problem16_49

University Physics with Modern Physics with Mastering Physics (11th Edition)

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16.49: a) Combining Eq. (16.14) and Eq. (16.15), 20 . 5 2 12 3 ) 10 ( 0 max 10 ) m W )(10 s m 344 )( m kg 20 . 1 ( 2 10 2 - = = β ρvI p Pa, 10 144 . 1 2 - × = or Pa, 10 14 . 1 2 - × = to three figures. b) From Eq. (16.5), and as in Example 16.1, ( 29 m. 10 51 . 7 Hz 587 Pa) 10 42 . 1 ( 2 s) m 344 ( Pa) 10 144 . 1 ( 2 9 5 2 max max - - × = × × = = = π πf B v p Bk p A c) The distance is proportional to the reciprocal of the square root of the intensity, and
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Unformatted text preview: hence to 10 raised to half of the sound intensity levels divided by 10. Specifically, m. 9 . 62 m)10 00 . 5 ( 2 3.00) (5.20 =-...
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  • Division, 9 m, 5.00 M, 1.20 kg, 2 Pa

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