PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm...
Transcript of PHYSICS 111 SPRING 2019 FINAL EXAM: May 8, 2019; 4:30pm...
PHYSICS111SPRING2019
FINALEXAM:May8,2019;4:30pm-6:30pmName(printed):______________________________________________RecitationInstructor:_________________________Section#_______INSTRUCTIONS:Thisexamcontains30multiple-choicequestion,eachworth3points,foranominalmaximumscoreof80pointsplus10extracreditpoints.Chooseoneansweronlyforeachquestion.Choosethebestanswertoeachquestion.Answerallquestions.Allowedmaterial:Beforeturningoverthispage,putawayallmaterialsexceptforpens,pencils,erasers,rulersandyourcalculator.Thereisaformulasheetattachedattheendoftheexam.Othercopiesoftheformulasheetarenotallowed.Calculator:Ingeneral,anycalculator,includingcalculatorsthatperformgraphing,ispermitted.Electronicdevicesthatcanstorelargeamountsoftext,dataorequations(likelaptops,e-bookreaders,smartphones)areNOTpermitted.CalculatorswithWiFitechnologyareNOTpermitted.Ifyouareunsurewhetherornotyourcalculatorisallowedfortheexam,askyourTA.Howtofillinthebubblesheet:
Useanumber2pencil.DoNOTuseink.Ifyoudidnotbringapencil,askforone.YouwillcontinuetousethesamebubblesheetyoualreadyusedforExam1-3.Bubbleanswers64-93onthebubblesheetforthisexam.Only,ifforsomereasonyouarestartinganewbubblesheet,writeandfillinthebubblescorrespondingto:- Yourlastname,middleinitial,andfirstname.- « « YourIDnumber(themiddle9digitsonyourISUcard)« « - SpecialcodesKtoLareyourrecitationsection.Alwaysusetwodigits(e.g.01,09,11,13).
Pleaseturnoveryourbubblesheetwhenyouarenotwritingonit.Ifyouneedtochangeanyentry,youmustcompletelyeraseyourpreviousentry.Also,circleyouranswersonthisexam.Beforehandinginyourexam,besurethatyouranswersonyourbubblesheetarewhatyouintendthemtobe.Youmayalsocopydownyouranswersonapieceofpapertotakewithyouandcomparewiththepostedanswers.Youmayusethetableattheendoftheexamforthis.Whenyouarefinishedwiththeexam,placeallexammaterials,includingthebubblesheet,andtheexamitself,inyourfolderandreturnthefoldertoyourrecitationinstructor.Nocellphonecallsallowed.Eitherturnoffyourcellphoneorleaveitathome.Anyoneansweringacellphonemusthandintheirwork;theirexamisover.
Bestofluck,Dr.SoerenPrell
64. Asshowninthefigure,abimetallicstrip,consistingofmetalGonthetopandmetalHonthebottom,isrigidlyattachedtoawallattheleft.ThecoefficientoflinearthermalexpansionformetalGisgreaterthanthatofmetalH.Ifthestripisuniformlyheated,itwill
A) Curvedownward.B) Remainhorizontal,butgetlonger.C) Neitherchangeshapenorlength.D) Remainhorizontal,butgetshorter.E) Curveupward.
SolutionBothmetalswillexpandwhenheated.MetalGwillexpandmorethanmetalHandthebimetallicstripwillcurvedownward.
65. Atemperaturechangeof20C°correspondstoaFahrenheittemperaturechangeof
A) 11FoB) 36FoC) 18FoD) 68FoE) 293Fo
Solution
TF =
95TC + 32 F!⇒ΔTF =TF −TF ,0 =
95TC + 32 F!
⎛
⎝⎜
⎞
⎠⎟−
95TC ,0 + 32 F!
⎛
⎝⎜
⎞
⎠⎟=
95ΔTC =
95
20 C!( ) = 36 C!
66. Aquantityofanidealgasiskeptinarigidcontainerofconstantvolume.Ifthegasisoriginallyatatemperatureof19°C,atwhattemperaturewillthepressureofthegasdoublefromitsoriginalvalue?
