Fundamentals of Physics Extended 10th Edition Solution Manual | Halliday, Resnick & Walker – 2025 Step-by-Step Physics Solutions

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Soluton_ManualFundamentals_of_Physics_Extended_10th_E dit g


ion


1. THINKgIngthisgproblemgwe’reggivengthegradiusgofgEarth,gandgaskedgtogcomputegi
tsgcircumference,gsurfacegareagandgvolume.


EXPRESSgAssuminggEarthgtogbegagspheregofgradius



REg =g(6.37gg106gm)(10−3gkm m)g=g6.37gg103g km,



thegcorrespondinggcircumference,gsurfacegareagandgvolumegare:
4g
Cg=g2gRg , Ag=g4gR2g,
g
Vg =g R3g.
E E E
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TheggeometricgformulasgareggivengingAppendixgE.




ANALYZE (a) Using the formulas given above, we find the circu
g g g g g g g g g g



mference to be g g




Cg=g2gRE =g2g(6.37gg103g km)g=g4.00104g km.


(b) Similarly,gthegsurfacegareagofgEarthgis

Ag=g4gRE =g4g(6.37gg103g km) =g5.10gg108g km2g,
2g

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(c) andgitsgvolumegi
s

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4g (6.37  103 km)
3g
4 =g1.08gg1012g km3.
 g =g R3g =g
g g g g g
V
E
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LEARNgFromgthegformulasggiven,gwegseegthatg C ,g A R2g,gandg V R3g.gThegratiosgofgvolume
RE E E

togsurfacegarea,gandgsurfacegareagtogcircumferencegareg Vg/gAg=gREg /g3g andg Ag/gCg=g2REg .



2. Thegconversiongfactorsgare:g1ggryg=g1/10 gline g,g1gline g=g1/12 g inch gandg1gpointg=g1/72ginc
h.gThegfactorsgimplygthat



1ggryg=g(1/10)(1/12)(72gpoints)g=g0.60gpoint.



Thus,g1ggry2g=g(0.60gpoint)2g=g0.36gpoint2,gwhichgmeansgthatg 0.50g gry2g=g 0.18g point 2 g.



3. Thegmetricgprefixesg(micro,gpico,gnano,g…)gareggivengforgreadygreferencegongtheginsidegfro
ntgcovergofgthegtextbookg(seegalsogTableg1–2).



(a) Sinceg1gkmg=g1gg103gmgandg1gmg=g1gg106gm,



1kmg=g103gmg=g(103gm)(10 gm m)g=g109 m.
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Theggivengmeasurementgisg1.0gkmg(twogsignificantgfigures),gwhichgimpliesgourgresultgshouldgbeg
writtengasg1.0gg109gm.



(b) Wegcalculategthegnumbergofgmicronsging1gcentimeter.gSinceg1gcmg=g10−2gm,


1cmg=g10−2gmg=g(10−2m)(106ggm m)g=g104 m.


Wegconcludegthatgthegfractiongofgonegcentimetergequalgtog1.0gmgisg1.0gg10−4.

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(c) Sinceg1gydg=g(3gft)(0.3048gm/ft)g=g0.9144gm,


1.0gydg=g(0.91m)(106ggm m)g=g9.1gg105g m.

4. (a)gUsinggthegconversiongfactorsg1ginchg=gg2.54 gcmgexactlygandg6gpicasg=g1ginch,g wegobtain
1ginch g  g6g picasgg
0.80gcmg=g (0.80g cm)gg  g  g1.9g picas.
2.54g cmg 1ginchg
  
(b)gWithg12gpointsg=g1gpica,gweghave

g 1ginchg  g6g picasg g12g pointsgg
0.80gcmg=g (0.80gcm)gg2.54gcmg g 1ginchg  g g 1gpica  g 23gpoints.
   


5. THINKgThisgproblemgdealsgwithgconversiongofgfurlongsgtogrodsgandgchains,gallgofgwhichgareg
unitsgforgdistance.


EXPRESSg Giveng thatg 1gfurlong =g 201.168gm,g1grodg=g5.0292g andg 1gchaing=g20.117gmg,g the
grelevantgconversiongfactorsgare m
1grod
1.0gfurlongg =g201.168gmg=g(201.168gmg =g40g rods,
) 5.0292 m

and
1gchain =10gchainsg.
1.0gfurlongg =g201.168gmg=g(201.168gmg) m
20.117

Notegthegcancellationgofgmg(meters),gthegunwantedgunit.



ANALYZEgUsinggthegabovegconversiongfactors,gwegfind


40grod
(a) thegdistancegdgingrodsgtogbeg dg =g 4.0gfurlongsg =(4.0gfurlongs)g =g160g rods,
s
1gfurlong
10gchainsg
(b) andgingchainsgtogb dg =g 4.0gfurlongsg =(4.0gfurlongs) =g 40g chains.
e 1gfurlong

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LEARNgSinceg4gfurlongsgisgaboutg800gm,gthisgdistancegisgapproximatelygequalgtog160grodsg(
1grodg g 5gmg)gandg40gchainsg(1gchaing g 20gmg).gSogourgresultsgmakegsense.



6. WegmakegusegofgTableg1-6.



(a) Weglookgatgthegfirstg(“cahiz”)gcolumn:g1gfanegagisgequivalentgtogwhatgamountgofgcahiz?gW
egnotegfromgthegalreadygcompletedgpartgofgthegtablegthatg1gcahizgequalsgagdozengfanega.g Thu
s,g1
fanegag=g121gg cahiz,gorg8.33gg10−2gcahiz.g Similarly,g“1gcahizg=g48gcuartilla”g(ingthegalready
g




completedgpart)gimpliesgthatg1gcuartillag=g481ggcahiz,gorg2.08gg10−2gcahiz.g Continuinggingthisgway
g




,gthegremaininggentriesgingthegfirstgcolumngareg6.94gg10−3gandg 3.4710−3g.



(b) Ingthegsecondg(“fanega”)gcolumn,gwegfindg0.250,g8.33gg10−2,gandg4.17gg10−2gforgthegla
stgthreegentries.



(c) Ingthegthirdg(“cuartilla”)gcolumn,gwegobtaing0.333gandg0.167gforgtheglastgtwogentries.



(d) Finally,gingthegfourthg(“almude”)gcolumn,gweggetg21g =g0.500gforgtheglastgentry.



(e) Sincegthegconversiongtablegindicatesgthatg1galmudegisgequivalentgtog2gmedios,gourgamountgof
7.00galmudesgmustgbegequalgtog14.0gmedios.



(f) Usinggthegvalueg(1galmudeg=g6.94gg10−3gcahiz)gfoundgingpartg(a),gwegconcludegthatg7.
00galmudesgisgequivalentgtog4.86gg10−2gcahiz.


(g) Sincegeachgdecimetergisg0.1gmeter,gtheng55.501gcubicgdecimetersgisgequalgtog0.055501gm3g
org55501gcm3.g Thus,g7.00galmudesg=g7.00g fanegag=g7.00g(55501gcm3)g=g3.24gg104gcm3.
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