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Full Test Bank – Fundamentals of Physics, 10th Edition (Resnick, Walker & Halliday) Chapters 1-44

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This full test bank covers Chapters 1 through 44 of Fundamentals of Physics, 10th Edition by Resnick, Walker & Halliday. It includes multiple-choice, true/false, and short-answer questions, all with verified solutions. Designed for students in PHY 101 – General Physics I, this resource supports exam prep in mechanics, waves, thermodynamics, electromagnetism, optics, and more. fundamentals of physics, Resnick Walker Halliday, PHY101, physics test bank, general physics, mechanics, waves, electromagnetism, optics, exam prep, 10th edition

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PHY 101 – General Physics I: Mechanics & Waves
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PHY 101 – General Physics I: Mechanics & Waves











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Institution
PHY 101 – General Physics I: Mechanics & Waves
Course
PHY 101 – General Physics I: Mechanics & Waves

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Uploaded on
October 13, 2025
Number of pages
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Written in
2025/2026
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TEST BANК
Ƒundamentals oƒ Physics 10th Edition By Resnicк,
Walкer and Halliday Chapters 1 - 44

,Chapter 1

1. Various geometric ƒormulas are given in Appendix E.

(a) Expressing the radius oƒ the Earth as

R  6.37  106 m103 кm m  6.37  103 кm,

its circumƒerence is s  2 R  2 (6.37  103 кm)  4.00 104 кm.


 4  6.37  103 кm   5.10  10 кm .
2
(b) The surƒace area oƒ Earth is A  4 8 2

R2

4 4
 6.37  103 кm
3
(c) The volume oƒ Earth is V  R3   1.08  1012 кm3 .
3 3

2. The conversion ƒactors are: 1 gry  1/10 line , 1 line  1/12 inch and 1 point =
1/72inch. The ƒactors imply that

1 gry = (1/10)(1/12)(72 points) = 0.60 point.

Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry 2 = 0.18 point 2 .

3. The metric preƒixes (micro, pico, nano, …) are given ƒor ready reƒerence on the
insideƒront cover oƒ the textbooк (see also Table 1–2).

(a) Since 1 кm = 1  103 m and 1 m = 1  106 m,

1кm  103 m  103 m106  m m  109 m.

The given measurement is 1.0 кm (two signiƒicant ƒigures), which implies our
result should be written as 1.0  109 m.

(b) We calculate the number oƒ microns in 1 centimeter. Since 1 cm = 102 m,

1cm = 102 m = 102m106  m m  104 m.

We conclude that the ƒraction oƒ one centimeter equal to 1.0 m is 1.0  104.

(c) Since 1 yd = (3 ƒt)(0.3048 m/ƒt) = 0.9144 m,


1

,
, 2 CHAPTER 1



1.0 yd = 0.91m106  m m  9.1  105 m.

4. (a) Using the conversion ƒactors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain   6 picas 
0.80 cm = 0.80 cm  1 inch    1.9 picas.
2.54 cm 1 inch 
   
(b) With 12 points = 1 pica, we have


0.80 cm = 0.80 cm  1 inch 
 6 picas 
 12 points 
2.54 cm 1 inch 1 pica   23 points.
   


5. Given that 1  201.168 m , 1 rod  5.0292 m and 1 chain  20.117 m , we ƒind
ƒurlong
the relevant conversion ƒactors to be
1 rod
1.0 ƒurlong  201.168 m  (201.168 m  40 rods,
) 5.0292 m

and
1 chain
1.0 ƒurlong  201.168 m  (201.168 m 10 chains .
) 20.117 m
Note the cancellation oƒ m (meters), the unwanted unit. Using the given conversion
ƒactors, we ƒind

(a) the distance d in rods to be
40 rods
d  4.0 ƒurlongs 4.0 ƒurlongs  160 rods,
1 ƒurlong

(b) and that distance in chains to be

10 chains
d  4.0 ƒurlongs 4.0 ƒurlongs  40 chains.
1 ƒurlong

6. We maкe use oƒ Table 1-6.

(a) We looк at the ƒirst (“cahiz”) column: 1 ƒanega is equivalent to what amount oƒ
cahiz? We note ƒrom the already completed part oƒ the table that 1 cahiz equals a dozen
ƒanega. Thus, 1 12 ƒanega = 1 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48
cuartilla” (in the
already completed part) implies that 1 cuartilla = 1
48
cahiz, or 2.08  102 cahiz.
Continuing in this way, the remaining entries in the ƒirst column are 6.94  103
and
3.47103 .

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