A) 273°CB) 38°CC) 122°CD) 91°CE) 311°C
Solution
T0 =19 !C = 19+273( )K = 292K; TP=T0
P0
⇒ T =T0PP0
= 2T0 = 2×292K = 584K = 584 −273( )!C = 311!C
67. Thefigureshowsagraphofthetemperatureofa
puresubstanceasafunctionoftimeasheatisaddedtoitataconstantrateinaclosedcontainer.IfLFisthelatentheatoffusionofthissubstanceandLVisitslatentheatofvaporization,whatisthevalueoftheratioLV/LF?
A) 7.2B) 3.5C) 4.5D) 5.0E) 1.5
SolutionThetimeittakestomeltthesubstanceis2unitsoftime(1sthorizontalsegment)inthegraphwhereasthetimeittakestoevaporateitisabout7units(2ndhorizontalsegment).Sinceheatisaddedataconstantratethetimesareproportionaltotherespectiveamountsofheatadded.SinceQ=mLandthemassmofthesubstanceisconstantLV/LF=3.5
68. Acamperisabouttodrinkhismorningcoffee.Hepours400gramsofcoffee,initiallyat75°Cintoa250-galuminumcup,initiallyat16°C.Whatistheequilibriumtemperatureofthecoffee-cupsystem,assumingnoheatislosttothesurroundings?Thespecificheatofaluminumis900J/kg•K,andthespecificheatofcoffeeisessentiallythesameasthatofwater,whichis4186J/kg•K.
A) 45oCB) 68oCC) 65oCD) 71oCE) 62oC
Solution
Qcoffee +Qcup = 0⇒ mcoffeeccoffee Tcoffee −T( ) + mcupccup Tcup −T( ) = 0
T =mcoffeeccoffeeTcoffee + mcupccupTcup
mcoffeeccoffee + mcupccup
= 68o C
69. Thethermodynamicprocessforanidealgasshownon
thepVdiagraminthefigureis
A) adiabaticB) isobaricC) isothermalD) isochoricE) notpossible
SolutionThere’snochangeinvolume.Thisprocesscouldbetheheatingupofafixedamountofgasinsideafixed-volumecontainer.Thepressureofthegasincreasesproportionaltotemperature.Constant-volumeprocessesarecalledisochoric.
70. Amonatomicidealgasundergoesanisothermalexpansionat300K,asthevolumeincreasesfrom0.010m3to0.040m3Thefinalpressureis130kPa.Whatisthechangeintheinternal(thermal)energyofthegasduringthisprocess?
A) 0.0kJB) –7.2kJC) +7.2kJD) +3.6kJE) –3.6kJ
SolutionTheinternalenergyofamonatomicidealgasisU=3/2nRT.Duringanisothermalprocessthetemperatureandtheinternalenergydonotchange.
71. ThefigureshowsapVdiagramforanidealgasthatiscarriedaroundacyclicprocess.Howmuchworkisdoneinonecycleifp0=4.00atmandV0=3.00liters?
A) 4850JB) 2420JC) 485JD) 2280JE) 1210J
Solution
W = (area enclosed by the path) = ΔpΔV = p0V0 = 4×1.01×105 Pa( ) 3.00×10−3 m3( ) = 1210 J
72. Anarchitectisinterestedinestimatingthe
rateofheatloss,ΔQ/Δt,throughasheetofinsulatingmaterialasafunctionofthethicknessofthesheet.Assumingfixedtemperaturesonthetwofacesofthesheetandsteadystateheatflow,whichoneofthegraphsshowninthefigurebestrepresentstherateofheattransferasafunctionofthethicknessoftheinsulatingsheet?
A)AB)BC)CD)DE)ESolution
Qt= kA
lΔT ∝ 1
l
73. Theprocessinwhichheatflowsbythemassmovementofmoleculesfromoneplacetoanotherisknownas
A)conductionB)radiationC)convectionD)sublimationE)dissipation
SolutionThethreeformsofheattransferareconduction,radiationandconvection.Nomatterisbeingtransportedinthefirsttwo.Onlyconvectioninvolvesthemovementofmolecules.
