## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 11 - Problems - Page 427: 3

#### Answer

(a) The cliff is $260~m$ away. (b) The intensity of the music arriving at the cliff is $1.5\times 10^{-10}~W/m^2$

#### Work Step by Step

(a) Let $r_c$ be the distance to the cliff. The sound takes $1.5~s$ to travel a distance of $2~r_c$. We can find $r_c$: $2~r_c = v~t$ $r_c = \frac{v~t}{2}$ $r_c = \frac{(343~m/s)(1.5~s)}{2}$ $r_c = 257.25~m$ $r_c = 260~m$ The cliff is $260~m$ away. (b) We can write an expression for the intensity 1.0-m away: $I_1 = \frac{P}{A}$ $I_1 = \frac{P}{4\pi~r_1^2}$ We can write an expression for the intensity at the cliff: $I_c = \frac{P}{A}$ $I_c = \frac{P}{4\pi~r_c^2}$ We can divide the two equations to find $I_c$: $\frac{I_c}{I_1} = \frac{\frac{P}{4\pi~r_c^2}}{\frac{P}{4\pi~r_1^2}}$ $I_c = \frac{r_1^2}{r_c^2}~\times I_1$ $I_c = \frac{(1.0~m)^2}{(257.25~m)^2}\times (1.0\times 10^{-5}~W/m^2)$ $I_c = 1.5\times 10^{-10}~W/m^2$ The intensity of the music arriving at the cliff is $1.5\times 10^{-10}~W/m^2$

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