Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 28
CHAPTER 28
Note:
At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in
nanometer
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 30
CHAPTER 30
Note:
A useful expression for the energy of a photon in terms of its wavelength is
E = hf = hc/ = (6.63 1034 J s
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 31
CHAPTER 31
Note:
1.
A factor that appears in the analysis of energies is
ke2 = (9.00 109 N m2/C2)(1.60 1019 C)2 = 2.30 1028
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 32
CHAPTER 32
Note:
1.
2.
3.
A useful expression for the energy of a photon in terms of its wavelength is
E = hf = hc/ = (6.63
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 33
CHAPTER 33
Note: A factor that appears in the analysis of energies is
ke2 = (9.00 109 N m2/C2)(1.60 1019 C)2 = 2.30 1028 J
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Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 23
CHAPTER 23
1.
For a flat mirror the image is as far behind the mirror as the object is in front, so the distance from objec
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 20
CHAPTER 20
1.
(a) The maximum force will be produced when the wire and the magnetic field are perpendicular,
so we have
Fma
CHAPTER 2
1.
2.
3.
We find the average speed from
average speed = d/t = (230 km)/(3.25 h) =
We find the time from
average speed = d/t;
25 km/h = (15 km)/t , which gives
70.8 km/h.
t = 0.60 h (36 min).
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 3
CHAPTER 3
1.
2.
3.
We choose the west and south coordinate system shown.
D2x
For the components of the resultant we have
W
R
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 6
CHAPTER 6
1.
Because there is no acceleration, the contact force must have the same magnitude as the weight. The
displacemen
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 10
CHAPTER 10
1.
2.
3.
4.
When we use the density of granite, we have
m = V = (2.7 103 kg/m3)(1 108 m3) =
2.7 1011 kg.
When we
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 11
CHAPTER 11
1.
We find the spring constant from the compression caused by the increased weight:
k = mg/x = (65 kg)(9.80 m/s2
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 12
CHAPTER 12
1.
Because the sound travels both ways across the lake, we have
L = !vt = !(343 m/s)(1.5 s) =
2.6 102 m.
2.
Beca
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 13
CHAPTER 13
1.
The number of atoms in a mass m is given by
N = m/Mmatom.
Because the masses of the two rings are the same, f
CHAPTER 14
1.
The required heat flow is
_Q = mc _T = (20.0 kg)(4186 J/kg C)(95C - 15C) =
6.7 106 J.
2.
For the work to equal the energy value of the food, we have
W = (750 Cal)(4186 J/Cal) =
3.14 106
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 16
CHAPTER 16
1.
2.
The number of electrons is
N = Q/e = ( 30.0 106 C)/( 1.60 1019 C/electrons) =
The magnitude of the Coulomb
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 17
CHAPTER 17
1.
We find the work done by an external agent from the work-energy principle:
W = KE + PE = 0 + q(Vb Va)
= ( 8.6
Solutions to Physics: Principles with Applications, 5/E, Giancoli
Chapter 18
CHAPTER 18
1.
The charge that passes through the battery is
1.4 105 C.
Q = I t = (5.7 A)(7.0 h)(3600 s/h) =
2.
The rate at
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