisabletoseparateionsofdifferentmasses.Assumingalloftheionshavethesamecharge,theonlyvariablethatisnotconstantintheequationistheionmass,m.Asmincreases,theradiusoftheion’spathalsoincreases.Thisresultsinseparatepathsforeachuniquemass.11.MagneticFieldAmodernmassspectrometercananalyzemoleculeshavingmassesofhundredsofprotonmasses.Ifthesinglychargedionsofthesemole-culesareproducedusingthesameacceleratingvoltage,howwouldthemassspectrometer’smagneticfieldhavetobechangedfortheionstohitthefilm?Sincer(1/B)2Vmq,asmincreases,sotoomustB.Ifmisraisedbyafactorofabout10,Bwouldhavetoincreasebyafactorofabout3becausetokeeprconstant,Bmustincreaseasm.12.PathRadiusAprotonmovesataspeedof4.2104m/sasitpassesthroughamagneticfieldof1.20T.Findtheradiusofthecircularpath.qvmBrvm(4.2104m/s)(1.671027kg)r3.7104mqB(1.601019C)(1.20T)13.MassAbeamofdoublyionized(2)oxygenatomsisacceleratedbyapotentialdifferenceof232V.Theoxygenthenentersamagneticfieldof75mTandfollowsacurvedpathwitharadiusof8.3cm.Whatisthemassoftheoxygenatom?q2VmB2r2qB2r2(2)(1.601019C)(75103T)2(8.3102m)2m2V(2)(232V)2.71026kgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual519

523Chapter26continued14.CriticalThinkingRegardlessoftheenergy20.Forlighttravelingthroughwater,theoftheelectronsusedtoproduceions,dielectricconstantis1.77.WhatistheJ.J.Thomsonnevercouldremovemorespeedoflighttravelingthroughwater?thanoneelectronfromahydrogenatom.c3.00108m/sWhatcouldhehaveconcludedaboutthevK1.77positivechargeofahydrogenatom?Itmustbeonlyasingleelementary2.25108m/scharge.21.Thespeedoflighttravelingthroughamaterialis2.43108m/s.WhatisthePracticeProblemsdielectricconstantofthematerial?c26.2ElectricandMagneticvKFieldsinSpacec23.00108m/s2pages705–713Kv2.43108m/s1.52page70615.Whatisthespeedinairofanelectromagneticwavehavingafrequencyof3.21019Hz?SectionReviewAllelectromagneticwavestravelthrough26.2ElectricandMagneticairoravacuumatc,3.00108m/s.FieldsinSpacepages705–71316.Whatisthewavelengthofgreenlighthavingafrequencyof5.701014Hz?page71322.WavePropagationExplainhowelectro-c3.00108m/s14Hz5.26107mmagneticwavesareabletopropagatef5.7010throughspace.17.AnelectromagneticwavehasafrequencyThechangingelectricfieldinducesaof8.21014Hz.Whatisthewavelengthchangingmagneticfield,andthechang-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofthewave?ingmagneticfieldinducesachangingc3.00108m/selectricfield.Thewavespropagateas14Hz3.7107mf8.210thesetwofieldsregenerateeachother.18.Whatisthefrequencyofanelectromagnetic23.ElectromagneticWavesWhataresomeofwavehavingawavelengthof2.2102m?theprimarycharacteristicsofelectromagneticcwaves?Doelectromagneticwavesbehavefdifferentlyfromthewaythatotherwaves,c3.00108m/ssuchassoundwaves,behave?Explain.f2m1.41010Hz2.210Electromagneticwavescanbedescribedbyfrequencyandwavelength.Theypage707behavesimilarlytootherwavesinthat19.Whatisthespeedofanelectromagnetictheyreflect,refract,diffract,interfere,andwavetravelingthroughtheair?UsecanbeDopplershifted.Thedifferencec299,792,458m/sinyourcalculation.betweentheelectromagneticwavesandc299,792,458m/sotherwaves,suchassoundwaves,isvK1.00054thatelectromagneticwavescantravelthroughavacuumandcanbepolarized.2.99712108m/s520SolutionsManualPhysics:PrinciplesandProblems

524Chapter26continued24.FrequencyAnelectromagneticwaveisscientistshavediscoveredthattheozonelayerfoundtohaveawavelengthof1.5105m.overbothAntarcticaandtheArcticOceanisWhatisthefrequencyofthewave?thinning.Usewhatyouhavelearnedaboutcelectromagneticwavesandenergytoexplainfwhysomescientistsareveryconcernedaboutthethinningozonelayer.c3.00108m/s13Hzf2.0101.5105mIftheentireozonelayeristhinning,theamountofUVradiationfromtheSun25.TVSignalsTelevisionantennasnormallythatisblockedbytheozonelayerwillhavemetalrodelementsthatareorienteddecrease,allowingmoreUVraystohorizontally.Fromthisinformation,whatreachthesurfaceofEarth.Thewave-canyoudeduceaboutthedirectionsofthelengthsofUVwavesareshortenough,electricfieldsintelevisionsignals?andtheirenergiesarehighenough,toTheyalsomustbehorizontal.damageskinmolecules.Thus,theresultingincreaseinUVraystowhich26.ParabolicReceiversWhyisitimportanthumanswouldbeexposedcouldforaparabolicdish’sreceivingantennatoincreasetheprevalenceofskincancer.beproperlyalignedwiththetransmitter?ParabolicdishantennasareonlyabletoreceivesignalswithinaverynarrowChapterAssessmentrangeofangles.Itis,therefore,neces-ConceptMappingsarytocarefullyalignthedishreceiverpage718withthetransmittingantennastomaxi-30.Completethefollowingconceptmapmizethereceivedsignalstrength.usingthefollowingtermandsymbols:E,c,magneticfield.27.AntennaDesignTelevisionchannels2through6havefrequenciesjustbelowtheFMradioband,whilechannels7throughElectromagneticwaves13havemuchhigherfrequencies.Whichsignalswouldrequirealongerantenna:thoseofchannel7orthoseofchannel6?magnetictravelattheProvideareasonforyouranswer.electricfieldfieldspeedoflightThesignalsofchannel6wouldrequirealongerantenna.Lower-frequencywaveswouldhavelongerwavelengths.EBc28.DielectricConstantThespeedoflighttravelingthroughanunknownmaterialis1.98108m/s.GiventhatthespeedoflightMasteringConceptsinavacuumis3.00108m/s,whatisthepage718dielectricconstantoftheunknownmaterial?31.Whatarethemassandchargeofancelectron?(26.1)vKThemassofanelectronisCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c23.00108m/s29.111031kg.Kv1.98108m/s2.30Itschargeis1.601019C.29.CriticalThinkingMostoftheUVradiation32.Whatareisotopes?(26.1)fromtheSunisblockedbytheozonelayerIsotopesareatomsofthesameelementinEarth’satmosphere.Inrecentyears,thathavedifferentmasses.Physics:PrinciplesandProblemsSolutionsManual521

525Chapter26continued33.ThedirectionofaninducedmagneticfieldApplyingConceptsisalwaysatwhatangletothechangingpage718electricfield?(26.2)38.TheelectronsinaThomsontubetravelfromAninducedmagneticfieldisalwaysatlefttoright,asshowninFigure26-14.rightanglestothechangingelectricfield.Whichdeflectionplateshouldbechargedpositivelytobendtheelectronbeamupward?34.WhymustanACgeneratorbeusedtopropagateelectromagneticwaves?IfaDCgeneratorwereused,whenwoulditcreateVUpperchargedelectromagneticwaves?(26.2)plateAnACgeneratorsuppliesthechangingelectricfield,whichinturngeneratesachangingmagneticfield.ADCgenera-Lowerchargedtorwouldonlygenerateachangingplateelectricfieldwhenturnedonoroff.35.Averticalantennawiretransmitsradio■Figure26-14waves.SketchtheantennaandtheelectricThetopplateshouldbechargedandmagneticfieldsthatitcreates.(26.2)positively.39.TheThomsontubeinquestion38usesamagneticfieldtodeflecttheelectronbeam.Whatwouldthedirectionofthemagneticfieldneedtobetobendthebeamdownward?Themagneticfieldwouldbedirected36.Whathappenstoaquartzcrystalwhenaintotheplaneofthepaper.voltageisappliedacrossit?(26.2)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.40.ShowthattheunitsofE/BarethesameasQuartzcrystalsbendordeformwhentheunitsforvelocity.voltageisplacedacrossthem.TheNquartzcrystalthenwillvibrateatasetCfrequency.EAmBNCAm37.Howdoesanantenna’sreceivingcircuitBecause1Ais1C/s,thisbecomesselectelectromagneticradiowavesofacertainfrequencyandrejectallothers?(26.2)ECmmBsCsByadjustingthecapacitanceoftheantennacircuit,theoscillationfrequency41.Thevacuumchamberofamassspectrome-ofthecircuitequalsthefrequencyoftheterisshowninFigure26-15.Ifasampleofdesiredradiowaves.Resonanceoccurs,ionizedneonisbeingtestedinthemasscausingtheelectronsinthecircuittospectrometer,inwhatdirectionmusttheoscillateatthatfrequency.magneticfieldbedirectedtobendtheionsintoaclockwisesemicircle?522SolutionsManualPhysics:PrinciplesandProblems

526Chapter26continuedChannel2waveshavealowerfrequencyandalongerwavelength,sochannel2requiresalongerantenna.ThelengthofVacuumchamberanantennaisdirectlyproportionaltoDeflectedwavelength.paths45.Supposetheeyesofanalienbeingaresensitivetomicrowaves.Wouldyouexpectsuchabeingtohavelargerorsmallereyesthanyours?Why?Theeyeswouldbemuchlarger,becausethewavelengthofmicrowaveradiationisFilmplatemuchlargerthanthatofvisiblelight.DualslitsMasteringProblemsIonbeam26.1InteractionsofElectricandMagneticFieldsandMatter■Figure26-15page719ThemagneticfieldisfoundbytheLevel1right-handruleandwouldbedirected46.Electronsmovingat3.6104m/spassoutfromandperpendiculartotheplanethroughanelectricfieldwithanintensityofthepaper.of5.8103N/C.Howlargeamagneticfield42.Ifthesignofthechargeontheparticlesinmusttheelectronsalsoexperiencefortheirquestion41ischangedfrompositivetopathtobeundeflected?negative,dothedirectionsofeitherorbothEvofthefieldshavetobechangedtokeeptheBparticlesundeflected?Explain.3E5.810N/CB0.16TYoucaneitherchangebothfieldsorv3.6104m/sneitherfield,butyoucannotchange47.Aprotonmovesacrossa0.36-Tmagneticonlyonefield.field,asshowninFigure26-16.Iftheprotonmovesinacircularpathwitharadius43.Foreachofthefollowingproperties,of0.20m,whatisthespeedoftheproton?identifywhetherradiowaves,lightwaves,orXrayshavethelargestvalue.XBXXXXXa.wavelengthPathofchargedparticleRadiowaveshavethelongestXXXXXXwavelengths.b.frequencyXXXrXXXXrayshavethehighestfrequencies.c.velocityAlltravelatthesamevelocity,whichXXXXXXisthespeedoflight.■Figure26-16Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.44.TVWavesThefrequencyoftelevisionq2vmBrwavesbroadcastonchannel2isabout58MHz.Thewavesbroadcastonchannel7Brq(0.36T)(0.20m)(1.601019C)vareabout180MHz.Whichchannelm1.671027kgrequiresalongerantenna?6.9106m/sPhysics:PrinciplesandProblemsSolutionsManual523

527Chapter26continued48.Aprotonentersa6.0102-Tmagneticfieldwithaspeedof5.4104m/s.Whatistheradiusofthecircularpathitfollows?mv(1.671027kg)(5.4104m/s)r9.4103mBq(6.0102T)(1.601019C)49.Anelectronisacceleratedbya4.5-kVpotentialdifference.Howstrongamagneticfieldmustbeexperiencedbytheelectronifitspathisacircleofradius5.0cm?12Vm1(2)(4.5103V)(9.111031kg)Brq0.050m1.601019C4.5103T50.Amassspectrometeryieldsthefollowingdataforabeamofdoublyionized(2)sodiumatoms:B8.0102T,q2(1.601019C),r0.077m,andV156V.Calculatethemassofasodiumatom.q2VmB2r2qB2r2(2)(1.601019C)(8.0102T)2(0.077m)2m3.91026kg2V(2)(156V)Level251.Analphaparticlehasamassofapproximately6.61027kgandhasachargeof2.Suchaparticleisobservedtomovethrougha2.0-Tmagneticfieldalongapathofradius0.15m.a.Whatspeeddoestheparticlehave?qvmBr19C)(0.15m)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Bqr(2.0T)(2)(1.6010vm6.61027kg1.5107m/sb.Whatisitskineticenergy?11Bqr2q2B2r2(2)(1.601019C)(2.0T)2(0.15m)2KEmv2m22m2m(2)(6.61027kg)7.01013Jc.Whatpotentialdifferencewouldberequiredtogiveitthiskineticenergy?KEqVKE7.01013J6VVq(2)(1.601019C)2.21052.Amassspectrometeranalyzescarbon-containingmoleculeswithamassof175103protonmasses.Whatpercentdifferentiationisneededtoproduceasampleofmoleculesinwhichonlycarbonisotopesofmass12,andnoneofmass13,arepresent?524SolutionsManualPhysics:PrinciplesandProblems

528Chapter26continuedThedifferencebetweencarbon-1226.2ElectricandMagneticFieldsinSpaceandcarbon-13isoneprotonmass.page719Todistinguishbetweenthese,apercentLevel1differentiationofoneprotonmassout54.RadioWavesTheradiowavesreflectedbyaof175103isneeded,orparabolicdishare2.0cmlong.Howlong11100percent.shouldtheantennabethatdetectsthewaves?175,0001750Theantennais,or1.0cmlong.53.SiliconIsotopesInamassspectrometer,2ionizedsiliconatomshavecurvatures,as55.TVAtelevisionsignalistransmittedonashowninFigure26-17.Ifthesmallercarrierfrequencyof66MHz.Ifthewiresonradiuscorrespondstoamassof28proton1areceivingantennaareplacedapart,masses,whatisthemassoftheothersili-4conisotope?determinethephysicaldistancebetweenthereceivingantennawires.11c44fVacuumchamber83.0010m/s(4)(66106Hz)r116.23cm1.1mr217.97cm56.Bar-CodeScannerAbar-codescannerr2usesalaserlightsourcewithawavelengthr1ofabout650nm.Determinethefrequencyofthelaserlightsource.Filmplatec3.00108m/s14Hzf4.610Dualslits650109mSiliconionbeam57.Whatistheoptimumlengthofareceivingantennathatistoreceivea101.3-MHzradiosignal?■Figure26-17Optimumantennalengthisq2V11cmB2r222fsomisproportionaltor28m/s3.0010mr2(2)(101.3106Hz)22mr2111.48mr22m2m1r1Level217.97cm258.AnEMwavewithafrequencyof100-MHz(28mp)16.23cm34mpistransmittedthroughacoaxialcablehav-m27kg)ingadielectricconstantof2.30.Whatisthe234mp(34)(1.6710velocityofthewave’spropagation?5.71026kgc8m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.v3.00101.98108m/sK2.30Physics:PrinciplesandProblemsSolutionsManual525

529Chapter26continued59.CellPhoneAcertaincellulartelephoneThefrequency,then,istransmitteroperatesonacarrierfrequencycfof8.00108Hz.Whatistheoptimallength(4)(0.083m)ofacellphoneantennadesignedtoreceive3.00108m/sthissignal?Notethatsingle-endedantennas,(4)(0.083m)suchasthoseusedbycellphones,generate9.0108HzpeakEMFwhentheirlengthisone-fourththewavelengthofthewave.63.AnunknownparticleisacceleratedbyaForasingle-endedantenna,optimumpotentialdifferenceof1.50102V.Theantennalengthisparticlethenentersamagneticfieldof11c50.0mT,andfollowsacurvedpathwitha44fradiusof9.80cm.Whatistheratioofq/m?3.00108m/sq2V(4)(8.00108Hz)mB2r20.0938m(2)(1.50102V)(50.0103T)2(9.80102m)2MixedReview1.25107C/kgpage720Level1ThinkingCritically60.Themassofadoublyionized(2)oxygenatomisfoundtobe2.71026kg.Ifthepage72064.ApplyConceptsManypolicedepartmentsmassofanatomicmassunit(amu)isequalto1.671027kg,howmanyatomicmassuseradargunstocatchspeedingdrivers.Aradargunisadevicethatusesahigh-fre-unitsareintheoxygenatom?quencyelectromagneticsignaltomeasure(2.71026kg)1amu16amuthespeedofamovingobject.Thefrequency1.671027kgoftheradargun’stransmittedsignalis61.RadioAnFMradiostationbroadcastsknown.ThistransmittedsignalreflectsoffCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.onafrequencyof94.5MHz.Whatistheofthemovingobjectandreturnstotheantennalengththatwouldgivethebestreceiverontheradargun.Becausethereceptionforthisstation?objectismovingrelativetotheradargun,thefrequencyofthereturnedsignalisdif-Optimumantennalengthisferentfromthatoftheoriginallytransmit-11c22ftedsignal.ThisphenomenonisknownastheDopplershift.Whentheobjectismov-3.00108m/s6Hz)ingtowardtheradargun,thefrequencyof(2)(94.510thereturnedsignalisgreaterthanthefre-1.59mquencyoftheoriginalsignal.IftheinitialtransmittedsignalhasafrequencyofLevel210.525GHzandthereturnedsignalshows62.AtwhatfrequencydoesacellphonewithaDopplershiftof1850Hz,whatisthean8.3-cm-longantennasendandreceivespeedofthemovingobject?Usethefollow-signals?Recallfromquestion59thatsingle-ingequation:endedantennas,suchasthoseusedbycellcfDopplervphones,generatepeakEMFwhentheirtarget2ftransmittedlengthisone-fourththewavelengthoftheWhere,wavetheyarebroadcastingorreceiving.vtargetvelocityoftarget(m/s)Theantennalengthiscspeedoflight(m/s)11c0.083m44f526SolutionsManualPhysics:PrinciplesandProblems

530Chapter26continuedfDopplerDopplershiftfrequency(Hz)ftransmittedfrequencyoftransmittedwave(Hz)cfDoppler(3.00108m/s)(1850Hz)vtarget2ftransmitted(2)(10.525109Hz)26.4m/s65.ApplyConceptsH.G.Wellswroteascience-fictionnovelcalledTheInvisibleMan,inwhichamandrinksapotionandbecomesinvisible,althoughheretainsallofhisotherfaculties.Explainwhyaninvisiblepersonwouldnotbeabletosee.Tosee,youmustdetectthelight,whichmeansthelightwillbeabsorbedorscattered.Essentially,aninvisiblepersonwouldbecompletelytransparentsolightwouldjustpassthroughtheeyewithouteverbeingabsorbedorscattered.66.DesignanExperimentYouaredesigningamassspectrometerusingtheprinciplesdiscussedinthischapter,butwithanelectronicdetectorreplacingthephotographicfilm.Youwanttodistinguishsinglyionized(1)moleculesof175protonmassesfromthosewith176protonmasses,butthespacingbetweenadjacentcellsinyourdetectoris0.10mm.Themoleculesmusthavebeenaccel-eratedbyapotentialdifferenceofatleast500.0Vtobedetected.WhataresomeofthevaluesofV,B,andrthatyourapparatusshouldhave?Thecharge-to-massratioforisotopesinamassspectrometerisq2V12Vmsotheradiusoftheisotope’spathisr.mB2r2BqThedifferenceintheradiiforthetwoisotopesis0.10103mr176r17512VBqm176m17512VBq176mp175mp2Vm1p176175BqThemagneticfield,then,is1761752VmpB0.10103mq176175(2)(500.0V)(1.671027kg)0.10103m1.601019C1.2TTheradiusfortheisotopewithamassequalto176protonmasses2V(176m(2)(5.00V)(176)(1.671027kg)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1p)1isr76Bq1.2T1.601019C3.6102mWhendesigningthespectrometer,youcanchooseanyvalueofVandB,providedVisatleast500.0V.However,sinceq/misconstant,VwillbeproportionaltoB2r2.Physics:PrinciplesandProblemsSolutionsManual527

531Chapter26continuedWritinginPhysicsThespheressharechargesequallywhentheyaretouchedtogether,sotheypage720eachcarry–1qofcharge.Theforce67.Composea1–2pagereportinwhichyouvarieswiththeproductofthecharges,outlinetheoperationofatypicaltelevision,sothenewforceistotheoldforceasDVD,orVCRinfraredremote-controlunit.1q1qisto4q2q,or1:8.Therefore,Explainwhythesimultaneoususeofmulti-thenewforceisF/8.Thedirectionofthepleremote-controlunitstypicallydoesnotnewforcewillberepulsive,ratherthancausetheunitstointerferewitheachother.attractive.Yourreportshouldincludeblockdiagramsandsketches.70.WhatistheelectricfieldstrengthbetweenRemotecontrolsuseabroadrangeoftwoparallelplatesspaced1.2cmapartifaIRfrequenciesthatarepulse-codemod-potentialdifferenceof45Visappliedtoulated.Eachbuttonontheremotepro-them?(Chapter21)ducesauniquesequenceofshortandV45V3V/morN/Clongpulses.BecauseofthewiderangeE3.810d0.012moffrequenciesusedbydifferentmanu-facturersandtheuniquepulsecodes71.Calculatethedailycostofoperatinganairused,itisveryunlikelythatdifferentcompressorthatrunsone-fourthofthetimeremotecontrolswillinterferewitheachanddraws12.0Afroma245-Vcircuitiftheother.costis$0.0950perkWh.(Chapter22)cost(E)(rate)CumulativeReviewpage720(IVt)(rate)68.AHe–Nelaser(633nm)isusedtoillu-(120A)(245V)(6h)minateaslitofunknownwidth,forminga$0.09501kWpatternonascreenthatislocated0.95mkWh1000Wbehindtheslit.Ifthefirstdarkbandis8.5mmfromthecenterofthecentralbright$1.68Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.band,howwideistheslit?(Chapter19)72.A440-cmlengthofwirecarrying7.7Aisxw1atrightanglestoamagneticfield.TheforceLonthewireis0.55N.WhatisthestrengthLofthefield?(Chapter24)wx1FBIL(633109m)(0.95m)BF8.5103mIL0.55N7.1105m(7.7A)(4.4m)0.016T69.Theforcebetweentwoidenticalmetalsphereswiththechargesshownin73.Anorth-southwireismovedtowardtheFigure26-18isF.Ifthespheresareeastthroughamagneticfieldthatispoint-touchedtogetherandreturnedtotheiringdown,intoEarth.Whatisthedirectionoriginalpositions,whatisthenewforceoftheinducedcurrent?(Chapter25)betweenthem?(Chapter20)north4q2q■Figure26-18528SolutionsManualPhysics:PrinciplesandProblems

532Chapter26continued3.00108m/s14Hzf5.2010ChallengeProblems5.77107mpage709For597nm,Visiblelightmakesuponlyaverysmallportionof3.00108m/s14Hzf5.0310theentireelectromagneticspectrum.Thewavelengths5.97107mforsomeofthecolorsofvisiblelightareshowninFor622nm,Table26-1.3.00108m/s14Hzf4.82106.22107mTable26-1For700nm,WavelengthsofVisibleLight3.00108m/s14HzColorWavelength(nm)f7m4.29107.0010Violet–Indigo390to45514Hztoonesignificantdigit)(410Blue455to492Thus,therangesareGreen492to577Violet:6.591014Hzto7.691014HzYellow577to597Blue:6.101014Hzto6.591014HzOrange597to622Green:5.201014Hzto6.101014HzYellow:5.031014Hzto5.201014HzRed622to700Orange:4.821014Hzto5.031014HzRed:4.291014Hzto4.821014Hz1.Whichcoloroflighthasthelongestwave-length?red2.Whichcolortravelsthefastestinavacuum?Theyalltravelatthesamespeed,c.3.Waveswithlongerwavelengthsdiffractaroundobjectsintheirpathmorethanwaveswithshorterwavelengths.Whichcolorwilldiffractthemost?Theleast?Redlightwilldiffractthemost.Violetlightwilldiffracttheleast.4.CalculatethefrequencyrangeforeachcoloroflightgiveninTable26-1.cThefrequencyisf.For390nm,3.00108m/s14Hzf7.69103.90107mFor455nm,3.00108m/s14HzCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.f7m6.59104.5510For492nm,3.00108m/s14Hzf6.10104.92107mFor577nm,Physics:PrinciplesandProblemsSolutionsManual529

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534CHAPTER27QuantumTheoryKEqVPracticeProblems0(1.601019C)(3.2J/C)27.1AParticleModel5.11019JofWavespages723–734page732page7306.Thethresholdwavelengthofzincis310nm.1.Anelectronhasanenergyof2.3eV.WhatisFindthethresholdfrequency,inHz,andtheenergyoftheelectroninjoules?theworkfunction,ineV,ofzinc.1.601019J19J(2.3eV)1eV3.710c3.00108m/s14Hzf9.7100310109m02.WhatistheenergyineVofanelectronwithWhf0avelocityof6.2106m/s?(6.631034J/Hz)12KEmv214Hz)1eV(9.7101.601019J1(9.111031kg)(6.2106m/s)224.0eV(1.751017J)1eV1.601019J7.Theworkfunctionforcesiumis1.96eV.1.1102eVWhatisthekineticenergy,ineV,ofphoto-electronsejectedwhen425-nmvioletlight3.Whatisthevelocityoftheelectroninfallsonthecesium?problem1?1240eVnmKEhf31kg,KE1mv2max0m9.111021240eVnm1.96eV2KE(2)(3.71019J)425nmvm9.111031kg0.960eV9.0105m/s8.Whenametalisilluminatedwith193-nm4.Thestoppingpotentialforaphotoelectricultravioletradiation,electronswithenergiescellis5.7V.Calculatethemaximumkineticof3.5eVareemitted.WhatistheworkenergyoftheemittedphotoelectronsineV.functionofthemetal?KEhfhfKEqV0019C)(5.7J/C)hc(1.6010hfKE0hfKE1eV1240eVnm1.601019JKE5.7eV1240eVnmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.5eV193nm5.Thestoppingpotentialrequiredtoprevent2.9eVcurrentthroughaphotocellis3.2V.Calculatethemaximumkineticenergyinjoulesofthephotoelectronsastheyareemitted.Physics:PrinciplesandProblemsSolutionsManual531

535Chapter27continued9.Ametalhasaworkfunctionof4.50eV.13.PhotoelectricandComptonEffectsWhatisthelongest-wavelengthradiationDistinguishthephotoelectriceffectfromthatwillcauseittoemitphotoelectrons?theComptoneffect.hcTheComptoneffectisthescatteringofhf4.50eV04.50eV,soaphotonbymatter,resultinginapho-01240eVnmtonoflowerenergyandmomentum.TheThus,04.50eV276nmphotoelectriceffectistheemissionofelectronsfromametalsamplewhenradi-ationofsufficientenergyisincidentonit.SectionReview27.1AParticleModel14.PhotoelectricEffectGreenlight(532nm)strikesanunknownmetal,ofWavescausingelectronstobeejected.Theejectedpages723–734electronscanbestoppedbyapotentialofpage7341.44V.Whatistheworkfunction,ineV,of10.PhotoelectricEffectWhyishigh-intensity,themetal?low-frequencylightunabletoejectelectronshc1240eVnmE2.33eVfromametal,whereaslow-intensity,high-greenlight532nmfrequencylightcan?Explain.KEejectedelectronqVLight,aformofelectromagneticradia-(1.601019C)tion,isquantizedandmassless,yetitdoeshavekineticenergy.Eachincident(1.44J/C)photoninteractswithasingleelectron.If1eVtheincidentphotondoesnothavesuffi-1.601019Jcientenergy,itcannotejectanelectron.1.44eVBecauseenergyisdirectlyrelatedtofre-quency,lowfrequencylightdoesnotWEgreenlightKEejectedelectronhavesufficientenergytoejectanelec-2.33eV1.44eVCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tron,whereashighfrequencylightdoes.0.89eV11.FrequencyandEnergyofHot-Body15.EnergyofaPhotonWhatistheenergy,inRadiationAsthetemperatureofabodyiseV,ofthephotonsproducedbyalaserincreased,howdoesthefrequencyofpeakpointerhavinga650-nmwavelength?intensitychange?Howdoesthetotalamountofradiatedenergychange?hc1240eVnmE1.9eV650nmBothfrequencyofpeakintensityandtotalenergyradiatedincrease.Thepeak16.PhotoelectricEffectAnXrayisabsorbedinfrequencyincreasesasT,whereastheaboneandreleasesanelectron.IftheXraytotalenergyincreasesasT4.hasawavelengthofapproximately0.02nm,estimatetheenergy,ineV,oftheelectron.12.PhotoelectricandComptonEffectsAnexperimentersendsanXrayintoatarget.hc1240eVnm4eVE6100.02nmAnelectron,butnootherradiation,emergesfromthetarget.Explainwhether17.ComptonEffectAnXraystrikesabone,thiseventisaresultofthephotoelectriccollideswithanelectron,andisscattered.effectortheComptoneffect.HowdoesthewavelengthofthescatteredItisaresultofthephotoelectriceffect,XraycomparetothewavelengthofthewhichisthecaptureofaphotonbyanincomingXray?electroninmatterandthetransferofthephoton’senergytotheelectron.532SolutionsManualPhysics:PrinciplesandProblems

536Chapter27continuedThescatteredXrayhasalongerwave-21.WhatvoltageisneededtoaccelerateanlengththantheincomingXray.electronsoithasa0.125-nmwavelength?hh18.CriticalThinkingImaginethatthecolli-p,sopsionoftwobilliardballsmodelstheinter-h2p2actionofaphotonandanelectronduring12KEmvtheComptoneffect.Supposetheelectronis22m2mreplacedbyamuchmoremassiveproton.6.631034JsWouldthisprotongainasmuchenergy0.125109mfromthecollisionastheelectrondoes?(2)(9.111031kg)Wouldthephotonloseasmuchenergyasit17J)1eVdoeswhenitcollideswiththeelectron?(1.544101.601019JTheanswertobothquestionsisno.A96.5eV,soitwouldhavetobetennisballcantransfermorekineticacceleratedthrough96.5V.energytoasoftballthanitcantoabowlingball.22.TheelectroninExampleProblem3hasadeBrogliewavelengthof0.14nm.Whatisthekineticenergy,ineV,ofaprotonPracticeProblems(m1.671027kg)withthesamewavelength?27.2MatterWaveshThedeBrogliewavelengthispages735–737mvhsothevelocityisvpage736m19.A7.0-kgbowlingballrollswithavelocityThekineticenergy,then,isof8.5m/s.121h2h2KE2mv2mm2m2a.WhatisthedeBrogliewavelengthofthebowlingball?(6.631034Js)2h6.631034Js(2)(1.671027kg)(0.14109m)2mv(7.0kg)(8.5m/s)1eV1.601019J1.11035mb.Whydoesthebowlingballexhibitno4.2102eVobservablewavebehavior?Thewavelengthistoosmalltoshowobservableeffects.SectionReview27.2MatterWaves20.WhatisthedeBrogliewavelengthandpages735–737speedofanelectronacceleratedbyapoten-page737tialdifferenceof250V?23.WavelikePropertiesDescribetheexperi-1mv2qV,somentthatconfirmedthatparticleshave2wavelikeproperties.2qV(2)(1.601019C)(250J/C)vm9.111031kgWhenabeamofelectronswasaimedatacrystal,thecrystalactedlikeadiffractionCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9.4106m/sgrating,causingtheelectronstoformahdiffractionpattern.Thediffractionofthemvelectrons(particles)issimilartothedif-6.631034Jsfractionoflight(waves)throughagrating.(9.111031kg)(9.4106m/s)7.71011mPhysics:PrinciplesandProblemsSolutionsManual533

537Chapter27continued24.WaveNatureExplainwhythewavenaturethebeampassedthrough,resultinginofmatterisnotobvious.thedistributionofphotonsoratomsThewavelengthsofmostobjectsareseenintheinterferencepattern.muchtoosmalltobedetected.28.CriticalThinkingPhysicistsrecentlymade25.DeBroglieWavelengthWhatistheadiffractiongratingofstandingwavesofdeBrogliewavelengthofanelectronlight.Atomspassingthroughthegratingacceleratedthroughapotentialdifferenceproduceaninterferencepattern.Ifthe1of125V?spacingoftheslitsinthegratingwere22qV(about250nm),whatwastheapproximatevmdeBrogliewavelengthoftheatoms?2(1.601019C)(125V)Fordiffractiongratings,dsin,9.111031kgwheredisthespacingoftheslits,and6.63106m/sistheangularseparationbetweenconsecutivepeaks.ThedeBrogliepmv(9.111031kg)(6.63106m/s)wavelength,then,is(250nm)sin.6.041024kgm/sIfweassumesinisaround0.1,thenh6.631034JsthedeBrogliewavelengthisafewtensp6.041024kgm/sofnanometers.1.101010m0.110nmChapterAssessment26.WavelengthsofMatterandRadiationConceptMappingWhenanelectroncollideswithamassivepage742particle,theelectron’svelocityandwave-29.Completethefollowingconceptmapusinglengthdecrease.Howisitpossibletotheseterms:dualnature,mass,waveproper-increasethewavelengthofaphoton?ties,momentum,diffraction.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.IfthephotonundergoesComptonscat-teringwithafixedtarget,thewave-dualnaturelengthofthephotonwillincrease.Note,however,thatthephoton’sspeedisnotchanged.Thephotonstilltravelsatc.particlewavepropertiesproperties27.HeisenbergUncertaintyPrincipleWhenlightorabeamofatomspassesthroughadou-bleslit,aninterferencepatternforms.Bothinterferencediffractionmassmomentumresultsoccurevenwhenatomsorphotonspassthroughtheslitsoneatatime.HowdoestheMasteringConceptsHeisenberguncertaintyprincipleexplainthis?page742TheHeisenberguncertaintyprinciple30.IncandescentLightAnincandescentlight-statesthatyoucannotsimultaneouslybulbiscontrolledbyadimmer.Whathappensknowtheprecisepositionandmomen-tothecolorofthelightgivenoffbythebulbastumofaparticle.Thus,ifyouknowthethedimmercontrolisturneddown?(27.1)precisepositionofaphotonoranThelightbecomesredder.atomasitpassesthroughtheslit,youcannotknowitsprecisemomentum.Becauseoftheunknownmomentum,youcannotbesurewhichoftheslits534SolutionsManualPhysics:PrinciplesandProblems

538Chapter27continued31.Explaintheconceptofquantizedenergy.37.HowdoestheComptoneffectdemonstrate(27.1)thatphotonshavemomentumaswellasQuantizedenergymeansthatenergyenergy?(27.1)canexistonlyinmultiplesofsomemin-Elasticcollisionstransferbothmomen-imumvalue.tumandenergy.Onlyifphotonshavemomentumcantheequationsbe32.WhatisquantizedinMaxPlanck’sinterpre-satisfied.tationoftheradiationofincandescentbodies?(27.1)38.Themomentum,p,ofaparticleofmatterisThevibrationalenergyoftheincandes-givenbypmv.Canyoucalculatethecentatomsisquantized.momentumofaphotonusingthesameequation?Explain.(27.2)33.Whatisaquantumoflightcalled?(27.1)No,usingtheequationyieldsaphotonaphotonmomentumofzerobecausephotonsaremassless.Thisresultisincorrect34.Lightabovethethresholdfrequencyshinesbecausemasslessphotonshavenon-onthemetalcathodeinaphotocell.Howzeromomenta.doesEinstein’sphotoelectriceffecttheoryexplainthefactthatasthelightintensity39.Explainhoweachofthefollowingelectronincreases,thecurrentofphotoelectronspropertiescouldbemeasured.(27.2)increases?(27.1)a.chargeEachphotonejectsaphotoelectron.BalancetheforceofgravityagainstLightwithgreaterintensitycontainstheforceofanelectricfieldonthemorephotonspersecond;thus,itcharge.causestheejectionofmorephoto-b.masselectronspersecond.Balancetheforceofanelectricfieldagainstthatofamagneticfieldto35.ExplainhowEinstein’stheoryaccountsforfindm/q,thenusethemeasuredthefactthatlightbelowthethresholdfre-valueofq.quencyofametalproducesnophotoelec-trons,regardlessoftheintensityofthec.wavelengthlight.(27.1)ScatterelectronsoffacrystalandPhotonsbelowthethresholdfrequencymeasuretheanglesofdiffraction.donothavesufficientenergytoejectanelectron.Iftheintensityofthelight40.Explainhoweachofthefollowingphotonincreases,thenumberofphotonspropertiescouldbemeasured.(27.2)increasesbuttheirenergydoesnot;a.energythephotonsarestillunabletoejectanMeasuretheKEoftheelectronselectron.ejectedfromametalforatleasttwodifferentwavelengths,ormeasurethe36.PhotographicFilmBecausecertaintypesKEoftheelectronsejectedfromaofblack-and-whitefilmarenotsensitiveknownmetalatonlyonewavelength.toredlight,theycanbedevelopedinab.momentumdarkroomthatisilluminatedbyredlight.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ExplainthisonthebasisofthephotonMeasurethechangeinwavelengththeoryoflight.(27.1)ofXraysscatteredbymatter.Redphotonsdonothaveenoughenergyc.wavelengthtocausethechemicalreactionthatMeasuretheangleofdiffractionexposesfilm.whenlightpassesthroughtwoslitsPhysics:PrinciplesandProblemsSolutionsManual535

