79. (a) With r = 10 m in Eq. 18-28, we have I=
P 4πr2
=⇒ P = 10 W .
(b) Using that value of P in Eq. 18-28 with a new value for r, we obtain I=
P W = 0.032 2 . 4π(5.0)2 m
Alternatively, a ratio I /I = (r/r ) could have been used. 2
(c) Using Eq. 18-29 with I = 0.0080 W/m2 , we have β = 10 log where I0 = 1 × 10−12 W/m2 .
I = 99 dB I0