Fundamentals of Physics Extended 10th Edition Solution Manual - Halliday & Resnick

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Soluton_ManualFundamentals_of_Physics_Extended_10th_Edit j


ion


1. THINKjInjthisjproblemjwe’rejgivenjthejradiusjofjEarth,jandjaskedjtojcomputejitsjcir
cumference,jsurfacejareajandjvolume.


EXPRESSjAssumingjEarthjtojbejajspherejofjradius



R =j(6.37jj106jm)(10−3jkm m)j=j6.37jj103j km,
E j




thejcorrespondingjcircumference,jsurfacejareajandjvolumejare:
j4j
Cj=j2jRj , Aj=j4jR2j, Vj =j R3j.
E E
3 E
ThejgeometricjformulasjarejgivenjinjAppendixjE.




ANALYZE (a) Using the formulas given above, we find the circum
j j j j j j j j j j



ference to be j j




Cj=j2jRE =j2j(6.37jj103j km)j=j4.00104j km.


(b) Similarly,jthejsurfacejareajofjEarthjis

Aj=j4jR2E =j4j(6.37jj103j km) =j5.10jj108j km2j,
2j


(c) andjitsjvolumejis

,2


4j j j4 (6.37  103 km) = 1.08  1012 km3.
3j

Vj =j R3jE =j j j j j j j j

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LEARNjFromjthejformulasjgiven,jwejseejthatj C ,j A R2j,jandj V R3j.jThejratiosjofjvolume
RE E E

tojsurfacejarea,jandjsurfacejareajtojcircumferencejarej Vj/jAj=jREj /j3j andj Aj/jCj=j2REj .



2. Thejconversionjfactorsjare:j1jgryj=j1 /10 j line j,j1jline j=j1 /12 j inch jandj1jpointj=j1/72jinch.jThejf
actorsjimplyjthat



1jgryj=j(1/10)(1/12)(72jpoints)j=j0.60jpoint.



Thus,j1j gry2j =j(0.60jpoint)2j =j 0.36jpoint2,j whichjmeansj thatj 0.50j gry2j=j 0.18j point 2 j.



3. Thejmetricjprefixesj(micro,jpico,jnano,j…)jarejgivenjforjreadyjreferencejonjthejinsidejfrontjco
verjofjthejtextbookj(seejalsojTablej1–2).



(a) Sincej1jkmj=j1jj103jmjandj1jmj=j1jj106jm,



1kmj=j103jmj=j(103jm)(106 jm m)j=j109 m.


Thejgivenjmeasurementjisj1.0jkmj(twojsignificantjfigures),jwhichjimpliesjourjresultjshouldjbejwritt
enjasj1.0jj109jm.



(b) Wejcalculatejthejnumberjofjmicronsjinj1jcentimeter.jSincej1jcmj=j10−2jm,


1cmj=j10−2jmj=j(10−2m)(106jjm m)j=j104 m.


Wejconcludejthatjthejfractionjofjonejcentimeterjequaljtoj1.0jmjisj1.0jj10−4.

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(c) Sincej1jydj=j(3jft)(0.3048jm/ft)j=j0.9144jm,


1.0jydj=j(0.91m)(106jjm m)j=j9.1jj105j m.

4. (a)jUsingjthejconversionjfactorsj1jinchj=j2.54 jcmjexactlyjandj6jpicasj=j1jinch,j wejobtain
j 1jinchj  j6j picasjj
0.80jcmj=j (0.80jcm)j j  j  j1.9j picas.
2.54j cmj 1jinchj
  
(b)jWithj12jpointsj=j1jpica,jwejhave

j 1jinchj  j6j picasj j12j pointsjj 
0.80jcmj=j (0.80jcm)j j2.54jcmj j 1jinchj  j j 1jpica j 23 j points.

   


5. THINKjThisjproblemjdealsjwithjconversionjofjfurlongsjtojrodsjandjchains,jalljofjwhichjarej units
jforjdistance.




EXPRESSj Givenj thatj 1jfurlongjr =j 201.168jm,j1jrodj=j5.0292jm andj 1jchainj=j20.117jmj,j the
elevantjconversionjfactorsjare
1jrod
1.0jfurlongj =j201.168jmj=j(201.168jmj) =j40j rods,
5.0292 m

and
1.0jfurlongj =j201.168jmj=j(201.168jmj) 1jchain =10jchainsj.
20.117 m

Notejthejcancellationjofjmj(meters),jthejunwantedjunit.



ANALYZEjUsingjthejabovejconversionjfactors,jwejfind


40jrods
(a) thejdistancejdjinjrodsjtojbej dj =j 4.0jfurlongsj =(4.0jfurlongs)j =j160j rods,
1jfurlong
10jchainsj
(b) andjinjchainsjtojbe dj =j 4.0j furlongsj =(4.0jfurlongs) =j 40j chains.
1jfurlong

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LEARNjSincej4jfurlongsjisjaboutj800jm,jthisjdistancejisjapproximatelyjequaljtoj160jrodsj(
1jrodj j 5jmj)jandj40jchainsj(1jchainj j 20jmj).jSojourjresultsjmakejsense.



6. WejmakejusejofjTablej1-6.



(a) Wejlookjatjthejfirstj(“cahiz”)jcolumn:j1jfanegajisjequivalentjtojwhatjamountjofjcahiz?jWejnot
ejfromjthejalreadyjcompletedjpartjofjthejtablejthatj1jcahizjequalsjajdozenjfanega.j Thus,j1
fanegaj=j 112jj cahiz,jorj8.33jj10−2jcahiz.j Similarly,j“1jcahizj=j48jcuartilla”j(injthejalready
j




jorj2.08jj10 jcahiz.j Continuingjinjthisjway,jthejr
completedjpart)jimpliesjthatj1jcuartillaj=j 1jjcahiz,
j −2
48
emainingjentriesjinjthejfirstjcolumnjarej6.94jj10−3jandj 3.4710−3j.



(b) Injthejsecondj(“fanega”)jcolumn,jwejfindj0.250,j8.33jj10−2,jandj4.17jj10−2jforjthejlastjthr
eejentries.



(c) Injthejthirdj(“cuartilla”)jcolumn,jwejobtainj0.333jandj0.167jforjthejlastjtwojentries.



(d) Finally,jinjthejfourthj(“almude”)jcolumn,jwejgetj 1j =2j0.500jforjthejlastjentry.



(e) Sincejthejconversionjtablejindicatesjthatj1jalmudejisjequivalentjtoj2jmedios,jourjamountjof
7.00jalmudesjmustjbejequaljtoj14.0jmedios.



(f) Usingjthejvaluej(1jalmudej=j6.94jj10−3jcahiz)jfoundjinjpartj(a),jwejconcludejthatj7.00jalm
udesjisjequivalentjtoj4.86jj10−2jcahiz.


(g) Sincejeachjdecimeterjisj0.1jmeter,jthenj55.501jcubicjdecimetersjisjequaljtoj0.055501jm3jorj55
501jcm3.j Thus,j7.00jalmudesj=j7.00j fanegaj=j7.00j(55501jcm3)j=j3.24jj104jcm3.
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