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

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

,Chapter 1 b




1. VariousbgeometricbformulasbarebgivenbinbAppendixbE.

(a) ExpressingbthebradiusbofbthebEarthbas

Rb b 6.37b b106b m103b kmb mb b 6.37b b103b km,

itsbcircumferencebisb sbb2bRb b2b(6.37bb103b km)bb4.00b104b km.

(b) ThebsurfacebareabofbEarthbisbAbb4bR2 b 4b 6.37bb103bbkm  b 5.10b b108 km2.
2




R bbbb  6.37bb103bbkm  b1.08bb1012bbkm 3 .
4bb 3 4b 3bbb
(c)bThebvolumebofbEarthbisb Vb b
3 3

2. Theb conversionb factorsb are:b1bgrybb1/10 b line b,b 1bline bb1/12 b inch bandb 1b pointb =b 1/72b
inch.bThebfactorsbimplybthat

1bgryb=b(1/10)(1/12)(72bpoints)b=b0.60bpoint.

Thus,b 1b gry2b =b (0.60b point)2b =b 0.36b point2,b whichb meansb thatb0.50 b gry 2 b=b 0.18 b point 2 b.

3. Thebmetricbprefixesb(micro,bpico,b nano,b …)bareb givenb forb readyb referenceb onb theb insideb
frontbcoverbofbthebtextbookb(seebalsobTableb1–2).

(a)bSinceb1bkmb=b1bb103bmbandb1bmb=b1bb106bm,

1kmb b 103b mb b 103b m106 bmb mbb109 m.

Theb givenb measurementb isb 1.0b kmb (twob significantb figures),b whichb impliesb ourb resultb
shouldbbebwrittenbasb1.0bb109bm.

(b)bWebcalculatebthebnumberbofbmicronsbinb1bcentimeter.bSinceb1bcmb=b102bm,

1cmb =b 102b m b =b 102m106b bmb mb b 104 m.

Webconcludebthatbthebfractionbofbonebcentimeterbequalbtob1.0bmbisb1.0bb104.b(

c)bSinceb1bydb=b(3bft)(0.3048bm/ft)b=b0.9144bm,


1

,
, 2 CHAPTERb1



1.0bydb =b 0.91m106b bmb mbb 9.1bb105 m.

4. (a)b Usingb theb conversionb factorsb 1b inchb =b 2.54b cmb exactlyb andb 6b picasb =b 1b inch,b web
obtain  b6b picasbb
0.80b cmb =b 0.80b cmb b 1binch  b  b1.9b picas.
2.54b cmb 1binchb 
   
(b)bWithb12bpointsb=b1bpica,bwebhave


0.80bcmb=b 0.80bcmb b 1binch   bbb12b pointsbb
bb6b picasb
2.54b cmb 1binchb 1bpica  b 23b points.
   


5. Givenbthatb1bfurlong b 201.168bmb,b1brodbb5.0292bm andb1bchainbb20.117b mb,b webfind
thebrelevantbconversionbfactorsbtobbe
1brod
1.0b furlongb b201.168bmbb(201.168bmb) b 40b rods,
5.0292 m
and
1bchain
1.0b furlongb b201.168b mb b (201.168b mb) 10b chainsb.
20.117bm
Noteb theb cancellationb ofb mb (meters),b theb unwantedb unit.b Usingb theb givenb conversionb
factors,bwebfind

(a) thebdistancebdbinbrodsbtobbe
40brods
db b 4.0bfurlongsb 4.0bfurlongsb b160b rods,
1bfurlong

(b) andbthatbdistancebinbchainsbtobbe

10bchainsb
db b 4.0b furlongsb 4.0bfurlongs b 40b chains.
1bfurlong

6. WebmakebusebofbTableb1-6.

(a) Weblookbatbthebfirstb(“cahiz”)bcolumn:b1bfanegabisbequivalentbtobwhatbamountbofbcahiz?b
Webnotebfrombthebalreadybcompletedbpartbofbthebtablebthatb1bcahizbequalsbabdozenbfanega.b
2
b cahiz,borb8.33bb10 bcahiz.b Similarly, b“1bcahizb =b48bcuartilla”b (inbthe
1
Thus,b1bfanegab=b 12
alreadybcompletedbpart)bimpliesbthatb1bcuartillab= 1bcahiz,48borb2.08bb 102b cahiz.bContinuin
b




gb inb thisb way,b theb remainingb entriesb inb theb firstb columnb areb 6.94b b 103b and
3.47103b.