74. Themotionofaparticleisdescribedinthevelocityvs.timegraphshowninthefigure.Overthenine-secondintervalshown,wecansaythatthespeedoftheparticle
A) RemainsconstantB) Decreasesandthenincreases.C) Onlydecreases.D) Onlyincreases.E) Increasesandthendecreases.SolutionInitiallythespeedishigh(about2.5m/s).Thenthespeedgoestozero,andthenincreasesagain(toabout4.0m/s).
75. Aballisthrownstraightup,reachesamaximumheight,thenfallstoitsinitialheight.Whichofthefollowingstatementsaboutthedirectionofthevelocityandaccelerationoftheballthehighestpointofitstrajectoryiscorrect?
A) Bothitsvelocityanditsaccelerationpointdownward.B) Bothitsvelocityanditsaccelerationpointupward.C) Itsvelocityiszeroanditsaccelerationpointsdownward.D) Itsvelocitypointsupwardanditsaccelerationiszero.E) Itsvelocityiszeroandanditsaccelerationiszero.SolutionAtthehighestpointthevelocityiszerosinceitchangesdirection.Theaccelerationatanytimeisg=9.8m/sanditpointsdownward(tothecenteroftheEarth).
76. Aboyjumpswithavelocityofmagnitude20.0m/satanangleof25.0°abovethe
horizontal.Whatisthehorizontalcomponentoftheboy'svelocity?
A)8.45m/sB)12.6m/sC)9.33m/sD)15.6m/sE)18.1m/sSolution
vx = vcosθ = 20.0 m/s( )cos 25.0!( ) = 18.1 m/s
77. Apersonwhonormallyweighs700Nisridinginaratherswiftelevatorthatismovingat
aconstantspeedof9.8m/s.Ifthispersonisstandingonabathroomscaleinsidetheelevator,whatwouldthescaleread?
A)morethan700NB)lessthan700N,butmorethanzeroC)0D)700NE)Itcouldbemoreorlessthan700N,dependingonwhethertheelevatorisgoingupordown.
SolutionTheelevatorismovingwithconstantspeed.Thus,thenormalforcebytheelevatoronthepersonequalstheirweightregardlessofthedirectionormagnitudeofthevelocity.
78. Inthefigure,whatdoesthespringscaleread?
Thepulleysareidealandthestringsandscalearealsomassless.
A)2.0NB)morethan19.6NC)0.0ND)1.0NE)0.5NSolutionThespringscalereadsthetensioninthestring,whichis1.0N.
79. A3.0-kgmassanda5.0-kgmasshangverticallyattheoppositeendsofaverylightrope
thatgoesoveranidealpulley(alsocalledAtwood’smachine).Ifthemassesaregentlyreleased,whatistheresultingaccelerationofthemasses?A)6.1m/s2B)3.7m/s2C)4.9m/s2D)2.5m/s2E)0.0m/s2
SolutionTheheaviermasswillacceleratedownwards.Choosethepositivedirectiontobeinthedirectionoftheacceleratio
(5.0 kg)g −T = (5.0 kg)a; T − (3.0 kg)g = (3.0 kg)a
⇒ (5.0 kg)g − (3.0 kg)g = (5.0 kg)a + (3.0 kg)a ⇒ a = 28
g = g4= 2.5 m/s
80. Aforceactsonanobject,causingittomoveparalleltotheforce.Thegraphinthefigure
showsthisforceasafunctionofthepositionoftheobject.Howmuchworkdoestheforcedoastheobjectmovesfrom6mto12m?
