Complete Solution Manual for Fundamentals of Physics Extended (10th Edition) – Halliday, Resnick, and Walker (Chapters 1–10)

1




Soluton_ManualFundamentals_of_Physics_Extended_10th_EVdi
tion


1. THINKVInVthisVproblemVwe’reVgivenVtheVradiusVofVEarth,VandVaskedVtoVcompu
teVitsVcircumference,VsurfaceVareaVandVvolume.


EXPRESSVAssumingVEarthVtoVbeVaVsphereVofVradius



RV =EV(6.37VV106Vm)(10−3Vkm m)V=V6.37VV103V km,



theVcorrespondingVcircumference,VsurfaceVareaVandVvolumeVare:
V4V
CV=V2VRV , AV=V4VR2V, VV =V R3V.
E E E
3
TheVgeometricVformulasVareVgivenVinVAppendixVE.




ANALYZEV(a)VUsingVtheVformulasVgivenVabove,VweVfindVtheVcirc
umferenceVtoVbe

CV=V2VRE =V2V(6.37VV103V km)V=V4.00104V km.


(b) Similarly,VtheVsurfaceVareaVofVEarthVis

AV=V4VE =V4V(6.37VV103V km) =V5.10VV108V km2V,
2V


(c) andVitsVvolumeVi R2
s

,2


(6.37VV103Vkm)
3V
V4V =V1.08VV1012V km3.
VVV4=V R3V =
V E
3 3


LEARNVFromVtheVformulasVgiven,VweVseeVthatV C ,V A R2V,VandV V R3V.VTheVratiosVofVvolume
RE E E

toVsurfaceVarea,VandVsurfaceVareaVtoVcircumferenceVareV VV/VAV=VREV /V3V andV AV/VCV=V2REV .



2. TheVconversionVfactorsVare:V1VgryV=V1/10 V line V,V1Vline V=V1/12 V inch VandV1VpointV=V1/
72Vinch.VTheVfactorsVimplyVthat



1VgryV=V(1/10)(1/12)(72Vpoints)V=V0.60Vpoint.



Thus,V1V gry2V=V(0.60Vpoint)2V =V 0.36Vpoint2,VwhichVmeansVthatV 0.50V gry2V=V 0.18V point 2 V.



3. TheVmetricVprefixesV(micro,Vpico,Vnano,V…)VareVgivenVforVreadyVreferenceVonVtheVinside
VfrontVcoverVofVtheVtextbookV(seeValsoVTableV1–2).




(a) SinceV1VkmV=V1VV103VmVandV1VmV=V1VV106Vm,



1kmV=V103VmV=V(103Vm) Vm m.
m)V=V109
(106

TheVgivenVmeasurementVisV1.0VkmV(twoVsignificantVfigures),VwhichVimpliesVourVresultVshould
VbeVwrittenVasV1.0VV109Vm.




(b) WeVcalculateVtheVnumberVofVmicronsVinV1Vcentimeter.VSinceV1VcmV=V10−2Vm,


1cmV=V10−2VmV=V(10−2m)(106VVm m.
m)V=V104


WeVconcludeVthatVtheVfractionVofVoneVcentimeterVequalVtoV1.0VmVisV1.0VV10−4.

, 3




(c) SinceV1VydV=V(3Vft)(0.3048Vm/ft)V=V0.9144Vm,


1.0VydV=V(0.91m)(106VVm m)V=V9.1VV105V m.

4. (a)VUsingVtheVconversionVfactorsV1VinchV=V2.54VcmVexactlyVandV6VpicasV=V1Vinch,V weVobtai
n V 1VinchV  V6V picasVV
0.80VcmV=V (0.80Vcm V)V  V  V1.9V picas.
2.54V cmV 1VinchV
  
(b)VWithV12VpointsV=V1Vpica,VweVhave

V 1VinchV  V6V picasV V12V pointsVV
0.80VcmV=V (0.80Vcm)VV2.54VcmV V 1VinchV  V V 1Vpica V 23Vpoints.
   


5. THINKVThisVproblemVdealsVwithVconversionVofVfurlongsVtoVrodsVandVchains,VallVofVwhi
chVareV unitsVforVdistance.


EXPRESSV GivenV thatV 1Vfurlo =V 201.168Vm,V1VrodV=V5.029 andV 1VchainV=V20.117V mV,V the
ngVrelevantVconversionVfactorsV 2Vm
are
1Vrod
1.0VfurlongV =V201.168VmV=V(201.168 =V40V rods,
VmV)
5.0292 m

and
1Vchai =10VchainsV.
1.0VfurlongV =V201.168VmV=V(201.168V nm
mV)
20.117

NoteVtheVcancellationVofVmV(meters),VtheVunwantedVunit.



ANALYZEVUsingVtheVaboveVconversionVfactors,VweVfind


40
(a) theVdistanceVdVinVrodsVtoVbeV dV =V 4.0VfurlongsV =(4.0Vfurlongs)V =V160V rods,
Vrods

1Vfurlong
10VchainsV
(b) andVinVchainsVtoV dV =V 4.0V furlongsV =(4.0Vfurlongs) =V 40V chains.
be 1Vfurlong

, 4


LEARNVSinceV4VfurlongsVisVaboutV800Vm,VthisVdistanceVisVapproximatelyVequalVtoV160VrodsV(
1VrodV V 5VmV)VandV40VchainsV(1VchainV V 20VmV).VSoVourVresultsVmakeVsense.



6. WeVmakeVuseVofVTableV1-6.



(a) WeVlookVatVtheVfirstV(“cahiz”)Vcolumn:V1VfanegaVisVequivalentVtoVwhatVamountVofVcahiz
?VWeVnoteVfromVtheValreadyVcompletedVpartVofVtheVtableVthatV1VcahizVequalsVaVdozenVfan
ega.V Thus,V1
fanegaV=V12V1VV cahiz,VorV8.33VV10−2Vcahiz.V Similarly,V“1VcahizV=V48Vcuartilla”V(inVtheValready

=VV1VVcahiz,VorV2.08VV10−2Vcahiz.V ContinuingVinVthis
completedVpart)VimpliesVthatV1VcuartillaV48
Vway,VtheVremainingVentriesVinVtheVfirstVcolumnVareV6.94VV10−3VandV 3.4710− V.
3




(b) InVtheVsecondV(“fanega”)Vcolumn,VweVfindV0.250,V8.33VV10−2,VandV4.17VV10−2VforVt
heVlastVthreeVentries.



(c) InVtheVthirdV(“cuartilla”)Vcolumn,VweVobtainV0.333VandV0.167VforVtheVlastVtwoVentries.



(d) Finally,VinVtheVfourthV(“almude”)Vcolumn,VweVgetV 1V =V0.500VforVtheVlastVentry.
2




(e) SinceVtheVconversionVtableVindicatesVthatV1ValmudeVisVequivalentVtoV2Vmedios,VourVamountVof
7.00ValmudesVmustVbeVequalVtoV14.0Vmedios.



(f) UsingVtheVvalueV(1ValmudeV=V6.94VV10−3Vcahiz)VfoundVinVpartV(a),VweVconcludeVtha
tV7.00ValmudesVisVequivalentVtoV4.86VV10−2Vcahiz.


(g) SinceVeachVdecimeterVisV0.1Vmeter,VthenV55.501VcubicVdecimetersVisVequalVtoV0.055501V
m3VorV55501Vcm3.V Thus,V7.00ValmudesV=V7.00V fanegaV=V7.00V(55501Vcm3)V=V3.24VV104Vcm3.
12 12