Physics 41 Homework Set 3 Chapter 17 Solutions from Serway 8th Edition

OC: 7, 8, 11, 13, 14 CQ: 3, 6, 8 Problems: 2, 3, 7, 11, 15, 18, 19, 21, 28, 30, 31, 35, 39, 45, 56, 59

OC: 7, 8, 11, 13, 14

CQ: 3, 6, 8

Problems

15. A cowboy stands on horizontal ground between two parallel vertical cliffs. He is not midway between the cliffs. He fires a shot, and hears its echoes. The second echo arrives 1.92 s after the first, and 1.47 s before the third. Consider only the sound traveling parallel to the ground and reflecting from the cliffs. Take the speed of sound as 340 m/s. (a) What is the distance between the cliffs? (b) What If? If he can hear a fourth echo, how long after the third echo does it arrive?

*P17.15 Let 1x represent the cowboy’s distance from the nearer canyon wall and 2x his distance from the farther cliff. The sound for the first echo travels distance 12x . For the second, 22x . For the third,

1 22 2x x+ . For the fourth echo, 1 2 12 2 2x x x+ + . Then 2 12 21.92 s

340 m sx x−

= and 1 2 22 2 21.47 s

340 m sx x x+ −

= .

Thus 11 340 m s 1.47 s 250 m2

x = = and 221.92 s 1.47 s

340 m sx

= + ; 2 576 mx = .

(a) So 1 2 826 mx x+ =

(b) ( )1 2 1 1 22 2 2 2 21.47 s

340 m sx x x x x+ + − +

=

P17.30 We presume the speakers broadcast equally in all directions.

(a) 2 23.00 4.00 m 5.00 mACr = + =

( )π π

β

β

−−

−

−

×= = = ×

×=

= =

36 2

22

6 2

12 2

1.00 10 W 3.18 10 W m4 4 5.00 m

3.18 10 W m10 dBlog

10 W m

10 dB 6.50 65.0 dB

Ir

P

(b) 4.47 mBCr =

( )

36 2

2

6

12

1.50 10 W 5.97 10 W m4 4.47 m

5.97 1010 dBlog10

67.8 dB

Iπ

β

β

−−

−

−

×= = ×

×=

=

(c) 2 23.18 W m 5.97 W mI µ µ= + 6

12

9.15 1010 dBlog 69.6 dB10

β−

−

×= =