539Chapter27continuedoradiffractiongrating,measuretheofincidentphotonsorthebrightnessofwidthofasingle-slitdiffractionpat-thelight,notthefrequencyofthelight.tern,ormeasuretheanglethelightisbentwhenitpassesthroughaprism.44.Potassiumemitsphotoelectronswhenstruckbybluelight,whereastungstenemitsApplyingConceptsphotoelectronswhenstruckbyultravioletradiation.page74241.Usetheemissionspectrumofanincandes-a.Whichmetalhasahigherthresholdcentbodyatthreedifferenttemperaturesfrequency?showninFigure27-1onpage724toBluelighthasalowerfrequencyandanswerthefollowingquestions.energythanUVlight.Thus,tungstena.Atwhatfrequencydoesthepeakemis-hasthehigherthresholdfrequency.sionintensityoccurforeachofthethreeb.Whichmetalhasalargerworkfunction?temperatures?tungsten4000K:~2.51014Hz,5800K:~3.51014Hz,8000K:~4.61014Hz45.ComparethedeBrogliewavelengthoftheb.Whatcanyouconcludeabouttherela-baseballshowninFigure27-11withthetionshipbetweenthefrequencyofpeakdiameterofthebaseball.radiationemissionintensityandtem-peratureforanincandescentbody?Thefrequencyofthepeakintensityincreaseswithincreasing21m/s0.10mtemperature.c.Bywhatfactordoestheintensityoftheredlightgivenoffchangeasthebody’stemperatureincreasesfrom4000Kto8000K?■Figure27-11Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.TheintensityintheredportionofThediameterofthebaseballisaboutthespectrumincreasesfromapprox-0.10m,whereasthedeBrogliewave-lengthis10–34m;thebaseballisaboutimately0.5to9.2,anincreasebya1033timeslargerthanthewavelength.factorofslightlygreaterthan18.42.Twoironbarsareheldinafire.OneglowsMasteringProblemsdarkred,whiletheotherglowsbright27.1AParticleModelofWavesorange.page742–743a.Whichbarishotter?Level1therodglowingbrightorange46.AccordingtoPlanck’stheory,howdoestheb.Whichbarisradiatingmoreenergy?frequencyofvibrationofanatomchangeifitgivesoff5.441019Jwhilechangingitstherodglowingbrightorangevalueofnby1?43.Willhigh-frequencylightejectagreaterEnhf,sonumberofelectronsfromaphotosensitiveE5.441019Jfsurfacethanlow-frequencylight,assumingnh(1)(6.631034Js)thatbothfrequenciesareabovethethresh-8.211014Hzoldfrequency?Notnecessarily;thenumberofejected47.Whatpotentialdifferenceisneededtostopelectronsisproportionaltothenumberelectronswithamaximumkineticenergyof4.81019J?536SolutionsManualPhysics:PrinciplesandProblems

540Chapter27continuedKEqV51.Thethresholdfrequencyofsodiumis0,so4.41014Hz.HowmuchworkmustbeKE4.81019CV0q(1.601019C)3.0Vdonetofreeanelectronfromthesurfaceofsodium?48.WhatisthemomentumofaphotonofvioletWorkhf0lightthathasawavelengthof4.0102nm?(6.631034J/Hz)(4.41014Hz)h6.631034Jsp4.0107m2.91019J1.71027kgm/s15Hz52.Iflightwithafrequencyof1.0010fallsonthesodiuminthepreviousprob-Level2lem,whatisthemaximumkineticenergyof49.Thestoppingpotentialofacertainmetalisthephotoelectrons?showninFigure27-12.Whatisthemaxi-KEhfhfmumkineticenergyofthephotoelectrons0inthefollowingunits?h(ff0)CathodeAnode(6.631034J/Hz)(1.001015Hz4.41014Hz)3.71019JLevel353.LightMeterAphotographer’slightmeterusesaphotocelltomeasurethelightfallingonthesubjecttobephotographed.What5.0Vshouldbetheworkfunctionofthecathodeifthephotocellistobesensitivetoredlight■Figure27-12(680nm)aswellastotheothercolorsa.electronvoltsoflight?KEqV01240eVnm1240eVnmW(1elementarycharge)(5.0V)680nm05.0eV1.8eVb.joules5.0eV1.601019J54.SolarEnergyAhomeusesabout41011Jofenergyeachyear.Inmanypartsofthe11eV8.01019JUnitedStates,thereareabout3000hofsunlighteachyear.50.Thethresholdfrequencyofacertainmetala.HowmuchenergyfromtheSunfallsonis3.001014Hz.Whatisthemaximumonesquaremetereachyear?kineticenergyofanejectedphotoelectronifEarthreceivesabout1000J/m2eachthemetalisilluminatedbylightwithasecond,sowavelengthof6.50102nm?3600s3000hE(1000J/m2s)KEhfhfhy0c11010J/m2peryearCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.hf0b.Ifthissolarenergycanbeconverted(6.631034Js)tousefulenergywithanefficiencyof3.00108m/s14Hz20percent,howlargeanareaof6.50107m3.0010converterswouldproducetheenergy1.071019Jneededbythehome?Physics:PrinciplesandProblemsSolutionsManual537

541Chapter27continued41011J6.631034J/HzArea(0.2)(11010J/m2)(9.111031kg)(4.2107m/s)2102m21.71011m0.017nm58.Aneutronisheldinatrapwithakinetic27.2MatterWavesenergyofonly0.025eV.page743a.Whatisthevelocityoftheneutron?Level11.601019J55.WhatisthedeBrogliewavelengthofanKE(0.025eV)eVelectronmovingat3.0106m/s?4.01021Jhmv12mv26.631034Js(9.111031kg)(3.0106m/s)2KE(2)(4.01021J)vm1.671027kg2.41010m2.2103m/s0.24nmb.FindthedeBrogliewavelengthofthe56.Whatvelocitywouldanelectronneedtoneutron.haveadeBrogliewavelengthofh3.01010m?mvh6.631034Js(1.671027kg)(2.2103m/s)mvh1.81010mvm6.631034J/Hz59.Thekineticenergyofahydrogenatom’s(9.111031kg)(3.01010m)electronis13.65eV.2.4106m/sa.Findthevelocityoftheelectron.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12KEmvLevel2257.Acathode-raytubeacceleratesanelectron2KEfromrestacrossapotentialdifferenceofvm3V.5.010(2)(13.65eV)(1.601019JeV)a.Whatisthevelocityoftheelectron?9.111031kg12qV2.19106m/smv2b.Calculatetheelectron’sdeBroglieqVwavelength.v1mh2mv(1.601019C)(5.0103V)v6.631034kgm/s131kg)(9.111031kg)(2.19106m/s)2(9.11100.332nm4.2107m/sc.Giventhatahydrogenatom’sradiusisb.Whatisthewavelengthassociatedwith0.519nm,calculatethecircumferenceoftheelectron?ahydrogenatomandcompareitwithhthedeBrogliewavelengthfortheatom’smvelectron.C2r538SolutionsManualPhysics:PrinciplesandProblems

542Chapter27continued(2)(0.519nm)3.26nmThecircumferenceisapproximatelyequaltotencompletewavelengths.Level360.AnelectronhasadeBrogliewavelengthof0.18nm.a.Howlargeapotentialdifferencediditexperienceifitstartedfromrest?hThedeBrogliewavelengthis,mvhwhichgivesavelocityofv.mThekineticenergy,then,is12KEmv21h2m2mh22m2Intermsofvoltage,thekineticenergyisKEqV.Combiningtheseandsolvingforvoltage,h2V2mq2(6.631034Js)2(2)(9.111031kg)(1.601019C)(0.1810947Vb.IfaprotonhasadeBrogliewavelengthof0.18nm,howlargeisthepotentialdifferencethatitexperiencedifitstartedfromrest?Usingthesamederivationasbefore,thevoltageish2V2mq2(6.631034Js)2(2)(1.671027kg)(1.601019C)(0.18109m)20.025VMixedReviewpage743–744Level161.Whatisthemaximumkineticenergyofphotoelectronsejectedfromametalthathasastoppingpotentialof3.8V?KEqV0(1elementarycharge)(3.8V)3.8eV62.Thethresholdfrequencyofacertainmetalis8.01014Hz.WhatistheworkCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.functionofthemetal?Whf0(6.631034J/Hz)(8.01014Hz)5.31019JPhysics:PrinciplesandProblemsSolutionsManual539

543Chapter27continued63.Iflightwithafrequencyof1.61015Hz348m/s)(6.6310Js)(3.0010fallsonthemetalinthepreviousproblem,1.601019Jwhatisthemaximumkineticenergyofthe(2.48eV)1eVphotoelectrons?5.01107mKEhfhf0501nm(6.631034J/Hz)(1.61015Hz)5.31019J67.AnelectronhasadeBrogliewavelengthof400.0nm,theshortestwavelengthofvisible5.31019Jlight.a.Findthevelocityoftheelectron.64.FindthedeBrogliewavelengthofadeuteron(nucleusof2Hisotope)ofmassh3.31027kgthatmoveswithaspeedofmv2.5104m/s.hvmhmv6.631034J/Hz34Js(9.111031kg)(400.0109m)6.6310(3.31027kg)(2.5104m/s)1.82103m/s8.01012mb.CalculatetheenergyoftheelectronineV.12Level2KEmv265.Theworkfunctionofironis4.7eV.131kg)(1.82103m/s)2a.Whatisthethresholdwavelengthof2(9.1110iron?eV1.601019Jhc1240eVnmW6eV009.43101240eVnm1240eVnm0W4.7eV68.ElectronMicroscopeAnelectronmicro-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.scopeisusefulbecausethedeBrogliewave-2.6102nmlengthsofelectronscanbemadesmallerthanb.Ironisexposedtoradiationofwave-thewavelengthofvisiblelight.Whatenergylength150nm.WhatisthemaximumineVhastobegiventoanelectronforittokineticenergyoftheejectedelectronshaveadeBrogliewavelengthof20.0nm?ineV?hThedeBrogliewavelengthis,hchc1240eVnmmvKE4.7eVh0150nmwhichgivesavelocityofv.m3.6eVThekineticenergy,then,is1266.Bariumhasaworkfunctionof2.48eV.KE2mvWhatisthelongestwavelengthoflight21hthatwillcauseelectronstobeemittedm2mfrombarium?h2hc2m2Workfunction2.48eVhf,so00(6.631034Js)2hc0(2)(9.111031kg)(20.0109m)22.48eV1eV1.601019J3.77103eV540SolutionsManualPhysics:PrinciplesandProblems

544Chapter27continuedLevel3ThinkingCritically69.Incidentradiationfallsontin,asshowninpage744Figure27-13.Thethresholdfrequencyof70.ApplyConceptsAhelium-neonlaseremitstinis1.21015Hz.photonswithawavelengthof632.8nm.CathodeAnodea.Findtheenergy,injoules,ofeachphotonemittedbythelaser.EachphotonhasenergyhcE167nm(6.631034Js)(3.00108m/s)632.8109m3.141019J■Figure27-13b.Atypicalsmalllaserhasapowerofa.Whatisthethresholdwavelengthoftin?0.5mW(equivalentto5104J/s).Howmanyphotonsareemittedeachcfsecondbythelaser?c3.00108m/s7mf1.21015Hz2.510P5104J/snE3.141019J/photonb.Whatistheworkfunctionoftin?21015photons/sWhf0(6.631034J/Hz)(1.21015Hz)71.ApplyConceptsJustbarelyvisiblelightwithanintensityof1.51011W/m28.01019Jentersaperson’seye,asshowninc.TheincidentelectromagneticradiationFigure27-14.hasawavelengthof167nm.WhatisthekineticenergyoftheejectedCorneaelectronsineV?KEhchf550nmLensmax0(6.631034J/Hz)(3.00108m/s)167109m8.01019JPupil3.91019J(diameter7.0mm)19J)1eV■Figure27-14(3.9101.601019J2.4eVa.Ifthislightshinesintotheperson’seyeandpassesthroughtheperson’spupil,whatisthepower,inwatts,thatenterstheperson’seye?Power(intensity)(area)(intensity)(r2)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(1.51011W/m2)((3.5103m)2)5.81016WPhysics:PrinciplesandProblemsSolutionsManual541

545Chapter27continuedb.Usethegivenwavelengthoftheinci-Fromthegraph,thethresholdfrequencydentlightandinformationprovidedinisf14Hz,whichgivesa04.9910cFigure27-14tocalculatethenumberofthresholdwavelengthof0fphotonspersecondenteringtheeye.8m/s03.0010601nmandawork4.991014HzEnergyperphotonfunctionofhcEWhf0(6.631034Js)(3.00108m/s)(6.631034J/Hz)(4.991014Hz)550109m3.311019J3.621019JWritinginPhysicsP5.81016J/spage744nE3.621019J/photon73.Researchthemostmassiveparticleforwhichinterferenceeffectshavebeenseen.1600photons/sDescribetheexperimentandhowtheinter-ferencewascreated.72.MakeandUseGraphsAstudentcompletedaphotoelectric-effectexperimentandrecordedAsof2003,thelargestisabuckyball,athestoppingpotentialasafunctionofwave-C60molecule.Nano-formedmetalliclength,asshowninTable27-1.Thephoto-gridswereusedasadiffractiongrating.cellhadasodiumcathode.Plotthedata(stoppingpotentialversusfrequency)anduseCumulativeReviewyourcalculatortodrawthebest-fitstraightpage744line(regressionline).Fromtheslopeand74.Thespringinapogostickiscompressedinterceptoftheline,findtheworkfunction,15cmwhenachildwhoweighs400.0Nthethresholdwavelength,andthevalueofstandsonit.Whatisthespringconstantofh/qfromthisexperiment.Comparethevaluethespring?(Chapter14)ofh/qtotheacceptedvalue.FkxCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Table27-1F400Nkx0.15mStoppingPotentialv.Wavelength3103N/m(nm)V(eV)02004.2075.Amarchingbandsoundsflatasitplayson3002.06averycoldday.Why?(Chapter15)4001.05Answer:Thepitchofawindinstrumentdependsonthespeedofsoundinthe5000.41airwithinit.Thecoldertheair,thelower6000.03thespeedofsoundandtheflatterthepitchofthesoundproduced.Convertwavelengthtofrequencyandplot.Determinethebeststraightline76.Achargeof8.0107Cexperiencesaforcethroughthedata.of9.0Nwhenplaced0.02mfromasecondSlope4.181015V/Hzcharge.Whatisthemagnitudeofthesec-4.181015J/HzCondcharge?(Chapter20)TheacceptedvalueisqAqBFKd2h(6.631034J/Hz)e(1.601019C)Fd2qBKq4.141015J/HzCA542SolutionsManualPhysics:PrinciplesandProblems

546Chapter27continued(9.0N)(0.02m)21.Findthemaximumkineticenergyofthe(9.0109Nm2C2)(8.0107C)vibratingobject.5107CKE1mv2277.Ahomeownerbuysadozenidentical120-V1(5.0103kg)(1.0102m/s)22lightsets.Eachlightsethas24bulbsconnect-edinseries,andtheresistanceofeachbulbis2.5107J6.0.Calculatethetotalloadinamperesifthehomeowneroperatesallthesetsfroma2.Thevibratingobjectemitsenergyinsingleexterioroutlet.(Chapter23)theformoflightwithafrequencyof5.01014Hz.IftheenergyisemittedinItotal12Isetasinglestep,findtheenergylostbytheVobject.(12)24REhf120V(12)(24)(6.0)(6.631034J/Hz)(5.01014Hz)1.0101A3.31019J78.Theforceona1.2-mwireis1.1103N.3.DeterminethenumberofequallysizedThewireisperpendiculartoEarth’smagneticenergy-stepreductionsthattheobjectfield.Howmuchcurrentisinthewire?wouldhavetomakeinordertoloseall(Chapter24)ofitsenergy.FBIL2.5107J11steps7.6103.31019J/stepF1.1103N1AI210BL(5105T)(1.2m)ChallengeProblempage731Supposeanickelwithamassof5.0gvibratesupanddownwhileitisconnectedtoaspring.Themaximumvelocityofthenickelduringtheoscil-lationsis1.0cm/s.Assumethatthevibratingnickelmodelsthequantumvibrationsoftheelectronswithinanatom,wheretheenergyofthevibrationsisgivenbytheequationEnhf.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Mass5.0gMaximumvelocity1.0cm/sPhysics:PrinciplesandProblemsSolutionsManual543

547

548CHAPTER28TheAtomr2(5.31011m)PracticeProblems2(2)2.11010mor0.21nm28.1TheBohrModelofr2(5.31011m)theAtom3(3)pages747–7594.81010mor0.48nmpage757r2(5.31011m)4(4)1.Calculatetheenergiesofthesecond,third,8.51010mor0.85nmandfourthenergylevelsinthehydrogenatom.5.Thediameterofthehydrogennucleusis13.6eV2.51015m,andthedistancebetweenEnn2thenucleusandthefirstelectronisabout13.6eV51011m.Ifyouuseaballwithadiame-E2(2)23.40eVterof7.5cmtorepresentthenucleus,how13.6eVfarawaywilltheelectronbe?E3(3)21.51eV11mx51013.6eV0.075m2.51015mE4(4)20.850eVx2103m2km,about1mile!2.CalculatetheenergydifferencebetweenE3andE2inthehydrogenatom.page758EE116.Findthewavelengthofthelightemittedin3E2(13.6eV)3222PracticeProblems2and3.Whichlinesin11Figure28-7correspondtoeachtransition?(13.6eV)1.89eV94hc3to2E3.CalculatetheenergydifferencebetweenE4andE(6.631034Js)(3.00108m/s)2inthehydrogenatom.(1.89eV)(1.601019J/eV)11EE4E2(13.6eV)42226.58107m658nm11hc(13.6eV)1642.55eV4to2E(6.631034Js)(3.00108m/s)4.Thetextshowsthesolutionoftheequation(2.55eV)(1.601019J/eV)h2n2rn42Kmq2forn1,theinnermost4.88107m488nmorbitalradiusofthehydrogenatom.Notethatwiththeexceptionofn2,allfactorsin7.Foraparticulartransition,theenergyofatheequationareconstants.Thevalueofmercuryatomdropsfrom8.82eVto6.67eV.r11m,or0.053nm.Usethis1is5.310a.Whatistheenergyofthephotonemit-informationtocalculatetheradiioftheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.tedbythemercuryatom?second,third,andfourthallowableenergyE8.82eV6.67eV2.15eVlevelsinthehydrogenatom.r2k,wherek5.31011mnn(Weareusingkforthecombinationofalltheconstantsintheequation.)Physics:PrinciplesandProblemsSolutionsManual545

549Chapter28continuedb.Whatisthewavelengthofthephoton11.BohrModelExplainhowenergyiscon-emittedbythemercuryatom?servedwhenanatomabsorbsaphotonofhclight.ETheinitialsumoftheenergyofthe(6.631034Js)(3.00108m/s)electronintheatomplustheenergy(2.15eV)(1.601019J/eV)oftheincidentphotonequalsthefinal5.78107m578nmenergyoftheelectronintheatom.12.OrbitRadiusAheliumionbehaveslikea8.Thegroundstateofaheliumionishydrogenatom.Theradiusoftheion’slow-54.4eV.Atransitiontothegroundstateestenergylevelis0.0265nm.Accordingtoemitsa304-nmphoton.WhatwastheBohr’smodel,whatistheradiusofthesec-energyoftheexcitedstate?ondenergylevel?hcE,soTheradiusdependsonn2,sothesecondlevelwouldhavearadiushc1240eVnmE4.08eVfourtimesthefirst,or0.106nm.304nmThereforeEexcitedEgroundE13.AbsorptionSpectrumExplainhowthe54.4eV4.08eVabsorptionspectrumofagascanbedeter-50.3eVmined.Describethereasonsforthespec-trum’sappearance.Toobtainanabsorptionspectrum,SectionReviewwhitelightispassedthroughasampleofgasandthenaspectroscope.28.1TheBohrModelofBecausethegasabsorbsspecificwave-theAtomlengths,thenormallycontinuousspec-pages747–759trumofwhitelightcontainsdarklines.page759Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.14.BohrModelHydrogenhasbeendetected9.Rutherford’sNuclearModelSummarizetransitioningfromthe101sttothe100ththestructureoftheatomaccordingtoenergylevels.Whatistheradiation’swave-Rutherford’snuclearmodel.length?Whereintheelectromagneticspec-InRutherford’snuclearmodel,allofantrumisthisemission?atom’spositivechargeandvirtuallyallEEofitsmassareconcentratedinatiny,101E100centrallylocatednucleusaroundwhich11(13.6eV)22negativelychargedelectronsorbit.1011002.68105eV,10.SpectraHowdotheemissionspectraofhc1240eVnmincandescentsolidsandatomicgasesdiffer?E2.68105eVInwhatwaysaretheysimilar?46.3106nm4.63cm,thewave-Incandescentsolidsproducespectralengthsindicatestheradiationisaconsistingofacontinuousbandofmicrowavecolors,whereasgasesproducespectramadeupofasetofdiscretelines.Allspectraarecreatedbyenergy-leveltransitionsinatoms.546SolutionsManualPhysics:PrinciplesandProblems

550Chapter28continued15.CriticalThinkingThenucleusoftheexplainwhythelawsofelectromagnet-hydrogenatomhasaradiusofaboutismdonotapplywithintheatom.1.51015m.Ifyouweretobuildamodelofthehydrogenatomusingasoftball(r19.QuantumModelExplainwhytheBohr5cm)torepresentthenucleus,wheremodeloftheatomconflictswiththewouldyoulocateanelectroninthen1Heisenberguncertaintyprinciple,whereasBohrorbit?Woulditbeinyourclassroom?thequantummodeldoesnot.Thescaleis5cm1.51015m,orTheuncertaintyprincipledoesn’tallow1cm3.01016m.TheBohrradiusaparticletohaveapreciselyknownis5.31011m.Inourmodelthiswouldposition,suchasaBohrorbit.Thebe(5.31011/3.01016)1cmquantummodelpredictsonlytheproba-1.8105cm,or1.8km.Thatwouldbebilitythattheradiusoftheelectronfarlargerthanmostschools.orbitwillhaveanygivenvalue.20.LasersExplainhowalasermakesuseofSectionReviewstimulatedemissiontoproducecoherentlight.28.2TheQuantumModelofWhenaphotonstrikesanatominthetheAtomexcitedstate,itstimulatestheexcitedpages760–765atomtoemitaphotonofthesameener-page765gyandinstepwiththeincidentphoton.16.LasersWhichofthelasersinTable28-1Theincidentphotonremainsunchangedemitsthereddestlight(visiblelightwiththeandthesetwophotonsinturnstrikelongestwavelength)?Whichofthelasersotherexcitedatoms,producingmoreemitbluelight?Whichofthelasersemitandmorein-step,coherentlight.beamsthatarenotvisibletothehumaneye?21.LaserLightWhatarethefourcharacteris-galliumaluminumarsenidelaserinred;ticsoflaserlightthatmakeituseful?argonionandindiumgalliumnitrideinblue.Krypton-fluorideexcimer,nitro-concentrated,highpower;directional;gen,galliumarsenide,neodymium,singlewavelength;coherentlightandcarbondioxidearenotvisibleto22.CriticalThinkingSupposethatanelec-thehumaneye.troncloudweretogetsosmallthatthe17.PumpingAtomsExplainwhethergreenatomwasalmostthesizeofthenucleus.lightcouldbeusedtopumparedlaser.UsetheHeisenberguncertaintyprincipletoWhycouldredlightnotbeusedtopumpexplainwhythiswouldtakeatremendousagreenlaser?amountofenergy.Yes.RedphotonshavelessenergythanThesmallertheelectroncloud,themoregreenones.Redphotonsdonothavepreciselyweknowthepositionoftheenoughenergytoputtheatomsinelectrons.Ifaparticle’spositioniswellenergylevelshighenoughtoenableknown,itsmomentummustbeuncer-themtoemitgreenphotons.tain.Theuncertaintyofthemomentumcanbelargeonlyifmomentumitselfis18.BohrModelLimitationsAlthoughitwaslarge.Therefore,thekineticenergyofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.abletoaccuratelypredictthebehavioroftheelectronalsomustbelarge,andithydrogen,inwhatwaysdidBohr’satomictakeslotsofenergytodothis.modelhaveseriousshortcomings?TheBohrmodelcouldnotpredictthebehaviorofanyotheratombesideshydrogen.ThemodelalsocouldnotPhysics:PrinciplesandProblemsSolutionsManual547

551Chapter28continuedAstheelectronsundergocentripetalChapterAssessmentacceleration,theywouldloseenergyConceptMappingandspiralintothenucleus.Inaddition,page770allatomsshouldradiateatallwave-23.Completethefollowingconceptmapusinglengths,notdiscretewavelengths.theseterms:energylevels,fixedelectronradii,27.AnalyzeandcritiquetheBohrmodeloftheBohrmodel,photonemissionandabsorption,atom.WhatthreeassumptionsdidBohrenergy-leveldifference.makeindevelopinghismodel?(28.1)stationarystates(quantizedenergyBohrmodellevels),atomemitsorabsorbsradiationhasonlywhenitchangesstates,angularmomentumisquantizedresults28.Gas-DischargeTubesExplainhowlinethatincludespectrafromgas-dischargetubesarepro-duced.(28.1)fixedelectronenergylevelphotonemissionradiiandabsorptionEnergyissuppliedtothegas,whichwithenergycausestheelectronstoexciteandmoveequaltotohigherenergylevels.Theelectronsenergy-leveldifferencethengiveoffthedifferenceinenergybetweenenergylevelsastheydropbackdowntoalessexcitedstate.TheMasteringConceptsenergydifferencesbetweenlevelscorre-page770spondstospectrallines.24.DescribehowRutherforddeterminedthatthepositivechargeinanatomisconcen-29.HowdoestheBohrmodelaccountforthetratedinatinyregion,ratherthanspreadspectraemittedbyatoms?(28.1)throughouttheatom.(28.1)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.PhotonwavelengthsaredeterminedbyHedirectedabeamofchargedparti-thedifferenceinenergiesofallowedclesatathinmetalsheetandmeasuredlevelsaselectronsjumpinwardtothenumberofparticlesdeflectedatvar-stationarystates.iousangles.Thesmallbutsignificantnumberdepletedatwideanglesindi-30.Explainwhylinespectraproducedbycatesaconcentratednucleus.hydrogengas-dischargetubesaredifferentfromthoseproducedbyheliumgas-dis-25.HowdoestheBohrmodelexplainwhythechargetubes.(28.1)absorptionspectrumofhydrogencontainsEachelementhasadifferentconfigura-exactlythesamefrequenciesasitsemissiontionofelectronsandenergylevels.spectrum?(28.1)Bohrsaidtheenergyofanemitted31.LasersAlaboratorylaserhasapowerofphotonoranabsorbedphotonisequalonly0.8mW(8104W).Whydoesittothechangeinenergyoftheatom,seemmorepowerfulthanthelightofawhichcanhaveonlyspecificvalues.100-Wlamp?(28.2)Lightisconcentratedintoanarrow26.Reviewtheplanetarymodeloftheatom.beam,ratherthanbeingspreadoveraWhataresomeoftheproblemswithawidearea.planetarymodeloftheatom?(28.1)548SolutionsManualPhysics:PrinciplesandProblems

552Chapter28continued32.AdevicesimilartoalaserthatemitsEnergyLevelDiagramforaMercuryAtommicrowaveradiationiscalledamaser.Whatwordslikelymakeupthisacronym?(28.2)Ionization(0.00)10.38E9(1.56)8.82MicrowaveAmplificationbyStimulatedE8(1.57)8.81EmissionofRadiationE7(2.48)7.90E6(2.68)7.7033.WhatpropertiesoflaserlightledtoitsuseE5(3.71)6.67inlightshows?(28.2)E4(4.95)5.43E3(5.52)4.86Lasersaredirectionalandsingle,pureE2(5.74)4.64colors.Energylevel(eV)ApplyingConceptsEnergyabovegroundstate(eV)page770E1(10.38)0.0034.Asthecomplexityofenergylevelschanges■ Figure28-21fromatomtoatom,whatdoyouthinkhappenstothespectrathattheyproduce?No,ittakes5.43eVtoraisetheelectrontotheEGenerally,thespectrabecomemore4leveland6.67eVtoE5.Theatomcanabsorbonlyphotonsthatcomplex.haveexactlytherightenergy.35.NorthernLightsThenorthernlightsare39.Acertainatomhasfourenergylevels,withcausedbyhigh-energyparticlesfromtheESunstrikingatomshighinEarth’satmos-4beingthehighestandE1beingthelowest.Iftheatomcanmaketransitionsbetweenphere.Ifyoulookedattheselightsthroughanytwolevels,howmanyspectrallinescanaspectrometer,wouldyouseeacontinuoustheatomemit?Whichtransitionproducesorlinespectrum?Explain.thephotonwiththehighestenergy?Linespectrum;thelightcomesfromSixlinesarepossible.Egasmadeofspecificelements.4→E1hasthelargestphotonenergy.36.IfwhitelightwereemittedfromEarth’ssur-40.Aphotonisemittedwhenanelectroninanfaceandobservedbysomeoneinspace,excitedhydrogenatomdropsthroughener-woulditsspectrumappeartobecontinu-gylevels.Whatisthemaximumenergythatous?Explain.thephotoncanhave?IfthissameamountNo,aswhitelightpassedthroughofenergyweregiventotheatomintheEarth’satmosphere,certainenergiesgroundstate,whatwouldhappen?wouldbeabsorbedbythegasescom-Themaximumenergyis13.6eV.Thisposingtheatmosphere.Itsspectrum,isalsotheionizationenergyforhydro-therefore,wouldhaveblacklinesonit.gen.Theelectronwouldhaveenoughenergytoleavethenucleus.37.Ismoneyagoodexampleofquantization?Iswater?Explain.41.ComparethequantummechanicaltheoryYes.Moneycomesinonlycertaindis-oftheatomwiththeBohrmodel.cretevalues.No,waterseemstocomeTheBohrmodelhasfixedorbitalradii.inanypossiblequantity.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thepresentmodelgivesaprobabilityoffindinganelectronatalocation.The38.RefertoFigure28-21.Aphotonwithener-Bohrmodelallowsforcalculationofgyof6.2eVentersamercuryatomintheonlyhydrogenatoms.Thepresentgroundstate.Willitbeabsorbedbythemodelcanbeusedforallelements.atom?Explain.Physics:PrinciplesandProblemsSolutionsManual549

553Chapter28continued42.Givenared,green,andbluelaser,which45.Acalciumatomisinanexcitedstateattheproducesphotonswiththehighestenergy?E6energylevel.HowmuchenergyisBluelighthasahigherfrequencyandreleasedwhentheatomdropsdowntothetherefore,higherenergy.E2energylevel?RefertoFigure28-22.E6E25.16eV2.93eV2.23eVMasteringProblems46.Aphotonoforangelightwithawavelength28.1TheBohrModeloftheAtomof6.00102nmentersacalciumatominpage771theE6excitedstateandionizestheatom.Level1Whatkineticenergywilltheelectronhave43.Acalciumatomdropsfrom5.16eVaboveasitisejectedfromtheatom?thegroundstateto2.93eVabovethehcgroundstate.WhatisthewavelengthoftheEphotonemitted?(6.631034J/Hz)(3.00108m/s)Ehfhc7m6.00101eVhc3.314J3.314J1.601019JE2.07eV1240eVnm5.16eV2.93eVEnergyneededtoionize6.08eV556nmE65.16eV0.92eV44.Acalciumatominanexcitedstate,E2,hasanenergylevel2.93eVabovethegroundPhotonenergyionizationenergystate.Aphotonofenergy1.20eVstrikesthekineticenergycalciumatomandisabsorbedbyit.To2.07eV0.92eV1.15eVwhatenergylevelisthecalciumatomraised?RefertoFigure28-22.Level247.CalculatetheenergyassociatedwiththeE7Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.EnergyLevelDiagramforaCalciumAtomandtheE2energylevelsofthehydrogen6.08(Ionization)atom.6E9E105.58E7E85.325.4915E6E713.6eVn2E55.165.18E44.62E34.5513.6eV10.278eV44.137213E22.93E213.6eVn21213.6eV3.40eV221Energyabovegroundstate(eV)48.CalculatethedifferenceinenergylevelsinE1thepreviousproblem.01■ Figure28-22E713.6eVn22.93eV1.20eV4.13eVE3113.6eV20.278eV7550SolutionsManualPhysics:PrinciplesandProblems

554Chapter28continued152.Usingthevaluescalculatedinproblem51,E213.6eVn2calculatethefollowingenergydifferences.1a.E13.6eV23.40eV6E52(0.378eV)(0.544eV)0.166eVE7E20.278eV(3.40eV)b.E6E33.12eV(0.378eV)(1.51eV)1.13eVRefertoFigure28-21forProblems49and50.c.E4E249.Amercuryatomisinanexcitedstateatthe(0.850eV)(3.40eV)2.55eVEd.E6energylevel.5E2a.Howmuchenergywouldbeneededto(0.544eV)(3.40eV)2.86eVionizetheatom?e.E5E3E67.70eV(0.544eV)(1.51eV)0.97eV10.38eV7.70eV2.68eV53.Usethevaluesfromproblem52todeter-b.Howmuchenergywouldbereleasedminethefrequenciesofthephotonsemit-iftheatomdroppeddowntothetedwhenanelectroninahydrogenatomE2energylevelinstead?makestheenergylevelchangeslisted.E24.64eVE7.70eV4.64eV3.06eVa.Ehf,sofhhfE50.Amercuryatominanexcitedstatehasan6E50.166eVenergyof4.95eV.Itabsorbsaphoton(0.166eV)(1.601019J/eV)fthatraisesittothenext-higherenergylevel.6.631034J/HzWhatistheenergyandthefrequencyofthe13Hz4.0110photon?b.hfEE6E31.13eV5E43.71eV(4.95eV)(1.13eV)(1.601019J/eV)1.24eVf6.631034J/HzEhf2.731014HzEfhc.hfE4E22.55eV1.601019J1.24eVeV(2.55eV)(1.601019J/eV)14Hzf6.631034Js2.99106.631034J/Hz6.151014Hz51.Whatenergiesareassociatedwithahydro-d.hfEgenatom’senergylevelsofE6E32.86eV2,E3,E4,E5,andE6?(2.86eV)(1.601019J/eV)f6.631034J/Hz13.6eV13.6eVE2n2(2)23.40eV6.901014Hz13.6eVe.hfEE6E30.97eV3(3)21.51eV(0.97eV)(1.601019J/eV)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.E13.6eVf6.631034J/Hz4(4)20.850eV2.31014Hz13.6eVE5(5)20.544eV54.DeterminethewavelengthsofthephotonsE13.6eVhavingthefrequenciesthatyoucalculated6(6)20.378eVinproblem53.Physics:PrinciplesandProblemsSolutionsManual551

555Chapter28continueda.cfc3.00108m/sf4.011013Hz7.48106m7480nmc3.00108m/sb.f2.731014Hz1.10106m1.10103nmc3.00108m/sc.f6.151014Hz4.88107m488nmc3.00108m/sd.f6.901014Hz4.35106m435nmc3.00108m/se.f2.31014Hz1.3106m1.3103nmLevel355.Ahydrogenatomemitsaphotonwithawavelengthof94.3nmwhenitsfallstothegroundstate.Fromwhatenergyleveldidtheelectronfall?cfc3.00108m/s9.43108m3.181015HzE34J/Hz)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.nE1(6.62610(3.181015Hz)2.111018JE2.111018JEnE1E2.171018J(2.111018J)61020J2.171018J20J610n2n236n656.Forahydrogenatominthen3Bohrorbital,findthefollowing.a.theradiusoftheorbitalh2n2r42Kmq2(6.631034Js)2(3)242(9.00109Nm2/C2)(9.111031kg)(1.601019C)24.771010m552SolutionsManualPhysics:PrinciplesandProblems

556Chapter28continuedb.theelectricforceactingbetweenthehc1240eVnmprotonandtheelectronE2.90eV2428nmKqFr2b.Inwhatpartofthespectrumisthislight?(9.00109Nm2/C2)(1.601019C)2blue(4.771010m)2Level21.01109N59.Acarbon-dioxidelaseremitsveryhigh-c.thecentripetalaccelerationofthepowerinfraredradiation.WhatistheenergyelectrondifferenceineVbetweenthetwolasingFmaenergylevels?ConsultTable28-1.F1.01109Ncfam9.111031kgc3.00108m/s1.111021m/s2f10,600109md.theorbitalspeedoftheelectron2.831013Hz(ComparethisspeedwiththespeedEhfoflight.)2(6.631034J/Hz)(2.831013Hz)var1.601019J/eV0.117eVvar(1.111021m/s2)(4.771010m)60.Thepowerinalaserbeamisequaltotheenergyofeachphotontimesthenumberof7.28105m/s,or0.24%ofcphotonspersecondthatareemitted.a.Ifyouwantalaserat840nmtohave28.2TheQuantumModeloftheAtomthesamepowerasoneat427nm,howpage771manytimesmorephotonspersecondLevel1areneeded?hc57.CDPlayersGalliumarsenidelasersareSinceEhf,theratioofenergycommonlyusedinCDplayers.Ifsucha427ineachphotonis0.508.840laseremitsat840nm,whatisthedifferenceTherefore,theratioofnumberofineVbetweenthetwolasingenergylevels?1photonspersecondis1.97cf0.508b.Findthenumberofphotonspersecondc3.00108m/sf9ina5.0-mW840-nmlaser.84010m14HzP(photons/s)(E/photon)nEso3.5710nP/E.EhfCalculatetheenergyofthephoton(6.631034J/Hz)(3.571014Hz)injoules.1.601019J/eVhcE1.5eV(1240eVnm)(1.601019J/eV)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.58.AGaInNilaserlasesbetweenenergylevels840nmthatareseparatedby2.90eV.19J;2.410a.Whatwavelengthoflightdoesitemit?5.0103J/shcSo,n2.41019J/photonE2.11016photons/sPhysics:PrinciplesandProblemsSolutionsManual553

