TEST BANK For Fundamentals of Physics 10th Edition By Resnick, Walker and Halliday Chapters 1 - 44 

TEST BANK For Fundamentals of Physics
10th Edition By Resnick, Walker and Halliday
Chapters 1 - 44

,Chapter 1 nj




1. Various njgeometric njformulas njare njgiven njin njAppendix njE.

(a) Expressing njthe njradius njof njthe njEarth njas

R n j  n j 6.37 n j  nj106 n j m103 nj km n j m nj  n j 6.37 n j  nj103 n j km,

its njcircumference njis n j s nj nj2njR n j nj2nj(6.37 nj nj103 n j km) nj nj4.00nj104 n j km.

(b) The njsurface njarea njof njEarth njis njA nj  n j 4 nj 6.37 nj nj103 nj  n j 5.10 nj km2.
2 8
nj10
km 
nj4 njR 2
nj



4 nj nj 3 4 n j
 6.37 nj nj103 njnjkm   nj1.08 nj nj1012 njnjkm3.
3 nj nj nj
(c) njThe njvolume njof njEarth njis nj V n j  n j R nj nj nj nj
3 3

2. The n j conversion n j factors n j are: nj1 njgry njn1j /10 n j line nj, nj 1 njline njnj1 /12 n j inch njand n j 1
n j point n j = n j 1/72njinch. njThe njfactors njimply njthat



1 njgry nj= nj(1/10)(1/12)(72 njpoints) nj= nj0.60 njpoint.

Thus, n j 1 n j gry2 n j = n j (0.60 n j point)2 n j = n j 0.36 n j point2, n j which n j means n j that nj0.50 n j gry 2 nj= n j 0.18
2
n j point nj .



3. The njmetric nj prefixes nj(micro, njpico, njnano, nj …) nj are njgiven nj for nj ready njreference njon
nj the nj insidenjfront njcover njof njthe njtextbook nj(see njalso njTable nj1–2).



(a) njSince nj1 njkm nj= nj1 nj nj103 njm njand nj1 njm nj= nj1 nj nj106 njm,

1km n j  nj 103 nj m n j  n j 103  njm n j m nj m.
m106
9
nj10
nj




The n j given n j measurement n j is n j 1.0 n j km n j (two n j significant nj figures), nj which
n j implies n j our n j result njshould njbe njwritten njas nj1.0 nj nj10 njm.
9



(b) njWe njcalculate njthe njnumber njof njmicrons njin nj1 njcentimeter. njSince nj1 njcm nj= nj102 njm,

1cm nj = n j 102 n j m n j = n j 102m106 n j  njm m.
n j m nj  nj 104

We njconclude njthat njthe njfraction njof njone njcentimeter njequal njto nj1.0 njm njis nj1.0

 nj104.nj(c) njSince nj1 njyd nj= nj(3 njft)(0.3048 njm/ft) nj= nj0.9144 njm,
nj

,1

, 2 CHAPTER 1



1.0 njyd n j = n j 0.91m106 nj  njm n j m nj m.
n j 9.1 nj nj10
5



4. (a) n j Using n j the n j conversion n j factors n j 1 n j inch n j = n j 2.54 n j cm n j exactly n j and n j 6 n j picas
n j = n j 1 n j inch, n j weno j btain
 nj6 n j picas nj n j
 n j = n j 0.80
0.80 n j cm 1  nj1.9 n j picas.
n j cm nj   nj njinch  n j 
2.54 n j cm nj 1 njinch n j 
   
(b) njWith nj12 njpoints nj= nj1 njpica, njwe
njhave


 nj= n j 0.80
0.80 njcm 1  nj6 n j picas nj nj12 n j points nj n j
njcm nj   nj njinch  n j  nj n j   n j 23 n j points.
2.54 n j cm nj 1 njinch n j 1 njpica 
   


5. Given njthat nj1  n j 201.168 njm nj, nj 1njrod nj and nj1ncj hain nj nj20.117 n j mnj, njwe njfind
njfurlong nj5.0292 nj m

the njrelevant njconversion njfactors njto njbe
1 njrod
1.0 nj furlong n j  nj201.168 njm nj nj(201.168  nj 40 n j rods,
njm nj)
5.0292 m

and
1 njchain
1.0 n j furlong n j  nj201.168 n j m n j  10 n j chains nj.
n j (201.168 n j m nj)
20.117
njm

Note n j the n j cancellation n j of n j m n j (meters), n j the n j unwanted n j unit. n j Using n j the
n j given n j conversionnjfactors, njwe njfind



(a) the njdistance njd njin njrods njto njbe
d n j  n j 4.0 nj furlongs n j 4.0 njfurlongs  nj160 n j rods,
40 njrods
nj

1 njfurlong

(b) and njthat njdistance njin njchains njto njbe

10 njchains n j
d n j  n j 4.0 nj furlongs n j 4.0 njfurlongs  n j 40 nj chains.
1 njfurlong

6. We njmake njuse njof njTable nj1-6.

(a) We njlook njat njthe njfirst nj(“cahiz”) njcolumn: nj1 njfanega njis njequivalent njto njwhat njamount
njof njcahiz? njWe njnote njfrom njthe njalready njcompleted njpart njof njthe njtable njthat nj1 njcahiz

n j cahiz, njor nj8.33 nj nj10
njequals nja njdozen njfanega. njThus, nj1 njfanega nj= n j
1 2 njcahiz.
12
n j Similarly, nj“1 njcahiz nj= nj48 njcuartilla” nj(in njthe


already njcompleted njpart) njimplies njthat nj1 njcuartilla nj48= nj
1
cahiz, njor nj2.08 nj
nj n j 102
n j cahiz. njContinuing nj in nj this nj way, nj the nj remaining nj entries nj in nj the nj first