OA02a Thermal Physics and 1D Kinematics PDF

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29/08/2016

OA02a:ThermalPhysicsand1DKinematics

OA02a:ThermalPhysicsand1DKinematics Due:11:59pmonWednesday,August24,2016 Youwillreceivenocreditforitemsyoucompleteaftertheassignmentisdue.GradingPolicy

PrelectureConceptQuestion17.06

PartA Ifboththepressureandvolumeofagivensampleofanidealgasaredoubled,whathappenstothetemperatureofthe gasinKelvins? ANSWER: Thetemperatureofthegasinincreasedbyfourtimesitsoriginalvalue. Thetemperatureofthegasisreducedtoonehalfitsoriginalvalue. Thetemperatureremainsconstant. Thetemperatureofthegasisincreasedbytwotimesitsoriginalvalue. Thetemperatureofthegasisreducedtoonefourthitsoriginalvalue.

Correct

PrelectureConceptQuestion17.04

PartA Whichofthefollowingstatementsaretrue? Checkallthatapply. ANSWER: Solidiceislessdensethanliquidwater. Watercontractsasitfreezesat0°C. Liquidwaterexpandswithincreasingtemperatureabove4°C. Liquidwaterexpandswithincreasingtemperaturebetween0°Cand4°C.

Correct

CalorimetryConceptualQuestion A0.5 blockofaluminum( insulatedtubofcoldwaterat0 ( blockaremeasuredtobe20 .

)isheatedto200 .Theblockisthenquicklyplacedinan )andsealed.Atequilibrium,thetemperatureofthewaterand

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Iftheoriginalexperimentisrepeatedwitha1.0 block?

aluminumblock,whatisthefinaltemperatureofthewaterand

Hint1.Heatenergyandtemperature Theamountofheatenergy neededtochangethetemperatureofasubstancedependsonthemassofthe substance ,thespecificheatofthesubstance ,andthechangeinthetemperature ofthesubstance. Thisissummarizedmathematicallyas .

Hint2.Conservationofenergy Thetotalamountofenergyinthesystemmustbeconserved.Oncetheblockisplacedincontactwiththecold water,heatenergywillflowfromthehottestblockintothecoldwater,loweringthetemperatureoftheblockand raisingthetemperatureofthewater.Alargerblockcantransferalargeramountofheatenergyforagiven changeintemperature.Thiswillresultinalargerequilibriumtemperature.

ANSWER: lessthan20 20 greaterthan20

Correct

PartB Iftheoriginalexperimentisrepeatedwitha1.0 temperatureofthewaterandblock?

copper(

)block,whatisthefinal

ANSWER: lessthan20 20 greaterthan20

Correct

PartC Iftheoriginalexperimentisrepeatedbut100 ofthe0 temperatureofthewaterandblock?

waterisreplacedwith100 of0

ice,whatisthefinal

Hint1.Latentheatoffusion Thelatentheatoffusionistheamountofheatenergyneededtomelt1.0 fusionoficeis334,000 .

ofasubstance.Thelatentheatof

ANSWER:

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lessthan20 20 greaterthan20

Correct

PartD Iftheoriginalexperimentisrepeatedbut100 ofthe0 temperatureofthewaterandblock?

waterisreplacedwithonly25 of0

ice,whatisthefina

Hint1.Findtheequilibriumtemperature Intheoriginalexperiment,thetemperatureof100 of0 waterwasraisedby20 .Theamountofheat requiredtodothiscanbefoundasfollows.Forspecificheat andchangeintemperature , , or . Therefore,

.

Ismoreorlessenergythanthisrequiredtomeltandheat25 oficeto20

?

Hint1.Howtoapproachtheproblem Theamountofenergyrequiredtomeltandheat25 oficeto20 canbebrokenintotwoparts:the energyrequiredtomelttheiceandtheenergyrequiredtoheatthewateraftertheicemelts.

ANSWER: more less

ANSWER: lessthan20 20 greaterthan20

Correct

Problem19.9

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Howmuchenergyisrequiredtobringa1.2 potofwaterat16

to100

?

Expressyouranswerusingtwosignificantfigures. ANSWER: = 4.2×105  

Correct

PartB Forhowlongcouldthisamountofenergyruna100

lightbulb?

