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

,Chapter651

1. Various65geometric65formulas65are65given65in65Appendix65E.

(a) Expressing65the65radius65of65the65Earth65as

R6 5 6 5 6.376 5 651066 5 m10365 km65 m65 6 5 6.376 5 651036 5 km,

its65circumference65is6 5 s6565265R6 5 65265(6.3765651036 5 km)65654.00651046 5 km.

(b) The65surface65area65of65Earth65is65A65 6 5 465  6.37656510365 6 5 5.1065 65 km2.
65465R2 108
65km 
2




46565 3 46 5
 6.3765651036565km  651.08656510126
3656565
(c)65The65volume65of65Earth65is65 V6 5 65 R 65656565
3
565km .
3 3

2. The6 5 conversion6 5 factors6 5 are:65165gry65651/10 6 5 line 65,65 165line 65651/12 6 5 inch 65an
d6 5 16 5 point6 5 =6 5 1/7265inch.65The65factors65imply65that

165gry65=65(1/10)(1/12)(7265points)65=650.6065point.

Thus,6 5 16 5 gry26 5 =6 5 (0.606 5 point)26 5 =6 5 0.366 5 point2,6 5 which6 5 means6 5 that650.50 6 5 gry 2 65=6 5 0.18 6 5 poi
nt 2 65.

3. The65 metric65 prefixes65 (micro,65 pico,65 nano,65 …)65 are65 given65 for65 ready65 reference6
5 on65 the65 inside65front65cover65of65the65textbook65(see65also65Table651–2).


(a)65Since65165km65=651656510365m65and65165m65=651656510665m,

1km6 5 65 10365 m6 5 6 5 10365 65m6 5 m656 m.
m106
9
510



The6 5 given6 5 measurement6 5 is6 5 1.06 5 km6 5 (two6 5 significant6 5 figures),6 5 which6 5 im
plies6 5 our6 5 result65should65be65written65as651.0656510965m.

(b)65We65calculate65the65number65of65microns65in65165centimeter.65Since65165cm65=6510265m,

1cm6 5 =6 5 1026 5 m 6 5 =6 5 102m1066 5 65m6 m.
5 m65 65 104

,We65conclude65that65the65fraction65of65one65centimeter65equal65to651.065m65is

651.0656510 .65(c)65Since65165yd65=65(365ft)(0.304865m/ft)65=650.914465m,
4




1

, 2 CHAPTER65
1


1.065yd6 5 =6 5 0.91m10665 65m6 5 m656 5 m.
9.16565105

4. (a)6 5 Using6 5 the6 5 conversion6 5 factors6 5 16 5 inch6 5 =6 5 2.546 5 cm6 5 exactly6 5 and6 5 66 5 pic
as6 5 =6 5 16 5 inch,6 5 we65obtain 65665 picas656 5
0.806 5cm65 =6 5 0.806 5 165inc 651.965 picas.
cm65  65 h
2.5465 cm65
 6 5
165inch6 5 

   
(b)65With651265points65=65165pica,65
we65have

0.8065cm65=6 5 0.8065c 165inc  6566 5 picas65 65126 5 points656 5
m65  65  6 5  656 5  6 5 236 5 points.
2.54h65 cm65 165inch6 5 165pica 
   


5. Given65that65 165furl 6 5 201.16865m65,65 165rod65655. and65 165chain656520.11765 m65,65 we65find
ong 029265m
the65relevant65conversion65factors65to65be
165rod
1.065 furlong6 5 65201.16865m6565(201.1 6 5 4065 rods,
6865m65) 5.0292 m

and
165chain
1.06 5 furlong6 5 65201.1686 5 m6 5 65 (201. 1065 chains65.
1686 5 m65) 20.11765
m
Note6 5 the6 5 cancellation6 5 of6 5 m6 5 (meters),6 5 the6 5 unwanted6 5 unit.6 5 Using6 5 the6 5
given6 5 conversion65factors,65we65find

(a) the65distance65d65in65rods65to65be
d6 5 65 4.065 furlongs6 5 4.065furlongs6 5 6516065 rods,
4065 rods
165furlong

(b) and65that65distance65in65chains65to65be

1065 chains6 5
d6 5 6 5 4.065 furlongs6 5 4.065furlongs 6 5 4065 chains.
165furlong

6. We65make65use65of65Table651-6.

(a) We65look65at65the65first65(“cahiz”)65column:65165fanega65is65equivalent65to65what65am
ount65of65cahiz?65We65note65from65the65already65completed65part65of65the65table65that65165
dozen65fanega.65Thus,65165fanega65=6 5 16 5 cahiz,65or658.33656510265ca
cahiz65equals65a6512
hiz.6 5 Similarly,65“165cahiz65=654865cuartilla”65(in65the
651
65cahiz,65or652.08656 5 10 6 5 c
2
already65completed65part)65implies65that65165cuartilla65=48
ahiz.65Continuing6 5 in6 5 this6 5 way,6 5 the6 5 remaining6 5 entries6 5 in6 5 the6 5 first6 5 column