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Complete Solution Manual Fundamentals of Physics Extended 10th Edition Halliday Questions & Answers with rationales $14.99   Add to cart

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Complete Solution Manual Fundamentals of Physics Extended 10th Edition Halliday Questions & Answers with rationales

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  • Fundamentals Of Physics, 10th Ed By Halliday

Fundamentals of Physics Extended 10th Edition Halliday Solutions Manual Complete Solution Manual Fundamentals of Physics Extended 10th Edition Halliday Questions & Answers with rationales PDF File All Pages All Chapters Grade A+ Fundamentals of Physics Extended 10th Edition Halliday Solutions Manua...

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  • November 2, 2024
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  • Fundamentals of physics, 10th ed by Halliday
  • Fundamentals of physics, 10th ed by Halliday
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TEST BANK For Fundamentals of Physics
10th Edition By Resnick, Walker and Halliday
Chapters 1 - 44

,Chapter 1 f




1. VariousfgeometricfformulasfarefgivenfinfAppendixfE.

(a) ExpressingfthefradiusfoffthefEarthfas

Rf f 6.37f f106f m103f kmf mf f 6.37f f103f km,

itsfcircumferencefisf sff2fRf f2f(6.37ff103f km)ff4.00f104f km.


f 4f 6.37ff103ffkm 
2
(b) ThefsurfacefareafoffEarthfisfAff4fR2 f 5.10f f108 km2.


R ffff  6.37ff103ffkm  f1.08ff1012ffkm 3 .
4ff 3 4f 3fff
(c)fThefvolumefoffEarthfisf Vf f
3 3

2. Thef conversionf factorsf are:f1fgry ff1 /10 f line f,f1fline ff1 /12 f inch fandf 1f pointf =f 1/72finc
h.fTheffactorsfimplyfthat

1fgryf=f(1/10)(1/12)(72fpoints)f=f0.60fpoint.

Thus,f 1f gry2f =f (0.60f point)2f =f 0.36f point2,f whichf meansf thatf0.50 f gry 2 f=f 0.18 f point 2 f.

3. Thefmetricfprefixesf(micro,fpico,f nano,f…)farefgivenf forf readyfreferencef onfthefinsideffro
ntfcoverfofftheftextbookf(seefalsofTablef1–2).

(a)fSincef1fkmf=f1ff103fmfandf1fmf=f1ff106fm,

1kmf f 103f mf f 103f m106 fmf mff109 m.

Thef givenf measurementf isf 1.0f kmf (twof significantf figures),f whichf impliesf ourf resultfsh
ouldfbefwrittenfasf1.0ff109fm.

(b)fWefcalculatefthefnumberfoffmicronsfinf1fcentimeter.fSincef1fcmf=f102fm,

1cmf =f 102f m f =f 102m106f fmf mf f 104 m.

Wefconcludefthatftheffractionfoffonefcentimeterfequalftof1.0fmfisf1.0ff104.f(c)fS

incef1fydf=f(3fft)(0.3048fm/ft)f=f0.9144fm,


1

,
, 2 CHAPTERf1



1.0fydf =f 0.91m106f fmf mff 9.1ff105 m.

4. (a)f Usingf thef conversionf factorsf 1f inchf =f 2.54f cmf exactlyf andf 6f picasf =f 1f inch,f wefobta
in  f6f picasff
0.80f cmf =f 0.80f cmf  f 1finch  f  f1.9f picas.
2.54f cmf 1finchf 
   
(b)fWithf12fpointsf=f1fpica,fwefhave


0.80fcmf=f 0.80fcmf f 1finch 
f f6f picas
f f12f pointsff
f 23f points.

ff
2.54f cmf 1finchf 1fpica
   


5. Givenfthatf1ffurlong f 201.168fmf,f1frodff5.0292fm andf1fchainff20.117f mf,fweffind
thefrelevantfconversionffactorsftofbe
1frod
1.0f furlongf f201.168fmff(201.168fmf) f 40f rods,
5.0292 m
and
1fchain
1.0f furlongf f201.168f mf f (201.168f mf) 10f chainsf.
20.117fm
Notef thef cancellationf off mf (meters),f thef unwantedf unit.f Usingf thef givenf conversionffa
ctors,fweffind

(a) thefdistancefdfinfrodsftofbe
40frods
df f 4.0f furlongsf 4.0ffurlongsf f160f rods,
1ffurlong

(b) andfthatfdistancefinfchainsftofbe

10fchainsf
df f 4.0ffurlongsf 4.0ffurlongs f 40f chains.
1ffurlong

6. WefmakefusefoffTablef1-6.

(a) Weflookfatftheffirstf(“cahiz”)fcolumn:f1ffanegafisfequivalentftofwhatfamountfoffcahiz?fW
efnoteffromfthefalreadyfcompletedfpartfofftheftablefthatf1fcahizfequalsfafdozenffanega.fThus,f
2
1ffanegaf=f 1f cahiz, forf8.33ff10 fcahiz.f Similarly, f “1fcahiz f=f48fcuartilla”f(infthe
12
2
alreadyfcompletedfpart)fimpliesfthatf1fcuartillaf= 1fcahiz,for
f
f2.08ff 10 f cahiz.fContinuingf i
48
nf thisf way,f thef remainingf entriesf inf thef firstf columnf aref 6.94f f 103fand
3.47103f.

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