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
nanometers. A useful expression for the energy of a photon in te
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)(3.00 108 m/s)(109 nm/m)/(1.60 1019 J/eV);
E = (1.24 1
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 J m = 1.44 MeV fm.
We find the product nucleus by bala
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 1034 J s)(3 108 m/s)(109 nm/m)/(1.60 1019 J/eV);
E = (
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 m = 1.44 MeV fm.
1.
2.
We find the distance from
D = 1/
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b
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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 object to
image is
do + di = 1.5 m + 1.5 m =
3.0 m.
2.
Becau
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).
We find the distance traveled from
average speed = d/t
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
RW = D1 + D2 cos 45
D2y
D2
= (125 km) + (65 km) cos 45 =
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
displacement in the direction of this force is the vertical displa
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 use the density of air, we have
m = V = LWH = (1.29 kg
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)/(0.028 m) = 2.28 104 N/m.
The frequency of vibration
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.
Because the sound travels down and back, we have
L = !vt =
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, for the ratio we have
NAu/NAg = MAg/MAu = 108/197 =
0.54
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 J.
3.
We find the temperature from
W = _Q = mc _T;
7700
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 force is
F = kQ1Q2/r2.
If we divide the expressions fo
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)
= ( 8.6 106 C)(+ 75 V 0)=
2.
6.5 104 J (done by the field).
W
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 which electrons pass any point in the wire is the curre
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
Fmax = ILB, or
Fmax/L = IB = (9.80 A)(0.80 T) =
7.8 N/m.
(
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