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**Unformatted text preview: **16.17: a) pmax = BkA = 2 πBfA v = 2 π (1.42×10 5 Pa) (150 Hz) (5.00×10 −6 m) (344 m s) = 1.95 Pa. b) From Eq. (16.14), 2 I = pmax 2 ρv = (1.95 Pa) 2 (2 × (1.2 kg m 3 )(344 m s)) = 4.58 × 10 −3 W m 2 . c) 10 × log ( 4.58×10 −3 10 −12 ) = 96.6 dB. ...

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