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Complete Solution Manual Fundamentals of Physics Extended 10th Edition Halliday Questions & Answers with rationales $17.39   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 Manu...

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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 R




1. VariousRgeometricRformulasRareRgivenRinRAppendixRE.

(a) ExpressingRtheRradiusRofRtheREarthRas

RR R 6.37R R106R m103R kmR mR R 6.37R R103R km,

itsRcircumferenceRisR sRR2RRR R2R(6.37RR103R km)RR4.00R104R km.

(b) TheRsurfaceRareaRofREarthRisRARR4R R 4R  6.37RR103RRkm R 5.10R R10 km2.
R2 8


2




R RRRR  6.37RR103RRkm  R1.08RR1012RRkm 3 .
4RR 3 4R 3RRR
(c)RTheRvolumeRofREarthRisR VR R
3 3

2. TheR conversionR factorsR are:R1RgryRR1/10 R line R,R 1Rline RR1/12 R inch RandR 1R pointR =R
1/72Rinch.RTheRfactorsRimplyRthat

1RgryR=R(1/10)(1/12)(72Rpoints)R=R0.60Rpoint.

Thus,R 1R gry2R =R (0.60R point)2R =R 0.36R point2,R whichR meansR thatR0.50 R gry 2 R=R 0.18 R point 2 R.

3. TheRmetricR prefixesR (micro,Rpico,R nano,R …)R areR givenR forR readyR referenceR onR theR ins
ideRfrontRcoverRofRtheRtextbookR(seeRalsoRTableR1–2).

(a)RSinceR1RkmR=R1RR103RmRandR1RmR=R1RR106Rm,

1kmR R 103R mR R 103R m1 RmR mRR10 m.
06 9



TheR givenR measurementR isR 1.0R kmR (twoR significantR figures),R whichR impliesR ourR res
ultRshouldRbeRwrittenRasR1.0RR109Rm.

,(b)RWeRcalculateRtheRnumberRofRmicronsRinR1Rcentimeter.RSinceR1RcmR=R102Rm,

1cmR =R 102R m R =R 102m106R RmR mR R 1 m.
04

WeRconcludeRthatRtheRfractionRofRoneRcentimeterRequalRtoR1.0RmRisR1.0RR10
4
.R(c)RSinceR1RydR=R(3Rft)(0.3048Rm/ft)R=R0.9144Rm,


1

, 2 CHAPTERR1



1.0RydR =R 0.91m106R RmR mRR 9.1RR1 m.
05

4. (a)R UsingR theR conversionR factorsR 1R inchR =R 2.54R cmR exactlyR andR 6R picasR =R 1R inch,R
weRobtain R6R picasRR
0.80R cmR =R 0.80R cmR 1Rinch R1.9R picas.
 R  R 
2.54R cmR 1RinchR 
   
(b)RWithR12RpointsR=R1Rpica,RweRha
ve


 RcmR=R 0.80RcmR 1Rinch  R6R picasR R12R pointsRR R 23R points.
0.80
 R  R  RR 
2.54R cmR 1RinchR 1Rpica 
   


5. GivenRthatR1Rfurlon R 201.168RmR,R 1RrodRR5.0292R andR1RchainRR20.117R mR,RweRfind
g m
theRrelevantRconversionRfactorsRtoRbe
1Rrod
1.0R furlongR R201.168RmRR(201.168R R 40R rods,
mR) 5.0292 m

and
1Rchain
1.0R furlongR R201.168R mR R (201.168R m 10R chainsR.
R)
20.117R
m
NoteR theR cancellationR ofR mR (meters),R theR unwantedR unit.R UsingR theR givenR conversi
onRfactors,RweRfind

(a) theRdistanceRdRinRrodsRtoRbe
40Rrod
dR R 4.0R furlongsR 4.0RfurlongsR R160R rods,
s
1Rfurlong

(b) andRthatRdistanceRinRchainsRtoRbe

10R chainsR
dR R 4.0R furlongsR 4.0Rfurlongs R 40R chains.
1Rfurlong

6. WeRmakeRuseRofRTableR1-6.

(a) WeRlookRatRtheRfirstR(―cahiz‖)Rcolumn:R1RfanegaRisRequivalentRtoRwhatRamountRofRcah
iz?RWeRnoteRfromRtheRalreadyRcompletedRpartRofRtheRtableRthatR1RcahizRequalsRaRdozenRfa
2
R =R R cahiz,R orR 8.33R R 10 Rcahiz.R Similarly,R ―1R cahizR =R 48R cuartill
1
nega.RThus,R1Rfanega
12
a‖R(inRthe
2
alreadyRcompletedRpart)RimpliesRthatR1RcuartillaR= 1Rcahiz,
R
RorR2.08RR 10 R cahiz.RContin
48
uingR inR thisR way,R theR remainingR entriesR inR theR firstR columnR areR 6.94R R 103Rand

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