557Chapter28continuedLevel361.HeNeLasersTheHeNelasersusedinmanyclassroomscanbemadetolaseatthreewavelengths:632.8nm,543.4nm,and1152.3nm.a.Findthedifferenceinenergybetweenthetwostatesinvolvedinthegenera-tionofeachwavelength.hc1240eVnmE,soE;substitutingthethreevaluesofgives1.96eV,2.28eV,and1.08eVb.Identifythecolorofeachwavelength.red,green,andinfrared,respectivelyMixedReviewpage772Level162.Aphotonwithanenergyof14.0eVentersahydrogenatominthegroundstateandionizesit.Withwhatkineticenergywilltheelectronbeejectedfromtheatom?Ittakes13.6eVtoionizetheatom,so14.0eV13.6eV0.4eVkineticenergy.63.CalculatetheradiusoftheorbitalassociatedwiththeenergylevelsE5andE6ofthehydrogenatom.h2n2(6.631034Js)2(5)2r542Kmq242(9.00109Nm2/C2)(9.111031kg)(1.601019C)21.33109m(6.631034Js)2(6)2r642(9.00109Nm2/C2)(9.111031kg)(1.601019C)2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.91109mLevel264.Ahydrogenatomisinthen2level.a.Ifaphotonwithawavelengthof332nmstrikestheatom,showthattheatomwillbeionized.13.6eV13.6eVE2n2(2)23.40eV,so3.40eVneededtoionizefromthislevel.hcEhf(6.631034J/Hz)(3.00108m/s)332109m5.991019J3.74eVYes,theatomisionized.554SolutionsManualPhysics:PrinciplesandProblems

558Chapter28continuedb.Whentheatomisionized,assumethatThinkingCriticallytheelectronreceivestheexcessenergypage772fromtheionization.Whatwillbethe67.ApplyConceptsTheresultofprojectingthekineticenergyoftheelectroninjoules?spectrumofahigh-pressuremercuryvapor3.74eV3.40eV0.340eVlampontoawallinadarkroomisshownin5.41020JFigure28-23.WhatarethedifferencesinenergylevelsforeachofthethreevisibleLevel3lines?65.Abeamofelectronsisdirectedontoasam-pleofatomichydrogengas.Whatminimumenergyoftheelectronsisneededforthehydrogenatomstoemittheredlightpro-ducedwhentheatomgoesfromthen3tothen2state?Theremustbeenoughenergytotransi-436nm546nm579nmtionastablehydrogenatomtothe■ Figure28-23n3state.436nm(2.84eV)fromE6toE3sinceEE1240eVnm3E1284eV;seeFig28-21to436nm(13.6eV)11findenergylevels.3212546nm(2.27eV)fromE6toE48(13.6eV)9579nm(2.14eV)fromE8toE512.1eV68.InterpretScientificIllustrationsAftertheemissionofthevisiblephotonsdescribedin66.Themostprecisespectroscopyexperimentsproblem67,themercuryatomcontinuestouse“two-photon”techniques.Twophotonsemitphotonsuntilitreachesthegroundwithidenticalwavelengthsaredirectedatstate.FromaninspectionofFigure28-21,thetargetatomsfromoppositedirections.determinewhetherornotanyofthesepho-Eachphotonhashalftheenergyneededtotonswouldbevisible.Explain.excitetheatomsfromthegroundstatetoNo.Thethreehighestenergylinesleavethedesiredenergylevel.Whatlaserwave-theatominstatesatleast4.64eVabovelengthwouldbeneededtomakeaprecisethegroundstate.Aphotonwiththisstudyoftheenergydifferencebetweenn1energyhasawavelengthof267nminandn2inhydrogen?theultraviolet.ThechangefromE4toE2EE2E1involvesanenergychangeofonly0.79eV,resultinginlightwithawavelength11(13.6eV)22of1570nmintheinfrared.21(13.6eV)369.AnaylzeandConcludeApositroniumatom4consistsofanelectronanditsantimatterrela-10.2eVtive,thepositron,boundtogether.Althoughthelifetimeofthis“atom”isveryshort—ontheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Foreachlaser,averageitlivesone-seventhofamicrosecond—itsenergylevelscanbemeasured.TheBohrhc1240eVnm243nmmodelcanbeusedtocalculateenergieswithE10.2eV22themassoftheelectronreplacedbyone-halfitsmass.Describehowtheradiioftheorbitsandtheenergyofeachlevelwouldbeaffected.Physics:PrinciplesandProblemsSolutionsManual555

559Chapter28continuedWhatwouldbethewavelengthoftheEatthepositionofthetestcharge?2toE1transition?(Chapter21)TheradiiwouldbetwiceaslargeF0.027N4N/CE9.010becausemappearsinthedenominatorq3.00107Coftheequation.Theenergieswouldbe73.Atechnicianneedsa4-resistorbutonlyhashalfaslargebecausemappearsinthe1-resistorsofthatvalue.Isthereawaytonumerator.Therefore,thewavelengthscombinewhatshehas?Explain.(Chapter23)wouldbetwiceaslarge.Thus,thelightemittedfromEYes.Putfour1resistorsinseries.2toE1wouldbe(2)(121nm)242nm.RTR1R2R3R4WritinginPhysics74.A1.0-m–longwireismovedatrightanglespage772toEarth’smagneticfieldwherethemagnetic70.Doresearchonthehistoryofmodelsoftheinductionis5.0105Tataspeedofatom.Brieflydescribeeachmodelandiden-4.0m/s.WhatistheEMFinducedinthetifyitsstrengthsandweaknesses.wire?(Chapter25)Studentsshoulddescribethe“raisinEMFBLvpudding”model,aclassicalplanetary(5.0105T)(1.0m)(4.0m/s)model,theBohrmodel,andthequantum2.0104V0.20mVmodel.Thefirstexplainshowatomscanhaveelectronsandmass,butfailsto75.TheelectronsinabeammoveatdescribetheresultsofRutherford’s2.8108m/sinanelectricfieldofexperiments.Theplanetaryexplains1.4104N/C.WhatvaluemustthemagneticelectronsandRutherford’sresults,butisfieldhaveiftheelectronspassthroughtheunstableandwouldcollapseinabout1crossedfieldsundeflected?(Chapter26)ns.Bohr’sexplainsknownspectraandEfitsRutherford’snuclearmodel,buthasvBunexplainedassumptionsandfailsthe4Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.E1.410N/Cuncertaintyprinciple,aswellasbeingBv2.8108m/sunabletodescribeatomswithmore5.0105T5.0101Tthanoneelectron.Thequantummodelcanexplainallknownfacts,butishard76.ConsiderthemodificationsthatJ.J.tovisualizeandrequirescomputerstoThomsonwouldneedtomaketohiscath-solvetheequations.ode-raytubesothatitcouldacceleratepro-tons(ratherthanelectrons),thenanswer71.Greenlaserpointersemitlightwithawave-thefollowingquestions.(Chapter26)lengthof532nm.Doresearchonthetypeoflaserusedinthistypeofpointeranda.Toselectparticlesofthesamevelocity,describeitsoperation.IndicatewhetherwouldtheratioE/Bhavetobechanged?thelaserispulsedorcontinuous.Explain.EItusesapulsedNdlaserat1064nm.No;V,soratioissameforaBTheIRisputintoa“frequencydoubling”givenV.crystal.Lightwithhalfthatwavelength,b.Forthedeflectioncausedbythemag-or532nm,results.neticfieldalonetoremainthesame,wouldtheBfieldhavetobemadeCumulativeReviewsmallerorlarger?Explain.page7722mv72.Theforceonatestchargeof3.00107CFormagneticfieldonly,Bqvris0.027N.Whatistheelectricfieldstrength556SolutionsManualPhysics:PrinciplesandProblems

560Chapter28continuedmvandrforabiggermass,BqBmustbebiggertokeepvconstant.77.Thestoppingpotentialneededtoreturnalltheelectronsejectedfromametalis7.3V.Whatisthemaximumkineticenergyoftheelectronsinjoules?(Chapter27)1.601019JKE(7.3eV)1eV1.21018JChallengeProblempage759AlthoughtheBohratomicmodelaccuratelyexplainedthebehaviorofahydrogenatom,itwasunabletoexplainthebehaviorofanyotheratom.VerifythelimitationsoftheBohrmodelbyanalyzinganelectrontransitioninaneonatom.Unlikeahydrogenatom,aneonatomhastenelectrons.Oneoftheseelectronsmakesatransi-tionbetweenthen5andthen3energystates,emittingaphotonintheprocess.1.Assumingthattheneonatom’selectroncanbetreatedasanelectroninahydrogenatom,whatphotonenergydoestheBohrmodelpredict?11EEiEf(13.6eV)52320.967eV2.Assumingthattheneonatom’selectroncanbetreatedasanelectroninahydrogenatom,whatphotonwavelengthdoestheBohrmodelpredict?hc1240eVnm1280nmE0.967eV3.Theactualwavelengthofthephotonemit-tedduringthetransitionis632.8nm.WhatisthepercenterroroftheBohrmodel’spre-dictionofphotonwavelength?PercenterrorCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.AcceptedvaluePredictedvalue100Acceptedvalue632.8nm1280nm100103%6328nmThecalculatedwavelengthisroughlytwiceaslargeastheactualwavelength.Physics:PrinciplesandProblemsSolutionsManual557

561

562CHAPTER29Solid-StateElectronicsPracticeProblems29.1ConductioninSolidspages775–783page7781.Zinc,withadensityof7.13g/cm3andanatomicmassof65.37g/mol,hastwofreeelectronsperatom.Howmanyfreeelectronsarethereineachcubiccentimeterofzinc?/cm32freee6.021023atoms1mol7.13gfreeeatommol65.37g3cm1.311023freee/cm32.Silverhas1freeelectronperatom.UseAppendixDanddeterminethenumberoffreeelectronsin1cm3ofsilver./cm31freee6.021023atoms1mol10.49gfreeeatommol107.87g3cm5.851022freee/cm33.Goldhas1freeelectronperatom.UseAppendixDanddeterminethenumberoffreeelectronsin1cm3ofgold./cm31freee6.021023atoms1mol19.32gfreeeatommol196.97g3cm5.901022freee/cm34.Aluminumhas3freeelectronsperatom.UseAppendixDanddeterminethenumberoffreeelectronsin1cm3ofaluminum./cm33e6.021023atoms1mol2.70gfreeeatommol26.982g3cm1.811023freee/cm35.ThetipoftheWashingtonMonumentwasmadeof2835gofaluminumbecauseitwasarareandcostlymetalinthe1800s.Useproblem4anddeterminethenumberoffreeelectronsinthetipoftheWashingtonMonument.freee(1.811023freee/cm3)2835g2.70g/cm31.901026freeeinthetipCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.page7806.Inpuregermanium,whichhasadensityof5.23g/cm3andanatomicmassof72.6g/mol,thereare2.251013freeelectrons/cm3atroomtemperature.Howmanyfreeelectronsarethereperatom?Physics:PrinciplesandProblemsSolutionsManual559

563Chapter29continued/atom1mol72.6gcm32.251013freeefreee6.021023atoms1mol5.23gcm35.191010freee/atom7.At200.0K,siliconhas1.89105freeelectrons/cm3.Howmanyfreeelectronsarethereperatomatthistemperature?WhatdoesthistemperaturerepresentontheCelsiusscale?/atom1mol28.09gcm31.89105freeefreee6.021023atoms1mol2.33gcm33.781018freee/atomTKTC273°TCTK273°200.0°273°73°C8.At100.0K,siliconhas9.231010freeelectrons/cm3.Howmanyfreeelectronsarethereperatomatthistemperature?WhatdoesthistemperaturerepresentontheCelsiusscale?/atom1mol28.09gcm39.231010freeefreee6.021023atoms1mol2.33gcm31.851032freee/atomTKTC273°TCTK273°100.0°273°Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.173°C9.At200.0K,germaniumhas1.161010freeelectrons/cm3.Howmanyfreeelectronsarethereperatomatthistemperature?/atom1mol72.6gcm31.161010freeefreee6.021023atoms1mol5.23gcm32.671013freee/atom10.At100.0K,germaniumhas3.47freeelectrons/cm3.Howmanyfreeelectronsarethereperatomatthistemperature?/atom1mol72.6gcm33.47freeefreee6.021023atoms1mol5.23gcm38.001023freee/atompage78311.Ifyouwantedtohave1104asmanyelectronsfromarsenicdopingasthermallyfreeelectronsinsiliconatroomtemperature,howmanyarsenicatomsshouldtherebepersiliconatom?FromExampleProblem3youknowthatthereare4.991022Siatoms/cm3,1.451010freee/cm3inSi,and1freee/Asatom.560SolutionsManualPhysics:PrinciplesandProblems

564Chapter29continuedefromAs(1104)(freeefromSi)However,theratioofatomsisneeded,notelectrons.efromAs(1104)(freeefromSi)Asatomsfreee/atomAsfreee/atomAsfreee/cm3SifreeefromSi(Siatoms)Siatoms/cm3SubstitutingintotheexpressionforAsatomsyields4)(Siatoms)freee/cm3Si(1103Siatoms/cmAsatoms/atomAsfreee4)freee/cm3Si(1103AsatomsSiatoms/cm/atomAsSiatomsfreee4)1.451010(1104.99102212.9110912.Ifyouwantedtohave5103asmanyelectronsfromarsenicdopingasthermallyfreeelectronsinthegermaniumsemiconductordescribedinproblem6,howmanyarsenicatomsshouldtherebepergermaniumatom?Usingthesolutioninproblem11asagoby3)freee/cm3Ge(5103AsatomsGeatoms/cmfreee/atomAsGeatomsFromproblem6,freee/cm3Ge2.25103FromExampleProblem3,freee/atomAs136.021023atoms1mol5.23g22atoms/cm3Geatoms/cm1mol72.6g34.3410cm3)2.251013(510Asatoms4.341022Geatoms12.5910613.Germaniumat400.0K,has1.131015thermallyliberatedcarriers/cm3.Ifitisdopedwith1Asatomper1millionGeatoms,whatistheratioofdopedcarrierstothermalcarriers?Usethesolutiontoproblem12asastartingpoint.dopedefreee/cm3GeGee3AsatomsGeatoms/cmGeatomsfreee/atomAsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.dopedeAsatomsGeatoms/cm3/atomAs)GeeGeatomsfreee/cm3Ge(freee14.34102211061.131015(1)38.4Physics:PrinciplesandProblemsSolutionsManual561

565Chapter29continued14.Siliconat400.0K,has4.541012thermallyliberatedcarriers/cm3.Ifitisdopedwith1Asatomper1millionSi,whatistheratioofdopedcarrierstothermalcarriers?Usingthesolutionproblem13asagobydopedeAsatomsSiatoms/cm3/atomAs)SieSiatomsfreee/cm3Si(freee14.99102211064.541012(1)1.1010415.Basedonproblem14,drawaconclusionaboutthebehaviorofgermaniumdevicesascomparedtosilicondevicesattemperaturesinexcessoftheboilingpointofwater.Germaniumdevicesdonotworkwellatsuchtemperaturesbecausetheratioofdopedcarrierstothermalcarriersissmallenoughthattemperaturehastoomuchinfluenceonconductivity.Siliconismuchbetter.SectionReview29.1ConductioninSolidspages775–783page78316.CarrierMobilityInwhichtypeofmaterial,aconductor,asemiconductor,oraninsulator,areelectronsmostlikelytoremainwiththesameatom?insulatorCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.17.SemiconductorsIfthetemperatureincreases,thenumberoffreeelectronsinanintrinsicsemiconductorincreases.Forexample,raisingthetemperatureby8°Cdoublesthenumberoffreeelectronsinsilicon.Isitmorelikelythatanintrinsicsemiconductororadopedsemiconductorwillhaveaconductivitythatdependsontemperature?Explain.Anintrinsiconebecauseallitsconductionisfromthermallyfreedelec-trons,whereasthedopedsemiconductordependsonthechargesfromthedopants,whichdependlittleontemperature.18.InsulatororConductor?Silicondioxideiswidelyusedinthemanufactureofsolid-statedevices.Itsenergy-banddiagramshowsagapof9eVbetweenthevalencebandandtheconductionband.Isitmoreusefulasaninsulatororaconductor?insulator19.ConductororInsulator?Magnesiumoxidehasaforbiddengapof8eV.Isthismaterialaconductor,aninsulator,orasemiconductor?insulator562SolutionsManualPhysics:PrinciplesandProblems

566Chapter29continued20.IntrinsicandExtrinsicSemiconductors24.DescribehowthediodesinthepreviousYouaredesigninganintegratedcircuitproblemshouldbeconnected.usingasinglecrystalofsilicon.YouwanttoTheanodeofoneconnectstothecath-havearegionwithrelativelygoodinsulat-odeoftheotherandthentheuncon-ingproperties.Shouldyoudopethisregionnectedanodemustbeconnectedtotheorleaveitasanintrinsicsemiconductor?positivesideofthecircuit.Leaveit.25.Describewhatwouldhappeninproblem21.CriticalThinkingSiliconproducesadou-23ifthediodeswereconnectedinseriesblingofthermallyliberatedcarriersforbutwithimproperpolarity.every8°Cincreaseintemperature,andger-Itwouldbeimpossibletoobtain2.5mAmaniumproducesadoublingofthermallyofcurrentwithanyreasonablepowerliberatedcarriersforevery13°Cincrease.Itsupplyvoltagebecauseoneofthewouldseemthatgermaniumwouldbediodeswouldbereverse-biased.superiorforhigh-temperatureapplications,buttheoppositeistrue.Explain.26.AgermaniumdiodehasavoltagedropofEventhoughrateofchangeforthermal0.40Vwhen12mApassesthroughit.Ifacarrierproductionisgreaterforsilicon,470-resistorisusedinseries,whatbatteryatanygiventemperaturesiliconshowsvoltageisneeded?farfewerthermallyliberatedcarriers.6.0VVbIRVdPracticeProblems(0.012A)(470)0.40V6.0V29.2ElectronicDevicespages784–789page786SectionReview22.Whatbatteryvoltagewouldbeneededtoproduceacurrentof2.5mAinthediodein29.2ElectronicDevicesExampleProblem4?pages784–7891.7Vpage78927.TransistorCircuitTheemittercurrentinaUsingFig.29-10,thediodehastransistorcircuitisalwaysequaltothesumVd0.50Vat2.5mAofthebasecurrentandthecollectorcur-VbIRVdrent:IEIBIC.Ifthecurrentgainfromthebasetothecollectoris95,whatisthe(0.0025A)(470)0.50Vratioofemittercurrenttobasecurrent?1.7VIGainC95IB23.Whatbatteryvoltagewouldbeneededtoproduceacurrentof2.5mAifanotherIEIBICidenticaldiodewereaddedinserieswithDividebothsidesbyIthediodeinExampleProblem4?B.IIVE1C19596bIRVdVdCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.IIBB(0.0025A)(470)0.50V0.50V96to12.2VPhysics:PrinciplesandProblemsSolutionsManual563

567Chapter29continued28.DiodeVoltageDropIfthediodecharac-ChapterAssessmentterizedinFigure29-10isforward-biasedbyabatteryandaseriesresistorsothatthereisConceptMappingmorethan10mAofcurrent,thevoltagepage794dropisalwaysabout0.70V.Assumethat33.Completetheconceptmapusingthefol-thebatteryvoltageisincreasedby1V.lowingterms:transistor,silicondiode,emitsa.Byhowmuchdoesthevoltageacrosslight,conductsbothways.thediodeorthevoltageacrosstheresis-torincrease?CircuitComponentsBecausethevoltageacrossthediodeisalways0.70V,thevoltageacrosstheresistorincreasesby1V.coppersiliconLEDtransistorb.Byhowmuchdoesthecurrentthroughwirediodetheresistorincrease?1VThecurrentincreasesbyI.conductsconductsemitsRamplifiesbothwaysonewaylight29.DiodeResistanceComparetheresistanceofapn-junctiondiodewhenitisforward-MasteringConceptsbiasedandwhenitisreverse-biased.page794Itconductsmuchbetterwhenforward-34.Howdotheenergylevelsinacrystalofanbiased,soitsresistanceismuchsmall-elementdifferfromtheenergylevelsinaerwhenforward-biasedthanwhensingleatomofthatelement?(29.1)reverse-biased.Theenergylevelsofasingleatomhavediscreteanduniquevalues.Theenergy30.DiodePolarityInalight-emittingdiode,levelsinacrystalhaveasmallrangewhichterminalshouldbeconnectedtothearoundthevaluesfoundinasinglep-endtomakethediodelight?atom.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.TheLEDmustbeforward-biased,sothepositiveterminalmustbeconnectedto35.Whydoesheatingasemiconductorincreasethep-end.itsconductivity?(29.1)Ahighertemperaturegiveselectrons31.CurrentGainThebasecurrentinatransis-additionalenergy,permittingmoreelec-torcircuitmeasures55Aandthecollectortronstoreachtheconductiveband.currentmeasures6.6mA.Whatisthecur-rentgainfrombasetocollector?36.Whatisthemaincurrentcarrierinap-typeIC6.6mAsemiconductor?(29.1)Gain120I0.055mABpositively-chargedholes32.CriticalThinkingCouldyoureplacean37.Anohmmeterisaninstrumentthatplacesanpn-transistorwithtwoseparatediodescon-potentialdifferenceacrossadevicetobenectedbytheirp-terminals?Explain.tested,measuresthecurrent,anddisplaysNo,thep-regionofannpn-transistortheresistanceofthedevice.Ifyouconnectmustbethinenoughtoallowelectronsanohmmeteracrossadiode,willthecur-topassthroughthebasetothecollector.rentyoumeasuredependonwhichendofthediodewasconnectedtothepositiveter-minaloftheohmmeter?Explain.(29.2)Yes,onewayyouforward-biasthediode;theotherwayyoureverse-biasit.564SolutionsManualPhysics:PrinciplesandProblems

568Chapter29continued38.Whatisthesignificanceofthearrowheadat43.Theresistanceofgraphitedecreasesastem-theemitterinatransistorcircuitsymbol?peraturerises.Doesgraphiteconductelec-(29.2)tricitymorelikecopperormorelikesiliconThearrowheadshowsthedirectionofdoes?theconventionalcurrent.morelikeSi39.Describethestructureofaforward-biased44.Whichofthefollowingmaterialswoulddiode,andexplainhowitworks.(29.2)makeabetterinsulator:onewithaforbiddenAforward-biaseddiodehasap-typegap8-eVwide,onewithaforbiddengapandann-typesemiconductorlayer,3-eVwide,oronewithnoforbiddengap?connectedtowiresoneitherendonewithan8-eVgapthroughmetalcaps.Thep-typelayerisconnectedtothepositiveterminalof45.Consideratomsofthethreematerialsinthebattery.Newholesarecreatedintheproblem44.Fromwhichmaterialwoulditp-typelayerandthoseholesmovebemostdifficulttoremoveanelectron?towardtheboundarybetweenthetwoonewithan8-eVgapsemiconductors.Newelectronsareaddedtothen-typelayer,andthose46.StatewhetherthebulbineachofthecircuitselectronsmovetowardtheboundaryofFigure29-17(a,b,andc)islighted.betweenthetwosemiconductors.Astheholesandelectronscombine,thecircuitiscompletedandthereiscur-rent.Thecurrentdirectionisfromthep-typesemiconductortothen-type.ApplyingConceptsabcpages794–795■Figure29-1740.Fortheenergy-banddiagramsshownincircuita:no,circuitb:no,circuitc:yesFigure29-16,whichonerepresentsamate-rialwithanextremelyhighresistance?47.InthecircuitshowninFigure29-18,statewhetherlampL1,lampL2,both,orneitherislighted.L1L2abc■Figure29-16■Figure29-1841.Fortheenergy-banddiagramsshowninFigure29-16,whichhavehalf-fullconduc-L1ison,L2isoff.tionbands?48.Usetheperiodictabletodeterminewhichaofthefollowingelementscouldbeaddedtogermaniumtomakeap-typesemicon-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.42.Fortheenergy-banddiagramsshownductor:B,C,N,P,Si,Al,Ge,Ga,As,In,Sn,inFigure29-16,whichonesrepresentorSb.semiconductors?B,Al,Ga,InbPhysics:PrinciplesandProblemsSolutionsManual565

569Chapter29continued49.DoesanohmmetershowahigherresistanceMasteringProblemswhenapn-junctiondiodeisforward-biased29.1ConductioninSolidsorreverse-biased?page795ThediodewillhavealowerresistanceLevel1whenitisforward-biased.53.Howmanyfreeelectronsexistinacubiccen-timeterofsodium?Itsdensityis0.971g/cm3,50.Iftheohmmeterinproblem49showstheitsatomicmassis22.99g/mol,andthereislowerresistance,istheohmmeterleadon1freeelectronperatom.thearrowsideofthediodeatahigherorlowerpotentialthantheleadconnectedtofreee/cm31etheotherside?atomhigherpotential,morepositive6.021023atoms0.971gmolmol322.99gcm51.Ifyoudopepuregermaniumwithgallium2.541022freee/cm3alone,doyouproducearesistor,adiode,oratransistor?Level2Youmakearesistorbecausethereisno54.Atatemperatureof0°C,thermalenergyjunction.frees1.55109e/cm3inpuresilicon.Thedensityofsiliconis2.33g/cm3,andthe52.Drawthetime-versus-amplitudewaveformatomicmassofsiliconis28.09g/mol.WhatforpointAinFigure29-19aassumingisthefractionofatomsthathavefreeaninputACwaveformasshowninelectrons?Figure29-19b.6.021023atoms2.33gmolmol328.09gcm1.55109e/cm3aA3.221013atom/eCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.29.2ElectronicDevicespage795Level1bTime55.LEDThepotentialdropacrossaglowingLEDisabout1.2V.InFigure29-20,theACvoltagepotentialdropacrosstheresistoristhedif-■Figure29-19ferencebetweenthebatteryvoltageandtheLED’spotentialdrop.Whatisthecurrentthrougheachofthefollowing?TimeVoltageR240OutputVoltageBattery1.2VV6.0VLEDPointAisnegativewithrespecttogroundandthegraphshowsthatthealternatingpolarityoftheinputwaveform■Figure29-20hasbeenrectifiedtoanegativepolarity.a.theLEDVbIRVd566SolutionsManualPhysics:PrinciplesandProblems

570Chapter29continuedVb.Determinethecollectorcurrent.IbVdRbydefinition:whenthebasecurrent6.0V1.2Viszero,soisthecollectorcurrent240c.Determinethevoltmeterreading.2.0101mA15Vb.theresistorWithnocurrentflow,thedrop2.0101mAacrossthecollectorresistoriszeroandthereisa15-Vdropacrossthe56.Jonwantstoraisethecurrentthroughthetransistor.LEDinproblem55upto3.0101mAsothatitglowsbrighter.Assumethatthe59.AssumethattheswitchshowninFigure29-potentialdropacrosstheLEDisstill1.2V.21ison,andthatthereisa0.70-VdropWhatresistorshouldbeused?acrossthebase-emitterjunctionandacur-VbVd6.0V1.2Vrentgainfrombasetocollectorof220.R160I3.0101mAa.Determinethebasecurrent.VILevel2R57.DiodeAsilicondiodewithI/Vcharacteris-3.5V0.70Vtics,asshowninFigure29-10,isconnected120,000toabatterythrougha270-resistor.The2.3105Abatteryforward-biasesthediode,andthediodecurrentis15mA.Whatisthebatteryb.Determinethecollectorcurrent.voltage?IC220IVbIRVdBIVC220IBd0.70V(fromfigure)V(220)(2.3105A)b(15mA)(270)0.70V5.1103A4.8Vc.Determinethevoltmeterreading.Level3Findthedropacrossthe150058.Assumethattheswitchshowninresistor:Figure29-21isoff.VresistorIRA(5.1103A)(1500)7.7V150015VThemeterisconnectedacrossthetransistor,120,000VbatteryVresistorVtransistorAVVmeterVtransistor3.5VVbatteryVresistor■Figure29-21Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.15V7.7Va.Determinethebasecurrent.7.3Vbyinspection,thebasecircuitisoff,soitiszeroPhysics:PrinciplesandProblemsSolutionsManual567

571Chapter29continuedMixedReviewbandtothevalenceband.FindthewidthoftheforbiddengapineVinthisdiode.pages795–7961240eVnmLevel1E2.25eV550nm60.Theforbiddengapinsiliconis1.1eV.Electromagneticwavesstrikingthesilicon64.RefertoFigure29-22.causeelectronstomovefromthevalencebandtotheconductionband.Whatisthelongestwavelengthofradiationthatcould220exciteanelectroninthisway?RecallthatE1240eVnm/.Alldiodesaresilicon.1240eVnmV1.1eV10.0V1100nminthenearinfraredA1A2Level261.SiDiodeAparticularsilicondiodeat0°C■Figure29-22showsacurrentof1.0nAwhenitisreverse-a.Determinethevoltmeterreading.biased.Whatcurrentcanbeexpectedifthe0.70Vtemperatureincreasesto104°C?Assumebyinspectionandtheapproximationthatthereverse-biasvoltageremainscon-thatasilicondiodewilldrop0.70Vstant.(Thethermalcarrierproductionofsil-whenitisforwardbiasedicondoublesforevery8°Cincreaseinb.DeterminethereadingofAtemperature.)1.104°C0ANumberof8°Cincreases8°Cbyinspection;0.70Visnotenough13toturnontwodiodesinseries.Thecurrentwilldouble13timesc.DeterminethereadingofA2.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Currentat104°C(1nA)(213)8.2AIV10.0V0.70V42mAR22062.GeDiodeAparticulargermaniumdiodeatThinkingCritically0°Cshowsacurrentof1.5Awhenitispage796reverse-biased.Whatcurrentcanbeexpected65.ApplyConceptsAcertainmotor,inifthetemperatureincreasesto104°C?Figure29-23,runsinonedirectionwithaAssumethatthereverse-biasingvoltagegivenpolarityappliedandreversesdirectionremainsconstant.(Thethermalcharge-withtheoppositepolarity.carrierproductionofgermaniumdoublesforevery13°Cincreaseintemperature.)MM104°CNumberof13°Cincreases813°CabThecurrentwilldouble8times.Currentat104°C(1.5A)(28)380AcLevel3M63.LEDAlight-emittingdiode(LED)producesgreenlightwithawavelengthof550nmwhenanelectronmovesfromtheconduction■Figure29-23568SolutionsManualPhysics:PrinciplesandProblems

572Chapter29continueda.Whichcircuit(a,b,orc)willallowthe9.0V2.5VRmotortoruninonlyonedirection?20.040A160a67.ApplyConceptsSupposethatthetwob.WhichcircuitwillcauseafusetoblowLEDsinproblem66arenowconnectediniftheincorrectpolarityisapplied?series.Ifthesamebatteryistobeusedandbacurrentof0.035Aisdesired,whatresistorc.Whichcircuitproducesthecorrectdirec-shouldbeused?tionofrotationregardlessoftheVRb(VD1VD2)appliedpolarity?Ic9.0V(1.75V2.5V)140d.Discusstheadvantagesanddisadvan-0.035Atagesofallthreecircuits.WritinginPhysicsThecircuitatahastheadvantageofsimplicity.Ithasthedisadvantageofpage796dropping0.70V,whichcanbe68.ResearchthePauliexclusionprincipleandimportantinlow-voltagecircuits.thelifeofWolfgangPauli.HighlighthisThecircuitatbhastheadvantageofoutstandingcontributionstoscience.notwasting0.70V.Ithasthedisad-Describetheapplicationoftheexclusionvantageofhavingtoreplacefuses.principletothebandtheoryofconduction,Thecircuitatchastheadvantageofespeciallyinsemiconductors.alwaysworking,regardlessofpolari-Studentanswerswillvary.Studentsty.Ithasthedisadvantageofwast-coulddiscusshiscontributionstoing1.4V.quantummechanics,thePaulimatrices,hissuggestionofaneutral,massless66.ApplyConceptsTheI/Vcharacteristicsofparticleinbetadecay(theneutrino),twoLEDsthatglowwithdifferentcolorsareandhisproofofthespin-statisticsshowninFigure29-24.Eachistobecon-theorem.nectedthrougharesistortoa9.0-Vbattery.Ifeachistoberunatacurrentof0.040A,69.Writeaone-pagepaperdiscussingtheFermiwhatresistorsshouldbechosenforeach?energylevelasitappliestoenergy-banddiagramsforsemiconductors.IncludeatLEDCurrentv.Voltageleastonedrawing.Studentanswerswillvary,butshouldindicatethattheFermienergyisthe0.04energyofthehighestoccupiedstateatabsolutezero.Itapplieswhenthereare0.02manyelectronsinasystem,andthus,Current(A)theelectronsareforcedintohigherenergylevelsduetothePauliexclusion00.5121.52.5principle.Voltage(V)■Figure29-24Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.VbIRVDVRbVDI9.0V1.75VR10.040A180Physics:PrinciplesandProblemsSolutionsManual569

573Chapter29continuedCumulativeReviewpage79670.Analphaparticle,adoublyionized(2)heliumatom,hasamassof6.71027kgandisacceleratedbyavoltageof1.0kV.Ifauniformmagneticfieldof6.5102Tismaintainedonthealphaparticle,whatwillbetheparticle’sradiusofcurvature?(Chapter26)12VmrBq1(2)(1.0103V)(6.71027kg)(6.5102T)(2)(1.601019C)0.010m71.Whatisthepotentialdifferenceneededtostopphotoelectronsthathaveamaximumkineticenergyof8.01019J?(Chapter27)KE8.01019J1eV1.601019J5.0eV,so5.0V72.CalculatetheradiusoftheorbitalassociatedwiththeenergylevelE4ofthehydrogenatom.(Chapter28)h2n2r42Kmq2(6.6271034Js)2(4)242(8.99109Nm2/C2)(9.111031kg)(1.601019C)28.491010m0.849nmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ChallengeProblempage786ApproximationsoftenareusedindiodeDiodeCurrentv.Voltagecircuitsbecausedioderesistanceisnotcon-stant.Fordiodecircuits,thefirstapproxima-14.0tionignorestheforwardvoltagedropacross12.0thediode.Thesecondapproximationtakesintoaccountatypicalvalueforthediodevolt-10.0agedrop.Athirdapproximationusesaddi-tionalinformationaboutthediode,oftenin8.0theformofagraph,asshownintheillustra-Graphicsolutiontiontotheright.Thecurveisthecharacteristic6.0Current(mA)current-volatagecurveforthediode.The4.0straightlineshowscurrent-voltageconditionsforallpossiblediodevoltagedropsfora180-2.0resistor,a1.8-Vbattery,andadiode,fromazerodiodevoltagedropand10.0mAatone0.00.20.61.01.41.8end,toa1.8-Vdrop,0.0mAattheotherend.Voltage(V)UsethediodecircuitinExampleProblem4withVb1.8V,butwithR180:570SolutionsManualPhysics:PrinciplesandProblems

574Chapter29continued1.Determinethediodecurrentusingthefirstapproximation.V1.8V1mAI1.010R1802.Determinethediodecurrentusingthesec-ondapproximationandassuminga0.70-Vdiodedrop.V1.8V0.70VI6.1mAR1803.Determinethediodecurrentusingthethirdapproximationbyusingtheaccompanyingdiodegraph.Thestraightlinerepresentsallthepos-sibleconditionsfora180-resistoranda1.8-Vbattery(fromadiodevolt-agedropof0.0Vat10.0mAononeendtoa1.8-Vdropat0.0mAontheotherend).ThesolutionistheintersectionofbothlinesandoccursatI6.3mA.4.Estimatetheerrorforallthreeapproxima-tions,ignoringthebatteryandresistor.Discusstheimpactofgreaterbatteryvolt-agesontheerrors.Forthefirstapproximation,theerroris(10.0mA6.1mA)/6.1mA,or64percent(whentheactualdiodedropis0.70V).Thiserrordecreasesforgreaterbatteryvoltages.Forthesecondapproximation,theerrorsourceisanydeviationfrom0.70Vastheactualdiodedrop.Itcannotbeexactlydetermined,butit’smuchlessthan64percent.Thiserroralsodecreaseswithhigherbatteryvoltages.Forthethirdapproximation,theerrorisduetograph-icresolutionandisnotaffectedbythebatteryvoltage.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual571

575CHAPTER30NuclearPhysicsPracticeProblems30.1TheNucleuspages799–805page8011.Threeisotopesofuraniumhavemassnumbersof234,235,and238.Theatomicnumberofuraniumis92.Howmanyneutronsareinthenucleiofeachoftheseisotopes?AZneutrons23492142neutrons23592143neutrons23892146neutrons2.Anisotopeofoxygenhasamassnumberof15.Howmanyneutronsareinthenucleusofthisisotope?AZ1587neutrons3.Howmanyneutronsareinthemercuryisotope20080Hg?AZ20080120neutrons4.Writethesymbolsforthethreeisotopesofhydrogenthathavezero,one,andtwoneutronsinthenucleus.1231H,1H,1Hpage805Usethesevaluestosolvethefollowingproblems:massofhydrogen1.007825u,massofneutron1.008665u,1u931.49MeV.5.Thecarbonisotope126Chasamassof12.0000u.a.Calculateitsmassdefect.Massdefect(isotopemass)(massofprotonsandelectrons)(massofneutrons)12.000000u(6)(1.007825u)(6)(1.008665u)0.098940ub.CalculateitsbindingenergyinMeV.Bindingenergy(massdefect)(bindingenergyof1u)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(0.098940u)(931.49MeV/u)92.161MeVPhysics:PrinciplesandProblemsSolutionsManual573