Expressyouranswerusingtwosignificantfigures. ANSWER: = 70 



Correct

Problem19.20 A60 icecubeatitsmeltingpointisdroppedintoaninsulatedcontainerofliquidnitrogen.

PartA Howmuchnitrogenevaporatesifitisatitsboilingpointof77 andhasalatentheatofvaporizationof200 Assumeforsimplicitythatthespecificheatoficeisaconstantandisequaltoitsvaluenearitsmeltingpoint.

?

Expressyouranswerusingtwosignificantfigures. ANSWER: = 0.12 



Correct

±AverageVelocityfromaPositionvs.TimeGraph LearningGoal: Tolearntoreadagraphofpositionversustimeandtocalculateaveragevelocity. Inthisproblemyouwilldeterminetheaveragevelocityofamovingobjectfromthegraphofitsposition asafunction oftime .Atravelingobjectmightmoveatdifferentspeedsandindifferentdirectionsduringanintervaloftime,butifwe askatwhat constant velocitytheobjectwouldhavetotraveltoachievethesamedisplacementoverthegiventime interval,thatiswhatwecalltheobject'saveragevelocity.Wewillusethenotation toindicateaveragevelocity overthetimeintervalfrom to .Forinstance, istheaveragevelocityoverthetimeintervalfrom to .

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PartA Consultingthegraphshowninthefigure,findtheobject'saveragevelocityoverthetimeintervalfrom0to1second. Answertothenearestinteger.

Hint1.Definitionofaveragevelocity Averagevelocityisdefinedastheconstantvelocityatwhichanobjectwouldhavetotraveltoachieveagiven displacement(differencebetweenfinalandinitialpositions,whichcanbenegative)overagiventimeinterval, fromtheinitialtime tothefinaltime .Theaveragevelocityisthereforeequaltothedisplacementdividedby thegiventimeinterval.Insymbolicform,averagevelocityisgivenby .

ANSWER: = 0 



Correct

PartB Findtheaveragevelocityoverthetimeintervalfrom1to3seconds. Expressyouranswerinmeterspersecondtothenearestinteger.

Hint1.Findthechangeinposition Thefinalandinitialpositionscanbereadoffthey axisofthegraph.Whatisthedisplacementduringthetime intervalfrom1to3seconds? Expressyouranswernumerically,inmeters ANSWER: = 40 



Hint2.Definitionofaveragevelocity https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4503427

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Averagevelocityisdefinedastheconstantvelocityatwhichanobjectwouldhavetotraveltoachieveagiven displacement(differencebetweenfinalandinitialpositions,whichcanbenegative)overagiventimeinterval, fromtheinitialtime tothefinaltime .Theaveragevelocityisthereforeequaltothedisplacementdividedby thegiventimeinterval.Insymbolicform,averagevelocityisgivenby .

ANSWER: = 20 



Correct Anoteaboutinstantaneousvelocity.Theinstantaneousvelocityatacertainmomentintimeisrepresentedbythe slopeofthegraphatthatmoment.Forstraightlinegraphs,the(instantaneous)velocityremainsconstantoverthe interval,sotheinstantaneousvelocityatanytimeduringanintervalisthesameastheaveragevelocityoverthat interval.Forinstance,inthiscase,theinstantaneousvelocityatanytimefrom1to3secondsisthesameasthe averagevelocityof .

PartC Nowfind

.

Giveyouranswertothreesignificantfigures.

Hint1.Anoteonthedisplacement Sincetheobject'spositionremainsconstantfromtime0totime1,theobject'sdisplacementfrom0to3isthe sameasinPartB.However,thetimeintervalhaschanged.

ANSWER: = 13.3 



Correct Notethat

isnotequaltothesimplearithmeticaverageof

and

,i.e.,

,becausetheyareaveragesfortimeintervalsofdifferentlengths.

PartD Findtheaveragevelocityoverthetimeintervalfrom3to6seconds. Expressyouranswertothreesignificantfigures.

Hint1.Determinethedisplacement Whatisthedisplacement? Answertothenearestinteger. ANSWER:

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OA02a:ThermalPhysicsand1DKinematics

= 40 



Hint2.Determinethetimeinterval Whatisthetimeinterval? Answertotwosignificantfigures. ANSWER: = 3.0  

ANSWER: = 13.3 



Correct

PartE Finally,findtheaveragevelocityoverthewholetimeintervalshowninthegraph. Expressyouranswertothreesignificantfigures.

