Chapter 13 Solutions - Mosinee School District
Chapter 13 Solutions - Mosinee School District
Chapter 13 Solutions - Mosinee School District
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<strong>Chapter</strong> <strong>13</strong><br />
f<br />
v boat wave<br />
, 5.0 m s<br />
20 m<br />
0.25 Hz<br />
<strong>13</strong>.49 The down and back distance is 4.00 m + 4.00 m = 8.00 m.<br />
The speed is then<br />
v<br />
d total<br />
t<br />
4 8.00 m<br />
0.800 s<br />
40.0 m s<br />
F<br />
Now,<br />
m<br />
L<br />
0.200 kg<br />
4.00 m<br />
5.00 10<br />
2<br />
kg m<br />
so<br />
F v<br />
5.00 10 kg m 40.0 m s 80.0 N<br />
2 2<br />
2<br />
<strong>13</strong>.50 The speed of the wave is<br />
v<br />
x<br />
t<br />
20.0 m<br />
0.800 s<br />
25.0 m s<br />
and the mass per unit length of the rope is m/ L 0.350 kg m . Thus, from v F , we obtain<br />
2<br />
2<br />
25.0 m s 0.350 kg m 219 N<br />
F v<br />
<strong>13</strong>.51 (a) The speed of transverse waves in the cord is v F , where mL is the mass per unit length. With<br />
the tension in the cord being F = 12.0 N, the wave speed is<br />
v<br />
F F FL<br />
m L m<br />
12.0 N 6.30 m<br />
0.150 kg<br />
22.4 m s<br />
(b)<br />
The time to travel the length of the cord is<br />
t<br />
L<br />
v<br />
6.30 m<br />
22.4 m s<br />
0.281 s<br />
<strong>13</strong>.52 (a) In making 6 round trips, the pulse travels the length of the line 12 times for a total distance of 144 m. The<br />
speed of the pulse is then<br />
Page <strong>13</strong>.20