576Chapter30continued6.Theisotopeofhydrogenthatcontainsoneprotonandoneneutroniscalleddeuterium.Themassoftheatomis2.014102u.a.Whatisitsmassdefect?Massdefect(isotopemass)(massofprotonandelectron)(massofneutron)2.014102u1.007825u1.008665u0.002388ub.WhatisthebindingenergyofdeuteriuminMeV?Bindingenergy(massdefect)(bindingenergyof1u)(0.002388u)(931.49MeV/u)2.2244MeV7.Anitrogenisotope,157N,hassevenprotonsandeightneutrons.Ithasamassof15.010109u.a.Calculatethemassdefectofthisnucleus.Massdefect(isotopemass)(massofprotonsandelectrons)(massofneutrons)15.010109u(7)(1.007825u)(8)(1.008665u)0.113986ub.Calculatethebindingenergyofthenucleus.Bindingenergy(massdefect)(bindingenergyof1u)(0.113986u)(931.49MeV/u)106.18MeV8.Anoxygenisotope,168O,hasanuclearmassof15.994915u.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Whatisthemassdefectofthisisotope?Massdefect(isotopemass)(massofprotonsandelectrons)(massofneutrons)15.994915u(8)(1.007825u)(8)(1.008665u)0.137005ub.Whatisthebindingenergyofitsnucleus?Bindingenergy(massdefect)(bindingenergyof1u)(0.137005u)(931.49MeV/u)127.62MeVSectionReview30.1TheNucleuspages799–805page8059.NucleiConsiderthesetwopairsofnuclei:121311116Cand6Cand5Band6C.Inwhichwayarethetwoalike?Inwhichwayaretheydifferent?574SolutionsManualPhysics:PrinciplesandProblems

577Chapter30continuedThefirstpairhasthesamenumberofprotons,butadifferentnumberofnucleons.Thesecondpairhasthesamenumberofnucleons,butadiffer-entnumberofprotons.10.BindingEnergyWhentritium,31H,decays,itemitsabetaparticleandbecomes32He.Whichnucleuswouldyouexpecttohaveamorenegativebindingenergy?Thetritiumnucleus,becausetritiumemitsaparticlewithmassandkinet-icenergyinitsdecay.11.StrongNuclearForceTherangeofthestrongnuclearforceissoshortthatonlynucleonsthatareadjacenttoeachotherareaffectedbytheforce.Usethisfacttoexplainwhy,inlargenuclei,therepulsiveelectromagneticforcecanovercomethestrongattractiveforceandmakethenucleusunstable.Theelectricforceislongrange,sothatallprotonsrepeleachother,eveninlargenuclei.Thestrongforceisshortrange,sothatonlyneighboringprotonsattract.Therepulsiveforcegrows,withincreasingnuclearsize,fasterthanthestrongforce.12.MassDefectWhichofthetwonucleiinproblem10hasthelargermassdefect?thetritiumnucleus13.MassDefectandBindingEnergyTheradioactivecarbonisotope146Chasamassof14.003074u.a.Whatisthemassdefectofthisisotope?Massdefect(isotopemass)(massofprotonsandelectrons)(massofneutrons)14.003074u(6)(1.007825u)(8)(1.008665u)0.113169ub.Whatisthebindingenergyofitsnucleus?Bindingenergy(massdefect)(bindingenergyof1u)(0.113196u)(931.49MeV/u)105.44MeV14.CriticalThinkingInoldstars,notonlyareheliumandcarbonproducedbyjoiningmoretightlyboundnuclei,butsoareoxygen(Z8)andsilicon(Z14).Whatistheatomicnumberoftheheaviestnucleusthatcouldbeformedinthisway?Explain.26,iron,becauseitsbindingenergyislargest.PracticeProblemsCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.30.2NuclearDecayandReactionspages806–814page80815.Writethenuclearequationforthetransmutationofaradioactiveuraniumisotope,23423092U,intoathoriumisotope,90Th,bytheemissionofanparticle.234230492U®90Th2HePhysics:PrinciplesandProblemsSolutionsManual575

578Chapter30continued16.Writethenuclearequationforthetransmu-21.Writethenuclearequationforthetransmu-tationofaradioactivethoriumisotope,tationofaseaborgiumisotope,263106Sg,into230arutherfordiumisotope,25990Th,intoaradioactiveradiumisotope,104Rf,bytheemis-226sionofanalphaparticle.88Ra,bytheemissionofanparticle.26325942302264106Sg®104Rf2He90Th®88Ra2He22.Aprotoncollideswiththenitrogenisotope17.Writethenuclearequationforthetransmu-15tationofaradioactiveradiumisotope,7N,forminganewisotopeandanalphaparticle.Whatistheisotope?Writethe22622288Ra,intoaradonisotope,86Rn,bydecay.nuclearequation.2262224151AX488Ra®86Rn2He7N1H®Z2HewhereZ712618.Aradioactiveleadisotope,21482Pb,canA151412changetoaradioactivebismuthisotope,214ForZ6,theelementmustbecarbon.83Bi,bytheemissionofaparticleandanThus,theequationisantineutrino.Writethenuclearequation.1511247N1H®6C2He.2142140e082Pb®83Bi1023.Writethenuclearequationsforthebeta19.Aradioactivecarbonisotope,14decayofthefollowingisotopes.6C,under-goesdecaytobecomethenitrogena.2101480Pbisotope7N.Writethenuclearequation.210AX0e080Pb®Z1014140e06C®7N10whereZ80(1)081A21000210.page809ForZ81,theelementmustbe20.Usetheperiodictableonpage916tocom-thallium.Thus,theequationispletethefollowing.2102100e080Pb®81Tl10.a.140e0Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6C®?10b.21083Bi14AX0e06C®Z10210AX0e0whereZ6(1)0783Bi®Z10whereZ83(1)084A140014A21000210ForZ7,theelementmustbe14ForZ84,theelementmustbenitrogen.Thus,theisotopeis7N.polonium.Thus,theequationisb.550e02102100e024Cr®?1083Bi®84Po10.550e0c.23424Cr®?1090Th234AX0e055AX0e090Th®Z1024Cr®Z10whereZ90(1)091whereZ24(1)025A23400234A550055ForZ91,theelementmustbeForZ25,theelementmustbeprotactinium.Thus,theequationismanganese.Thus,theisotopeis552342340e025Mn.90Th®91Pa10.576SolutionsManualPhysics:PrinciplesandProblems

579Chapter30continuedd.23927.Tritium,393Np1H,oncewasusedinsomewatches239AX0e0toproduceafluorescentglowsothatthe93Np®Z10watchescouldbereadinthedark.IfthewhereZ93(1)094brightnessoftheglowisproportionaltoA23900239theactivityofthetritium,whatwouldbeForZ94,theelementmustbethebrightnessofsuchawatch,incompari-plutonium.Thus,theequationissontoitsoriginalbrightness,whenthe2392390e0watchissixyearsold?93Np®94Pu10.Sixyearsisabouthalfoftritium’shalf-lifeof12.3years.Thus,thepage810117brightnessis2,oraboutoftheRefertoFigure30-4andTable30-2tosolvethe210originalbrightness.followingproblems.24.Asampleof1.0goftritium,31H,ispro-duced.WhatwillbethemassofthetritiumSectionReviewremainingafter24.6years?24.6years(2)(12.3years),whichis30.2NuclearDecayand2half-lives.Reactions1tremainingoriginalpages806–8142page81412(1.0g)28.BetaDecayHowcananelectronbe2expelledfromanucleusinbetadecayifthe0.25gnucleushasnoelectrons?25.Theisotope238Inthenucleusaneutronchangesintoa93Nphasahalf-lifeof2.0days.If4.0gofneptuniumisproducedprotonandemitsanelectronandanonMonday,whatwillbethemassofantineutrino.neptuniumremainingonTuesdayofthe29.NuclearReactionsThepoloniumisotopenextweek?2108.0days(4)(2.0days),whichis84Poundergoesalphadecay.Writetheequationforthereaction.4half-livesremaining210AX484Po®Z2He1t(original)whereZ84282214A2104206(4.0g)2ForZ82,theelementmustbelead.0.25gThus,theequationis210206484Po®82Pb2He.26.Asampleofpolonium-210ispurchasedforaphysicsclassonSeptember1.Itsactivityis30.Half-LifeUseFigure30-4andTable30-2to2106Bq.Thesampleisusedinanexperi-estimateinhowmanydaysasampleof13153ImentonJune1.Whatactivitycanbewouldhavethree-eighthsitsoriginalactivity.expected?Fromthegraph,3/8areleftataboutThehalf-lifeof21084Pois138days.1.4halflives.Fromthetable,thehalf-Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thereare273daysorabout2half-liveslifeis8.07days,soitwouldtakeaboutbetweenSeptember1andJune1.11days.Sotheactivity6decays12210s25105BqPhysics:PrinciplesandProblemsSolutionsManual577

580Chapter30continued31.NuclearReactorLeadoftenisusedasaradiationshield.Whyisitnotagoodchoiceforamoderatorinanuclearreactor?Leadisusedasaradiationshieldbecauseitabsorbsradiation,includingneutrons.Amoderatorshouldonlyslowtheneutronssotheycanbeabsorbedbythefissilematerial.32.FusionOnefusionreactioninvolvestwodeuteriumnuclei,21H.Adeuteriummoleculecontainstwodeuteriumatoms.Whydoesn’tthismoleculeundergofusion?Thenucleiinthemoleculemustbemovingveryfasttoundergofusion.33.EnergyCalculatetheenergyreleasedinthefirstfusionreactionintheSun,1120e01H1H®1H10.TheenergyreleasedisE((initialmass)(finalmass))(931.49MeV/u)(2(massofprotium)(massofdeuterium)(massofpositron))(931.49MeV/u)2(1.007825u)2.014102u(9.111031kg)1u1.66051027kg(931.49MeV/u)0.931MeV34.CriticalThinkingAlphaemittersareusedinsmokedetectors.Anemitterismountedononeplateofacapacitor,andtheparticlesstriketheotherplate.Asaresult,thereisapotentialdifferenceacrosstheplates.Explainandpredictwhichplatehasthemorepositivepotential.TheplatebeingstuckbytheparticleshasthemorepositivepotentialbecausethepositivelychargedparticlesmovepositivechargefromtheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.emitterplatetothestruckplate.PracticeProblems30.3TheBuildingBlocksofMatterpages815–823page82135.Themassofaprotonis1.671027kg.a.Findtheenergyequivalentoftheproton’smassinjoules.Emc2(1.671027kg)(3.00108m/s)21.501010Jb.ConvertthisvaluetoeV.1.501010JE1.602171019J/eV9.36108eV578SolutionsManualPhysics:PrinciplesandProblems

581Chapter30continuedc.Findthesmallesttotal-rayenergythatcouldresultinaproton-antiprotonpair.Theminimumenergyis(2)(9.36108eV)1.87109eV36.Apositronandanelectroncanannihilateandformthreegammas.Twogammasaredetected.Onehasanenergyof225keV,theother357keV.Whatistheenergyofthethirdgamma?Asshowninthissection,theenergyequivalentofthepositronandelec-tronis1.02MeV.Thus,theenergyofthethirdgammais1.02MeV0.225MeV0.357MeV0.438MeV.37.Themassofaneutronis1.008665u.a.Findtheenergyequivalentoftheneutron’smassinMeV.E(neutronmassinu)(931.49MeV/u)(1.008665u)(931.49MeV/u)939.56MeVb.Findthesmallesttotal-rayenergythatcouldresultintheproductionofaneutron-antineutronpair.Thesmallestpossible-rayenergywouldbetwicetheneutronenergy.Etotal2En(2)(939.56MeV)1879.1MeV38.Themassofamuonis0.1135u.Itdecaysintoanelectronandtwoneutrinos.Whatistheenergyreleasedinthisdecay?Energyreleased(massofmuonmassofelectron)(931.49MeV/u)31kg)1u0.1135u(9.109101.66051027kg(931.49MeV/u)105.2MeVSectionReview30.3TheBuildingBlocksofMatterpages815–823page82339.NucleusBombardmentWhywouldaprotonrequiremoreenergythananeu-tronwhenusedtobombardanucleus?Theprotonandnucleusbothhavepositivecharge,sotheyrepeleachother.Therefore,theprotonmusthaveenoughkineticenergytoovercomethepotentialenergycausedbytherepulsion.TheunchargedneutrondoesCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.notexperiencethisrepulsiveforce.40.ParticleAcceleratorProtonsintheFermiLaboratoryaccelerator,Figure30-11,movecounterclockwise.Inwhatdirectionisthemagneticfieldofthebendingmagnets?down,intoEarthPhysics:PrinciplesandProblemsSolutionsManual579

582Chapter30continued41.PairProductionFigure30-18showstheChapterAssessmentproductionoftwoelectron/positronpairs.WhydoesthebottomsetoftrackscurvelessConceptMappingthanthetoppairoftracks?page828Theelectron/positronpairatthebottom44.Organizethefollowingtermsintothehasmorekineticenergy,andtherefore,conceptmap:StandardModel,quarks,agreatervelocity.gammarays,forcecarriers,protons,neutrons,leptons,Wbosons,neutrinos,electrons,gluons.42.TheStandardModelResearchthelimita-tionsoftheStandardModelandpossibleStandardreplacements.ModelAnswersinclude:TheStandardModelhasseveralparametersthatcanbeIncludesobtainedonlyfromexperiment.TheforceHiggsparticlethatwouldsettheenergyquarksleptonscarriersscalehasnotbeenfound.Itisnotreallyatheorythatexplains,andsoisnotWhichExamplesExamplescomposecomplete.Supersymmetryandstringgammatheoryaretwopossiblereplacements.protonselectronsraysgluons43.CriticalThinkingConsiderthefollowingneutronsneutrinosWbosonsequations.u®dWandW®eHowcouldtheybeusedtoexplaintheMasteringConceptsradioactivedecayofanucleonthatresultspage828intheemissionofapositronandaneutri-45.Whatforceinsideanucleusactstopushtheno?Writetheequationinvolvingnucleonsnucleusapart?Whatforceinsidethenucle-ratherthanquarks.usactstoholdthenucleustogether?(30.1)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.AnucleoncontainingauquarkcouldTherepulsiveelectricforcebetweenthedecaybythisprocesstoanucleonwithpositiveprotonsactstopushthenucle-onelessuandonemoredquark.Ausapart.Thestrongnuclearforceactsprotonisuudandaneutronisuud,sotoholdthenucleustogether.thefollowingequationwouldwork:110046.Definethemassdefectofanucleus.To1p®0n1e0whatisitrelated?(30.1)Themassdefectisthedifferencebetweenthesumofthemassesoftheindividualparticlesofthenucleusandthemassofthenucleus.ItisrelatedtobindingenergythroughEmc2.47.Whicharegenerallymoreunstable,smallorlargenuclei?(30.1)Largenucleiaregenerallymoreunstable.Thegreaternumberofprotonsmakestherepulsiveelectricforceovercometheattractivestrongforce.580SolutionsManualPhysics:PrinciplesandProblems

583Chapter30continued48.Whichisotopehasthegreaternumberofpro-fuseintoalargernucleus,themassoftons,uranium-235oruranium-238?(30.1)themoretightlybound,largenucleusisTheybothhavethesamenumberofpro-less,andtheextramassappearsastons,butdifferentnumbersofneutrons.energy.49.Definethetermtransmutationasusedin55.High-EnergyPhysicsWhywouldalinearnuclearphysicsandgiveanexample.(30.2)acceleratornotworkwithaneutron?(30.3)Transmutationisthechangeofoneele-Itacceleratesparticlesusingelectricmentintoanotherbymeansofaforce,andtheneutronhasnocharge.nuclearreactioncausedbyradioactive56.ForcesInwhichofthefourinteractionsdecayornuclearbombardment.For(strong,weak,electromagnetic,andgravita-example,uranium-238decaysintothori-tional)dothefollowingparticlestakepart?um-234andanparticle.(30.3)50.RadiationWhatarethecommonnamesa.electronforanparticle,particle,andradiation?electromagnetic,weak,gravitational(30.2)b.protonAnaparticleisaheliumnucleus,astrong,electromagnetic,gravitationalparticleisanelectron,andradiationc.neutrinoisahigh-energyphoton.weak51.Whattwoquantitiesmustalwaysbecon-servedinanynuclearequation?(30.2)57.Whathappenstotheatomicnumberandmassnumberofanucleusthatemitsatheatomicnumber,toconservecharge;positron?(30.3)themassnumber,toconservethenum-berofnucleonsTheatomicnumberdecreasesby1,andthereisnochangeinthemassnumber.52.NuclearPowerWhatsequenceofeventsZ®Z1;A®Amustoccurforachainreactiontotakeplace?(30.2)58.AntimatterWhatwouldhappenifamete-Manyneutronsmustbereleasedbyafis-oritemadeofantiprotons,antineutrons,sionednucleusandabsorbedbyneigh-andpositronslandedonEarth?(30.3)boringnuclei,causingthemtoundergoItwouldannihilatewithanequivalentfission.amountofmatter,producinganextremelylargeamountofenergy.53.NuclearPowerWhatroledoesamodera-torplayinafissionreactor?(30.2)ApplyingConceptsThemoderatorslowsthefastneutrons,page828increasingtheirprobabilityofbeing59.FissionAWebsiteclaimsthatscientistsabsorbed.havebeenabletocauseironnucleitoundergofission.Istheclaimlikelytobe54.Fissionandfusionareoppositeprocesses.true?Explain.Howcaneachreleaseenergy?(30.2)Itisnottrue.IronisthemosttightlyCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Whenalargeatom,suchasuranium,boundelement,andthusthemoststa-undergoesfission,themassoftheblenucleus.Itcannotdecaybyeitherproductsislessthanthatoftheoriginalfissionorfusion.nucleus.Anamountofenergyequiva-lenttothedifferenceinmassisreleased.However,whensmallnucleiPhysics:PrinciplesandProblemsSolutionsManual581

584Chapter30continued60.UsethegraphofbindingenergypernucleoninFigure30-2todeterminewhetherthereaction2131H1H®2Heisenergeticallypossible.Theinitialbindingenergyislessthanthefinalbindingenergyand,there-fore,thereactionisenergeticallypossible.61.IsotopesExplainthedifferencebetweennaturallyandartificiallyproducedradioactiveisotopes.Anaturalradioactivematerialisonethatisfoundtoexistnormallyinoresandthatexperiencesradioactivedecay.Artificially,radioactivematerialexperiencesradioactivedecayonlyafterbeingsubjectedtobombardmentbyparticles.62.NuclearReactorInanuclearreactor,waterthatpassesthroughthecoreofthereactorflowsthroughoneloop,whilethewaterthatproducessteamfortheturbinesflowsthroughasecondloop.Whyaretheretwoloops?Thewaterthatpassesthroughthecoreisathighpressure,soitdoesn’tboil.Asecondloopcarrieswateratlowerpressure,producingsteam.63.Thefissionofauraniumnucleusandthefusionoffourhydrogennucleitopro-duceaheliumnucleusbothproduceenergy.a.Whichproducesmoreenergy?AsshowninSection30-2,theenergyproducedbythefissionofoneuraniumnucleusis200MeV.Theenergyproducedbythefusionoffourhydrogennucleiis24MeV.b.Doesthefissionofakilogramofuraniumnucleiorthefusionofakilogramofhydrogenproducemoreenergy?Themassnumberoffissionableuraniumis238.ThemassnumberCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofhydrogenis1.Foranequalmassoftheelements,youwouldhave238timesmorehydrogennucleithanuraniumnuclei.Thefissionofeachuraniumnucleuswouldproduce200MeVofenergy.Thefusionof238nucleiofhydrogentoproduceheliumnucleiwouldproduce238(25MeV),orabout1500MeVofenergy.4c.Whyareyouranswerstopartsaandbdifferent?Althoughthefissionofoneuraniumnucleusproducesmoreenergythanthefusionoffourhydrogennucleitoproducehelium,therearemorethan200timesasmanyhydrogennucleiin1kgasthereareura-niumnuclei.MasteringProblems30.1TheNucleuspages828–829Level164.Whatparticles,andhowmanyofeach,makeupanatomof10947Ag?47electrons,47protons,62neutrons582SolutionsManualPhysics:PrinciplesandProblems

585Chapter30continued65.Whatistheisotopicsymbol(theoneusedinnuclearequations)ofazincatomcomposedof30protonsand34neutrons?6430Zn66.Thesulfurisotope3216Shasanuclearmassof31.97207u.a.Whatisthemassdefectofthisisotope?massdefect(isotopemass)(massofprotonsandelectrons)(massofneutrons)31.97207u(16)(1.007825u)(16)(1.008665u)0.29177ub.Whatisthebindingenergyofitsnucleus?Bindingenergy(massdefect)(bindingenergyof1u)(0.29177u)(931.5MeV/u)271.78MeVc.Whatisthebindingenergypernucleon?Bindingenergypernucleonis271.8MeV8.494MeV/nucleon32nucleonsLevel267.Anitrogenisotope,127N,hasanuclearmassof12.0188u.a.Whatisthebindingenergypernucleon?Bindingenergy(massofdefect)(bindingenergyfor1u)((isotopemass)(massofprotonsandelectrons)(massofneutrons))(bindingenergyfor1u)((12.0188u)(7)(1.007825u)(5)(1.008665u))(931.49MeV/u)73.867MeV73.867MeVThebindingenergypernucleonis6.1556MeV/nucleon12nucleonsb.Doesitrequiremoreenergytoseparateanucleonfroma147Nnucleusorfroma12147Nnucleus?7Nhasamassof14.00307u.For147N,thebindingenergyisBindingenergy(massdefect)(bindingenergyfor1u)((isotopemass)(massofprotonsandelectrons)(massofneutrons))(bindingenergyfor1u)((14.00307u(7)(1.007825u)(7)(1.008665u))(931.49MeV/u)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.104.66MeV104.66MeVThebindingenergypernucleonis7.4757MeV/nucleon14nucleonsThus,removinganucleonfrom147Nrequiresmoreenergy.Physics:PrinciplesandProblemsSolutionsManual583

586Chapter30continued68.Thetwopositivelychargedprotonsinaheliumnucleusareseparatedbyabout2.01015m.UseCoulomb’slawtofindtheelectricforceofrepulsionbetweenthetwoprotons.Theresultwillgiveyouanindicationofthestrengthofthestrongnuclearforce.qAqBFK2d(1.601019C)(1.601019C)(9.0109Nm2/C2)58N2.01015m69.Thebindingenergyfor42Heis28.3MeV.Calculatethemassofaheliumiso-topeinatomicmassunits.massoftheisotope(massdefect)(massofprotonsandelectrons)(massofneutrons)bindingenergybindingenergyof1u(massofprotonsandelectrons)(massofneutrons)28.3MeV(2)(1.007825u)(2)(1.008665u)931.49MeV/u4.00u30.2NuclearDecayandReactionspage829Level170.Writethecompletenuclearequationforthealphadecayof22286Rn.222AX486Rn®Z2HeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.whereA222-4218Z86-284ForZ84,theelementmustbepolonium.Thus,theequationis222218486Rn®84Po2He71.Writethecompletenuclearequationforthebetadecayof8936Kr.89AX0e036KrZ10whereZ36(1)037A890089ForZ37,theelementmustberubidium.Thus,theequationis89890e036Kr37Rb11072.Completeeachnuclearreaction.a.225489Ac®2He_____225422189Ac®2He87Frb.2270e__________88Ra®12270e227088Ra®189Ac0584SolutionsManualPhysics:PrinciplesandProblems

587Chapter30continuedc.651175.Whenaboronisotope,1129Cu0n®_____®1p_____5B,isbombardedwithprotons,itabsorbsaprotonandemits6516616529Cu0n®29Cu®1p28Nianeutron.d.2351961a.Whatelementisformed?92U0n®40Zr3(0n)_____2351961136carbon92U0n®40Zr30n52Teb.Writethenuclearequationforthis73.Anisotopehasahalf-lifeof3.0days.Whatreaction.percentoftheoriginalmaterialwillbeleftafter111A15B1p®6C0na.6.0days?whereA1111116.0d2.0half-livesThus,theequationis3.0d1111111t5B1p®6C0npercentremaining1002c.Theisotopeformedisradioactiveand12.0decaysbyemittingapositron.Write1002thecompletenuclearequationforthis25%reaction.b.9.0days?11A0e06C®ZX109.0d3.0half-liveswhereZ61053.0dA1100111tpercentremaining100ForZ5,theelementmustbe2boron.Thus,theequationis13.0210011110e06C®5B1013%76.Thefirstatomicbombreleasedanenergyc.12days?equivalentof2.0101kilotonsofTNT.One12d4half-lives12J.kilotonofTNTisequivalentto5.0103.0dUranium-235releases3.211011J/atom.percentremaining1t100Whatwasthemassoftheuranium-235that2underwentfissiontoproducetheenergyof14.0thebomb?1002E(2.0101kton)(5.01012J/kton)6.3%1atom1mol3.211011J6.021023atom74.Inanaccidentinaresearchlaboratory,aradioactiveisotopewithahalf-lifeofthree0.235kgdaysisspilled.Asaresult,theradiationismoleighttimesthemaximumpermissible1.2kgamount.Howlongmustworkerswaitbeforetheycanentertheroom?77.Duringafusionreaction,twodeuterons,12Fortheactivitytofalltoitspresent1H,combinetoformaheliumisotope,83amount,youmustwaitthreehalf-lives,2He.Whatotherparticleisproduced?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.or9days.223A1H1H®2HeZXwhereZ1120A2231Thus,theparticlemustbe10n,aneutron.Physics:PrinciplesandProblemsSolutionsManual585

588Chapter30continuedLevel2b.du78.2091284Pohasahalf-lifeof103years.Howlongdu1chargewouldittakefora100-gsampletodecayso33c.ddthatonly3.1gofPo-209wasleft?111tud0chargeremainingoriginal332LetRremainingmassandIorigi-82.Baryonsareparticlesthataremadeofthreenalmassquarks.Findthechargeoneachofthefol-1tIlowingbaryons.RI2t2a.neutron:ddutI122Rddu2033log(2t)logIb.antiproton:uudR221Iuud333tlog2logR1IlogRtLevel2log283.ThesynchrotronattheFermiLaboratory100glog3.1ghasadiameterof2.0km.Protonscirclinginitmoveatapproximatelythespeedoflog2lightinavacuum.5half-livesa.Howlongdoesittakeaprotontocom-Therefore,thesamplewouldtakeaboutpleteonerevolution?500yearstodecay.dvt30.3TheBuildingBlocksofMatterwheredisthecircumferenceofthepage829synchrotron,Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Level1d(2.0103m)79.Whatwouldbethechargeofaparticlesotv8m/s3.010composedofthreeupquarks?2.1105s2Eachuquarkhasachargeof.3b.Theprotonsentertheringatanenergy2of8.0GeV.Theygain2.5MeVeachuuu32elementarychargesrevolution.Howmanyrevolutionsmust3theytravelbeforetheyreach400.0GeV80.Thechargeofanantiquarkisoppositethatofenergy?ofaquark.Apioniscomposedofanupquarkandananti-downquark,ud.What400.0109eV8.00109eV2.5106eV/revwouldbethechargeofthispion?1.6105rev21ud1elementary33c.Howlongdoesittaketheprotonstobechargeacceleratedto400.0GeV?81.Pionsarecomposedofaquarkandananti-t(1.6105rev)(2.1105s/rev)quark.Findthechargeofapionmadeup3.4softhefollowing.d.Howfardotheprotonstravelduringa.uuthisacceleration?22du0chargedvt(3.00108m/s)(3.4s)331.0109m,orabout1millionkm586SolutionsManualPhysics:PrinciplesandProblems

589Chapter30continued84.Figure30-21showstracksinabubbleLevel2chamber.Whataresomereasonsonetrack87.Oneofthesimplestfusionreactionsmightcurvemorethananother?involvestheproductionofdeuterium,21H(2.014102u),fromaneutronandapro-ton.Writethecompletefusionreactionandfindtheamountofenergyreleased.Theequationforthereactionis1120n0p®1HTheenergyreleasedisE((initialmass)(finalmass))(931.49MeV/u)((massofneutron)(massofproton)(massofdeuterium))(931.49MeV/u)■Figure30-21((1.008665u)(1.007276u)Thepathsoffaster-movingparticles(2.014102u))(931.49MeV/u)wouldcurveless.1.7130MeVMixedReview88.A232page83092Unucleus,mass232.0372u,decaysto228Level190Th,mass228.0287u,byemittinganparticle,mass4.0026u,witha85.Eachofthefollowingnucleicanabsorbankineticenergyof5.3MeV.Whatmustbeparticle.Assumethatnosecondaryparti-thekineticenergyoftherecoilingthoriumclesareemittedbythenucleus.Completenucleus?eachequation.a.144Thetotalkineticenergyofthedecay7N2He®_____productsis144AX7N2He®ZKEtotalKEthoriumKEwhereZ729Thus,A14418KEthoriumKEtotalKEForZ9,theelementmustbe(massdefect)fluorine.Thus,theequationis(931.49MeV/u)KE144187N2He®9F((massof23292U)b.274(massof22813Al2He®_____90Th)(massof4274A2He))13Al2He®ZX(931.49MeV/u)KEwhereZ13215((232.0372u)(228.0287u)A27431(4.0026u))ForZ15,theelementmustbe(931.49MeV/u)5.3MeVphosphorus.Thus,theequationis0.2MeV27431Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13Al2He®15P86.21186Rnhasahalf-lifeof15h.Whatfractionofasamplewouldbeleftafter60h?60h1414half-lives,soisleft.15h216Physics:PrinciplesandProblemsSolutionsManual587

590Chapter30continuedThinkingCriticallynumberofreactionsoccurringeachsecondispage83041026J/s89.InferGammarayscarrymomentum.The1038reactions/s4.01012J/reactionmomentumofagammarayofenergyEisequaltoE/c,wherecisthespeedoflight.92.InterpretDataAnisotopeundergoingWhenanelectron-positronpairdecaysintoradioactivedecayismonitoredbyaradia-twogammarays,bothmomentumandtiondetector.Thenumberofcountsineachenergymustbeconserved.Thesumofthefive-minuteintervalisrecorded.Theresultsenergiesofthegammaraysis1.02MeV.IfareshowninTable30-3.Thesampleisthepositronandelectronareinitiallyatrest,thenremovedandtheradiationdetectorwhatmustbethemagnitudeanddirectionrecords20countsresultingfromcosmicofthemomentumofthetwogammarays?raysin5min.Findthehalf-lifeoftheBecausetheinitialmomentumiszero,isotope.Notethatyoushouldfirstsubtractthismustbethefinalmomentum.Thus,the20-countbackgroundreadingfromeachthetwogammaraysmusthaveequalresult.Thenplotthecountsasafunctionandoppositemomentum.Themagni-oftime.Fromyourgraph,determinethetudeofthemomentaishalf-life.1Ep2CTable30-3(1.02106eV)(1.601019J/eV)RadioactiveDecay(2)(3.00108m/s)Measurements2.721022kgm/sTime(min)Counts(per5min)Theymoveinoppositedirections.098790.InferAnelectron-positronpair,initiallyat5375rest,alsocandecayintothreegammarays.10150Ifallthreegammarayshaveequalenergies,1570Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.whatmustbetheirrelativedirections?Make2040asketch.2525Thequestionbecomesthefollowing:3018howcanthreeparticles,eachwiththesamemomentum,havezerototalabout4minmomentum?Thethreegammaraysleavewithanglesof120°betweenthemWritinginPhysicsonaplane.page83093.Researchthepresentunderstandingofdark91.EstimateOnefusionreactionintheSunmatterintheuniverse.Whyisitneededbyreleasesabout25MeVofenergy.Estimatecosmologists?Ofwhatmightitbemade?thenumberofsuchreactionsthatoccurAbout25percentoftheuniverseisdarkeachsecondfromtheluminosityofthematter.ItisneededtoexplaingalacticSun,whichistherateatwhichitreleasesenergy,41026W.rotationandtheexpansionoftheuni-verse.Accordingtoonetheory,dark1eV1.60221019J,somatterisnotmadeofordinarymatter25MeV(25106eV/reaction)thatiscoveredbytheStandardModel.(1.60221019J/eV)Itmayinteractwithordinarymatteronly4.01012J/reactionthroughgravityandweaknuclearThetotalpoweris41026J/s,sotheforces.588SolutionsManualPhysics:PrinciplesandProblems

591Chapter30continued94.Researchthehuntforthetopquark.Whyb.CalculatetheenergyoftheelectronineV.didphysicistshypothesizeitsexistence?12KEmvTheoristshadpredictedtheexistence2ofdifferentflavorsofquarks.Theyreal-1(9.111031kg)(1.82103m/s)2izedquarksoccurinpairs.Theupand2downquarkpairandthecharmedand1eVstrangepairhadbeenfoundbyexperi-1.601019Jments.Whenthebottomquarkwas9.43106eVfoundin1977,theyfeltitmusthaveapartner.Theshortlifetimeandhigh98.Aphotonwithanenergyof14.0eVentersamassofthetopquarkmadeitdifficulthydrogenatominthegroundstateandion-tofind.ItfinallywasfoundataFermilabizesit.Withwhatkineticenergywilltheexperimentin1995.electronbeejectedfromtheatom?(Chapter28)CumulativeReviewKE(photonenergy)(energyofpage830electroninthegroundstate)95.Anelectronwithavelocityof1.7106m/s14.0eV(13.6eV)isatrightanglestoa0.91-Tmagneticfield.Whatistheforceontheelectronproduced0.4eVbythemagneticfield?(Chapter24)99.Asilicondiode(V0.70V)thatiscon-FBqvducting137mAisinserieswitharesistor(0.91T)(1.601019C)anda6.67-Vpowersource.(Chapter29)(1.7106m/s)a.Whatisthevoltagedropacrossthe2.51013Nresistor?V96.AnEMFof2.0mVisinducedinawirethatRVsourceVdiodeis0.10mlongwhenitismovingperpen-6.67V0.70Vdicularlyacrossauniformmagneticfieldat5.97Vavelocityof4.0m/s.Whatisthemagneticb.Whatisthevalueoftheresistor?inductionofthefield?(Chapter25)V5.97VEMFBLvR137mA43.6IEMF2.0103VBLv(0.10m)(4.0m/s)5.0103T5.0mT97.AnelectronhasadeBrogliewavelengthof400.0nm,theshortestwavelengthofvisiblelight.(Chapter27)a.Findthevelocityoftheelectron.hmvhvCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.m6.631034J/Hz(9.111031kg)(400.0109m)1.82103m/sPhysics:PrinciplesandProblemsSolutionsManual589

592Chapter30continuedChallengeProblempage82123892Udecaysbyemissionandtwosuccessiveemissionsbackintouraniumagain.1.Showthethreenucleardecayequations.Fortheemission,238AX492U®Z2HewhereZ92–290A2384234ForZ90,theelementmustbethorium.Thus,theequationis238234492U®90Th2HeForthefirstemission,234AX0e090Th®Z10whereZ90(1)091A23400234ForZ91,theelementmustbeprotactinium.Thus,theequationis2342340e090Th®91Pa10Forthesecondemission,234AX0e091Pa®Z10whereZ91(1)092A23400234ForZ92,theelementmustbeuranium.Thus,theequationis2342340e091Pa®92U102.Predicttheatomicmassnumberoftheuraniumformed.A234Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.590SolutionsManualPhysics:PrinciplesandProblems

593AppendixBAdditionalProblemsChapter1........................593Chapter16.......................641Chapter2........................595Chapter17.......................642Chapter3........................598Chapter18.......................644Chapter4........................602Chapter19.......................646Chapter5........................607Chapter20.......................649Chapter6........................612Chapter21.......................651Chapter7........................616Chapter22.......................655Chapter8........................618Chapter23.......................658Chapter9........................622Chapter24.......................662Chapter10.......................626Chapter25.......................664Chapter11.......................628Chapter26.......................665Chapter12.......................631Chapter27.......................669Chapter13.......................635Chapter28.......................672Chapter14.......................638Chapter29.......................674Chapter15.......................640Chapter30.......................677

594AppendixBfolicacid:Chapter1(400mcg)(0.001mg/mcg)0.4mgpage858vitaminB12:1.Thedensity,,ofanobjectisgivenbythe(6mcg)(0.001mg/mcg)0.006mgratiooftheobject’smass,m,andvolume,V,biotin:accordingtotheequationm/V.Whatis(30mcg)(0.001mg/mcg)0.03mgthedensityofacubethatis1.2cmoneachsideandhasamassof25.6g?6.Whattypeofrelationshipisshowninthem25.6g3scatterplot?Writeanequationtomodel315g/cmV(1.2cm)thedata.2.Anobjectthatismovinginastraightlinewithspeedvcoversadistance,dvt,in1.2timet.Rewritetheequationtofindtintermsofdandv.Howlongdoesittake1.0aplanethatistravelingat350km/hto0.8travel1750km?0.6d1750kmt5.0hv350km/h0.43.Convert523kgtomilligrams.0.21000g1000mg523kg(523kg)01kg1g1234565.23108mg1.2inverserelationship;yx4.Theliquidmeasuremilliliter,mL,isthesameas1cm3.Howmanymillilitersof7.Howmanysignificantdigitsarethereinliquidcanbeheldina2.5-m3container?eachofthefollowingmeasurements?3(2.5m3)100cm31mL2.5m3a.100m1m1cm12.5106mLb.0.0023m/s5.Partofthelabelfromavitamincontainer2isshownbelow.Theabbreviation“mcg”c.100.1mstandsformicrograms.Convertthevalues4tomilligrams.d.2.00235EachTabletContains%DVCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FolicAcid400mcg100%VitaminB126mcg100%Biotin30mcg10%Physics:PrinciplesandProblemsSolutionsManual593