Hint1.Determinethedisplacement Whatisthedisplacement? Answertothenearestinteger. ANSWER: = 0 



ANSWER: = 0 



Correct Notethatthoughtheaveragevelocityiszeroforthistimeinterval,theinstantaneousvelocity(i.e.,theslopeofthe graph)hasseveraldifferentvalues(positive,negative,zero)duringthistimeinterval. Noteaswellthatsinceaveragevelocityoveratimeintervalisdefinedasthechangeinposition(displacement)in thegivenintervaldividedbythetime,theobjectcantravelagreatdistance(here80meters)andstillhavezero averagevelocity,sinceitendedupexactlywhereitstarted.Therefore,zeroaveragevelocitydoesnotnecessarily meanthattheobjectwasstandingstilltheentiretime!

WhatVelocityvs.TimeGraphsCanTellYou

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Acommongraphicalrepresentationofmotionalongastraightlineisthevvs.tgraph,thatis,thegraphof(instantaneous) velocityasafunctionoftime.Inthisgraph,time isplottedonthehorizontalaxisandvelocity ontheverticalaxis.Note thatbydefinition,velocityandaccelerationarevectorquantities.Instraightlinemotion,however,thesevectorshaveonly singlenonzerocomponentinthedirectionofmotion.Thus,inthisproblem,wewillcall thevelocityand the acceleration,eventhoughtheyarereallythecomponentsofthevelocityandaccelerationvectorsinthedirectionofmotion respectively. Hereisaplotofvelocityversustimeforaparticlethattravelsalongastraightlinewithavaryingvelocity.Refertothisplot toanswerthefollowingquestions.

PartA Whatistheinitialvelocityoftheparticle,

?

Expressyouranswerinmeterspersecond.

Hint1.Initialvelocity Theinitialvelocityisthevelocityat

.

Hint2.Howtoreadavvs.tgraph Recallthatinagraphofvelocityversustime,timeisplottedonthehorizontalaxisandvelocityonthevertical axis.Forexample,intheplotshowninthefigure, at .

ANSWER: = 0.5 



Correct

PartB Whatisthetotaldistance

traveledbytheparticle?

Expressyouranswerinmeters.

Hint1.Howtoapproachtheproblem Recallthattheareaoftheregionthatextendsoveratimeinterval underthevvs.tcurveisalwaysequalto thedistancetraveledin .Thus,tocalculatethetotaldistance,youneedtofindtheareaoftheentireregion underthevvs.tcurve.Inthecaseathand,theentireregionunderthevvs.tcurveisnotanelementary geometricalfigure,butratheracombinationoftrianglesandrectangles.

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OA02a:ThermalPhysicsand1DKinematics

Hint2.Findthedistancetraveledinthefirst20.0seconds Whatisthedistance

traveledinthefirst20secondsofmotion,between

and

?

Expressyouranswerinmeters.

Hint1.Areaoftheregionunderthevvs.tcurve Theregionunderthevvs.tcurvebetween and dimensions by ,andatriangleofbase figure.

canbedividedintoarectangleof andheight ,asshowninthe

ANSWER: = 25 



Hint3.Findthedistancetraveledinthesecond20.0seconds Whatisthedistance

traveledinthesecond20secondsofmotion,from

to

?

Expressyouranswerinmeters.

Hint1.Areaoftheregionunderthevvs.tcurve Theregionunderthevvs.tcurvebetween by ,asshowninthefigure.

and

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isarectangleofdimensions

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OA02a:ThermalPhysicsand1DKinematics

ANSWER: = 40 



Hint4.Findthedistancetraveledinthelast10.0seconds Whatisthedistance

traveledinthelast10secondsofmotion,from

to

?

Expressyouranswerinmeters.

Hint1.Areaoftheregionunderthevvs.tcurve Theregionunderthevvs.tcurvebetween height ,asshowninthefigure.

and

isatriangleofbase

and

ANSWER: = 10 



ANSWER: = 75 



Correct

PartC Whatistheaverageacceleration

oftheparticleoverthefirst20.0seconds?

