507_Physics ProblemsTechnical Physics

507_Physics ProblemsTechnical Physics - Chapter 17 P17.33 =...

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Chapter 17 509 P17.33 β = F H G I K J 10 10 12 log I I = 10 10 10 12 bg ej Wm 2 I 120 100 dB 2 af = .; I 100 2 100 10 dB 2 I 10 11 dB 2 . (a) ℘= 4 2 π rI so that rI rI 1 2 12 2 2 = rr I I 21 1 2 2 300 30 0 = F H G I K J = × = . . . . m m a f (b) I I 1 2 11 5 949 10 = F H G I K J = × . . . . m a f P17.34 (a) Et r I t =℘ = = × = 4 4 100 7 00 10 0 200 1 76 2 2 2 ππ W m s k J 2 a f a f .. . (b) = × × F H G I K J = 10 700 10 108 2 12 log . . dB P17.35 (a) The sound intensity inside the church is given by = F H G I K J = F H G I K J == = −− 10 101 10 10 10 10 10 0 012 6 0 12 10 1 12 1.90 ln ln . . I I I I dB dB 2 222 We suppose that sound comes perpendicularly out through the windows and doors. Then, the radiated power is = = IA 00126 220 0277 . m W 22 e j . Are you surprised by how small this is? The energy radiated in 20.0 minutes is F H G I K J = 200 60 0 332 . Js m in s 1.00 min J . (b) If the ground reflects all sound energy headed downward, the sound power,
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