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Instructor’s Solutions Manual – College Physics (2025 Evergreen Release) | Alan Giambattista | Complete Worked Solutions

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This listing contains the complete Instructor’s Solutions Manual for College Physics, 2025 Evergreen Release by Alan Giambattista. It includes detailed, step-by-step worked solutions for all textbook problems, providing clear explanations and accurate calculations. Perfect for students needing guided support, exam preparation, and homework assistance. Aligned with the 2025 Evergreen edition to ensure up-to-date physics problem solving.

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College Physics
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College Physics











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Institución
College Physics
Grado
College Physics

Información del documento

Subido en
4 de diciembre de 2025
Número de páginas
2260
Escrito en
2025/2026
Tipo
Examen
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INSTRUCTOR'S SOLUTIONS MANUALS

COLLEGE PHYSICS

2025 EVERGREEN RELEASE

CHAPTER NO. 01: INTRODUCTION

CONCEPTUAL QUESTIONS

1. Knowledge of physics is important for a full understanding of many scientific disciplines,
such as chemistry, biology, and geology. Furthermore, much of our current technology can
only be understood with knowledge of the underlying laws of physics. In the search for more
efficient and environmentally safe sources of energy, for example, physics is essential. Also,
many people study physics for the sense of fulfillment that comes with learning about the
world we inhabit.

2. Without precise definitions of words for scientific use, unambiguous communication of
findings and ideas would be impossible.

3. Even when simplified models do not exactly match real conditions, they can still provide
insight into the features of a physical system. Often a problem would become too
complicated if one attempted to match the real conditions exactly, and an approximation can
yield a result that is close enough to the exact one to still be useful.

4. After solving a problem, it is a good idea to check that the solution is reasonable and makes
intuitive sense. Do the units work out correctly? In the symbolic version of the answer,
before numbers are substituted, would the expression change in a reasonable way if each
parameter were made larger? Smaller? Very much larger or smaller? It may also be useful to
explore other possible methods of solution as a check on the validity of the first. A good
student thinks of a framework of ideas and skills that they are constructing for themself. The
problem solution may extend or strengthen this framework

5. Scientific notation eliminates the need to write many zeros in very large or small numbers,
and to count them. Also, the number of significant digits is indicated unambiguously when a
quantity is written this way.

,6. In scientific notation the decimal point is often placed after the first (leftmost) numeral. The
number of digits written equals the number of significant figures.

7. Not all of the significant digits are known definitely. The last (rightmost) digit, called the
least significant digit, is an estimate and is less definitely known than the others.

8. It is important to write a quantity with the correct number of significant figures so that we
can indicate how precisely a quantity is known and so that we do not mislead the reader by
writing digits that are not at all known to be correct.

9. The kilogram, meter, and second are three of the base units used in SI, the international
system of units.

10. The international system SI uses a well-defined set of internationally agreed upon standard
units and makes measurements in terms of these units, their combinations, and their powers
of ten. The U.S. customary system contains units that are primarily of historical origin and
are not based upon powers of ten. As a result of this international acceptance and of the ease
of manipulation that comes from dealing with powers of ten, scientists around the world
prefer to use SI.

11. Fathoms, kilometers, miles, and inches are units with the dimension length. Grams and
kilograms are units with the dimension mass. Years, months, and seconds are units with the
dimension time.

12. The first step toward successfully solving almost any physics problem is to thoroughly read
the question and obtain a precise understanding of the scenario. The second step is to
visualize the problem, often making a quick sketch to outline the details of the situation and
the known parameters.

13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment.
Such trends are more apparent when data is plotted graphically rather than listed in numerical
tables.

, 14. The statement gives a number for the speed of sound in air, but fails to indicate the units used
for the measurement. Without units, the reader cannot relate the speed to one given in
familiar units such as km/s.

MULTIPLE-CHOICE QUESTIONS

1. (b) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (b) 8. (d) 9. (b) 10. (c)

PROBLEMS

1. Strategy The New fence will be 100% + 37% = 137% of the height of the old fence.
1.37 × 1.8 m = 2.5 m
Solution Find the height of the new fence.
Discussion. Long ago you were told that 37% of 1.8 is 0.37 times 1.8.

2
2. Strategy. Relate the surface area A to the radius r using A = 4π r .
Solution. Find the ratio of the new radius to the old.
A1 4π r12 and
= = π r22 2.0
A2 4= = A1 2.0(4π r12 ).
4π r22 = 2.0(4π r12 )
r22 = 2.0r12
2
 r2

  = 2.0
 r1

r2
= =2.0 1.4
r1

The radius of the balloon increases by a factor of 1.4.

2
3. Strategy Relate the surface area A to the radius r using A = 4π r .
Solution Find the ratio of the new radius to the old.
A1 = 4π r12 and A2 = 4π r22 = 1.160 A1 = 1.160(4π r12 ).

4π r22 = 1.160(4π r12 )
r22 = 1.160r12
2
 r2 
  = 1.160
 r1 
r2
= 1.160 = 1.077
r1

The radius of the balloon increases by 7.7%.

, Discussion. Because the surface area is proportional to the square of the radius, the
percentage change in radius is smaller than the percentage change in area—in fact, a bit less
than half as large. The factor of 4π divides out and plays no part in determining the answer.
The answer just comes from the proportionality of area to the square of the radius. The
circumference also increases by 7.7%.



4. Strategy To find the factor by which the child’s height increased, divide their new height by
their old height. Subtract 1 from this value and multiply by 100 % to find the percentage
increase.
Solution Find the factor.
1.65 m
= 1.10
1.50 m
Find the percentage.
1.10 − 1 = 0.10, so the percent increase is 10 % .


5. Strategy To find the factor by which the metabolic rate of a 70 kg human exceeds that of a
5.0 kg cat, use a ratio.
3/ 4 3/ 4
 mh   70 
  = = 7.2
 mc   5.0 
Solution Find the factor:

Discussion. The proportionality could be written into an equation as (Metabolic rate) = K
(Body mass)3/4 where K is a proportionality constant (with very odd units). If we write down
this equation for a human and again for a cat, and then divide the two, the K divides out and
we obtain the quantity (mh/mc)3/4 that we evaluated. Get used to using your calculator to follow
the order of operations without your having to re-enter any numbers. On my calculator I type
70  5 = ^ 0.75 = .
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