595Chapter1continued8.Thebuoyant(upward)forceexertedbywaterb.Arethedatalinear?onasubmergedobjectisgivenbytheformulanoFVg,whereisthedensityofwaterc.Plottheperiodversusthesquarerootof(1.0103kg/m3),Visthevolumeofthethelength.objectinm3,andgistheaccelerationduetogravity(9.80m/s2).Theforceisinnewtons.ITRewritetheequationtofindVintermsofF.0.30.6Usethistofindthevolumeofasubmerged0.40.9barrelifthebuoyantforceonitis9200N.0.61.3FVg0.81.69200N30.91.80.94m(1.0103kg/m3)(9.80m/s2)1.02.09.Whatis2.3kg0.23g?2.3kg0.23g2.3103g0.23g2.02.3103g2.3kg10.Solvethefollowingproblems.1.0a.15.5cm12.1cmPeriod(s)188cm20.00.20.40.60.81.014.678mb.3.2m/slm4.6sd.Whatistherelationshipbetweentheperiodandthesquarerootofthelength?11.AnexperimentwasperformedtodeterminetheperiodofapendulumasafunctionofThegraphislinear,sotheperiodisthelengthofitsstring.ThedatainthetableproportionaltothesquarerootofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.belowweremeasured.thelengthofthestring.Length(m)Period(s)12.Basedonthepreviousproblem,whatshould0.10.6betheperiodofapendulumwhoselengthis0.7m?0.20.9around1.7s0.41.30.61.60.81.81.02.0a.Plotperiod,T,versuslength,l.2.01.0Period(s)0.00.20.40.60.81.0Length(m)594SolutionsManualPhysics:PrinciplesandProblems

596a.WhatisthepositionofthecaratChapter20.00s?pages859–86040.0m1.Aposition-timegraphforabicycleisshownb.Whatisthepositionoftheautomobileinthefigurebelow.after2.00shaselapsed?40.020.0mc.Howfardidtheautomobiletravel30.0betweenthetimes1.00sand3.00s?distancetraveled30.0m10.0m20.020.0mPosition(m)10.03.Ajoggerrunsataconstantrateof10.0mevery2.0s.Thejoggerstartsattheorigin,and0.001.002.003.004.005.00runsinthepositivedirectionfor3600.0s.Time(min)Thefigurebelowisaposition-timegraphshowingthepositionofthejoggerfroma.Whatisthepositionofthebicycleattimet0.0stotimet20.0s.Whereis1.00min?therunnerattimet5.0s?t15.0s?35.0mb.Whatisthepositionofthebicycleat100.03.50min?35.0m75.0c.Whatisthedisplacementofthebicyclebetweenthetimes1.00min50.0and5.00min?Position(m)0.00m25.0d.Describethemotionofthebicycle.Thebicycleisnotmoving.0.04.08.012.016.020.02.ThepositionofanautomobileisplottedasaTime(s)functionoftimeintheaccompanyingfigure.Att5.0sthejoggerisat25.0m.50.0Att15.0sthejoggerisat75.0m.40.030.020.0Position(m)10.00.0Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.002.003.004.005.00Time(s)Physics:PrinciplesandProblemsSolutionsManual595

597Chapter2continued4.Twotrainssimultaneouslyleavethesametrainstationatnoon.Onetraintravels2.0CarAnorthandtheothertravelssouth.Theposition-timegraphforbothtrainsis1.5CarBshownintheaccompanyingfigure.1.0North1000.0Position(km)0.50.01.02.03.04.05.00.0Time(h)Time(min)4.06.08.010.0Position(km)a.HowfarapartarethetwocarswhencarBstartsoutatt1.0min?0.75kmb.Atwhattimedothecarsmeet?South1000.02.0mina.Whatisthepositionofthetrainc.Howfarapartarethecarsattimetravelingnorthat6.0h?t3.0min?600.0km0.75kmb.Whatisthepositionofthetraintravelingsouthat6.0h?6.Theposition-timegraphfortwojoggers,AandB,isshownintheaccompanyingfigure.300.0kmc.Whatisthedistancebetweenthetrains2.00at6.0h?Whatisthedistanceat10.0h?JoggerB900.0km,1500.0kmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.50d.Atwhattimearethetrains600.0kmapart?1.004.0hPosition(km)e.Whichtrainismovingmorequickly?0.50JoggerAnorthboundtrain5.Twocarsheadoutinthesamedirection.0.0010.020.030.040.050.0CarAstarts1.0minbeforecarB.TheTime(min)position-timegraphsforbothcarsareshownintheaccompanyingfigure.a.Howfarapartarethetworunnersat10.0min?1.50kmb.Atwhattimearethey1.00kmapart?20.0minc.Howfarapartaretheyat50.0min?0.50kmd.Atwhattimedotheymeet?40.0min596SolutionsManualPhysics:PrinciplesandProblems

598Chapter2continuede.WhatdistancedoesjoggerBcovera.Whatistheaveragevelocityofthebetween30.0minand50.0min?airplane?0.00kmvelocityslopeoflinef.WhatdistancedoesjoggerAcover800.0km0.0kmbetween30.0minand50.0min?5.0h1.0h1.00km2.0102km/hb.Whatistheaveragespeedofthe7.Achild’stoytrainmovesataconstantairplane?speedof2.0cm/s.speedabsolutevalueofthevelocitya.Drawtheposition-timegraphshowing2.0102km/hthepositionofthetoyfor1.0min.9.Theposition-timegraphforahot-airballoonthatisinflightisshownin120theaccompanyingfigure.90Time(h)600.01.02.03.04.05.0Position(cm)30010.0102030405060Time(s)20.0b.Whatistheslopeofthelinerepresentingthemotionofthetoy?Position(km)30.0rise120cmsloperun60s2.0cm/s,thesameasthe40.0train’sspeeda.Whatistheaveragevelocityofthe8.Thepositionofanairplaneasafunctionballoon?oftimeisshowninthefigurebelow.velocityslopeTime(h)0.00km(40.0km)5.0h0.0h0.01.02.03.04.05.08.0km/hb.Whatistheaveragespeedofthe200.0balloon?speedabsolutevalueofslope400.08.0km/hPosition(km)600.0800.0Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual597

5993.Acartravelingat21m/smissestheturnoffonChapter3theroadandcollidesintothesafetyguardrail.pages861–862Thecarcomestoacompletestopin0.55s.1.Jasonandhissister,Tara,areridingbicycles.a.WhatistheaverageaccelerationofJasontriestocatchuptoTara,whohasathecar?10.0-sheadstart.vvfi0.00m/s21m/satt0.55sfi8.0Jason38m/s2Tarab.Ifthesafetyrailconsistedofasectionof6.0rigidrail,thecarwouldstopin0.15s.Whatwouldbetheaccelerationinthis4.0case?vvVelocity(m/s)fi0.00m/s21m/satt0.15s2.0fi1.4102m/s20.010.020.030.040.04.Onthewaytoschool,JamalrealizesthatTime(s)helefthisphysicshomeworkathome.Hiscarwasinitiallyheadingnorthat24.0m/s.a.WhatisJason’sacceleration?Ittakeshim35.5stoturnhiscararound8.0m/s0.0m/sandheadsouthat15.0m/s.IfnorthisaJ30.0s10.0sdesignatedtobethepositivedirection,0.40m/s2whatistheaverageaccelerationofthecarb.WhatisTara’sacceleration?duringthis35.5-sinterval?a8.0m/s0.0m/svfvi15.0m/s24.0m/sT40.0s0.0satt35.5sfi0.20m/s21.10m/s2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.Atwhattimedotheyhavethesamevelocity?5.Acheetahcanreachatopspeedof27.8m/sin5.2s.Whatisthecheetah’saverage20.0sacceleration?2.Adragsterstartsfromrestandacceleratesvvfi27.8m/s0.0m/sfor4.0satarateof5.0m/s2.Itthentravelsatt5.2sfiataconstantspeedfor2.5s.Aparachute5.3m/s2opens,stoppingthevehicleataconstantratein2.0s.Plotthev-tgraphrepresenting6.Afterbeinglaunched,arocketattainsaspeedtheentiremotionofthedragster.of122m/sbeforethefuelinthemotoriscompletelyused.Ifyouassumethattheaccel-20.0erationoftherocketisconstantat32.2m/s2,15.0howmuchtimedoesittakeforthefueltobecompletelyconsumed?10.0vaVelocity(m/s)5.0tv122m/s0.00m/s0.0t23.79s2.04.06.08.010.0a32.2m/sTime(s)598SolutionsManualPhysics:PrinciplesandProblems

600Chapter3continued7.Anobjectinfreefallhasanaccelerationofc.Whatisthedisplacementoftherunner9.80m/s2assumingthatthereisnoairresis-betweent50.0sandt60.0s?tance.Whatisthespeedofanobjectdropped1d(8.0m/s)(60.0s50.0s)fromthetopofatallcliff3.50safterithas2beenreleased,ifyouassumetheeffectofair4.0101mresistanceagainsttheobjectisnegligible?v10.Drawthev-tgraphofanautomobilethatatacceleratesuniformlyfromrestatt0.00svatgt(9.80m/s2)(3.50s)andcoversadistanceof180.0min12.0s.34.3m/sSincethecaracceleratesuniformly,thev-tgraphisastraightline.Startingfrom8.Atrainmovingwithavelocityof51m/seasttheorigin,theareaistriangular.undergoesanaccelerationof2.3m/s2asit1Thus,dvapproachesatown.Whatisthevelocityofthe2maxttrain5.2safterithasbeguntodecelerate?2d(2)(180.0m)v30.0m/svmaxt12.0svfviatt30.0vtvfai(2.3m/s2)(5.2s)51m/s39m/s15.0Velocity(m/s)9.Thev-tgraphofarunnerisshownintheaccompanyingfigure.0.02.04.06.08.010.012.0Time(s)8.011.Thev-tgraphofacarisshownintheaccom-6.0panyingfigure.Whatisthedisplacementofthecarfromt0.00stot15.0s?4.030.0Velocity(m/s)2.015.00.0010.020.030.040.050.060.0Time(s)0.004.008.0012.0016.00ThedisplacementistheareaundertheTime(s)Velocity(m/s)graph.15.0a.Whatisthedisplacementoftherunnerbetweent0.00sandt20.0s?d130.02(8.0m/s)(20.0s)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8.0101mdisplacementareaunderv-tgraphb.WhatisthedisplacementoftherunnerThetotaldisplacementisthesumofbetweent20.0sandt50.0s?thedisplacementsfrom0.00sto4.00s,4.00sto6.00s,6.00sto8.00s,8.00stod(8.0m/s)(50.0s20.0s)10.0s,and10.0sto15.0s:240mPhysics:PrinciplesandProblemsSolutionsManual599

601Chapter3continueddtotald1d2d3d4d5vfvi1.23m/s3.24m/stfa2.32m/s21(30.0m/s)(4.00s)1.93s2(30.0m/s)(2.00s)15.Acheetahcanacceleratefromresttoaspeed1of27.8m/sin5.20s.Thecheetahcanmain-(30.0m/s)(2.00s)2tainthisspeedfor9.70sbeforeitquickly1runsoutofenergyandstops.Whatdistance(30.0m/s)(2.00s)2doesthecheetahcoverduringthis14.9-srun?(30.0m/s)(5.00s)Duringtheaccelerationperiod:30.0mvfviatfvv12.Supposeacarrollsdowna52.0-m-longafi27.8m/s0.00m/st5.20sinclinedparkinglotandisstoppedbyaffence.Ifittookthecar11.25storolldown5.35m/s2thehill,whatwastheaccelerationoftheWiththisaccelerationthedistancecarbeforestrikingthefence?duringaccelerationcanbedetermined.12dfdivitf2atfv2v22a(ddfifi)vi0.00m/ssincethecarstartsv2v2fifromrestdfdi2a2(dfdi)(2)(52.0m)2(27.8m/s)2(0.00m/s)2at2(11.25s)20.823m/s(2)(5.35m/s2)f72.3m13.AskydiverinfreefallreachesaspeedofDuringtheconstantspeedperiod,65.2m/swhensheopensherparachute.a0.00m/s2.Theparachutequicklyslowsherdownto2.7.30m/sataconstantrateof29.4m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.dvt(27.8m/s)(9.70s)Duringthisperiodofacceleration,howfar2.70102mdoesshefall?v2v22a(ddThetotaldistanceisthenthesumoffifi)thetwo.v2v2distance2.70102m72.3mfiddfi2a342m(7.30m/s)2(65.2m/s)2(2)(29.4m/s2)16.Acabdriverinahurryissittingataredlight.71.4mWhenthelightturnsgreensherapidlyaccel-eratesfor3.50sat6.80m/s2.ThenextlightShehasfallen71.4mduringtheaccel-isstillred.Shethenslamsonthebrakes,erationperiod.acceleratingatarateof9.60m/s2beforecomingtorestatthestoplight.Whatwasher14.Achildrollsaballupahillat3.24m/s.totaldistanceforthistrip?Iftheballexperiencesanaccelerationof2.32m/s2,howlongwillittakefortheballDuringthefirstpartofthetrip,tohaveavelocityof1.23m/sdownthehill?12dfdivitf2atfLetthepositivedirectionbeupthehill.vd1212)(3.50s)2fviatffdi2atf2(6.80m/s41.6m600SolutionsManualPhysics:PrinciplesandProblems

602Chapter3continuedForthesecondpart,firstdetermine19.Arockisthrownupwardwithaspeedofthespeedofthecabattheendofthe26m/s.Howlongafteritisthrownwillthefirstaccelerationperiod.rockhaveavelocityof48m/stowardthevground?fviatf(0.00m/s)(6.80m/s2)(3.50s)Takeupwardasthepositivedirection.v23.8m/sfviatfwhereagvDuringthesecondaccelerationperiod,fvi48m/s26m/st2v22a(dfg9.80m/s27.6svdfifi)v2v220.Thehigh-diveboardatmostpoolsisfiddfi2a3.00mabovethewater.Adivinginstructor2divesofftheboardandstrikesthewater(0.00m/s)(23.8m/s)(2)(9.60m/s2)1.18slater.29.5ma.Whatwastheinitialvelocityofthediver?Thetotaldistancetraveledisthenthed12whereagsumofthetwo.fdivitf2atfdistance41.6m29.5m71.1m1(d(g)t2fdi)2f17.Acyclistridesataconstantspeedof12.0m/svitffor1.20minandthencoaststoastopwith(3.00m)12)(1.18s)2uniformacceleration21.2slater.Ifthetotal2(9.80m/sdistancetraveledis1321m,thenwhatisthe1.18saccelerationwhilethebikecoaststoastop?3.24m/sDuringthefirstpartofthemotion,b.Howhighabovetheboarddidthetheridermovesatconstantspeed,diverrise?ora0.00m/s2.Atthehighestpoint,thespeedisThedistancecoveredisthen0.00m/s.dvt(12.0m/s)(72s)864m.v2v22a(ddThedistancecoveredduringthecoast-fifi)whereagingperiodisthen1321864457m.vf2vi2dd2v22a(dfi2(g)vdfifi)(0.00m/s)2(3.24m/s)2vf2vi2(0.00m/s)2(12.0m/s)2(2)(9.80m/s2)a2(d(2)(457m)fdi)0.536m0.158m/s218.Ahikertossesarockintoacanyon.Hehearsitstrikewater4.78slater.Howfardownisthesurfaceofthewater?Assumethatdownwardisthepositivedirection.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12dfdivitf2atf0.00m(0.00m/s)(4.78s)1(9.80m/s2)(4.78s)22112mPhysics:PrinciplesandProblemsSolutionsManual601

6033.Drawafree-bodydiagramforasubmergedChapter4beachballasitrisestowardthesurfacejustpages862–864afterbeingreleased.Identifytheforcesacting1.Drawafree-bodydiagramforthespaceonthebeachballandindicatethedirectionshuttlejustafteritleavestheground.ofthenetforceandtheacceleration.Identifytheforcesactingontheshuttle.Makesurethatyoudonotneglectairresis-tance.Alsobesurethatyouindicatethedirectionoftheacceleration,aswellasthenetforce.FthrustFbuoyantFairresistanceFgThenetforceandaccelerationarebothFnetaupward.FgFwater2.Drawafree-bodydiagramforagoldfishresistancethatismotionlessinthemiddleofafish-Thenetforceandaccelerationvectorsbowl.Identifytheforcesactingontheareupward.fish.Indicatethedirectionofthenetforceonthefishandthedirectionoftheacceler-4.Muturiisrearrangingsomefurniture.Heationofthefish.pushesthedresserwithaforceof143N,andthereisopposingfrictionalforceofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.112N.Whatisthenetforce?Fnet143N112N31N5.OneofthefloatsinaThanksgivingDayparaderequiresfourpeoplepullingonropestomaintainaconstantspeedof3.0km/hforthefloat.Twopeoplepullwithaforceof210Neach,andtheothertwopullwithaforceof140Neach.FbuoyantFgThenetforceandtheaccelerationarea.Drawafree-bodydiagram.bothzero.210N(2)Ff140N(2)602SolutionsManualPhysics:PrinciplesandProblems

604Chapter4continuedb.Whatistheforceoffrictionbetweenthec.Ifthebonehasamassof2.5kg,whatisfloatandtheground?itsacceleration?FFnet210N210N140Nnetma140NFfFnet7N2a3m/s0sincespeedisconstantm2.5kgF2Nf7.0108.Alargemodelrocketenginecanproduce6.Fivepeopleareplayingtug-of-war.Andersathrustof12.0Nuponignition.ThisandAlysonpulltotherightwith45Nandengineispartofarocketwithatotalmass35N,respectively.CalidandMarisolpulltoof0.288kgwhenlaunched.theleftwith53Nand38N,respectively.WithwhatforceandinwhatdirectiondoesBenitopullifthegameistied?Assignthecoordinatesystemsothatpositiveforcesaretotheright.Sincetheropeisnotaccelerating:FnetFAndersFAlysonFCalidFMarisolFBenitoa.Drawafree-bodyFthrust0N45N35N53N38NdiagramoftherocketFBenitojustafterlaunch.0N11NFBenitoFgb.WhatisthenetforcethatisactingonFBenito11NthemodelrocketjustafteritleavestheBenitopullstotheright.ground?F7.Twodogsfightoverabone.ThelargerofnetFthrustFgthetwopullsonthebonetotherightwith12.0N(0.288kg)(9.80m/s2)aforceof42N.Thesmalleronepullsto9.2Nupwardtheleftwithaforceof35N.c.Whatistheinitialaccelerationoftherocket?FnetmaFnetam1.20N(0.288kg)(9.80m/s2)Dog1Dog20.288kga.Drawthefree-bodydiagramforthebone.31.8m/s2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.35N42N9.Erikaisonanelevatorandpressesthebuttontogodown.Whentheelevatorb.Whatisthenetforceactingonthebone?firststartsmoving,ithasanaccelerationofF2downward.Erikaandtheelevatornet42N35N7Ntotheright2.5m/shaveacombinedmassof1250kg.Physics:PrinciplesandProblemsSolutionsManual603

605Chapter4continueda.Drawafree-bodydiagramforthe12.Duringaspacelaunch,anastronauttypicallyelevator.undergoesanaccelerationof3gs,whichFmeansheexperiencesanaccelerationthatTisthreetimesthatofgravityalone.Whatwouldbetheapparentweightofa205-kgastronautthatexperiencesa3-gliftoff?FgLetupbethenegativedirection.b.WhatisthetensioninthecablethatFnetFg,apparentFgmaprovidestheupwardforceontheelevatorcar?Fg,apparentFgmaFnetFTFgma(205kg)(9.80m/s2)F(205kg)(3)(9.80m/s2)TmaFg8040N(1250kg)(2.5m/s2)(1250kg)(9.80m/s2)13.AlfonsoandSarahliketogoskydivingtogether.Alfonsohasamassof88kg,and9100NupwardSarahhasamassof66kg.Whileinfreefall10.Nganhasaweightof314.5NonMarsandtogether,AlfonsopushesSarahhorizontallyaweightof833.0NonEarth.withaforceof12.3N.a.WhatisNgan’smassonMars?Fg,EarthmgEarthFg,Earth833.0Na.WhatisAlfonso’shorizontalacceleration?m85.0kgg9.80m/s2EarthFonSarahmSarahaSarahb.WhatistheaccelerationduetogravityFonSarahonMars,gMars?aSarahmSarahFCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.g,MarsmgMars12.3N20.19m/sF314.5N66kgg,MarsgMarsm85.0kgb.WhatisSarah’shorizontalacceleration?3.70m/s2FonAlphonsoFonSarahmAlphonsoaAlphonso11.AlexisonthewrestlingteamandhasamassFof85.3kg.Beingawhizatphysics,hereal-onAlphonso12.3Naizesthatheisoverthe830.0-NcutoffforhisAlphonsom88kgAlphonsoweightclass.Ifhecanconvincethetrainers20.14m/stomeasurehisweightintheelevator,whatmustbetheaccelerationoftheelevatorso14.A7.25-gbulletisfiredfromagun.Thethathejustmakeshisweightclass?muzzlevelocityofthebulletis223m/s.FnetFg,apparentFgmaAssumethatthebulletacceleratesataconstantratealongthebarrelofthegunFg,apparentFgabeforeitemergeswithconstantspeed.mThebarrelofthegunis0.203mlong.830.0kg(85.3kg)(9.80m/s2)Whataverageforcedoesthebulletexert85.3kgonthegun?0.0696m/s2byNewton’sthirdlawa0.0696m/s2downFonbulletFongun604SolutionsManualPhysics:PrinciplesandProblems

606Chapter4continuedFonbulletmbulletabulletv2v22afibullet(dfdi)27.5kg(223m/s)a5m/s2bullet(2)(0.203m)1.22102.5kgFonbulletmbulletabullet(0.00725kg)(1.22105m/s2)888Nso,888Nonthegun15.A15.2-kgpolicebatteringramexertsanaverageforceof125Nona10.0-kgdoor.a.Whatistheaverageaccelerationofthea.Whichlengthofstringwillbreakfirst?door?Afree-bodydiagramcanbedrawnByNewton’sthirdlaw,forthe2.5-kgmass,m1,andthetwomassestakentogether,mF1m2.LetdoorFramFT1bethetensioninthetopstring,FdoormdooradoorandFT2thetensioninthemiddleFstring,andFT3thetensionapplieddoor125N2adoorm10.0kg12.5m/sbythephysicsteacher.doorb.WhatistheaverageaccelerationoftheFT1batteringram?FrammramaramFram125N2FT2aramm15.2kg8.22m/sram16.Asademonstration,aphysicsteacherFgFT3attachesa7.5-kgobjecttotheceilingbyanearlymasslessstring.Thisobjectthensup-FT3Fgportsa2.5-kgobjectbelowitbyanother2.5kgm110.0kgm1m2pieceofstring.Finally,anotherpieceofFstringhangsoffthebottomofthelowerT2FT3Fg0objecttobepulledwitheverincreasingforceF2)FT2(2.5kg)(9.80m/sT3untilthestringbreakssomewhere.Thestring24.5NFwillbreakwhenthetensionreaches156N.T3FT1FT3Fg0F2)FT1(10.0kg)(9.80m/sT398.0NFT3RegardlessoftheappliedforceCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.FT1FT2FT3.Therefore,theupperstringwillbreakfirst.b.Whatisthemaximumdownwardforcethephysicsteachercanapplybeforethestringbreaks?Physics:PrinciplesandProblemsSolutionsManual605

607Chapter4continuedThestringbreakswhenFb.Whatisthenormalforceonthesculp-T1156N.FturewhenthetensioninthecableisT198.0NFT319,250N?FT3FT198.0N156N98.0NWhatisthenormalforceonthe58Nsculpturewhenthetensioninthecableis19,250N?17.A10.0-kgobjectisheldupbyastringFnetFTFNFg0thatwillbreakwhenthetensionexceeds1.00102N.AtwhatupwardaccelerationFNFgFTwillthestringbreak?(2225kg)(9.80m/s2)FnetFTFgma19,250NF2560NTFgam2560Nupward1.00102N(10.0kg)(9.80m/s2)10.0kg0.200m/s218.Alargesculptureisloweredintoplacebyacrane.Thesculpturehasamassof2225kg.Whenthesculpturemakescontactwiththeground,thecraneslowlyreleasestheten-sioninthecableasworkersmakefinaladjustmentstothesculpture’spositionontheground.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.a.Drawafree-bodydiagramofthesculp-turewhenitisincontactwiththeground,andthereisstilltensioninthecablewhiletheworkersmakethefinaladjustments.FTFNFg606SolutionsManualPhysics:PrinciplesandProblems

608Chapter5pages865–8671.Asoccerballiskickedfroma22-m-tallplatform.Itlands15mfromthebaseoftheplatform.Whatisthenetdisplacementoftheball?Sincethetwodistancesareperpendicular,usethePythagoreantheorem.R2A2B2RA2B2(22m)2(15m)227m2.Ifthenetdisplacementis32mforthesamesituation,asdescribedinproblem1,howfarfromtheplatformbasemusttheballland?R2A2B2BR2A2(32m)2(22m)223m3.Foranysingleforcevector,thereisonlyoneangleforwhichitsx-andy-componentsareequalinsize.a.Whatisthatangle?RsinRcossinRcosRtan1Rtan1(1)45°Rb.Howmanytimesbiggerisavectoratthisparticularanglethaneitherofitscomponents?R11Rsinsinsin45°1.4timesbigger4.Acueballonabilliardstabletravelsat1.0m/sfor2.0s.Afterstrikinganotherball,ittravelsat0.80m/sfor2.5satanangleof60.0°fromitsoriginalpath.a.Howfardoesthecueballtravelbeforeandafteritstrikes120.0°theotherball?Infirstlegitgoesxbeforevbeforetbefore(1.0m/s)(2.0s)2.0mandinsecondlegCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.xaftervaftertafter(0.80m/s)(2.5s)2.0mPhysics:PrinciplesandProblemsSolutionsManual607

609Chapter5continuedb.Whatisthenetdisplacementofthecueballfortheentire4.5-stimeinterval?Notethattheanglebetweenthetwolegsis120.0°,not60.0°.UsingthelawofcosinesR2x2y22xy(cos)Rx2y22xy(cos)(2.0m)2(2.0m)2(2)(2.0m)(2.0m)(cos120.0°)3.5m5.Thetablebelowrepresentsasetofforcevectors.Thesevectorsbeginattheoriginofacoordinatesystem,andendatthecoordinatesgiveninthetable.yVector#x-value(N)y-value(N)A0.06.0BB5.00.0C0.010.0Aa.Whatisthemagnitudeoftheresultantofthesumofthesethreevectors?CRxAxBxCxx0.0N5.0N0.0N5.0NRRyAyByCy6.0N0.0N(10.0N)4.0NRRx2Ry2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(5.0N)2(4.0N)26.4Nb.Whatisthesizeoftheangle,,thattheresultantmakeswiththehorizontal?Rtan1ytan14.0NR5.0Nx39°c.Intowhichquadrantdoestheresultantpoint?Becausethex-componentispositiveandthey-componentisnegative,theresultantpointsintoquadrant4.6.A9.0-kgcratesitsonalevel,roughfloor.A61-Nforceisneededjusttostartitmoving.Whatisthesizeofthecoefficientofmaximumstaticfriction?ifthereisnomotion,FappliedFfsFN;therefore,FFappliedappliedsFmgN61N0.69(9.0kg)(9.80m/s2)608SolutionsManualPhysics:PrinciplesandProblems

610Chapter5continued7.Giventhegraphbelow,answerthefollowingIfvisconstant,FappliedFfkFNquestions.Fmappliedbucketleadg4kFFNN(0.255kg)(9.80m/s2)312N(N)0.21fF2b.Ifsomeextrapelletsareadded,describe1thebehavioroftheblock.SinceFappliednowexceedsFf,0Fnet0,theblockwillaccelerate2468overtime.FN(N)a.Whatisthevalueof9.Aboattravels75kmsoutheast,then56kmkforthissystem?dueeast,then25km30.0°northofeast.kistheslopeoftheline.a.SketchthevectorsetonaN-E-S-Wgrid.4N0N0.5N8N0Nb.Ifthefrictionalforceis1.5N,whatisFN?WEFromthegraph,FN3Nc.DoestriplingFNtripleFapplied?SYes,theyaredirectlyproportional.b.FinditsnetE-Wcomponentofdisplace-d.DoFappliedandFNactinthesamementandnetN-Scomponentofdirection?Explainwhyorwhynot.displacement.No,thefrictionforceactshorizontallyREWAEWBEWCEWandoppositetoappliedforce,whilethenormalforceactsperpendicular(75km)(cos(45°))tothefixedsurface.(56km)(cos0.0°)(25km)(cos30.0°)8.Awoodenblocksitsonalevellabtable.Astringdrapedoverapulleyconnectstoa1.3102kmbucketthatcanbefilledwithleadpellets.RNSANSBNSCNSMaggiewantstomeasurehowmuchappliedmass(pelletsbucket)isneededtomove(75km)(sin(45°))theblockalongthetableataconstantspeed.(56km)(sin0.0°)(25km)(sin30.0°)41kmc.Finditsnetdisplacement.RREW2RNS2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(1.3102km)2(41km)2a.Iftheappliedmassofthebucketand136kmtheleadpelletsis0.255kg,andtheblockhasaweightof12N,whatisthevalueofk?Physics:PrinciplesandProblemsSolutionsManual609

611Chapter5continuedd.FinditsnetanglerelativetoanE-Waxis.FfFnetma(NonegativesignRherebecausethevalueofashouldtan1NSRincorporatethat.)EW141kmThus,tan21.310kmkFNma18°kmgma18°southofeastakg10.A1.25-kgboxisbeingpulledacrossalevelsurfacewherev2v2kis0.80.Ifthestringissud-v2v22ad,soafidenlycut,causingtheappliedforcetofi2dimmediatelygoto0.0N,whatistherateofv2v2fiaccelerationoftheblock?2dkgLetthedirectionoftheappliedforcebepositive.(v2v2)fiThenetforceontheboxisdueto2dgfrictionandisinthenegativedirection.((0.0m/s)2(21m/s)2)0.36FnetmaFfkFNkmg(2)(62m)(9.80m/s2)a2)b.Willtheanswerchangeifapuckofkg(0.80)(9.80m/sdifferentmass,butsamematerialand7.8m/s2inthenegativedirectionshapeisused?Explainwhyorwhynot.11.Ifthevelocityoftheboxinproblem10wasNo,theresultisindependentof5.0m/sattimezero,whatwillbeitsspeedmass.Notehowitcancelsinaboveafter0.50s?Howfarwillittravelinthatsolution.timeinterval?13.A105-Nsuitcasesitsonarubberrampatvfviatanairportcarouselata25°angle.TheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.v2)(0.50s)f5.0m/s(7.8m/sweightvectorcanbebrokenintotwo1.1m/sperpendicularcomponents.v2v22(5.0m/s)2fi(1.1m/s)105Nd2a(2)(7.8m/s2)1.5mFxFy12.A0.17-kghockeypuckleavesthestickonaslapshottraveling21m/s.Ifnootherforcesactonthepuck,frictioneventuallywillFgbringittorest62maway.a.Whatisthemagnitudeofthecompo-a.Basedonthesedata,determinethenentparalleltotherampsurface?valueofkforthishockeypuckonice.FLetthedirectionofmotionbeposi-xFwsin(105N)(sin25°)tive.Thus,theforceoffrictionisin44Nthenegativedirection.b.Whatisthemagnitudeofthecompo-FnentatarightangletotherampfkFN(Remember:frictionisinthenegativedirection.)surface?FyFwcos(105N)(cos25°)95N610SolutionsManualPhysics:PrinciplesandProblems

612Chapter5continuedc.Whichofthosecomponentsisthemgsinmagsinanormalforce?kmgcosgcosFyvfgsint14.Inproblem13,whichforceoffsetsthegcoskineticfrictionforce?IfFxexceedsFf,will(9.80m/s2)(sin42°)15m/sthesuitcaseacceleratedowntheramp?3.0s(9.80m/s2)(cos42°)Thecomponentparalleltoramp,Fx,offsetsthefrictionforce.SinceFnet0.21alongrampnolongerequalszero,thesuitcasewillaccelerate.17.A64-Nboxispulledbyaropeataconstantspeedacrossaroughhorizontalsurface.15.DaliladecidestodeterminethecoefficientIfthecoefficientofkineticfrictionis0.81,ofmaximumstaticfrictionforwoodagainstwhatisthemagnitudeoftheappliedforcewoodbyconductinganexperiment.Firstifthatforceisdirectedparalleltothefloor?sheplacesablockofwoodonasmallFappliedFfsinceFnet0plank;next,sheslowlyliftsoneendoftheplankupward.ShenotestheangleatwhichFfkFN(0.81)(64N)52Ntheblockjustbeginstoslide,andclaimsthat18.Supposethatinproblem17theappliedstan.Sheiscorrect.Showwhy.forceisdirectedbytheropeatanangle,,tothefloor.a.ThenormalforceisnolongersimplyFg.Showhowtocomputenetverticalforce.Fnet,yFg(Fappliedsin)b.Thefrictionalforcestillopposesthehorizontalmotionofthebox.Showhowtocomputenethorizontalforce.Ifthereisnomotion,FnetalongtheFnet,xFappliedcos(Ff)ramp0andFFappliedcoskFNxFwsinFfsFNsFwcosFwsinFappliedcoskFnet,yThus,tansFcoswThisworks,regardlessoftheweightoftheblock.16.Jonathan,withamassof81kg,startsatrestfromthetopofawaterslideangledat42°fromtheground.Heexitsatthebottom3.0slatergoing15m/s.Whatisk?vfviatwherevi0Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.vfatFnetmaFgFfFgkFnmgsinkmgcosPhysics:PrinciplesandProblemsSolutionsManual611

613Chapter6yElectronbeamdpages867–8681.Afootballplayerkicksafieldgoalfromadis-xtanceof45mfromthegoalpost.Thefootballdislaunchedata35°angleabovethehorizon-Deflectorplatestal.WhatinitialvelocityisrequiredsothatthefootballjustclearsthegoalpostcrossbarthatLocatetheoriginofthecoordinatesys-is3.1mabovetheground?Ignoreairresis-temmidwaybetweentheplatesatthetanceandthedimensionsofthefootball.leftedge.Therebyyi0andvyi0.TheoriginofthecoordinatesystemisThepositionofanelectronintheregionatthepointofthekick.oftheplatesisgivenby:vat2xivicosxvxit,y2vyivisinEliminatingtgivesthetrajectory:Lettbethetimewhenthefootballax2a2crossestheverticalplaneofthey22xvxi2vxiuprights.Then:Werequirethetrajectorytopassthroughxvxitthepointjustbeyondthedeflectionplate:xsota2vxiy2v2xxigt2yvThen,yit26m/s)20.010mx22v2y(2)(3.0102gaxixvxix2(0.100m)2vyiv2xi9.01012m/s2xgx2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(visin)vicos2(v23.Askateboardtrackhasahorizontalsegmenticos)2followedbyarampthatdeclinesata45°gx(tan)(x)2v2(cos)2angle,asshown.igx2vi2(cos)2((tan)(x)y)(9.80m/s2)(45m)2(2)(cos35°)2((tan35°)(45m)3.1m)23m/s45°2.Inacertaincathode-raytube,abeamofelectrons,movingataconstantvelocity,entersaregionofconstantelectricforcea.Howlongwouldtherampneedtobetomidwaybetweentwoparallelplatesthatareprovidealandingforaskateboarderwho10.0cmlongand1.0cmapart.Inthislaunchesfromthehorizontalsegmentatregion,theelectronsexperienceanaccelera-avelocityof5.0m/s?tion,a,towardtheupperplate.Iftheelec-Theskateboarder’strajectoryistronsenterthisregionatavelocityofgivenby:3.0106m/s,whatistheaccelerationthatg2,for0xxyxneedstobeappliedsothattheelectrons2v2Lxijustmisstheupperdeflectionplate?612SolutionsManualPhysics:PrinciplesandProblems

614Chapter6continuedTherampcanberepresentedbytheIfthecoordinatesystemischosenequation:sothattheoriginisontheflooratyxthepointtheballisthrown,thentheheightoftheballovertimeis:Byeliminatingy,wecanfindtheval-gt2uesofxthatsatisfybothequations.yyivyit2gx2xgx1x022Attimet,themaximumheightis2v2vxixi2vyigvyi2x0,x2vxixymaxyivyig2ggLThesecondsolutionrepresentsthev2gv2yiyiypointwheretheskateboarderreturnsig2g2totheramp.v2v2(sin)2yii(2)(5.0m/s)2yiyix2g2gL9.80m/s25.1m(10.0m/s)2(sin68°)2yLxL5.1m1.50m2)(2)(9.80m/sx5.9mLLrampcosxL5.1mb.Ifthebasketrimheightis3.05m,howLrampcoscos45°7.2mfarabovetherimistheball?ThenegativesignisbecauseoftheThehorizontalpositionoftheballcoordinatesystemassigned.Theovertimeis:lengthoftherampisapositivexvxit(vicos)(t)value:7.2m.xsotb.Iftheinitialvelocityisdoubled,whatvicoshappenstotherequiredlengthoftheTheheightoftheballthenis:ramp?gt2Notethattherequiredramplengthisyrimyivyit2proportionaltothesquareoftheini-xtialvelocity.Therefore,therequiredyivisinvicosramplengthisquadrupledto28.8m.gx22v4.Inanattempttomakea3-pointshotfromicosadistanceof6.00m,abasketballplayergx2yloftstheballatanangleof68°abovetheitanxv2(cos)2ihorizontal.Theballhasaninitialvelocityof10.0m/sandaninitialdistancefromthe1.50m(tan68°)(6.00m)floorof1.50m.(9.80m/s2)(6.00m)2a.Whatisthemaximumheightreached(2)(10.0m/s)2(cos68°)2bytheball?3.78mTheverticalcomponentoftheveloc-heightaboverim3.78m3.05mityovertimeis:0.73mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.vyvyigtTheballis0.73mabovetherim.Themaximumheightoccursattimet5.Howfardoesabaseballthatisthrownhori-1whentheverticalvelocityiszero.zontallyat42.5m/sdropoverahorizontalvyitdistanceof18.4m?gPhysics:PrinciplesandProblemsSolutionsManual613