A) 70JB) 0JC) 20JD) 30JE) 40J
Solution
W = 1
212 m − 6 m( ) 10 N( ) = 30 J
81. Agirlthrowsastonefromabridge.Considerthefollowingwaysshemightthrowthestone.Thespeedofthestoneasitleavesherhandisthesameineachcase.CaseA:Thrownstraightup.CaseB:Thrownstraightdown.CaseC:Thrownstraightouthorizontally.Inwhichcasewillthespeedofthestonebegreatestwhenithitsthewaterbelowifthereisnosignificantairresistance?
A)caseAB)caseBC)casesAandBD)caseCE)Thespeedwillbethesameinallcases.SolutionMechanicalenergyisconservedduringfreefallwithnoairresistance.Inallcasesaretheinitialkineticandpotentialenergiesthesame.Thefinalpotentialenergiesarealsothesame.Therefore,thefinalkineticenergiesandthusspeedshavetobethesameaswell.
82. Asshowninthefigure,agivenforceisappliedtoarodinseveraldifferentways.InwhichcaseisthetorqueaboutthepivotpointPduetothisforcethegreatest?
A)1B)2C)3D)4E)5SolutionForcase1,theforceisperpendiculartothelineconnectingthepointPandthepointwheretheforceapplies.Inaddition,thedistancebetweenPandthepointwheretheforceappliesisthelargest.Therefore,case1hasthelargestlevelarm.
83. Astoneinitiallymovingat8.0m/sonalevelsurfacecomestorestduetofrictionafterittravels11m.Whatisthecoefficientofkineticfrictionbetweenthestoneandthesurface?
A)0.13B)0.43C)0.80D)0.30E)0.25Solution
ΔKE =W = −Fs = − fks = −µmgs = −KE0 = − 1
2mv2 ⇒ µ = v2
2gs= 0.3
84. Ameterstickbalancesatthe50.0-cmmark.Ifamassof50.0gisplacedatthe90.0-cmmark,thestickbalancesatthe61.3-cmmark.Whatisthemassofthemeterstick?
A)89.7gB)178gC)32.6gD)73.4gE)127g
Solution
61.3 cm −50 cm( )m = 90 cm − 61.3 cm( ) 50 g( )⇒ m = 127g
85. Twoiceskaterssuddenlypushoffagainstoneanotherstartingfromastationaryposition.The45-kgskateracquiresaspeedof0.375m/srelativetotheice.Whatspeeddoesthe60-kgskateracquirerelativetotheice?A)0.38m/sB)0.50m/sC)0.28m/sD)0.75m/sE)0.00m/sSolution
ΔPtotal = 0⇒ m1v1 + m2v2 = 0⇒ v2 =
m1
m2
v1 = 0.28 m/s
86. Saltwaterhasgreaterdensitythanfreshwater.Ablockofwoodfloatsinbothfreshwaterandinsaltwaterandablockofironsinksinbothfreshwaterandinsaltwater.ThebuoyantforceonthewoodblockinsaltwateriscalledFW,SandinfreshwateritiscalledFW,F.ThebuoyantforceontheironblockinsaltwateriscalledFI,SandinfreshwateritiscalledFI,F.Whichoftherelationsbetweenthebuoyantforcesiscorrect?A)FW,S>FW,FandFI,S>FI,FB)FW,S=FW,FandFI,S>FI,FC)FW,S>FW,FandFI,S=FI,FD)FW,S=FW,FandFI,S=FI,FE)FW,S<FW,FandFI,S<FI,FSolutionThebuoyantforceonafloatingobjectisalwaysequaltoitsweight.Thebuoyantforceofanyfullyorpartiallysubmergedobjectisequaltotheweightofthedisplacedfluid.
87. TheintensityofthewavesfromapointsourceatadistancedfromthesourceisI.Whatis
theintensityatadistance2dfromthesource?
A)2IB)I/4C)I/√2D)I/2E)4I
Solution
I(d) = P4πd 2 ⇒ I(2d) = P
4π 2d( )2 = I(d)4
88. Anobjectofmassm=8.0kgisattachedtoanidealspringandallowedtohangintheearth'sgravitationalfield.Thespringstretches2.2cmbeforeitreachesitsequilibriumposition.Ifitwerenowallowedtooscillatebythisspring,whatwouldbeitsfrequency?
