## College Physics (4th Edition)

Sound energy is produced by the loudspeaker at a rate of $0.099~W$
) We can find the intensity of the sound: $\beta = 10~log\frac{I}{I_0}$ $71 = 10~log\frac{I}{I_0}$ $7.1 = log\frac{I}{I_0}$ $10^{7.1} = \frac{I}{I_0}$ $I = (10^{7.1})~I_0$ $I = (10^{7.1})~(1.0\times 10^{-12}~W/m^2)$ $I = 1.26\times 10^{-5}~W/m^2$ We can find the power of the source: $P = I~A$ $P = I~4\pi~r^2$ $P = (1.26\times 10^{-5}~W/m^2)~(4\pi)~(25~m)^2$ $P = 0.099~W$ Sound energy is produced by the loudspeaker at a rate of $~0.099~W$