615Chapter6continuedThetimeneededbythebaseballtotravel(2)(vRicos)(visin)18.4misgxvx(v2)(2)(sin)(cos)xt,sotvixgInthistimeintervaltheballdropsandusethetrigonometricidentitygt2gx2y2sincossin2toget22vx2vRisin29.80m/s218.4m2g242.5m/sb.Whatlaunchangle,,resultsinthe0.918mlargestrange?sinehasamaximumvalueat90°so,6.Aprojectileislaunchedfromzeroheight290°withaninitialangle,,abovethehorizon-talandwithaninitialvelocity,v45°i.a.Showthattherangeoftheprojectile—Rismaximumwhen45°thedistancefromthelaunchpointat7.Ifaringwereconstructedaspartofaspacewhichtheheightisagainzero—isstation,howfastmusta50.0-m-radiusringgivenby:rotatetosimulateEarth’sgravity?v2Risin2Forobjectsinuniformcircularmotion,gv2thecentripetalaccelerationisaLocatetheoriginofthecoordinatecr.systematthepointoflaunch.TheTosimulateEarth’sgravity,settheprojectilecoordinatesovertimearecentripetalaccelerationtog.givenby:v2acrg12xvxit,yvyit2gtv2grCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Eliminatingtgivesthetrajectoryvgrpath.vyig2(9.80m/s2)(50.0m)22.1m/syxxvxi2v2xiTherequiredperiodTisvyig(2)(50.0m)xxT2r14.2svxi2v2v22.1m/sxiSettingy0givesthetworootsof8.Aturntableforvinylrecordsworksbycon-thisequation.strainingtheneedletotrackinsideagroovevyiginaverycloseapproximationofuniformx0,x0vxi2v2xicircularmotion.IftheturntablerotationalThesecondisthesolutionofinterest.1speedis33rpm,whatistheneedle’s3SolveforxandrenameitasR.centripetalaccelerationwhenitis14.6cm2vxivyifromthecenter?Rg1Substitute:(33rev/min)0.55rev/s3vxivicos,vyivisin,T11.80s0.556rev/s42r42(0.146m)2acT21.80s3.20m/s614SolutionsManualPhysics:PrinciplesandProblems

616Chapter6continued9.AparkrideisdesignedsothattheriderisThedirectionis:ontheedgeofarevolvingplatform,3.5mvRP3.7m/sfromtheplatform’scenter.Thisplatformistan1tan1v4.2m/sPQmounted8.0mfromthecenterofalarger41°northofeastrevolvingplatform.Thesmallerplatformmakesonerevolutionin6.0sandcom-10.Twoobjectsareplacedonaflatturntablepletes2.0revforeachrevolutionoftheat10.0cmand20.0cmfromthecenter,largerplatform.Allrotationsarecounter-respectively.Thecoefficientofstaticfrictionclockwise.Attheinstantshown,whatarewiththeturntableis0.50.Theturntable’sthemagnitudeandthedirectionoftherotationalspeedisgraduallyincreased.rider’svelocitywithrespecttotheground?a.Whichofthetwoobjectswillbegintoslidefirst?Why?Theouterobjectslidesfirst.The42rcentripetalaccelerationacT2isQproportionaltotheradius,r.Thecentripetalaccelerationisprovidedbythefrictionalforceuptothelimit8.0mvRPspecifiedbythecoefficientofstaticPfriction.Therefore,atanygivenrota-vPQRtionalspeed,thefrictionalforceisgreaterfortheobjectatthelarger3.5mradius,andtheforceonthatobjectreachesthelimitfirst.LetthevelocityoftheriderRwithrespecttopointPbevb.AtwhatrotationalspeeddoestheinnerRP.TheperiodofplatformPis6.0s,sothemagnitudeobjectbegintoslide?anddirectionofvRPisLetvbethevelocityinm/s.Then,2rv22r(2)(3.5m)v,andavRPTcr.T6.0sAtthemaximumangularvelocity,3.7m/s,norththefrictionalforceFfsFNTheperiodofplatformQis(2.0)(6.0s)smgmustbeequaltomac.12s,sothemagnitudeanddirectionmv2ofvsmgrPQisv2r(2)(8.0m)vsgrPQT12s4.2m/s,east(0.50)(9.80m/s2)(0.10m)Thevelocityoftheriderwithrespectto0.70m/sthegroundis11.Anairplane’sairspeedis2.0102km/hvRQvRPvPQdueeast.BecauseofawindblowingtovRQvRP2vPQ2thenorth,itisapproachingitsdestination15°northofeast.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.vRQ(3.7m/s)2(4.2m/s)2a.Whatisthewindspeed?5.6m/svwindtanvairPhysics:PrinciplesandProblemsSolutionsManual615

617Chapter6continuedvwindvairtanChapter7(2.0102km/h)(tan15°)pages868–86954km/h1.Ayearisdefinedasthetimeittakesforab.Howfastistheairplaneapproachingitsplanettotravelonefullrevolutionarounddestination?theSun.OneEarthyearis365days.UsingthedatainTable7-1,calculatethenumbervcosairofEarthdaysinoneNeptuneyear.IfvgroundNeptunetakesabout16htocompleteonevvairofitsdays,howmanyNeptuniandayslonggroundcosisNeptune’syear?2.0102km/h23TNrNcos15°TrEE2.1102km/hT2r3TENNr312.Ariverisflowing4.0m/stotheeast.AEboateronthesouthshoreplanstoreacha(365days)2(4.501012m)3dockonthenorthshore30.0°downriverby(1.501011m)3headingdirectlyacrosstheriver.6.00104EarthdaysForevery16Earthdays,therewillbe24Neptuniandays,4Earthdays)24Neptuniandays(6.001016Earthdays30.0°4Neptuniandays9.00102.SupposeanewplanetwasdiscoveredtoorbittheSunwithaperiod5timesthatCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ofPluto.UsingthedatainTable7-1,calcu-latetheaveragedistancefromtheSuna.Whatshouldbetheboat’sspeedrelativeforthisnewplanet.tothewater?T2r3AALetvb/wbetheboat’sspeedrelativeTPrPtothewater.Then,LetTPperiodforPlutovtanwvb/wTAperiodfornewplanet5TPvw4.0m/s3T2r3vrAPb/wtantan30.0°AT2P6.9m/s3(5T2(5.871012m)3b.Whatistheboat’sspeedrelativetotheP)T2dock?P221.721013mvb/dvwvb/w(4.0m/s)2(6.9m/s)23.IfameteoritehittheEarthandmovedit2.411010mclosertoVenus,howmany8.0m/sdayswouldtherebeinanEarthyear?UsethedatainTable7-1.616SolutionsManualPhysics:PrinciplesandProblems

618Chapter7continuedNewaveragedistancefromtheSunT2r3AATrBBT2r3TBAAr3B(365days)2(1.501011m2.411010m)3(1.501011m)3281days4.TheMoonisasatelliteoftheEarth.TheMoontravelsonefullrevolutionaroundtheEarthin27.3days.GiventhatthemassoftheEarthis5.971024kg,whatistheaveragedistancefromtheEarthtotheMoon?Lookatunitsfirst:Theperiodisgiventousindays,whichmustbeconvertedtosecondsduetotheunitsofG.24h60min60s6s(27.3days)2.36101day1h1minr3T2GMET2GM3Er243(2.36106s)2(6.671011Nm2/kg2)(5.971024kg)423.83108m5.Asatellitetravels7.18107mfromthecenterofoneoftheplanetsinoursolarsystemataspeedof4.20104m/s.UsingthedatainTable7-1,identifytheplanet.GMvrv2r(4.20104m/s)2(7.18107m)27kg,JupiterM1.9010G6.671011Nm2/kg26.Earth’satmosphereisdividedintofourlayers:thetroposphere(0–10km),thestratosphere(10–50km),themesosphere(50–80km),andthethermosphere(80–500km).Whatistheminimumvelocityanobjectmusthavetoenterthethermosphere?rhr4m6.38106mE8.010GMvEr(6.671011Nm2/kg2)(5.971024kg)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8.0104m6.38106m7.8103m/sPhysics:PrinciplesandProblemsSolutionsManual617

6196.A92-kgmanusesa3.05-mboardtoattemptChapter8tomoveaboulder,asshowninthediagrampages869–870below.Hepullstheendoftheboardwitha1.Therotationalvelocityofamerry-go-forceequaltohisweightandisabletomoveroundisincreasedataconstantratefromitto45°fromtheperpendicular.Calculate1.5rad/sto3.5rad/sinatimeof9.5s.thetorqueapplied.Whatistherotationalaccelerationofthemerry-go-round?(3.5rad/s)(1.5rad/s)t9.5s0.21rad/s245°2.Arecordplayer’sneedleis6.5cmfromthecenterofa45-rpmrecord.Whatisthevelocityoftheneedle?Firstconvertrpmtorad/s.45rev2rad1minFrsin4.71rad/s1min1rev60smgrsin1mvr(6.5cm)(4.71rad/s)(92kg)(9.80m/s2)(3.05m)(sin45°)100cm0.31m/s1.9103Nm3.Supposeabaseballrolls3.2macrossthe7.Ifa25-kgchildtriestoapplythesametorquefloor.Iftheball’sangulardisplacementisasinthepreviousquestionusingonlyhisor82rad,whatisthecircumferenceoftheball?herweightfortheappliedforce,whatwouldthelengthoftheleverarmneedtobe?drd3.2mLrsinr82rad0.039m1.9103NmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Fmg(25kg)(9.80m/s2)c2r2(0.039m)0.25m7.8m4.Apainterusesa25.8-cmlongscrewdrivertoprythelidoffofacanofpaint.Ifaforceof8.Logan,whosemassis18kg,sits1.20m85Nisappliedtomovethescrewdriver60.0°fromthecenterofaseesaw.IfShiromustfromtheperpendicular,calculatethetorque.sit0.80mfromthecentertobalanceFrsin(85N)(0.258m)(sin60.0°)Logan,whatisShiro’smass?19NmFLrLFSrSmLgrLmSgrS5.Aforceof25Nisappliedverticallyatthemendofawrenchhandlethatis45cmlongmLrLSrtotightenaboltintheclockwisedirection.SWhattorqueisneededbythebolttokeep(18kg)(1.20m)thewrenchfromturning?0.80mInordertokeepthewrenchfromturn-27kging,thetorqueontheboltmustbeequal9.Twoforces—55-Nclockwiseand35-Ninmagnitudebutoppositeindirectionofcounterclockwise—areappliedtoamerry-thetorqueappliedbythewrench.go-roundwithadiameterof4.5m.WhatisFrsin(25N)(0.45m)(sin90.0°)thenettorque?11Nmcounterclockwise618SolutionsManualPhysics:PrinciplesandProblems

620Chapter8continued12.Amerry-go-roundintheparkhasaradiusof2.6mandamomentofinertiaof1773kgm2.Whatisthemassofthemerry-go-round?4.512mImr22I(2)(1773kgm2)35Nmr2(2.6m)255N5.2102kg13.Themerry-go-rounddescribedintheprevi-Thenettorqueisthesumoftheousproblemispushedwithaconstantforceindividualtorques.of53N.Whatistheangularacceleration?netnetccwcwIFccwrsinFcwrsinFrsin1(FccwFcw)rsin(53N)(2.6m)(sin90.0°)(35N55N)4.5m(sin90.0°)1773kgm227.8102rad/s245Nm45Nmclockwise14.Whatistheangularvelocityofthemerry-go-rounddescribedinproblems12and1310.Astudentsitsonastoolholdinga5.0-kgafter85s,ifitstartedfromrest?dumbbellineachhand.Heextendshisarmsfisuchthateachdumbbellis0.60mfromthettaxisofrotation.Thestudent’smomentofinertiais5.0kgm2.Whatisthemomentofi0sinceitstartedfromrestinertiaofthestudentandthedumbbells?ftI2(5.0kg)(0.60m)2(7.8102rad/s2)(85s)singledumbbellmr1.8kgm26.6rad/sItotal2IsingledumbbellIstudent15.Anice-skaterwithamomentofinertiaof(2)(1.8kgm2)5.0kgm21.1kgm2beginstospinwithherarmsextended.After25s,shehasanangular8.6kgm2velocityof15rev/s.Whatisthenettorqueactingontheskater?11.Abasketballplayerspinsabasketballwitharadiusof15cmonhisfinger.ThemassofnetItheballis0.75kg.WhatisthemomentofI(fi)inertiaaboutthebasketball?tI2mr222(1.1kgm2)(15rev/s0rev/s)55(0.75kg)(0.15m)255Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6.8103kgm22radrev4.1NmPhysics:PrinciplesandProblemsSolutionsManual619

621Chapter8continued16.Aboardthatis1.5mlongissupportedintwoplaces.Iftheforceexertedbythefirstsupportis25Nandtheforcedexertedbythesecondis62N,whatisthemassoftheboard?FnetF1F2(Fg)Sincethesystemisinequilibrium,Fnet0.0F1F2FgFgF1F2mgF1F2F1F225N62Nmg9.80m/s28.9kg17.Achildbeginstobuildahouseofcardsbylayingan8.5-cm-longplayingcardwithamassof0.75gacrosstwootherplayingcards:supportcardAandsupportcardB.IfsupportcardAis2.0cmfromtheendandexertsaforceof1.5103N,howfarfromtheendissupportcardBlocated?LettheaxisofrotationbeatthepointsupportcardAcomesincontactwiththetopcard.FnetFAFB(Fg)Sincethesystemisinequilibrium,Fnet0.0FAFBFgFBFgFACopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mgFASincetheaxisofrotationisaboutsupportcardA,A0.sonetBgThesystemisinequilibrium,sonet0.0BgBgBrBFBandgrgFgrBFBrgFgrrgFggmgrBFmgFBA1(0.085m)0.020m(7.5104kg)(9.80m/s2)2(7.5104kg)(9.80m/s2)1.5103N2.8102m28cm620SolutionsManualPhysics:PrinciplesandProblems

622Chapter8continued18.IfsupportcardAinthepreviousproblemwasmovedsothatitnowis2.5cmfromtheend,howfarfromtheotherenddoessupportcardBneedtobetoreestablishequilibrium?rgmgrBmgFA1(0.085m)0.025m(7.5104kg)(9.80m/s2)2(7.5104kg)(9.80m/s2)1.5103N2.2102m22cmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual621

623Chapter9pages870–8711.Aballwithaninitialmomentumof6.00kgm/sbouncesoffawallandtravelsintheoppositedirectionwithamomentumof4.00kgm/s.Whatisthemagnitudeoftheimpulseactingontheball?Choosethedirectionawayfromthewalltobepositive.pf4.00kgm/spi6.00kgm/sImpulsepfpi(4.00kgm/s)(6.00kgm/s)10.0kgm/s2.Iftheballinthepreviousprobleminteractswiththewallforatimeintervalof0.22s,whatistheaverageforceexertedonthewall?ImpulseFtImpulse10.0kgm/sF45Nt0.22s3.A42.0-kgskateboardertravelingat1.50m/shitsawallandbouncesoffofit.Ifthemagnitudeoftheimpulseis150.0kg·m/s,calculatethefinalvelocityoftheskateboarder.Choosethedirectionawayfromthewalltobepositive.Impulsepfpimvf(mvi)m(vfvi)ImpulsevCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fmvi150.0kgm/s1.50m/s42.0kg2.07m/s4.A50.0-gtoycartravelingwithavelocityof3.00m/sduenorthcollideshead-onwithan180.0-gfiretrucktravelingwithavelocityof0.50m/sduesouth.Thetoyssticktogetherafterthecollision.Whatarethemagnitudeanddirectionoftheirvelocityafterthecollision?Choosenorthtobepositive.pipfpciptipfmcvcimtvti(mcmt)vfmvcvcimtvtifmcmt(50.0g)(3.00m/s)(180.0g)(0.50m/s)50.0g180.0g0.26m/s,duenorth622SolutionsManualPhysics:PrinciplesandProblems

624Chapter9continued5.A0.040-kgbulletisfiredintoa3.50-kgThesecondskatermoveswestwithablockofwood,whichwasinitiallyatrest.velocityof1.3m/s.Thebulletremainsembeddedwithintheblockofwoodafterthecollision.The8.Supposea55.0-kgice-skater,whowasbulletandtheblockofwoodmoveatainitiallyatrest,firesa2.50-kggun.Thevelocityof7.40m/s.Whatwastheoriginal0.045-kgbulletleavesthegunatavelocityvelocityofthebullet?of565.0m/s.Whatisthevelocityoftheice-skateraftershefiresthegun?pipfppipfbipwipf0pmfspfbbvbimwvwi(mbmw)vf(mwherevsmg)vfsmbvfbwi0.Thus,becausethefinalmassoftheskater(0.040kg3.50kg)(7.40m/s)includesthemassofthegunheldbyvbi0.040kgtheskater.Then,6.5102m/smvbvfbfsmsmg6.BallA,withamassof0.20kg,strikesballB,(0.045kg)(565.0m/s)withamassof0.30kg.Theinitialvelocity55.0kg2.50kgofballAis0.95m/s.BallBisinitiallyatrest.Whatarethefinalspeedanddirection0.44m/sofballAandBafterthecollisionifthey9.A1200-kgcannonisplacedatrestonsticktogether?anicerink.A95.0-kgcannonballisshotpipffromthecannon.IfthecannonrecoilsatmAvAi(mAmB)vfaspeedof6.80m/s,whatisthespeedofmthecannonball?vAvAifmpAmBipf(0.20kg)(0.95m/s)Sincethecannonandcannonballareat0.20kg0.30kgrestbeforetheblast,pi0.00kgm/s.0.38m/sinthesamedirectionasSopballA’sinitialvelocityfcpfbmcvfcmbvfb7.Anice-skaterwithamassof75.0kgpushesmvcvfcoffagainstasecondskaterwithamassoffbmb42.0kg.Bothskatersareinitiallyatrest.(1200kg)(6.80m/s)Afterthepush,thelargerskatermovesoff95.0kgwithaspeedof0.75m/seastward.Whatisthevelocity(magnitudeanddirection)of86m/sthesmallerskaterafterthepush?10.An82-kgreceiver,moving0.75m/snorth,pipfistackledbya110.0-kgdefensivelinemanm1vf1m2vf2moving0.15m/seast.Thefootballplayershitthegroundtogether.CalculatetheirfinalmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.v1vf1f2mvelocity(magnitudeanddirection).2p(75.0kg)(0.75m/s)rimrvri,y(82kg)(0.75m/s)42.0kg62kgm/snorth1.3m/spdimdvdi,x(110.0kg)(0.15m/s)16kgm/s,eastPhysics:PrinciplesandProblemsSolutionsManual623

625Chapter9continuedLawofconservationofmomentumstatespipfpf,xpi,x16kgm/spf,ypi,y62kgm/s1p2p2)2f(pf,xf,y1((16kgm/s)2(62kgm/s)2)264kgm/spf64kgm/svfm82kg110.0kgrmd0.33m/sp1f,ytanpf,xm1rvri,ytanmdvdi,x1(82kg)(0.75m/s)tan(110.0kg)(0.15m/s)75°11.A985-kgcartravelingsouthat29.0m/shitsatrucktraveling18.0m/swest,asshowninthefigurebelow.Afterthecollision,thevehiclessticktogetherandtravelwithafinal29.0m/smomentumof4.0104kgm/satCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.anangleof45°.Whatisthemass18.0m/softhetruck?45°p2p2p2ffxfyp2p2p2fixiyp2m2v2m2v2fccitti12m2v224pp4.010kgm/smfccitv2ti14kgm/s)(985kg)2(29.0m/s)2(4.0102(18.0m/s)21.6103kg624SolutionsManualPhysics:PrinciplesandProblems

626Chapter9continued12.A77.0-kgwomaniswalking0.10m/seastinthegym.Amanthrowsa15.0-kgballsouthandaccidentallyhitsthewoman.Thewomanandtheballmovetogetherwithavelocityof0.085m/s.Calculatethedirectionthewomanandtheballmove.pcos1fxpfpcos1ixpfmcos1wviw(mwmb)vf(77.0kg)(0.10m/s)cos1(77.0kg15.0kg)(0.085m/s)1.0101degreessouthofeastCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual625

6275.Ifthetwo-blocksystemdescribedintheChapter10previousproblemwasinitiallyatrest,whatpage871isthefinalvelocity?1.Atoytruckispushedacrossatable0.80m12KEmvnorth,andpulledbackacrossthetable20.80msouth.Ifaconstanthorizontalforce12m(vof15Nwasappliedinbothdirections,2fvi)whatisthenetwork?vi0sinceblockswereinitiallyatrest.Choosenorthtobethepositivedirection.12SoKEmv2fWnetWnorthWsouth2KE(2)(77J)(Fd)(Fd)vfm15kg(15N)(0.80m)(15N)(0.80m)3.2m/s0.0J6.Atoycarwithamassof0.75kgispulled2.A15-kgchildexperiencesanaccelerationof3.2macrossthefloorwithaconstantforce0.25m/s2assheispulled1.7mhorizontallyof110N.If67Jofworkisdone,whatistheacrossthefloorbyhersister.Calculatetheupwardangleoftheforce?changeinthechild’skineticenergy.WFdcosKEWFdmad1Wcos(15kg)(0.25m/s2)(1.7m)Fd6.4J167Jcos79°(110N)(3.2m)3.Amanpushesacouchadistanceof0.75m.7.Ifa75-Wlightbulbisleftonfor2.0h,howIf113Jofworkisdone,whatisthemagni-muchworkisdone?tudeofthehorizontalforceapplied?WFd3600s3s(2.0h)7.2101.0hCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.W113J2NF1.510Wd0.75mPt4.TwoblocksaretiedtogetherbyahorizontalWPt(75W)(7.2103s)stringandpulledadistanceof2.7macross5.4105Janairhockeytablewithaconstantforceof35N.Theforceisdirectedatanupward8.A6.50-horsepower(hp)self-propelledangleof35°fromthe9.0-kgblock,aslawnmowerisabletogofrom0.00m/sshowninthefigure.Whatisthechangeinto0.56m/sin0.050s.Ifthemassofthekineticenergyinthetwo-blocksystem?lawnmoweris48.0kg,whatdistancedoesthelawnmowertravelinthistime?35N(Use1hp746W.)35°746W3W6.0kg9.0kg(6.50hp)1hp4.8510vatWKEFdcos(35N)(2.7m)(cos35°)77J626SolutionsManualPhysics:PrinciplesandProblems

628Chapter10continuedW11.Aworkerusesapulleytolifta45-kgobject.PtIfthemechanicaladvantageofthepulleyisFdmad5.2,whatistheeffortforceexertedbythettworker?vFrmtdm(vMAfvi)dFett2Frmg(45kg)(9.80m/s2)Pt2FedMAMA5.2m(vfvi)85N(4.85103W)(0.050s)2(48.0kg)(0.36m/s0.00m/s)12.Whenthechainonabicycleispulled0.45m0.95cm,therearwheelrimmovesadistanceof14cm.Ifthegearhasaradius9.Awinchthat’spoweredbya156-Wmotorof3.5cm,whatistheradiusoftherearliftsacrate9.8min11s.Whatisthemasswheel?ofthecrate?drIMAeed9.8mdrvrravgt11s0.89m/sredr(3.50cm)(14.0cm)Ifthecrateisinitiallyatrest:rrd0.95cmevvvivffavg2252cmvf2vavg(2)(0.89m/s)1.8m/sWKEmv2Ptt2t2Pt2Ptm22v(vfvi)(2)(156W)(11s)(1.8m/s)21.1103kg10.Amanexertsaforceof310Nonalevertoraiseacratewithamassof910kg.Iftheefficiencyoftheleveris78percent,whatisthelever’sIMA?FmgrFeFee(100)e(100)(910kg)(9.80m/s2)310N(100)7837Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual627

629Chapter11KEWFd(67N)(5.5m)3.7102Jpage8721.Acratewithamassof210kgishorizontallySincethecratewasinitiallyatrest,acceleratedbyaforceof95N.TheforceKEKEfisdirectedatanupwardanglesothattheKE2Jverticalpartoftheforceis47.5Nandf3.710thehorizontalpartoftheforceis82.3N4.A150-kgroller-coastercarclimbstothe(seethediagrambelow).Ifthecrateistopofa91-m-highhill.Howmuchworkpulled5.5macrossthefloor,whatistheisdoneagainstgravityasthecarisliftedchangeinkineticenergyofthecrate?tothetopofthehill?47.5N95NWFdFmg82.3NWmgd210kg(150kg)(9.80m/s2)(91m)Smoothfloor1.3105JWKEFd5.Apendulumbobwithamassof0.50kg(82.3N)(5.5m)swingstoamaximumheightof1.0m.What4.5102Jisthekineticenergywhenthependulumbobisataheightof0.40m?2.Assumingthatthecratedescribedinprob-EtotKEPElem1wasinitiallyatrest,whatisthefinalSinceenergyisconserved,thetotalvelocityofthecrate?energyisconstant.KEKEfKEiWhenthependulumbobisatthetopofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1mv2theswing,EKEtotPEatthetopofthef2fswing.12KEi2mviPEmgh(0.50kg)(9.80m/s2)(1.0m)Sincethecratewasinitiallyatrest,4.9JKEi0.Whenthependulumbobisat0.40m,So,KEKE1mv2PEmgh(0.50kg)(9.80m/s2)(0.40m)f2f2.0J2KE(2)(4.5102J)vfm210kgKEEtotPE4.9J2.0J2.1m/s2.9J6.Asledanditsrider,withatotalmassof3.Ifthecratedescribedinproblem195kg,areperchedontopofa25-m-tallexperiencedafrictionalforceof15N,hill.Asecondhillis12mtall(seethedia-whatisthefinalkineticenergyofthecrate?grambelow).TheriderisgivenaninitialLetthedirectionofthemotionofthepushproviding3674Jofkineticenergy.cratebepositive.Neglectingfriction,whatisthevelocityattheWKEFdtopofthetopofthe12-m-tallhill?FFHFf82.3N(15N)67N628SolutionsManualPhysics:PrinciplesandProblems

630Chapter11continuedPEFghPE3.0104J2NF8.61025mgh35m12m9.Thebighillonaroller-coasterrideis91mtall.Ifthemassoftheroller-coastercaranditstworidersis314kgandthemaximumEKEPEvelocityreachedbythisroller-coasterrideAtthetopofthe25-m-highhill:is28m/s,howmuchenergywaslosttofriction?KE3674J2)(25m)Atthetopofthe91-m-highhill:PEmgh(95kg)(9.80m/sEPEmgh2.3104J(314kg)(9.80m/s2)(91m)EKEPE3674J2.3104J2.8105J2.7104JAtthebottomofthehill:Atthetopofthe12-m-highhill:122)(12m)EKE2mvPEmgh(95kg)(9.80m/s1.1104J2KE(2)(2.8105J)vm314kgKEEPE2.7104J1.1104J42m/sifnoenergyislosttofriction1.6104JTheactualKEatthebottomofthehill:2KE(2)(1.6104J)vKE1mv212m95kg(314kg)(28m/s)2218m/s5J1.210EKEE7.A35-kgchildisridingonaswingthatrisesftoamaximumheightof0.80m.NeglectingE5J1.2105JfEKE2.810friction,whatisthechild’sgravitationalpotentialenergyatthetopoftheswing?1.6105JWhatisthechild’skineticenergyatthetopoftheswing?10.Adartboardwithamassof2.20kgissuspendedfromtheceilingsuchthatitisPEmgh(35kg)(9.80m/s2)(0.80m)initiallyatrest.A0.030-kgdartisthrown2.7102Jatthedartboardwithavelocityof1.3m/s.Afterthedarthitsthedartboard,thedartWhentheswingisatitsmaximumandtheboardinitiallymovewithavelocityheight,thevelocityiszero.of0.025m/s.HowmuchkineticenergyisKE1mv20Jlostinthesystem?2KE1m212i2dvd2(0.030kg)(1.3m/s)8.AsledanditsriderareperchedatthetopofCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.ahillthatis35mtall.Ifthegravitational2.5102Jpotentialenergyis3.0104J,whatisthe12weightofthesledandrider?KEf2(mdmb)vfPEmgh12(0.030kg2.20kg)(0.025m/s)2Fgmg6.97104JPhysics:PrinciplesandProblemsSolutionsManual629

631Chapter11continuedE12.A0.150-kgballthatisthrownatavelocityoflostKEiKEf30.0m/shitsawallandbouncesbackinthe(2.5102J)(6.97104J)oppositedirectionwithaspeedof25.0m/s.2.4102JHowmuchworkwasdonebytheball?Conservationofenergy:11.Inaphysicslaboratory,studentscrashcartsWtogetheronafrictionlesstrack.AccordingballKE0tothefollowingdata,waskineticenergyWballKEKEfKEiconserved?KE1mv21(0.150kg)(30.0m/s)2i2i2Mass(kg)v(m/s)v(m/s)ifCartA0.250.180.2167.5JCartB0.360.200.11KE1mv21(0.150kg)(25.0m/s)2f2f21246.9JKEmiA2AviAWballKEfKEi122(0.25kg)(0.18m/s)46.9J67.5J4.0103J20.6J12KEmiB2BviB12(0.36kg)(0.20m/s)27.2103J12KEmfA2AvfA122(0.25kg)(0.21m/s)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5.5103J12KEmfB2BvfB12(0.36kg)(0.11m/s)22.2103JIfKEisconserved:KEiKEfKEiKEiAKEiB4.0103J7.2103J11.2103JKEfKEfAKEfB5.5103J2.2103J7.7103JKEisnotconserved.630SolutionsManualPhysics:PrinciplesandProblems

632e.298KChapter12TCTK27329827325°Cpages872–874f.316K1.ConvertthefollowingCelsiustemperaturestoKelvintemperatures.TCTK27331627343°Ca.196°C4.A9.8-gleadbulletwithamuzzlevelocityTKTC27319627377Kof3.90102m/sisstoppedbyawoodenb.32°Cblock.WhatisthechangeintemperatureofTthebulletifone-fourthofitsoriginalkineticKTC27332273305Kenergygoesintoheatingthebullet?c.212°Cv3.90102m/sTKTC273212273485Kd.273°CWoodenBulletTblockKTC2732732730Km9.8ge.273°C12TKEmvKTC273273273546K2f.27°C1andKEQmCTT4KTC27327273300K11mv2mCT422.FindtheCelsiusandtheKelvintempera-turesforthefollowingobjects.1v28a.averagebodytemperatureTC98.6°Fisabout37°C,310K12m/s)28(3.9010b.hotcoffee130J/kgKabout70°C,343Kc.icedtea1.5102Kabout0°C,273K5.Whatisthechangeintemperatureof2.2kgofd.boilingwaterthefollowingsubstancesif8.5103Jofther-100°C,373Kmalenergyisaddedtoeachofthesubstances?a.ice3.ConvertthefollowingKelvintemperaturesQtoCelsiustemperatures.TmCa.4K3J8.510T(2.2kg)(2060J/kgK)CTK2734273269°Cb.25K1.9KTb.waterCTK27325273248°CQc.272KTmCCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.TCTK2732722731°C8.5103Jd.373K(2.2kg)(4180J/kgK)T0.92KCTK273373273100°CPhysics:PrinciplesandProblemsSolutionsManual631

633Chapter12continuedc.steamQTmC8.5103J(2.2kg)(2020J/kgK)1.9Kd.aluminumQTmC8.5103J(2.2kg)(897J/kgK)4.3Ke.silverQTmC8.5103J(2.2kg)(235J/kgK)16Kf.copperQTmC8.5103J(2.2kg)(300J/kgK)1.0101K6.A2350-kggranitetombstoneabsorbs2.8107JofenergyfromtheSuntoCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.changeitstemperaturefrom5.0°Catnightto20.0°Cduringanautumnday.Determinethespecificheatofgranitefromthisinformation.QmCTQ2.8107J2J/kgKC7.910mT(2350kg)(20.0°C5.0°C)7.A2.00103-gsampleofwaterat100.0°Cismixedwitha4.00103-gsampleofwaterat0.0°Cinacalorimeter.Whatistheequilibriumtemperatureofthemixture?mACA(TfTAi)mBCB(TfTBi)0SinceCACBinthiscase,thespecificheatcancelsout.m(2.00kg)(100.0°C)(4.00kg)(0.0°C)TATAimBTBifm2.00kg4.00kg33°CAmB632SolutionsManualPhysics:PrinciplesandProblems

634Chapter12continued8.A220-gironhorseshoeisheatedto825°Candthenplungedintoabucketfilledwith20.0kgofwaterat25.0°C.Assumingthatnoenergyistransferredtothesurroundings,whatisthefinaltemperatureofthewateratequilibrium?WatermT25.0°CACA(TfTAi)mBCB(TfTBi)0m20.0kgmTACATAimBCBTBiHorseshoefmT825°CACAmBCB(0.22kg)(450J/kgK)(825°C)(20.0kg)(4180J/kgK)(25.0°C)26°C(0.22kg)(450J/kgK)(20.0kg)(4180J/kg°K)9.Acookadds1.20kgofsoupbonesto12.5kgofhotbrothforflavoring.TheBonestemperaturesofthebonesandthebrothm1.20kgBrotht20.0CT90.0Care20.0°Cand90.0°C,respectively.Assumethatthespecificheatofthebrothisthesameasthatofwater.Alsoassumethatnoheatislosttotheenvironment.Iftheequilibriumtemperatureis87.2°C,whatisthespecificheatofthebones?mACA(TfTAi)mBCB(TfTBi)0mCBCB(TfTBi)AmA(TfTAi)(12.5kg)(4180J/kgK)(87.2°C90.0°C)(1.20kg)(87.2°C20.0°C)1.81103J/kgK10.A150.0-gcoppercylinderat425°Cisplacedonalargeblockoficeat0.00°C.Assumethatnoenergyistransferredtothesurroundings.Whatisthemassoftheicethatwillmelt?Theenergyformeltingtheicecomesfromthecoolingofthecopper.QmCTThisquantityofenergyisavailabletomelttheice,soQmiceHfQmCTmiceHHff(0.1500kg)(385J/kg°C)(425°C0.00°C)3.34105J/kgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.0735kgoficePhysics:PrinciplesandProblemsSolutionsManual633

635Chapter12continued11.Howmuchenergyisneededtomeltonetroy13.Acylindercontaining1.00gofwaterattheounce,31.1g,ofgoldatitsmeltingpoint?boilingpointisheateduntilallthewaterturnsintosteam.TheexpandingsteamQmHfpushesapiston0.365m.Thereisa215-N(0.0311kg)(6.30104J/kg)frictionalforceactingagainstthepiston.Whatisthechangeinthethermalenergy1.96103Jofthewater?UQW12.Tryingtomakeaninterestingpattern,anartistslowlypours1.00cm3ofliquidQmHvgoldontoalargeblockoficeat0.00°C.(1.00103kg)(2.26106J/kg)Whilebeingpoured,theliquidgoldisatitsmeltingpointof1064°C.Thedensityof2.26103Jgoldis19.3g/cm3,anditsspecificheatisWFd(215N)(0.365m)78.5J128J/kgK.WhatmassoficewillmeltafterUQWallthegoldcoolsto0.00°C?Theenergytomelttheicecomesfrom2.26103J–78.5Jtheenergyreleasedwhenthegoldsolid-2.18103Jifiesandthenreleasedasthegoldcoolsfromitsmeltingtemperatureto0.00°C.14.Whatisthechangeintemperatureofwaterafterfallingovera50.0-m-tallwaterfall?QtotQsolidificationQcoolingAssumethatthewaterisatrestjustbeforeQsolidificationmHffallingoffandjustafteritreachesthe(0.0193kg/cm3)(1.00cm3)bottomofthefalls.(6.30104J/kg)UQ–WU0andWKEPEmgh1216JQmCTQcoolingmCTThus,mCTmghCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(0.0193kg/cm3)(1.00cm3)gh(9.80m/s2)(50.0m)(128J/kg°C)TC4180J/kgK(1064°C0.00°C)0.117K2628JQ15.Two6.35-kgleadbricksarerubbedagainsttot1216J2628J3844JeachotheruntiltheirtemperaturerisesbyQicemHf1.50K.HowmuchworkwasdoneontheQbricks?miceHfUQWandU03844J1.15102kgWQmCT3.34105J/kg(12.70kg)(130J/kg)(1.50K)2500J634SolutionsManualPhysics:PrinciplesandProblems

6364.Asshowninthefigurebelow,abubbleofChapter133isreleasedgaswithavolumeof1.20cmpages874–875underwater.Asitrisestothesurface,the1.UseTable13.1toestimatethepressuretemperatureofthebubbleincreasesfrominatmospheres(atm)onaclimberstand-27°Cto54°C,andthepressureiscuttoingatopMt.Everest.Isthismoreorlessone-thirdofitsinitialvalue.Whatisthethanhalfstandardatmosphericpressurevolumeofthebubbleatthesurface?(1.0atm)?4Pa)1.0atm(3105Pa0.3atm1.010Thisislessthanhalfofstandardatmosphericpressure.2.Awomanwearinghighheelsbrieflysupportsallherweightontheheels.Ifhermassis45kgandeachheelhasanareaof1.2cm2,whatpressuredoessheexertonthefloor?T127273300KFPTA254273327KmgP1V1P2V2AT1T2(45kg)(9.80m/s2)100cm2PV1V1T2(2)(1.2cm2)1m2T1P21.8103kPaP3)(327K)1(1.20cm13.Atypicalbrickhasdimensionsof(300K)3P120.0cm10.0cm5.0cmandweighs33.9cm20.0N.Howdoesthepressureexertedbyatypicalbrickwhenitisrestingon5.Asampleofethanegas(molarmassitssmallestsidecomparetothepressure30.1g/mol)occupies1.2102m3at46°Cexertedwhenitisrestingonitslargestside?and2.4105Pa.HowmanymolesofethaneFarepresentinthesample?WhatisthemasspsmallAsmallofthesample?20.0N100cm2PVn(10.0cm)(5.0cm)1mRT4.0103Pa(2.4105Pa)(1.2102m3)(8.31Pam3/molK)(319K)FplargeAlarge1.1molofethane20.0N100cm2mMn(20.0cm)(10.0cm)1m(30.1g/mol)(1.1mol)1.00103Pa33gCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thepressureisfourtimesaslargewhenthebrickrestsonitssmallestside.Physics:PrinciplesandProblemsSolutionsManual635

