SOLUTIONS MANUAL
Physics
ST
John D. Cutnell, Kenneth W. Johnson, David Young, and Shane Stadler
──────────────────────────────────────────────────
12th Edition
UV
IA
26
_A
PP
RO
VE
D?
?
, TABLE OF CONTENTS
Solutions Manual: Physics, 12th Edition
ST
By John D. Cutnell, Kenneth W. Johnson, David Young, and Shane Stadler
Chapter 1 Introduction and Mathematical Concepts
UV
Chapter 2 Kinematics in One Dimension
Chapter 3 Kinematics in Two Dimensions
Chapter 4 Forces and Newton’s Laws of Motion
Chapter 5 Dynamics of Uniform Circular Motion
IA
Chapter 6 Work and Energy
Chapter 7 Impulse and Momentum
26
Chapter 8 Rotational Kinematics
Chapter 9 Rotational Dynamics
Chapter 10 Simple Harmonic Motion and Elasticity
_A
Chapter 11 Fluids
Chapter 12 Temperature and Heat
Chapter 13 The Transfer of Heat
PP
Chapter 14 The Ideal Gas Law and Kinetic Theory
Chapter 15 Thermodynamics
Chapter 16 Waves and Sound
RO
Chapter 17 The Principle of Linear Superposition and Interference Phenomena
Chapter 18 Electric Forces and Electric Fields
Chapter 19 Electric Potential Energy and the Electric Potential
Chapter 20 Electric Circuits
VE
Chapter 21 Magnetic Forces and Magnetic Fields
Chapter 22 Electromagnetic Induction
Chapter 23 Alternating Current Circuits
D?
Chapter 24 Electromagnetic Waves
Chapter 25 The Reflection of Light: Mirrors
Chapter 26 The Refraction of Light: Lenses and Optical Instruments
?
Chapter 27 Interference and the Wave Nature of Light
1
, Chapter 28 Special Relativity
Chapter 29 Particles and Waves
Chapter 30 The Nature of the Atom
ST
Chapter 31 Nuclear Physics and Radioactivity
Chapter 32 Ionizing Radiation, Nuclear Energy, and Elementary Particles
UV
IA
26
_A
PP
RO
VE
D?
?
2
, CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
ST
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
UV
1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the
last vector.
2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to
the shortest distance between the tail of A and the head of B. Thus, R is less than the
IA
magnitude (length) of A plus the magnitude of B.
3. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are
not reversed.
26
4. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector
A is not reversed.
_A
5. (c) When the two vector components Ax and Ay are added by the tail-to-head method,
the sum equals the vector A. Therefore, these vector components are the correct ones.
6. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
PP
component along the x axis (Ax = 0 m) and its scalar component along the y axis is
negative.
7. (e) The scalar components are given by Ax = (450 m) sin 35.0 = 258 m and
RO
Ay = (450 m) cos 35.0 = 369 m.
8. (d)
VE
9. Rx = 0 m, Ry = 6.8 m
10. R = 7.9 m, degrees
D?
11. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides
are known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
4.0 km
?
12. (b) The angle is found by using the inverse tangent function, tan1 53 .
3.0 km
Physics
ST
John D. Cutnell, Kenneth W. Johnson, David Young, and Shane Stadler
──────────────────────────────────────────────────
12th Edition
UV
IA
26
_A
PP
RO
VE
D?
?
, TABLE OF CONTENTS
Solutions Manual: Physics, 12th Edition
ST
By John D. Cutnell, Kenneth W. Johnson, David Young, and Shane Stadler
Chapter 1 Introduction and Mathematical Concepts
UV
Chapter 2 Kinematics in One Dimension
Chapter 3 Kinematics in Two Dimensions
Chapter 4 Forces and Newton’s Laws of Motion
Chapter 5 Dynamics of Uniform Circular Motion
IA
Chapter 6 Work and Energy
Chapter 7 Impulse and Momentum
26
Chapter 8 Rotational Kinematics
Chapter 9 Rotational Dynamics
Chapter 10 Simple Harmonic Motion and Elasticity
_A
Chapter 11 Fluids
Chapter 12 Temperature and Heat
Chapter 13 The Transfer of Heat
PP
Chapter 14 The Ideal Gas Law and Kinetic Theory
Chapter 15 Thermodynamics
Chapter 16 Waves and Sound
RO
Chapter 17 The Principle of Linear Superposition and Interference Phenomena
Chapter 18 Electric Forces and Electric Fields
Chapter 19 Electric Potential Energy and the Electric Potential
Chapter 20 Electric Circuits
VE
Chapter 21 Magnetic Forces and Magnetic Fields
Chapter 22 Electromagnetic Induction
Chapter 23 Alternating Current Circuits
D?
Chapter 24 Electromagnetic Waves
Chapter 25 The Reflection of Light: Mirrors
Chapter 26 The Refraction of Light: Lenses and Optical Instruments
?
Chapter 27 Interference and the Wave Nature of Light
1
, Chapter 28 Special Relativity
Chapter 29 Particles and Waves
Chapter 30 The Nature of the Atom
ST
Chapter 31 Nuclear Physics and Radioactivity
Chapter 32 Ionizing Radiation, Nuclear Energy, and Elementary Particles
UV
IA
26
_A
PP
RO
VE
D?
?
2
, CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
ST
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
UV
1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the
last vector.
2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to
the shortest distance between the tail of A and the head of B. Thus, R is less than the
IA
magnitude (length) of A plus the magnitude of B.
3. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are
not reversed.
26
4. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector
A is not reversed.
_A
5. (c) When the two vector components Ax and Ay are added by the tail-to-head method,
the sum equals the vector A. Therefore, these vector components are the correct ones.
6. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
PP
component along the x axis (Ax = 0 m) and its scalar component along the y axis is
negative.
7. (e) The scalar components are given by Ax = (450 m) sin 35.0 = 258 m and
RO
Ay = (450 m) cos 35.0 = 369 m.
8. (d)
VE
9. Rx = 0 m, Ry = 6.8 m
10. R = 7.9 m, degrees
D?
11. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides
are known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
4.0 km
?
12. (b) The angle is found by using the inverse tangent function, tan1 53 .
3.0 km
