Page 659 - 9780077418427.pdf

Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefiles
tiL12214_appd_633-642.indd Page 636 09/10/10 8:35 AM user-f463
tiL12214_appd_633-642.indd Page 636 09/10/10 8:35 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

Example 5.5, p. 124 Example 6.6, p. 152
T F = 86.0° 5 _ I = 0.5 A P = IV

T C = (T F – 32°)

v = ? 9 V = 120 V = (0.5 A)(120 V)
5 _ P = ?

= (86.0° – 32°) C _ _
J
9 = (0.5)(120) ×

s C
5 _
= (54.0°) _ J

9 = 60
s
270.0°
_

= = 60 W
9
= 30.0° Example 6.8, p. 154
I = 0.5 A
)
(
2.00 ft∕s

(

v T = v 0 + _ T p ) V = 120 V
p °C
p = IV = 60 W
_
2.00 ft∕s

= 1,087 ft/s + (30.0°C) Rate = $0.10/kWh

°C
Cost = ?
= 1,087 ft/s + 60.0 ft/s
(watts)(time)(rate)
__

= 1,147 ft/s Cost =
_
W

1,000
kW
Example 5.7, p. 125 _
W

(60 W)(1.00 h)$0.10
t = 1.00 s d _ ___
kW

v = ∴ d = vt =

t
W
v = 1,147 ft/s ft _ 1,000 _

kW
d = ? = (1,147 ) (1.00 s) __ h $ kW
s
(60)(1.00)(0.10) W _ _ _ _

×

ft _ s _ = 1,000 × × kWh W
1
1
= (1,147)(1.00) ×

s
1
= $0.006 or 0.6 of a cent per hour
= 1,147 ft
Sound traveled from the source to a reflecting surface, then back to
the source, so the distance is CHAPTER 7
1 _

1,147 ft × = 574 ft Example 7.4, p. 194
2 14
f = 7.00 × 10 Hz
–34
h = 6.63 × 10 J . s
Example 5.9, p. 130
E = ?
L = 0.5 m nv _

f n = E = hf
v = 400 m/s 2L m _ 14 1 _
(2) (400 )
–34
s

__ = (6.63 × 10 J . s) (7.00 × 10 )
F 2 = ? = s
(2)(0.5 m)
1 _
–34
14
(2)(400) m _
_ 1 _ = (6.63 × 10 )(7.00 × 10 ) J . s ×
s

= ×

(2)(0.5) s m –19
= 4.64 × 10 J
_
800 1 _

=
1 s
= 800 Hz CHAPTER 8
CHAPTER 6 Example 8.2, p. 208
14
f = 7.30 × 10 Hz
Example 6.4, p. 151 h = 6.63 × 10 J . s
–34
V = 120 V E = ?
R = 30 Ω
E = hf
I = ?
14 1 _
–34

V _ = (6.63 × 10 J . s) (7.30 × 10 )
s

V = IR ∴ I =
R 1 _
14
–34
_ = (6.63 × 10 )(7.30 × 10 ) J . s ×
120 V
s
= V _
30 = 4.84 × 10 –19 J

A
120 V _ A _
_

= ×
30 1 V
= 4 A
636 APPENDIX D Solutions for Follow-Up Example Exercises D-4