637Chapter13continued6.TheconstantRequals8.31Pam3/molKasitispresentedintheidealgaslawinthischapter.Onemoleofanidealgasoccupies22.4Latstandardtemperatureandpressure,STP,whichisdefinedas0.00°Cand1.00atm.Giventhisinformation,deducethevalueofRinLatm/molK.PVRnT(1.00atm)(22.4L)(1mol)(273K)0.0821Latm/molK7.Supposethattwolinkedpistonsarebothcylindricalinshape.Showthattheratioofforcesgeneratedisdirectlyproportionaltothesquareoftheradiiofthetwocross-sectionalcircularareas.FF12AA12FF12r2r212Fr211,adirectproportionFr2228.Thecross-sectionalareasofthepistonsFinthesystemshownbelowhavearatioof25to1.Ifthemaximumforcethatcanbeappliedtothesmallpistonis12N,whatisthemaximumweightthatcanbelifted?F2A1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.F1A2(12N)(25)13.0102N9.Acarofmass1.35103kgsitsonalargepistonthathasasurfaceareaof1.23m2.Thelargepistonislinkedtoasmallpistonofarea144cm2.Whatistheweightofthecar?Whatforcemustamechanicexertonthesmallpistontoraisethecar?F3kg)(9.80m/s2)gmg(1.35101.32104NFF1A22A1mgA2A1(1.35103kg)(9.80m/s2)(1.44102m2)1.23m2155N636SolutionsManualPhysics:PrinciplesandProblems

638Chapter13continued10.Atwhatdepthinfreshwaterdoesthewater13.Aglassmirrorusedinamountaintopexertapressureof1.00atm(1atmtelescopeissubjecttotemperaturesranging1.013105Pa)onascubadiver?from15°Cto45°C.Atthelowesttemper-Pghature,ithasadiameterof5.1m.IfthecoefficientoflinearexpansionforthisglassPis3.0106°C1,whatisthemaximumhgchangeindiameterthemirrorundergoes1.013105N/m2duetothermalexpansion?(1.00103kg/m3)(9.80m/s2)LLT10.3m(3.0106°C1)(5.1m)(45°C(15°C))11.Anicebergfloatsinseawater,partlyunder9.2104mwaterandpartlyexposed.ShowthatthevalueofVsubmerged/Vtotalequalsice/seawater.14.Asshowninthefigurebelow,abimetallicWhatpercentageoftheicebergisexposed?stripismadefromapieceofcopperUse1.03103kg/m3forthedensityofsea-(16106°C1)andapieceofsteelwater,and0.92103kg/m3forthedensity(8106°C1).Thetwopiecesareofice.equalinlengthatroomtemperature.Becausetheicefloats,CopperFbuoyantFgseawaterVsubmergedgiceVtotalgWoodenVsubmergedicehandleSteelVtotalseawatera.Asthestripisheatedfromroomtem-Thisisageneralresultforfloatingperature,whatistheratioofthechangesolids.inthelengthofthecoppertothatof0.92103kg/m3thesteel?iceseawater1.03103kg/m3LcoppercopperL1T0.89LsteelsteelL1TSo,89percentissubmergedandcopper11percentisexposed.steel12.Aconcreteblock(36106°C1)of16106°C1volume0.035m3at30.0°Ciscooledto8106°C110.0°C.Whatisthechangeinvolume?21VVTb.Howwillthestripbendwhenitisheated(36106°C1)(0.035m3)aboveroomtemperature?Whenitis(10.0°C30.0°C)cooledbelowroomtemperature?5.0105m3Whenheated,itbendswithcopperontheoutsideofthearc.Whenitiscooled,thecopperwillbeontheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.insideofthebend.Physics:PrinciplesandProblemsSolutionsManual637

639Chapter13continued15.Afishingline’sleadsinkerhasavolumeof1.40105m3.ThedensityofleadisChapter141.2104kg/m3.Whatistheapparentpages875–876weightofthesinkerwhenimmersedin1.Whatisthemassofthewatermelonshownfreshwater?Seawaterisslightlydenserthanbelow,ifthespringconstantis128N/m?freshwater.Isthesinker’sapparentweightbiggerorsmallerinseawater?Explain.FapparentFgFbuoyant0.48mleadgVwatergVgV(leadwater)(9.80m/s2)(1.40105m3)(1.2104kg/m31.00103kg/m3)Fkxmgkx(128N/m)(0.48m)1.5Nmg9.80m/s2Thebuoyantforceisslightlybiggerinseawater,duetoitsgreaterdensity.6.3kgThus,theapparentweightisslightly2.Howmanycentimeterswillaspringstretchsmallerthaninfreshwater.whena2.6-kgblockishungverticallyfromaspringwithaspringconstantof89N/m?Fkxmgmg(2.6kg)(9.80m/s2)xk89N/m0.29mor29cm3.HowmuchelasticpotentialenergydoesCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.aspringwithaconstantof54N/mhavewhenitisstretched18cm?PE1kx21(54N/m)(0.18m)2sp220.87J4.Whatistheperiodofthependulumoftheclockbelow?XIIXIIlXIIT2IXIIIVIIIIVgVIIVVI0.62m29.80m/s21.6s62cm0.52kg638SolutionsManualPhysics:PrinciplesandProblems

640Chapter14continued5.Aclockpendulumhasaperiodof0.95s.9.GeoisstandingonabreakwaterandheHowmuchlongerwillithavetobetohavenoticesthatonewavegoesbyevery4.2s.aperiodof1.0s?Thedistancebetweencrestsis12.3m.a.Whatisthefrequencyofthewave?lT2g11f0.24HzT4.2sgT2l2b.Whatisthespeedofthewave?4g11l2T2)vfT(12.3m)4.2s2l142(T2122.9m/s9.80m/s((1.0s)2(0.95s)2)420.024m2.4cmThependulumwillhavetobe2.4cmlonger.6.Whatisthelengthofapendulumwithaperiodof89.4ms?lT2ggT2(9.80m/s2)(89.4103s)2l42421.98103m1.98mm7.AlischoppingwoodacrossaclearingfromSu.Suseestheaxecomedownandhearsthesoundoftheimpact1.5slater.Howwideistheclearing?dvt(343m/s)(1.5s)5.1102m8.Whenanorchestraistuningup,thefirstviolinistplaysanoteat256Hz.a.Whatisthewavelengthofthatsoundwaveifthespeedofsoundinthecon-certhallis340m/s?v340m/s1.3mf256Hzb.Whatistheperiodofthewave?11Tf256HzCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.00391sor3.91msPhysics:PrinciplesandProblemsSolutionsManual639

641Chapter15(85Hz)1129.6m/spage876343m/s1.Soundwavesarebeingusedtodetermine93Hzthedepthofafreshwaterlake,asshowninthefigurebelow.Ifthewateris25°Cand5.Ifthespeedofawaveona78-cm-longittakes1.2sfortheechotoreturntotheguitarstringisknowntobe370m/s,whatsensor,howdeepisthelake?isthefundamentalfrequency?12L(2)(0.78m)1.56mvf12L370m/sDetectorSound1.56msource240Hz6.BoatAistravelingat4.6m/s.BoatBismovingawayfromboatAat9.2m/s,as1.2sdepthvt(1493m/s)showninthefigurebelow.Thecaptainof22mboatBblowsanairhornwithafrequency9.010of550Hz.WhatfrequencydoesboatA2.Findthewavelengthofan8300-Hzwaveinhear?Use343m/sforthespeedofsound.copper.vdvsvf3560m/s8300HzBoatABoatB0.43mvvfddfsvvCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.s3.Janetisstanding58.2mfromNinovan.IfNinovanshouts,howlongwillittake343m/s4.6m/s(550Hz)343m/s(9.2m/s)Janettohearher?Use343m/sforthespeedofsound.5.3102Hzd58.2mt7.Asubmarineistravelingtowardastationaryv343m/sdetector.Ifthesubmarineemitsa260-Hz0.170ssoundthatisreceivedbythedetectorasa4.Theengineonamotorcyclehumsat85Hz.262-Hzsound,howfastisthesubmarineIfthemotorcycletravelstowardarestingtraveling?Use1533m/sforthespeedofobserveratavelocityof29.6m/s,whatfre-soundinseawater.quencydoestheobserverhear?Use343m/svvfdandvforthespeedofsound.dfsvvd0,sosvvfdandvfvdfsvvd0,sovvsssfd1fdfsvfs1svsv1fvd260Hz(1533m/s)1262Hz12m/s640SolutionsManualPhysics:PrinciplesandProblems

642Chapter15continued8.Theendofapipeisinsertedintowater.Chapter16Atuningforkisheldoverthepipe.Ifthepiperesonatesatlengthsof15cmandpages876–87735cm,whatisthefrequencyofthetuning1.Whatisthedistance,r,betweenthelight-fork?Use343m/sforthespeedofsound.bulbandthetableinthefigurebelow?closedpipe:2(LBLA)(2)(0.35m0.15m)0.40mP2152lmv343m/sf0.40mr8.6102HzE180lx9.A350-Hztuningforkisheldovertheendofapipethatisinsertedintowater.Whatisthespacingbetweentheresonancelengthsofthepipeifthespeedofsoundis348m/s?vandfPE4r2Lforaclosedpipe,soBLA2P2152lmv348m/sr0.98mL4E4(180lx)BLA2f(2)(350Hz)0.50m2.Whatistheluminousfluxofaflashlightthatprovidesanilluminanceof145lxtothesurfaceofwaterwhenheld0.50mabovethewater?PE4r2P4r2E4(0.50m)2(145lx)4.6102lm3.Anoverheadlightfixtureholdsthreelight-bulbs.Eachlightbulbhasaluminousfluxof1892lm.Thelightfixtureis1.8mabovethefloor.Whatistheilluminanceonthefloor?P(3)(1892lm)2lxE1.4104r24(1.8m)24.Whatisthewavelengthinairoflightthathasafrequencyof4.61014Hz?c3.00108m/sCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.f4.61014Hz6.5107m650nmPhysics:PrinciplesandProblemsSolutionsManual641

643Chapter16continued5.Aheliumatomisinagalaxytravelingat6m/sawayfromtheChapter17aspeedof4.8910Earth.AnastronomeronEarthobservespage877afrequencyfromtheheliumatomof1.Alightrayisreflectedoffaplanemirrorat6.521014Hz.Whatfrequencyoflightanangleof25°fromthenormal.Alightisemittedfromtheheliumatom?rayfromanothersourceisreflected54°vfromthenormal.Whatisthedifferenceinfobsf1ctheanglesofincidenceforthetwolightTheyaremovingawayfromeachothersources?sousetheplusform.Difference54°25°v29°ffobs1c14Hz)4.89106m/s2.Alightrayisreflectedoffaplanemirror,as(6.521013.00108m/sshown.Whatistheangleofreflection?6.631014Hz13°6.AnastronomerobservesthatamoleculeinagalaxytravelingtowardEarthemitsi90°angletomirror90°13°77°lightwithawavelengthof514nm.Theastronomeridentifiesthemoleculeasonerithatactuallyemitslightwithawavelength77°of525nm.Atwhatvelocityisthegalaxymoving?3.Theanglebetweenanincidentandareflectedrayis70.0°.Whatistheangleofreflection?obs5.14107m5.25107mriv70.0°c2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.35.0°vc8m/s)4.Animageisproducedbyaconcavemirror,(3.0010asshown.Whatistheobject’sposition?5.14107m5.25107m5.25107m25cm6.29106m/s15cmCImagerf215cm27.5cm642SolutionsManualPhysics:PrinciplesandProblems

644Chapter17continuedfdi9.Aballis6.5mfromaconvexmirrorwithadodmagnificationof0.75.Iftheimageis0.25mifindiameter,whatisthediameterofthe(7.5cm)(25cm)actualball?25cm7.5cmh1.0101cmmiho5.WhatisthemagnificationoftheobjectinThevariablehcanbeusedforanyproblem4?dimensioninplaneandsphericaldmirrors.Inthisproblem,hodiametermidoftheball.oh25cmhiom11cm0.25m2.30.756.Iftheobjectinproblem4is3.5cmtall,0.33mhowtallistheimage?hmihohimho(2.3)(3.5cm)8.2cmTheimageis8.2cmtall.Thenegativesignmeansitisinverted.7.A6.2-m-tallobjectis2.3mfromaconvexmirrorwitha0.8-mfocallength.Whatisthemagnification?fddoidof(0.8m)(2.3m)2.3m(0.8m)0.6mdmido(0.6m)2.3m0.38.Howtallistheimageinproblem7?hiCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.mhohimho(0.3)(6.2m)2mPhysics:PrinciplesandProblemsSolutionsManual643

645Atthewaterandflintglassboundary:Chapter18n12sin2pages877–8781sinn11.ApieceofflintglassAir(1.62)(sin38.1°)islyingontopofacon-sin148.7°Flintglass1.33tainerofwater(seefigurebelow).WhenaredbeamWater4.Anobjectthatis24cmfromaconvexoflightinairisincidentlensproducesarealimagethatis13cmupontheflintglassatanfromthelens.Whatisthefocallengthofangleof28°,whatisthethelens?angleofrefractionintheflintglass?dido(13cm)(24cm)fd13cm24cmidon1sin1n2sin28.4cmn11sin12sinn25.A5.0-cm-tallobjectisplaced16cmfroma(1.0003)(sin28°)convexlenswithafocallengthof8.4cm.sin117°1.62Whataretheimageheightandorientation?fd(8.4cm)(16cm)do2.Whentheangleofrefractionintheflintid16cm8.4cmofglassofproblem1is22°,whatistheangleofrefractioninthewater?17.7cmn1sin1n2sin2hidimhdoon11sin12sinndiho(17.7cm)(5.0cm)2hid16cmo(1.62)(sin22°)sin127°5.5cm1.33Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Theimageisinvertedrelativetothe3.Whenthebeamoflightinproblem1entersobject.theflintglassfromthewater,whatisthemaximumangleofincidenceinthewater6.Anobjectis185cmfromaconvexlenssuchthatlightwilltransmitintotheairabovewithafocallengthof25cm.Whenantheflintglass?Hint:Useanangleofrefractioninvertedimageis12cmtall,howtallisofthelightbeaminairthatisalmost90°.theassociatedobject?Thiswilloccurwhentheangleoffd(25cm)(185cm)dorefractionofthelightbeamintheairisid185cm25cmofalmost90°.Use289.9°.29cmn1sin1n2sin2hidimhdAttheflintglassandairboundary:oond(185cm)(12cm)12sin2hohi1sinnod29cm1i(1.0003)(sin89.9°)77cmsin138.1°1.62Theobjectis77cmtall.644SolutionsManualPhysics:PrinciplesandProblems

646Chapter18continued7.Whataretheimageheightandorientation9.Aconcavelenswithafocallengthofproducedbythesetupinthefollowingfigure?220cmproducesavirtualimagethatis36cmtall.Whentheobjectisplaced128cmfromthelens,whatistheObjectmagnification?fd(220cm)(128cm)do5.0cmid128cm(220cm)of81cmFd(81cm)mi0.6310.0cmd128cmo15.0cmfd(15.0cm)(10.0cm)doid10.0cm15.0cmof3.0101cmhdmiihdood(3.0101cm)hihoid10.0cmo15cmTheimageisuprightrelativetotheobject.8.Aconvexlenscanbeusedasamagnifyingglass.Whenanobjectthatis15.0cmfromthelenshasanimagethatisexactly55timesthesizeoftheobject,whatisthefocallengthofthelens?dmidodimdo(55)(15.0cm)825cmdfidodido(825cm)(15.0cm)(825cm)(15.0cm)15.3cmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual645

6473.Lightwithawavelengthof454.5nmpassesChapter19throughtwoslitsthatare95.2cmfromapages878–879screen.Thedistancebetweenthecenters1.Aphysicsstudentperformsadouble-slitofthecentralbandandthefirst-orderbanddiffractionexperimentonanopticalbench,is15.2mm.Whatistheslitseparation?asshowninthefigurebelow.Thelightxdfromahelium-neonlaserhasawavelengthLof632.8nm.ThelaserlightpassesthroughLdtwoslitsseparatedby0.020mm.Whatisxthedistancebetweenthecentersofthe(4.545107m)(0.952m)centralbandandthefirst-orderband?0.0152mScreen2.85105mLaserDiffractiongrating4.Whatcoloroflightwouldbereflectedfromthesoapywaterfilmshowninthefigurebelow?50.0cmAir150.0cmSoapywaterL150.0cm50.0cm76.1nm100.0cmnSW1.33xdAirL1L2dmx2nswd(6.328107m)(1.000m)Form0,2d1.5m2nsw2.0103.2102m3.2cm4dnswCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(4)(76.1nm)(1.33)2.Alaserofunknownwavelengthreplaces405nmthehelium-neonlaserintheexperimentdescribedinproblem1.ToobtainthebestThelightisvioletincolor.diffractionpattern,thescreenhadtobe5.A95.7-nmfilmofanunknownsubstanceismovedto104.0cm.Thedistancebetweenabletoprevent555-nmlightfrombeingthecentersofthecentralbandandthefirst-reflectedwhensurroundedbyair.Whatisorderbandis1.42cm.Whatwavelengthoftheindexofrefractionofthesubstance?lightisproducedbythelaser?1L104.0cm50.0cm2dm2nsub54.0cm1Form0,2d.xd2nsubL(1.42102m)(2.0105m)nsub4d0.540m555nm5.3107m530nm(4)(95.7nm)1.45646SolutionsManualPhysics:PrinciplesandProblems

648Chapter19continued6.Anoilfilm(n1.45)onthesurfaceofa2L2x1wpuddleofwateronthestreetis118nmthick.Whatfrequencyoflightwouldbe(2)(6.85107m)(1.1m)reflected?2.5105m10.063m2dm2noil9.Thewidthofthecentralbrightbandofa1Form0,2d.diffractionpatternis2.9cm.Laserlight2noilofunknownwavelengthpassesthrougha4dnoilsingleslitthatis0.042mmwideandonto(4)(118nm)(1.45)ascreenthatis1.5mfromtheslit.Whatisthewavelengthofthelight?685nmc2L2x1wfc2x1wf2L3.00108m/s(0.029m)(4.2105m)6.85107m(2)(1.5m)4.381014Hz4.1107m7.Violetlight(415nm)fallsonaslitthat10.Adiffractiongratinghas13,400linesperis0.040mmwide.Thedistancebetweeninch.Whatistheslitseparationdistance?thecentersofthecentralbrightbandand(Use1in2.54cm)thethird-orderdarkbandis18.7cm.What1in2.54cm1mdisthedistancefromtheslittothescreen?13,400lines1in100cmmL1.90106m/linexmwxmw11.Lightfromahelium-neonlaser(Lm632.8nm)passesthroughthediffraction(0.187m)(4.0105m)gratingdescribedinproblem10.Whatis(3)(4.15107m)theanglebetweenthecentralbrightlineandthefirst-orderbrightline?6.0mdsin8.Redlight(685nm)fallsonaslitthatis1sin0.025mmwide.Thedistancebetweenthedcentersofthecentralbrightbandandthe16.328107msecond-orderdarkbandis6.3cm.Whatissin1.90106mthewidthofthecentralband?19.5°mLxmwxLmwmCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(0.063m)(2.5105m)(2)(6.85107m)1.1mPhysics:PrinciplesandProblemsSolutionsManual647

649Chapter19continued12.Adiffractiongratingwithslitsseparatedby3.40106misilluminatedbylightwithawavelengthof589nm.Theseparationbetweenlinesinthediffractionpatternis0.25m.Whatisthedistancebetweenthediffractiongratingandthescreen?dsin1sind15.89107msin3.40106m9.98°xtanLxLtan0.25mtan9.98°1.4mCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.648SolutionsManualPhysics:PrinciplesandProblems

650Chapter20page8791.Themagnitudeofacharge,q,istobedeter-minedbytransferringthechargeequallytotwopithballs.Eachofthepithballshasamassofm,andissuspendedbyaninsulatingthreadoflengthl.Whenthechargeistransferred,thepithballsseparatetoformanequilibriuminlwhicheachthreadformsanangle,,withthevertical.a.Drawaforcediagramshowingtheforcesthatareactingontherightmostpithball.ddFTFEFgwhere:FTthreadtensionFEelectrostaticforceFggravitationalforceb.Deriveanexpressionforqasafunctionof,m,andl.Sincethesystemisinequilibrium,thenethorizontalandverticalforcesmustbezero.FTsinFE0FTcosFg0RearrangeandsubstituteFEandFg.q2q2q2FAqBTsinFEK(2d)2K4d2K16d2FTcosFgmgCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.DividethefirstequationbythesecondtoeliminateFT.q2sintanKcos16mgd2Physics:PrinciplesandProblemsSolutionsManual649

651Chapter20continuedFromthediagram,dlsin.Substitutefordandsolveforq.16mgl2sin2tanq2Kmgtanq4lsinKc.Useyourderivedexpressiontodeterminethevalueofqwhen5.00°,m2.00g,andl10.0cm.(2.00103kg)(9.80m/s2)(tan5.0°)q(4)(10.0102m)(sin5.0°)(9.0109Nm2/C2)1.52108C2.Asshowninthefigurebelow,fourcharges,eachwithchargeq,aredistributedsym-metricallyaroundtheorigin,O(0,0),atA(1.000,0),B(0,1.000),C(1.000,0),andD(0,1.000).Findtheforceonafifthcharge,qT,locatedatT(5.000,0).y(m)B0CATx(m)DBecauseofsymmetry,theforcecomponentsonchargeTinthedirectionofthey-axiscancel,andonlythex-directionforcesneedtobecomputed.FT,xFAT,xFBT,xFCT,xFDT,xCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1cosB1cosDKqqTAT2BT2CT2DT25125211KqqT16(2)2523610.1657KqqTN3.Considerthatthefourchargesinthepreviousproblemarenowcombinedintoasinglecharge,4q,locatedattheorigin.WhatistheforceonchargeqT?(4q)(qT)4K52KqqT250.1600KqqTNAchargeuniformlyspreadoverasphericalsurfacemaybetreatedasifallofthechargeisconcentratedatthesphere’scenter.Thepreviousproblemisatwo-dimensionalapproximationofthiscondition.650SolutionsManualPhysics:PrinciplesandProblems

652Chapter21FqE(0.5107C)(6.2104N/C)pages879–8801.Whatarethemagnitudeandthesignofa3.1103NpointchargethatexperiencesaforceofThedirectionisoppositethatof0.48Neastwhenplacedinanelectrictheelectricfieldbecausethechargefieldof1.6105N/Cwest?3N,isnegative.So,E3.110qF0.48N3.0106C15°southofwestE1.6105N/CBecausetheforceexertedonthecharge4.Bywhatpercentmustthedistancefromaisoppositethedirectionoftheelectricpointchargeincreaseinordertohaveafield,thechargemustbenegative.reductionintheelectricfieldstrengthby40percent?2.Atestchargeof1.0106C,locatedatTheelectricfieldstrengthofapointthepointT(0,1)m,experiencesaforceofcharge,0.19Ndirectedtowardtheorigin,alongtheqy-axis,duetotwoidenticalpointchargesEK2,rlocatedatA(1,0)mandB(1,0)m.isinverselyproportionaltothesquareyofthedistancefromthecharge.Er212qT(0,1)Er2T21Eqqr21r2AB2E1x2A(1,0)B(1,0)Er1r2E12Inthiscase,E2(10.40)E10.60E11Then,rr20.6011.2r1a.WhatisthesignofthechargesatAandB?Thedistancemustincreaseby30percent.ThechargesatAandBmustbepositivetocauseanattractiveforce5.Aparticleofmassm2.0106kgisonthetestcharge.inacircularorbitaboutapointcharge.b.WhatarethemagnitudeanddirectionThecharge,q,is3.0105CandisataoftheelectricfieldatT?distanceofr20.0cm.F0.19NEq1.0106CE1.9105N/CFEThedirectionisoppositetheforcebecauseitisanegativecharge.So,20.0cmqE1.9105N/C,awayfromtheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.origin.3.Atestchargeof0.5107Cisplacedinanelectricfieldof6.2104N/C,directeda.Whatistheelectricfieldstrengthat15°northofeast.Whatistheforceexperi-allpointsontheorbitaroundtheencedbythecharge?pointcharge?Physics:PrinciplesandProblemsSolutionsManual651

653Chapter21continuedSincetheorbitiscircular,theTheelectricfieldstrengthisthesumdistanceoftheparticlefromtheoftwocomponents,oneduetoeachpointchargeisconstant.Fromcharge,bothpointingtowardtheExampleProblem2,theelectricnegativecharge.Atthemidpoint,fieldstrengthdependsonlyonthethemagnitudesofthecomponentdistanceandthecharge,andisfieldstrengthsareequal.givenby2Kq8KqqEEqEqr2EKr2r22SolveforqintermsofE.9Nm2/C2)3.0105C(9.010(0.200m)22(4.8104N/C)(0.50m)2Erq8K(8)(9.0109Nm2/C2)6.8106N/C,radiallyoutwardatallpointsontheorbit1.7107Cb.Consideringonlyelectrostaticforces,7.Apithballweighing3.0102Ncarriesawhatcharge,q,ontheparticleischargeof1.0106C.Thepithballisrequiredtosustainanorbitalperiodof3.0103s?placedbetweentwolarge,parallel,metalplates,thatareseparatedby0.050m.WhatTheelectrostaticforceFEmustbeisthepotentialdifference,V,thatmustbetowardthepointchargetoaccountappliedinordertosuspendthepithballforthecentripetalaccelerationofbetweentheplates?theparticleinorbit.Therefore,theparticlechargemustbenegative.VEdTheforcerequiredtokeeptheFEdparticleinorbitis(seeChapter6,qMotioninTwoDimensions):(3.0102N)(0.050m)42r1.0106CF,radiallyinward.EmacmT23VCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1.510Thechargeontheparticlemustthenbe8.Theelectricfieldstrength,definedasaforceFEperunitcharge,hastheunitsnewtonsperqEcoulomb,N/C.Theformulaforelectric42rmpotentialdifference,VEd,however,ET2suggeststhatEalsocanbeexpressedasvoltspermeter,V/m.42(0.200m)(2.0106kg)(6.8106N/C)(3.0103s)2a.Byanalysisoftheunits,showthatthesetwoexpressionsfortheunitsofelectric2.6107Cfieldstrengthareequivalent.6.TwochargesofequalmagnitudeandVJ/C(Nm)/CNmmmCoppositesignareplaced0.50mapart.Theelectricfieldstrengthmidwaybetweenthemb.SuggestareasonwhyV/misoftenais4.8104N/Ctowardthenegativecharge.moreusefulwaytoexpresselectricWhatisthemagnitudeofeachcharge?fieldstrength.Inpractice,electricpotentialdiffer-qEqencesanddistancesaremucheasierrrtomeasurethanforcesoncharges.22652SolutionsManualPhysics:PrinciplesandProblems

654Chapter21continued9.Supposeyouhavetwoparallel,metalplates13.Anoildropweighing7.51015Ncarriesthathaveanelectricfieldbetweenthemofthreeexcesselectrons.strength3.0104N/C,andare0.050ma.Whatpotentialdifferenceisrequiredapart.Considerapoint,P,located0.030mtosuspendthedropbetweenparallelfromplateA,thenegativelychargedplate,platesseparatedby2.3cm?whenansweringthefollowingquestions.Va.WhatistheelectricpotentialatPFgFEqEneEnedrelativetoplateA?FdgV4V/m)(0.030m)VPA(3.010ne9.0102V(7.51015N)(2.3102m)(3)(1.601019C)b.WhatistheelectricpotentialatPrelativetoplateB,thepositivelychargedplate?3.6102VV4V/m)(0.020m)b.Iftheoildroppicksupanotherelectron,PB(3.010byhowmuchmustthepotentialdiffer-6.0102Vencebetweentheplatesbereducedtomaintaintheoildropinsuspension?10.Acertain1.5-Vsize-AAbatteryhasastorageFcapacityof2500C.HowmuchworkcangdVthisbatteryperform?neWqV(2.5103C)(1.5V)Sothepotentialdifferenceis3Jinverselyproportionaltothe3.810numberofexcesselectrons.11.Inavacuumtube,electronsacceleratefromV2n1thecathodeelementtotheplateelement,V1n2whichismaintainedatapositivepotentialnV1Vwithrespecttothecathode.Iftheplatevolt-2n12ageis240V,howmuchkineticenergy3(3.6102V)2.7102Vhasanelectronacquiredwhenitreaches4theplate?Reductionof90VThekineticenergygainedbyanelectronisthesameasthepotentialenergylost14.Achargeof2.00106Cismovedagainstbytheelectronasitmovesthroughtheaconstantelectricfield.If4.50104Jofpotentialdifferenceof240V.workisdoneonthecharge,whatistheKEWeVpotentialdifferencebetweentheinitial(1.601019C)(240V)andthefinallocationsofthecharge?3.81017JWq4.50104JVq2.00106C12.Anoildropwithfiveexcesselectronsissus-225Vpendedinanelectricfieldof2.0103N/C.Whatisthemassoftheoildrop?15.CapacitorsC1220FandC2470FareFgmgFEqEneEconnectedacrossa48.0-VelectricpotentialCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Fdifference.gneEmgga.Whatarethecharges,q1andq2,oneachofthecapacitors?(5)(1.601019C)(2.0103N/C)2q9.80m/sCV1.61016kgPhysics:PrinciplesandProblemsSolutionsManual653

655Chapter21continuedq6F)(48.0V)e.Basedonyourresponsestotheabove,1C1V(22010makeaconjectureconcerningthe1.1102Cequivalentcapacitanceofasystemofq6F)(48.0V)capacitors—allofwhichareconnected2C2V(47010acrossthesamepotentialdifference.2.3102CTheequivalentcapacitanceisb.Whatisthetotalcharge,qT,onboththesumofalloftheindividualcapacitors?capacitances.qTq1q216.Anew90.0-VbatterywithastoragecapacityC1VC2Vof2.5104Cchargesa6800-Fcapacitor(CwiththeswitchinpositionA.Thenthe1C2)VswitchisthrowntopositionBtodischarge(220106F470106F)thecapacitor.(48.0V)3.3102CBAc.Repeatstepsaandbforanewpotentialdifference,v96.0V.Dischargecircuit90.0V6F)(96.0V)q1C1V(220102.1102Cq6F)(96.0V)a.Howmanytimescanthiscyclebe2C2V(470104.5102Crepeatedbeforethebatteryiscompletelydischarged?qTq1q2Whenthecapacitoriscompletely(Ccharged,itcontains1C2)V(220106F470106F)q6F)(90.0V)cCV(680010(96.0V)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.61C6.6102CThenumberofcyclesNisd.Nowconsiderthetwocapacitorsasasystem.Whatwouldbeasingleequiva-2.5104CNlentcapacitor,C0.61C/cycleeq,thatcouldreplaceC1andC2,andbecapableofyieldingthe4.1104cyclessameresults?b.IfthecapacitordischargeoccursinLetCeqC1C2690F120ms,whatistheaveragepowerdissipatedinthedischargecircuitThen,duringonecycle?qTCeqVTheenergystoredinthefully(690106F)(48.0V)chargedcapacitorequalstheworkperformedtostorethecharge.3.3102CWqandcVTheaveragepowerdissipatedqTCeqVequalsthetotalenergydissipated(690106F)(96.0V)dividedbythetimerequired.q6.6102CPEWcVtttyieldingsimilarresultsaspreviously.654SolutionsManualPhysics:PrinciplesandProblems

656Chapter21continued(0.61C)(90.0V)Chapter220.12s2Wpages881–8824.6101.Adecorativelightbulbratedat7.50Wdraws60.0mAwhenlit.Whatisthe17.A0.68-Fcapacitorcarriesachargeononeplateof1.36105C.Whatisthepotentialvoltagedropacrossthebulb?differenceacrosstheleadsofthiscapacitor?PVIqP(7.50W)CV125VVI(60.0103A)q1.36105C2.A1.2-Vnickel-cadmiumbatteryhasaVC0.68106Fratedstoragecapacityof4.0103mAh2.0101V(milliamp-hours).a.Whatisthebatterychargecapacityincoulombs?Hint:1C1As.3mAh)1A3600sq(4.0101000mA1h1.4104As1.4104Cb.Howlongcanthisbatterysupplyacurrentof125mA?qIt4.0103mAht32h125mA3.Aheatingelementofanelectricfurnaceconsumes5.0103Wwhenconnectedacrossa240-Vsource.Whatcurrentflowsthroughtheelement?PVIP5.0103WI21AV240V4.Thecoldfilamentresistanceofalightbulbis20.0.Thebulbconsumes75Wwhenitisoperatingfroma120-Vsource.Bywhatfactordoesthestart-upcurrentinthebulbexceedtheoperatingcurrent?VIstartRPIoperateVVIRV2(120V)2startCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.IoperatePRP(20.0)(75W)V9.6Thestart-upcurrentis9.6timeslargerthantheoperatingcurrent.Physics:PrinciplesandProblemsSolutionsManual655

657Chapter22continued5.Inthecircuitshownbelow,apotentiometer8.Modifythediagramofthepreviousproblemisusedtovarythecurrenttothelamp.Iftoincludeanammeterandavoltmetertotheonlyresistanceisduetothelamp,whatmeasurethevoltagedropacrossthe680-isthecurrentinthecircuit?resistor.II22012V90.0VRlamp20.0AVV12VII0.60AR20.0lamp6806.Anelectrictoasterconsumes1875Winoperation.Ifitispluggedintoa125-V9.Ifthetotalresistanceinthecircuitsource,whatisitsresistance?inproblem8is9.0102,whatwouldtheammeterandthevoltmeterindicate?V2PAmmeterreadingisRV2(125V)2VR8.33IP1875WR90.0V0.10A7.Drawacircuitdiagramtoincludea90.0-V9.0102battery,a220-resistor,anda680-Voltmeterreadingisresistorinseries.ShowthedirectionofVIRconventionalcurrent.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.(0.10A)(680)68VI22010.Anelectricmotorwithaloaddelivers5.2hptoitsshaft(1hp746W).Undertheseconditions,itoperatesat82.8percent90.0Vefficiency.(Efficiencyisdefinedastheratioofpoweroutputtopowerinput.)a.Howmuchcurrentdoesthemotordrawfroma240-Vsource?IPoutputEfficiency0.828Pinput680PoutputPinputEfficiencyIVV(5.2hp)(746W/hp)0.828240V2.0101A656SolutionsManualPhysics:PrinciplesandProblems

658Chapter22continuedb.Whathappenstotheremainingb.Assumeabulbhasalifetimeof17.2percentoftheinputpower?1.0103handanelectricenergycostItisprimarilydissipatedasI2Rof$0.12/kWh.Howmuchlessdoeslossesinthemotorwindings.the55-WbulbcosttooperateoverUltimately,thisenergyisconverteditslifetime?tothermalenergyandmustbeThe60.0-Wbulbcostsremovedfromthemotorbyfans(0.0600kW)(1.0103h)($0.12/kWh)orothermeans.$7.20The55-Wbulbcosts11.Anindustrialheatingprocessusesacurrent(0.055kW)(1.0103h)($0.12/kWh)of380Asuppliedat440-Vpotential.$6.60a.WhatistheeffectiveresistanceoftheThedifferenceincostisheatingelement?$7.20$6.60$0.60RV440V1.2c.IsmostofthecostsavingsduetoI380Ahigherbulbefficiency,orduetotheb.Howmuchenergyisusedbythisconsumer’swillingnesstoacceptlowerprocessduringan8-hshift?lightoutput?EPtVItMostofthesavingsisduetoa3600swillingnesstoacceptlowerlight(440V)(380A)(8h)1houtput.Ifthe60.0-Wbulbcould5109Jbemadetooutput800.0Im,itscostwouldbe800.012.An8.0-electricheateroperatesfrom($7.20)$6.86,resultingina840.0a120-Vsource.savingsof$0.26forthe55-Wbulb.a.HowmuchcurrentdoestheheaterSoofthe$0.60savingsoverthebulbrequire?lifetime,only$0.26,or43percent,isduetohigherefficiency.V120VI15AR8.014.BywhatfactorwouldtheI2Rlossintrans-b.Howmuchtimedoestheheaterneedto4Jofthermalenergy?missionwiresbereducedifthetransmissiongenerate2.010voltagewereboostedfrom220Vto22kV?EPtVItAssumethattherateofenergydeliveredis4unchanged.E2.010Jt11sVI(120V)(15A)Thepowerrequiredisthesame.Thecurrent,however,wouldbereducedby13.Amanufactureroflightbulbsadvertisesthatthesamefactorthatthevoltagewasits55-Wbulb,whichproduces800.0lmofboosted,102.TheI2Rlosswouldbelightoutput,providesalmostthesamelightreducedbyafactorof104becausetheasa60.0-Wbulbwithanenergysavings.currentissquared.The60.0-Wbulbproduces840.0lm.15.Theelectric-utilityinvoiceforahouseholda.Whichbulbmostefficientlyconvertsshowsausageof1245kWhduringaelectricenergytolight?certain30-dayperiod.WhatistheaverageCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.800.0lmThe55-Wbulbproduces,powerconsumptionduringthisperiod?55Wor14.5lm/W.The60.0-WbulbE1.245106Wh1d840.0lmPproduces,or14.0lm/W.t30d24h60.0WThe55-Wbulbismoreefficient.2103WPhysics:PrinciplesandProblemsSolutionsManual657