Expressyouranswerinmeterspersecondpersecond.

Hint1.Definitionandgraphicalinterpretationofaverageacceleration Theaverageacceleration ofaparticlethattravelsalongastraightlineinatimeinterval thechangeinvelocity experiencedbytheparticletothetimeinterval ,or

istheratioof

.

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OA02a:ThermalPhysicsand1DKinematics

Inavvs.tgraph,then,theaverageaccelerationequalstheslopeofthelineconnectingthetwopoints representingtheinitialandfinalvelocities.

Hint2.Slopeofaline Theslope ofalinefrompointA,ofcoordinates "rise"overthe"run,"or

,topointB,ofcoordinates

,isequaltothe

.

ANSWER: = 0.075 



Correct Theaverageaccelerationofaparticlebetweentwoinstantsoftimeistheslopeofthelineconnectingthetwo correspondingpointsinavvs.tgraph.

PartD Whatistheinstantaneousacceleration oftheparticleat

?

Hint1.Graphicalinterpretationofinstantaneousacceleration Theaccelerationofaparticleatanygiveninstantoftimeoratanypointinitspathiscalledtheinstantaneous acceleration.Ifthevvs.tgraphoftheparticle'smotionisknown,youcandirectlydeterminetheinstantaneous accelerationatanypointonthecurve.Theinstantaneousaccelerationatanypointisequaltotheslopeofthe linetangenttothecurveatthatpoint.

Hint2.Slopeofaline Theslope ofalinefrompointA,ofcoordinates "rise"overthe"run,"or

,topointB,ofcoordinates

,isequaltothe

.

ANSWER: 1 0.20 =

0.20 0.022 0.022

Correct Theinstantaneousaccelerationofaparticleatanypointonavvs.tgraphistheslopeofthelinetangenttothe curveatthatpoint.Sinceinthelast10secondsofmotion,between and ,thecurveisa straightline,thetangentlineisthecurveitself.Physically,thismeansthattheinstantaneousaccelerationofthe particleisconstantoverthattimeinterval.Thisistrueforanymotionwherevelocityincreaseslinearlywithtime. Inthecaseathand,canyouthinkofanothertimeintervalinwhichtheaccelerationoftheparticleisconstant?

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Nowthatyouhavereviewedhowtoplotvariablesasafunctionoftime,youcanusethesametechniqueanddrawan accelerationvs.timegraph,thatis,thegraphof(instantaneous)accelerationasafunctionoftime.Asusualinthesetypes ofgraphs,time isplottedonthehorizontalaxis,whiletheverticalaxisisusedtoindicateacceleration .

PartE Whichofthegraphsshownbelowisthecorrectaccelerationvs.timeplotforthemotiondescribedintheprevious parts?

Hint1.Howtoapproachtheproblem Recallthatwhenevervelocityincreaseslinearlywithtime,accelerationisconstant.Intheexamplehere,the particle'svelocityincreaseslinearlywithtimeinthefirst20.0 ofmotion.Inthesecond20.0 ,theparticle's velocityisconstant,andthenitdecreaseslinearlywithtimeinthelast10 .Thismeansthattheparticle's accelerationisconstantovereachtimeinterval,butitsvalueisdifferentineachinterval.

Hint2.Findtheaccelerationinthefirst20 Whatis

,theparticle'saccelerationinthefirst20 ofmotion,between

and

?

Expressyouranswerinmeterspersecondpersecond.

Hint1.Constantacceleration Sincewehavealreadydeterminedthatinthefirst20 ofmotiontheparticle'saccelerationisconstant, itsconstantvaluewillbeequaltotheaverageaccelerationthatyoucalculatedinPartC.

ANSWER: = 0.075 



Hint3.Findtheaccelerationinthesecond20 Whatis

,theparticle'saccelerationinthesecond20 ofmotion,between

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and

? 12/20

29/08/2016

OA02a:ThermalPhysicsand1DKinematics

Expressyouranswerinmeterspersecondpersecond.

Hint1.Constantvelocity Inthesecond20 ofmotion,theparticle'svelocityremainsunchanged.Thismeansthatinthistime interval,theparticledoesnotaccelerate.

ANSWER: = 0 
...