A)0.52HzB)3.4HzC)0.20×10−3HzD)1.6HzE)21HzSolution
F = kΔx = mg; f = ω
2π= 1
2πkm
= 12π
gΔx
= 3.4 Hz
89. A0.588mlongstringwithamassof0.031kg,istightlyclampedatbothends.Iftheloweststandingwavefrequencyofthestringis326Hz,howfastdowavestravelonthisstring?
A)724m/sB)383m/sC)582m/sD)475m/sE)Impossibletotellwithoutknowingthetensioninthestring.
Solution
v = f λ = 326 Hz( ) 2× 0.588 m( ) = 383 m/s
Laboratoryfinalexamquestions
90. Inacertainexperiment,wemeasuretwoquantities:a=60unitsandb=30units,andweuse
thosevaluestoestimatetheratiobetweenthem:
2.0ab=
However,weknowthatourmeasurementscouldbeeachoffby3units.Whatisthentheassociatedmaximumerrorofourestimationofa/b?
A. Exactly0,becausetheerrorscancelout.B. 0.1C. 0.2D. 0.3E. 0.4
Theminimumandmaximumvaluesofthisratioare:
min
min max
max
max min
60 3 1.730 360 3 2.330 3
aab b
aab b
−= = =++= = =−
Therefore,thevaluesoftheratiocanbe0.3aboveorbelowthecentralvalueof2.0.
91. Theplotbelowshowsthepositionofacaralongastraightstreet.Asyoucansee,thedataisperfectforalinearfit.
Theslopeofthelinearfitrepresentsthe______________ofthecar,anditsintersectionwiththeverticalaxisrepresentsthe_________________ofthecar.A. acceleration;initialpositionB. acceleration;finalpositionC. velocity;initialpositionD. velocity;finalpositionE. velocity;initialtime
Theslopeofapositionversustimegraphisalwaysthevelocity.Theintersectionwiththevertical(position)axiscorrespondstothepositionvalueatt=0,sothisistheinitialpositionofthecar.
92. StudentstakingPHYS111onplanetXperformthefollowingexperimenttomeasuretheaccelerationofgravityonthesurfaceofthatplanet:theydropasmallrockfromdifferentheightshandmeasurethetimetherocktakestohittheground.Theythenplotthedatainthegraphsbelow.
Basedontheseresults,whatisthevalueofgonplanetX?A. 4.5m/s2B. 7.5m/s2C. 8.2m/s2D. 9.0m/s2E. 18m/s2
Forany1Dmotionwithconstantacceleration, 20 0
12y yy y v t a t= + + .Ifwetakey=0atgroundlevel
andpositivesup,thenwehavey0=h,y=0,v0y=0,anday=−g,whichyields:21
2h gt=
Indeed,thehvs.t2plotshowslinearbehaviorandintersectstheverticalaxisveryclosetotheorigin,soitlookslikeitfitsthisequation.Wecanthenidentifytheslopeofthelinearfitwithg/2.Thelinearfit(showninredonthegraph)containspoints(0,0)and(4s2,18m),sotheslopeis:
( )( )
22
18 0 m4.5m/s
4 0 sslope
−= =
−
Therefore, 22 9.0m/sg slope= × =
93. A0.50-kgpuckmovesatconstantspeedontheplaneofthepaperandleavesamarkonthesurfacewithafrequencyof60Hz.Eachofthesemarksisacircleinthefigurebelow.Thedistancebetweenthefirstandlastcircleis35cm.Findthelinearmomentumofthepuck.
A. 1.2kgm/sB. 1.3kgm/sC. 1.7kgm/sD. 2.2kgm/sE. 2.9kgm/s
( )( )
= = = =
= = =
35cmseparationbetweentwomarks 8 262.5cm/s 2.625m/s
1timebetweentwomarks s60
0.50kg 2.625m/s 1.3kgm/s
v
p mv
35cm
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