659Chapter23b.Whatisthevoltagedropinthe400-resistor?pages882–884V1.Aseriescircuitcontainsa47-resistor,an400IR40082-resistor,anda90.0-Vbattery.What(0.1000A)(400.0)resistance,R3,mustbeaddedinseriesto40.00Vreducethecurrentto350mA?V4.ShowthatthetotalpowerdissipatedinIsourceacircuitofseries-connectedresistorsisRPI2R,whereRistheequivalentresistance.VRsourceIThetotaldissipatedpowerisequaltothesumofthepowerdissipationsin90.0V0.350Aeachoftheresistors.257PP1P2…Foraseriescircuit,RR1R2R3I2R2R1I2…R3RR1R2I2(R1R2…)2574782I2R1285.Inanexperiment,threeidenticallightbulbs2.Whatistheminimumnumberof100.0-areconnectedinseriesacrossa120.0-Vresistorsthatmustbeconnectedinseriessource,asshown.Whentheswitchisclosed,togetherwitha12.0-batterytoensureallofthebulbsareilluminated.However,thatthecurrentdoesnotexceed10.0mA?whentheexperimentisrepeatedthenextVsourceday,bulbsAandBareilluminatedbrighterI,whereI0.0100ARthannormal,andbulbCisdark.Avolt-Letnequalthenumberof100.0-meterisusedtomeasurethevoltagesintheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.resistors.Thus,forthisseriescircuit,circuitasshown.Thevoltmeterreadingsare:R100.0nV1120.0VV12.0V260.0V0.0100AVR30.0V12.0V0.0100A(BulbA)(BulbB)(BulbC)100.0nRARBRC12.0n12.0(0.0100)(100.0)Theremustbeatleast12resistors.120.0VV1V2V33.A120.0-generatorisconnectedinserieswitha100.0-resistor,a400.0-resistor,anda700.0-resistor.a.Howmuchcurrentisflowinginthea.Whathashappenedinthiscircuit?circuit?BulbChasdevelopedashortcircuitVIsourceinitsbaseandthus,nocurrentflowsRthroughitsfilamentanditremains120.0Vdark.Theshortactsasazeroresis-100.0400.0700.0tanceandallowsthecircuitcurrent0.1000A658SolutionsManualPhysics:PrinciplesandProblems

660Chapter23continuedtoflow.Themeasurementsconfirm7.Tworesistorsareconnectedinseriesacrossthattheavailablevoltageisdroppeda12.0-Vbattery.Thevoltagedropacrossacrosstheremainingtwobulbs.oneoftheresistorsis5.5V.b.ExplainwhybulbsAandBarebrightera.Whatisthevoltagedropacrossthethannormal.otherresistor?Thevoltagedropacrosseachbulb12.0V5.5V6.5V120.0Vincreasedfrom40.00V,b.Ifthecurrentinthecircuitis5.0mA,3120.0Vwhatarethetworesistorvalues?to60.00V.Thepower22VVdissipatedineachbulb,PR,RIhasincreasedbecauseofthe5.5VincreasedV.R10.0050Ac.Istheremoreorlesscurrentflowing1100nowinthecircuit?6.5VMorecurrentflowsthroughtheR20.0050Acircuit.Theequivalentresistanceof1300twobulbsandashortcircuitislowerthanthatofthreebulbs.This8.Inproblem7,thespecificationsforthelowerequivalentresistanceresultsresistorsstatethattheirresistancemayvaryinanincreasedcurrent.fromthelistednominalvalue.Ifthepossi-blerangesofactualresistancevaluesare6.Astringofholidaylightshas25identicalasfollows,1050Rbulbsconnectedinseries.Eachbulbdissi-11160,and1240Rpates1.00Wwhenthestringisconnected21370,whatisthepossi-bleminimumandmaximumvalueofthetoa125-Voutlet.nominal5.5-Vvoltagedrop?a.HowmuchpowermustbesuppliedbyConstructatabletodeterminethevolt-the125-Vsource?agedropsresultingfromtheextremeP(25bulbs)(1.00W/bulb)valuesoftheresistors.Thevoltagedrop25.0Wiscalculatedusingtheequation12Rb.Whatistheequivalentresistanceof1.RRthiscircuit?12V2PRResistorR1ResistorR2VoltageDropkk(V)V2(125V)2RP25.0W6251.051.245.5c.Whatistheresistanceofeachbulb?1.051.375.2Becausethebulbsareinseries,1.161.245.8RR62525.01.161.375.5bulb2525Thevoltagedropwillrangefrom5.2Vd.Whatisthevoltagedropacrosseachto5.8V.bulb?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Becausethebulbsareinseries,V125VV5.00Vbulb2525Physics:PrinciplesandProblemsSolutionsManual659

661Chapter23continued9.Avoltage-dividercircuitisconstructedwithb.Whatistheoperatingresistanceofeachapotentiometer,asshownbelow.Thereareofthebulbs?twofixedresistancesof113kand294k.SincethebulbsareconnectedTheresistanceofthepotentiometercaninparallel,thesamevoltageisrangefrom0.0to100.0k.connectedacrosseachofthem.V2PR113kV2RP100.0V100.0k2(125V)R2087575.0WVout294k(125V)2R6252525.0Wc.Thebulbsarerewiredinseries.Whatisa.WhatisVoutwhenthepotentiometerthedissipatedpowerineachbulb?isatitsminimumsettingof0.0?Whichbulbisthebrightest?VoutVminThecurrentinthecircuitisnow:294kVsourceI294k100.0k113kR(100.0V)VRR257558.0V125Vb.WhatisVoutwhenthepotentiometer833isatitsmaximumsettingof100.0k?0.150AVoutVmaxThepowerdissipatedineachbulbis:294k100.0kP2R2(208)75I75(0.150A)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.294k100.0k113k(100.0V)4.68WP2R2(625)77.7V25I25(0.150A)c.Whatpotentiometersettingisrequired14.1WtoadjustVouttoexactly65.0?The25-Wbulbisthebrightest.ThesettingisthesameastheratiooftherangefromV11.A10.0-Vbatteryhasaninternalresistanceminto65.0Vtothetotalavailablerange.of0.10.Theinternalresistancecanbemodeledasaseriesresistor,asshown.(65.0V)Vmin65.0V58.0VVmaxVmin77.7V58.0VRI0.1035.5%10.Twolightbulbs,oneratedat25.0Wandoneratedat75.0W,areconnectedinparallelacrossa125-Vsource.10.0VVRLa.Whichbulbisthebrightest?The75.0-Wbulbdissipatesthemostpowerandthereforeburnsthebrightest.BatteryLoad660SolutionsManualPhysics:PrinciplesandProblems

662Chapter23continueda.Deriveanexpressionforthebattery211V…terminalvoltage,V,asafunctionofR1R2thecurrent,I.V2voltagerisesumofvoltagedropsR10.0IRIV13.AholidaylightstringoftenbulbsisV10.0IRIequippedwithshuntsthatshortoutthe10.0–0.10Ibulbswhenthevoltagedropincreasestolinevoltage,whichhappenswhenabulbb.Createagraphofvoltageversuscurrentburnsout.Eachbulbhasaresistanceforacurrentrangeof0.0to1.0A.of200.0.Thestringisconnectedtoa(1.0A,9.9V)householdcircuitat120.0V.Ifthestring10.0isprotectedbya250.0-mAfuse,howmany9.9bulbscanfailwithoutblowingthefuse?Thestringwilloperateaslongasthecurrentislessthan250.0mA.VIVoltage(V)RV0.250AnRbulb000.51.0120.0V0.250ACurrent(Amps)(n)(200.0)c.WhatvalueofloadresistorRn120.0VLneedsto(0.250A)(200.0)beplacedacrossthebatteryterminalstogiveacurrentof1.0A?n2.4VVThestringwilloperatewithatleastIsourcesourceRRRthreegoodbulbs,ornomorethanILsevenfailedbulbs.Vsource10.0VRR0.10LII1.0A14.A60.0-Wlightbulbanda75.0-Wlightbulb9.9areconnectedinserieswitha120-Vsource.d.HowdoestheV-Ifunctiondifferfroma.Whatistheequivalentresistanceofthatofanidealvoltagesource?thecircuit?AnidealvoltagesourceprovidesV2PthespecifiedpotentialdifferenceRregardlessoftheamountofcurrent2V2(120V)drawnfromit.RP60.0W75.0W12.Showthatthetotalpowerdissipatedina1.1102circuitofparallel-connectedresistorsis:b.Supposean1875-WhairdryerisnowV2P,whereRistheequivalentresistance.pluggedintotheparallelcircuitwiththeRlightbulbs.WhatisthenewequivalentThetotaldissipatedpowerisequaltoresistanceofthecircuit?Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.thesumofthepowerdissipationsinV2(120V)2eachoftheresistors.RP60.0W75.0W1875WPP1P2…7.2V2V2P…RR12Physics:PrinciplesandProblemsSolutionsManual661

663Chapter23continued15.WhatistheequivalentresistanceoftheChapter24followingresistornetwork?pages884–88529.4k25.5k1.ThemagneticfieldofEarthresemblesthefieldofabarmagnet.Thenorthpoleofacompassneedle,usedfornavigation,gen-6.34kerallypointstowardthegeographicnorthpole.WhichmagneticpoleoftheEarthis47.5kthecompassneedlepointingtoward?7.87kthesouthmagneticpoleofEarth2.AmagnetisusedtocollectsomespilledRpaperclips.Whatisthemagneticpoleat129.4ktheendofthepaperclipthatisindicatedR236.34k7.87kinthefigure?14.21kRN4525.5k47.5k73.0k11111SRRR14.21k73.0kp2345Rp11.9kRR1Rp29.4k11.9k41.3knorthCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.Jingdanturnedascrewdriverintoamagnetbyrubbingitwithastrongbarmagnettopickupascrewthathasfallenintoaninaccessiblespot.Howcouldhedemagnetizehisscrewdriverafterpickingupthescrew?Hecoulddropit,heatit,hititwithahammer,orsomehowotherwisejigglethemagneticdomainsintorandomness.4.Along,straight,current-carryingwirecarriesacurrentfromwesttoeast.Acompassisheldabovethewire.a.Whichdirectiondoesthenorthpoleofthecompasspoint?southb.Whenthecompassismovedunderneaththewire,inwhichdirectiondoesthenorthpolepoint?north662SolutionsManualPhysics:PrinciplesandProblems

664Chapter24continued5.Considerthesketchoftheelectromagnetin7.Howlongisawireina0.86-Tfieldthatthefigurebelow.Whichcomponentscouldcarriesacurrentof1.4Aandexperiencesyouchangetoincreaseordecreasetheaforceof13N?strengthoftheelectromagnet?ExplainyourFILBanswer.F13NL11mIB(1.4A)(0.86T)8.A6.0-Tmagneticfieldbarelypreventsa0.32-mlengthofcopperwirewithacurrentof1.8Afromdroppingtotheground.Whatisthemassofthewire?FILBmgILB(1.8A)(0.32m)(6.0T)Toincreasethestrengthofthemagnet,mg20.35kg9.80m/sthevoltageofthepowersourcecouldbeincreased,thenumberofthewraps9.Howmuchcurrentwillbeneededtocouldbeincreased,orferromagneticproduceaforceof1.1Nona21-cm-longcorescouldbeadded.pieceofwireatrightanglestoa0.56-Tfield?Todecreasethestrengthofthemagnet,F1.1NI9.4AthevoltageofthepowersourcecouldLB(0.21m)(0.56T)bedecreased,fewerwrapscouldbeused,orhollowornonmagneticcores10.Alphaparticles(particlescontainingtwocouldbeused.protons,twoneutrons,butnoelectrons)aretravelingatrightanglestoa47-Tfield6.Sketchagraphshowingtherelationshipwithaspeedof36cm/s.Whatistheforcebetweenthemagneticfieldaroundaoneachparticle?straight,current-carryingwire,andtheFqvBdistancefromthewire.(3.201019C)(0.36m/s)(4.7105T)5.41024N11.Aforceof7.11012NisexertedonsomeAl3ions(anatommissingthreeelectrons)thataretravelingat430km/sperpendiculartoamagneticfield.Whatisthemagneticfield?F7.11012NBqv(4.801019C)(4.3105m/s)Magneticfieldstrength34TDistancefromwire12.ElectronstravelingatrightanglestoamagneticfieldexperienceaforceofThisisagraphofy1/x.Theunits/num-8.31013Nwhentheyareinamagneticbersdonotmatter.They-axisshouldbeCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.fieldof6.2101T.Howfastarethelabeledmagneticfieldstrengthandtheelectronsmoving?x-axisdistancefromwire.F8.31013NvqB(1.601019C)(6.2101T)8.4106m/sPhysics:PrinciplesandProblemsSolutionsManual663

665b.Whatisthepeakvoltageofthegenerator?Chapter25VeffVpage885max92V21.A72-cmwireismovedatanangleof72°throughamagneticfieldof1.7102T5.WhatistheRMSvoltageofahouseholdandexperiencesanEMFof1.2mV.Howoutletifthepeakvoltageis125V?fastisthewiremoving?Veff2Vmax(2)(165V)117VEMFv6.Anoutlethasapeakvoltageof170V.LBsin1.2103Va.Whatistheeffectivevoltage?(0.72m)(1.7102T)(sin72°)Veff2Vmax0.10m/s(2)(170V)120V2.A14.2-mwiremoves3.12m/sperpendicularb.Whateffectivecurrentisdeliveredtoantoa4.21-Tfield.11-toaster?a.WhatEMFisinducedinthewire?V120VI11AEMFBLvsinR11(4.21T)(14.2m)(3.12m/s)7.Astep-uptransformerhasaprimarycoil(sin90.0°)consistingof152turnsandasecondarycoilwith3040turns.Theprimarycoilreceivesa187Vpeakvoltageof98V.b.Assumethattheresistanceinthewireisa.Whatistheeffectivevoltageinthe0.89.Whatistheamountofcurrentprimarycoil?inthewire?IV187V2.1102AVeff2Vmax(2)(98V)69VR0.89b.WhatistheeffectivevoltageintheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.A3.1-mlengthofstraightwirehasasecondarycoil?resistanceof3.1.Thewiremovesatn2304026cm/satanangleof29°throughaV2nV1152(69V)1.4kV1magneticfieldof4.1T.Whatistheinducedcurrentinthewire?8.Astep-downtransformerhas9000turnsEMFBLvsinontheprimarycoiland150turnsonthesecondarycoil.TheEMFintheprimary(4.1T)(3.1m)(0.26m/s2)(sin29°)coilis16V.Whatisthevoltagebeing1.6Vappliedtothesecondarycoil?EMF1.6Vn1150IR3.10.52AV1nV29000(16V)0.27V24.Ageneratordeliversaneffectivecurrent9.Atransformerhas124turnsontheof75.2Atoawirethathasaresistanceprimarycoiland18,600turnsontheof0.86.secondarycoil.a.Whatistheeffectivevoltage?a.Isthisastep-downorastep-uptransformer?VeffIeffR(75.2)(0.86)65Vstep-up664SolutionsManualPhysics:PrinciplesandProblems

666Chapter25continuedb.IftheeffectivevoltageinthesecondaryChapter26coilis3.2kV,whatisthepeakvoltagebeingdeliveredtotheprimarycoil?pages885–886n1.Astreamofsinglyionized(1)fluorineV1V1,effn2,effatomspassesundeflectedthroughamag-2neticfieldof2.5103Tthatisbalanced124(3.2kV)21V3V/m.Thebyanelectricfieldof3.51018,600massofthefluorineatomsis19timesthatV1,max2V1,eff(2)(21V)ofaproton.1a.Whatisthespeedofthefluorineions?3.010Theelectricforceandmagneticforcesbalanceeachother.BqvEqEvB3.5103V/m2.5103T1.4106m/sb.Iftheelectricfieldisswitchedoff,whatistheradiusofthecircularpathfollowedbytheions?Themagneticforceactsasacen-tripetalforce.mv2BqvrmvrBq(19)(1.671027kg)(1.4106m/s)(1.601019C)(2.5103T)110m2.AnelectronmovesperpendiculartoEarth’smagneticfieldwithaspeedof1.78106m/s.IfthestrengthofEarth’smagneticfieldisabout5.00105T,whatistheradiusoftheelectron’scircularpath?mv2BqvrmvrBq(9.111031kg)(1.78106m/s)(5.00105T)(1.601019C)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.0.203mPhysics:PrinciplesandProblemsSolutionsManual665

667Chapter26continued3.Aprotonwithavelocityof3.98104m/sperpendiculartothedirectionofamagneticfieldfollowsacircularpathwithadiameterof4.12cm.Ifthemassofaprotonis1.671027kg,whatisthestrengthofthemagneticfield?mv2BqvrmvBrq(1.671027kg)(3.98104m/s)(2.06102m)(1.601019C)2.02102T4.Abeamofdoublyionized(2)calciumatomsisanalyzedbyamassspectrome-ter.IfB4.5103T,r0.125m,andthemassofthecalciumionsis6.681026kg,whatisthevoltageofthemassspectrometer?q2VmB2r2calciumqB2r2V2m(3.201019C)(4.5103T)2(0.125m)2(2)(6.681026kg)0.76V5.Thespeedoflightincrownglassis1.97108m/s.Whatisthedielectricconstantofcrownglass?cvKc2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Kv3.00108m/s21.97108m/s2.326.Thedielectricconstantofdiamondis6.00.Whatisthespeedoflightindia-mond?cvK3.00108m/s6.001.22108m/s666SolutionsManualPhysics:PrinciplesandProblems

668Chapter26continued7.Bycurvingdifferentisotopesthroughpathswithdifferentradii,amassspectrom-etercanbeusedtopurifyasampleofmixeduranium-235anduranium-238iso-topes.AssumethatB5.00103T,V55.0V,andthateachuraniumisotopehasa5ionizationstate.Uranium-235hasamassthatis235timesthatofapro-ton,whileuranium-238hasamassthatis238timesthatofaproton.Bywhatdis-tancewillthetwoisotopesbeseparatedbythemassspectrometer?R1Uranium-235R2Uranium-238R2R1Uraniumionbeamq2VmB2r22VmrU-235qB2(2)(55.0V)(235)(1.671027kg)(5)(1.601019kg)(5.00103T)21.47m2VmrU-238qB2(2)(55.0V)(238)(1.671027kg)(5)(1.601019kg)(5.00103T)21.48mSeparationrU-238rU-2351.48m1.47m0.01m8.AmassspectrometeroftenisusedincarbondatingtodeterminetheratioofC-14isotopestoC-12isotopesinabiologicalsample.Thisratiothenisusedtoestimatehowlongagotheonce-livingorganismdied.Becauseamassspectrometerissensi-tivetothecharge-to-massratio,itispossibleforacontaminantparticletoalterthevaluemeasuredfortheC-14/C-12ratio,andthus,yielderroneousresults.Whenion-ized,C-14formsanionwitha4charge,andthemassofC-14is14timesthatofaproton.Consideracontaminantlithiumparticle.Ifthemostcommonlithiumiso-topehasamassthatisseventimesthatofaproton,whatmustbethechargeoftheCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.lithiumionneededtocontaminateacarbon-14experiment?qqC2VLimB2r2mCLimLiqLiqCmCPhysics:PrinciplesandProblemsSolutionsManual667

669Chapter26continued7mp(4)14mp29.Whatisthewavelengthofaradiowavewithafrequencyof90.7MHz?cf3.00108m/s90.7106Hz3.31m10.Whatisthefrequencyofamicrowavewithawavelengthof3.27mm?cfcf3.00108m/s3.27103m9.171010Hz11.WhatisthefrequencyofanXraywithawavelengthof1.001010m?cfcfCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3.00108m/s1.001010m3.001018Hz12.Inrecentyears,physicistshaveslowedthespeedoflightpassingthroughamaterialtoabout1.20mm/s.Whatisthedielectricconstantofthismaterial?cvKc2Kv3.00108m/s21.20103m/s6.251022668SolutionsManualPhysics:PrinciplesandProblems

670Chapter27pages886–8871.Ifthemaximumkineticenergyofemittedphotoelectronsis1.41018J,whatisthestoppingpotentialofacertainphotocell?KEqV0KEV0q1.41018J1.601019C8.8V2.Thestoppingpotentialofaphotocellis2.3V.Whatistheinitialvelocityofanemittedphotoelectronthatisbroughttoastopbythephotocell?12KEqVmv022qV0vm(2)(1.601019C)(2.3V)9.111031kg9.0105m/s3.Ifaphotoelectrontravelingat8.7105m/sisstoppedbyaphotocell,whatisthephotocell’sstoppingpotential?12KEqVmv02mv2V02q(9.111031kg)(8.7105m/s)2(2)(1.601019C)2.2V4.Lightwithafrequencyof7.51014Hzisabletoejectelectronsfromthemetalsurfaceofaphotocellthathasathresholdfrequencyof5.21014Hz.Whatstoppingpotentialisneededtostoptheemittedphotoelectrons?KEhfhf0h(ff0)andKEqV0Thus,KEh(ff(6.6261034J/Hz)(7.51014Hz5.21014Hz)V0)0qq1.601019C0.95VCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual669

671Chapter27continued5.Ametalhasaworkfunctionof4.80eV.Willultravioletradiationwithawave-lengthof385nmbeabletoejectaphotoelectronfromthemetal?Firstcalculatetheenergyofthephoton.1240eVnmE1240eVnm385nm3.22eVToejectaphotoelectronfromthemetal,theenergyoftheincidentradiationmustbegreaterthantheworkfunctionofthemetal.Becausetheenergyoftheincidentradiation,3.22eV,islessthantheworkfunction,4.80eV,aphotoelectronwillnotbeejected.6.WhenametalisilluminatedwithradiationCathodeAnodewithawavelengthof152nm,photoelectronsareejectedwithavelocityof7.9105m/s.Whatistheworkfunction,ineV,ofthemetal?v7.9105m/sFirstcalculatetheenergyofthephotoninjoules.152nmhcE(6.6261034J/Hz)(2.998108m/s)7Incident1.5210mradiation1.311018JNext,calculatethekineticenergy.12KEmv2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1(9.111031kg)(7.9105m/s)222.81019JKEEWWEKE1.311018J2.81019J1.01018J(1.01018J)1eV1.601019J6.2eV670SolutionsManualPhysics:PrinciplesandProblems

672Chapter27continued7.ThedeBrogliewavelengthforanelectrontravelingat9.6105m/sis7.61010m.Whatisthemassoftheelectron?hmvhmv6.6261034J/Hz(7.61010m)(9.6105m/s)9.11031kg8.WhatisthedeBrogliewavelengthofa68-kgmanmovingwithakineticenergyof8.5J?12KEmv22KEvmThedeBrogliewavelengthisthen,hhmmvm2KE1h(2mKE)21(6.6261034J/Hz)((2)(68kg)(8.5J))21.91035m9.AnelectronhasadeBrogliewavelengthof5.21010m.Whatpotentialdifferenceisresponsibleforthiswavelength?hmvhvmand12KE–qVmv02mv2V02qmh22q2m2h22q2m(6.6261034J/Hz)2(2)(1.601019C)(5.21010m)2(9.111031kg)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5.6VPhysics:PrinciplesandProblemsSolutionsManual671

673Chapter28page8871.AnelectroninahydrogenatommakesatransitionfromE3toE1.Howmuchenergydoestheatomlose?1En13.6eVn2EE1E313.613.6123212.1eV2.Anelectroninthehydrogenatomloses3.02eVasitfallstoenergylevelE2.Fromwhichenergyleveldidtheatomfall?EEfEi113.02eV13.6n2n2yx13.6eV3.02eV3.40eVn2xnx63.Whichenergylevelinahydrogenatomhasaradiusof7.63109m?h2n2r42Kmq242Kmq2rnh2Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.42(9.0109Nm2/C2)(9.111031kg)(1.601019C)2(7.63109m)(6.6261034Js)212Alternatively,rr2wherer0m0istheBohrradiusinhydrogen,530mm.r7.63109mThus,m5.301011m14412r04.WhenanelectronfallsfromE4toE1,whatisthefrequencyoftheemittedphoton?EE1E4(2.171018J)118J)112(2.1710422.041018JEhfEfh2.041018J6.6261034Js3.081015Hz672SolutionsManualPhysics:PrinciplesandProblems

674Chapter28continued5.IfanelectronmovesfromE3toE5,whatisthewavelengthofthephotonabsorbedbytheatom?EE5E3(2.171018J)1(2.171018J)152321.541019JhcEhfdhcdE(6.6261034Js)(3.00108m/s)1.541019J1.29106m6.Ifahydrogenatominitsgroundstateabsorbsaphotonwithawavelengthof93nm,itjumpstoanexcitedstate.Whatisthevalueoftheenergyinthatexcitedstate?hcEhf(6.6261034Js)(3.00108m/s)9.3108m2.141018J(2.141018J)1eV1.601019J13.3eVabsorbedEfEiE13.6eV13.3eV0.27eVCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Physics:PrinciplesandProblemsSolutionsManual673

675Chapter29pages887–8881.Indiumhas3freeelectronsperatom.UseAppendixDanddeterminethenum-beroffreeelectronsin1.0kgofindium.freee/atom)(N11000g(freeekgA)M1kgSubstitute:freee/atom3freee/1atom,N23atoms/mol,M114.82g/molA6.0210freee3freee6.021023atoms1mol1000gkg1atom1mol114.82g1kg1.571025freee/kgforindium2.Cadmiumhas2freeelectronsperatom.UseAppendixDanddeterminethenumberoffreeelectronsin1.0dm3ofCd.freee/atom)(N11000cm3(freee()dm3A)M1dm3Substitute:freee/atom2freee/1atom,N23atoms/mol,M112.41g/mol,A6.02108.65g/cm3freee2freee6.021023atoms1mol8.65g1000cm3dm31atom1mol112.41g1cm31dm39.261025freee/dm3incadmium3.Copperhas1freeelectronperatom.Whatlengthof1.00-mmdiametercopperCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.wirecontains7.811024freeelectrons?UseAppendixDforphysicalconstants.1)atoms11Lwirer2(numberoffreeefreeeN(M)ASubstitute:N23atoms/mol,A6.0210M63.546g/mol,8.92g/cm3,numberoffreee7.811024freee,rwire1.00mm/20.50mm0.050cmL1(7.811024freee)1atomwire21freee0.100cm21mol63.546g1cm36.0221023atoms1mol8.92g1.18104cm118m674SolutionsManualPhysics:PrinciplesandProblems

676Chapter29continued4.At400.0K,germaniumhas1.131015freeelectrons/cm3.HowmanyfreeelectronsperGeatomarethereatthistemperature?freee11(M)(freeecm3forGe)atomNASubstitute:N23atoms/mol,A6.0210M72.63g/mol,5.23g/cm3,freee/cm3forGe1.131015freee/cm3freee1mol72.63g1cm31.131015freeeatom6.021023atoms1mol5.23gcm3freee/atomofGeat400.0K2.611085.At400.0K,siliconhas4.541012freeelectrons/cm3.HowmanyfreeelectronsperSiatomarethereatthistemperature?freee11(M)(4.541012freee/cm3forSi)atomNASubstitute:N23atoms/mol,M28.09g/mol,2.33g/cm3,A6.0210freee/cm3forSi4.541012freee/cm3at400.0Kfreee1mol28.09g1cm34.541012freeeatom6.021023atoms1mol2.33gcm3freee/atomofSiat400.0K9.0910116.At200.0K,siliconhas3.791018freeelectronsperatom.Howmanyfreeelectrons/cm3arethereinsiliconatthistemperature?freeefreee1(N()cm3atomA)MSubstitute:freee/atom3.791018freee/1atom,N23atoms/mol,M28.09g/mol,2.33g/cm3A6.0210freee3.791018e6.021023atoms1mol2.33gcm31atom1mol28.09g1cm31.89105freee/cm3insiliconat200K7.Siliconhas1.451010freeelectrons/cm3atroomtemperature.Ifyouwantedtohave3106asmanyelectronsfromarsenicdopingasthermalfreeelectronsfromsiliconatroomtemperature,howmanyarsenicatomsshouldtherebeforeachsiliconatom?Eacharsenicatomprovides1freeelectron.UseAppendixDforphysicalconstants.AsatomsnumberAsatomsnumberofAseSiatomsAsethermaleCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.numberofthermale11(M)cm3NASubstitute:M28.09g/mol,N23atoms/mol,2.33g/cm3,A6.02101Asatom1Ase,3106Ase/thermale,numberofthermale/cm31.451010e/cm3Physics:PrinciplesandProblemsSolutionsManual675

677Chapter29continuedAsatoms1Asatoms3106Ase1.451010thermale28.09gSiatoms1Asethermalecm3molmolcm36.021023Siatoms2.33g8.711078.At200.0K,germaniumhas1.161010thermallyliberatedchargecarriers/cm3.Ifitisdopedwith1Asatomto525,000Geatoms,whatistheratioofdopedcarrierstothermalcarriersatthistemperature?SeeAppendixDforphysicalconstants.dopedcarriersdopedcarriers11(Nthermalcarriers5.25105atomsA)M()thermalcarriersSubstitute:M72.6g/mol,N23atoms/mol,5.23g/cm3A6.0210thermalcarriers1.161010thermalcarriers/cm3dopedcarriers1dopedcarrier6.021023atomsmol5.23gthermalcarriers5.25105atomsmol72.6gcm3cm31.161010thermalcarriers7.121069.At200.0K,siliconhas1.89105thermallyliberatedchargecarriers/cm3.Ifitisdopedwith1Asatomto3.75millionSiatoms,whatistheratioofdopedcarri-erstothermalcarriersatthistemperature?SeeAppendixDforphysicalconstants.dopedcarriersdopedcarriers11(Nthermalcarriers3.75106atomsA)M()thermalcarriersCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Substitute:M28.09g/mol,N23atoms/mol,2.33g/cm3A6.0210thermalcarriers1.89105thermalcarriers/cm3dopedcarriers1dopedcarrier6.021023atomsmol2.33gthermalcarriers3.75106atomsmol28.09gcm3cm31.89105thermalcarriers7.05101010.Thediodeshownbelowhasavoltagedrop,Vd,Rof0.45VwhenI11mA.Ifa680-resistor,R,isconnectedinseries,whatpowersupplyvoltage,IVbisneeded?VbIRVd(0.011A)(680)0.45VVb7.9VDiodeVd676SolutionsManualPhysics:PrinciplesandProblems

678Chapter29continued11.AdiodeinacircuitsimilartotheoneinChapter30Figure29-26hasavoltagedrop,Vd,of0.95VwhenI18mA.Ifa390-resistor,pages888–889R,isconnectedinseries,whatpowersupply1.Carbon-14,or14C,isanisotopeofthevoltage,Vcommon12bisneeded?6C,andisusedindatingancientVartifacts.WhatisthecompositionofitsbIRVdnucleus?(0.018A)(390)0.95Vsixprotons(equaltoZ)andeight8.0Vneutrons(equaltoAZ)12.Whatpowersupplyvoltagewouldbeneeded2.Anisotopeofiodine(Z53)isusedtotoproduceacurrentof27mAinthecircuittreatthyroidconditions.Itsmassnumberisinproblem10?Assumethediodevoltageis131.Howmanyneutronsareinitsnucleus?uncharged.1315378neutronsVbIRVd3.Theonlynonradioactiveisotopeoffluorine(0.027A)(680)0.45Vhasnineprotonsandtenneutrons.19Va.Whatisitsmassnumber?massnumber(A)19b.Theatomicmassunit,u,isequalto1.661027kg.Whatisfluorine-19’sapproximatemassinkilograms?(19nucleons)(1.661027kg/nucleon)3.1510–26kgc.Writethefullsymbolofthisatom.199F4.Themagnesiumisotope2512Mghasamassof24.985840u.a.Calculateitsmassdefect.massdefect(isotopemass)(massofprotonsandelectrons)(massofneutrons)24.985840u(12)(1.007825u)(13)(1.008665u)0.220705ub.CalculateitsbindingenergyinMeV.bindingenergy(massdefect)(bindingenergyof1u)(0.220705u)(931.49MeV/u)Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.205.58MeVPhysics:PrinciplesandProblemsSolutionsManual677

679Chapter30continued5.Theisotope10b.Identifytheparticlethatisejected.5Bhasamassof10.012939u.a.Calculatethemassdefect.anparticlemassdefect(isotopemass)2228.Theradioisotope84Poundergoesalpha(massofprotonsandelectrons)decaytoformanisotopeoflead(leadhas(massofneutrons)atomicnumber82).Determinewhatthemassnumberofthatisotopemustbeby10.012939u(5)(1.007825u)writinganuclearequation.(5)(1.008665u)222A40.069511u84Po→ZPb2Heb.CalculateitsbindingenergyinMeV.whereZ84282bindingenergy(massdefect)A2224218(bindingenergyof1u)Thustheequationis222218484Po→82Pb2He(0.069511u)(931.49MeV/u)Themassnumberis218.64.749MeV9.Thegraphbelowshowsasequenceofalphac.Calculateitsbindingenergypernucleon.andbetadecays,labeled1,2,3,and4.bindingenergypernucleonisConsultTable30-1asneeded.64.744MeV6.4749MeV/nucleonA10nucleons2186.Themoststableisotopeofallis5626Fe.Itsbindingenergypernucleonis12168.75MeV/nucleon.a.Whatisthebindingenergyofthis3214isotope?2(56nucleons)(–8.75MeV/nucleon)Massnumber2124–4.90102MeVCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.Whatisthemassdefectofthisisotope?210bindingenergymassdefectbindingenergyof1u808284Z4.90102MeVAtomicnumber931.49MeV/ua.Whichrepresentalphadecays,and0.526uwhichrepresentbetadecays?7.Theisotope239alphadecays:1and4,inwhich94Pucanbetransmutedtoanisotopeofuranium,235massnumberdecreasesby4and92U.atomicnumberby2;betadecays:2a.Writethenuclearequationforthisand3,inwhichatomicnumbertransmutation.increasesby1andmassnumber239Pu→235A9492UZXstaysthesamewhereZ94922b.WhatistheoverallchangeinmassintheA2392354sequence?Inthenumberofneutrons?ForZ2,theelementmustbeOverallchangeinmassis218uhelium.Thustheequationis210u8ulower.Overallchangein2392354Zis84–822lower.Thus,overall94Pu→92U2Hechangeinthenumberofneutronsis6lower(8nucleonsless,2protonsless,so6neutronsless).678SolutionsManualPhysics:PrinciplesandProblems

680Chapter30continued10.UseAppendixDtocompletethetwo13.Iodine-131hasahalf-lifeof8.0days.Ifnuclearequations.Includecorrectthereare60.0mgofthisisotopeattimesubscriptsandsuperscriptsforeachzero,howmuchremains24dayslater?oftheparticles.24days3.0half-livesa.328.0days/half-life15P®?antineutrino1t32A0e0remainingoriginal15P→ZX102whereZ15(1)01613.0(60.0mg)2A3200327.5mgForZ16,theelementmustbesulfur.Thus,theequationis14.RefertoTable30-2.Givenasampleof32320e015P→16S10cobalt-60,b.235a.howlongatimeisneededforittogo92U®?throughfourhalf-lives?235A492U→ZX2He(4)(30years)120yearswhereZ92290b.whatfractionremainsattheendofthatA2354231time?1tForZ90,theelementmustberemainingoriginal2thorium.Thus,theequationis23523141492U→90Th2Heoriginal211.Writethecompletenuclearequationforthe1original3416betadecayof16S.34A0e015.Thegraphshowstheactivityofacertain16S→ZX10radioisotopeovertime.Deduceitshalf-lifewhereZ16(1)017fromthisdata.A340034ForZ17,theelementmustbe48chlorine.Thus,theequationis34340e016S→17Cl10(Bq)3651012.Apositronisidenticaltoabetaparticle,24exceptthatitschargeis1insteadof1.3719KundergoesspontaneouspositronActivitydecaytoformanisotopeofargon.Identify12theisotopeofargonbywritinganuclearequation.03691237A0e019K→ZAr10Time(h)whereZ191018Every3.0htheactivityishalfofthepre-A370037viousactivity,sothehalf-lifeis3.0h.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Thustheequationis37370e019K→18Ar10Theisotopeisargon37.Physics:PrinciplesandProblemsSolutionsManual679

681Chapter30continued16.Themassofaprotonandofanantiproton18.When1mol(235g)ofuranium-235is1.00728u.Recallthattheconversionofundergoesfission,about2.01010kJofexactly1uintoenergyyields931.5MeV.energyarereleased.When4.0gofhydro-a.Calculatethemassusedupwhenagenundergoesfusion,about2.0109kJareprotonandanantiprotonannihilatereleased.oneanother.a.Foreach,calculatetheenergyyieldper(2)(1.00728u)2.01456ugramoffuel.2.0109kJ8b.Calculatetheenergyreleasedhere.5.0104.0g(2.01456u)(931.5MeV/u)1876.5MeVkJ/gforfusionofhydrogen;17.Onesourceofastar’senergyisthefusionof2.01010kJ78.510235gtwodeuteronstoformanalphaparticleplusagammaray.kJ/gforfissionofuraniumb.WhichprocessproducesmoreenergyParticleMassinupergramoffuel?2H2.01361Fusionreleasesaboutsixtimesas4He4.00262muchenergypergramoffuelas00.00000fission.a.Writethenuclearequationforthisfusion.22401H1H→2He0b.Calculatethemass“lost”inthisprocess.massbeforemassafter2(2.0136u)4.0026u0.0246uCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.c.CalculatetheenergyreleasedinMeV.(0.0246u)(931.49MeV/u)22.9MeV680SolutionsManualPhysics:PrinciplesandProblems