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Halliday 2 Cap 18 - P18 079

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    December 2018
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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