problem16_04

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

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16.4: The values from Example 16.8 are Hz, 1000 Pa, 10 16 . 3 4 = × = f B m. 10 2 . 1 8 - × = A Using Example 16.5, , s m 295 s m 344 293K K 216 = = v so the pressure amplitude of this wave is Pa). 10 16 . 3 ( 2 4 max × = = = A v πf B BkA p Pa. 10 8.1 m) 10 2 . 1 ( s m 295 Hz) (1000 2 3 8 - - × = × π This is 0.27 Pa) 10 (3.0 Pa) 10 1 . 8 ( 2 3
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Unformatted text preview: × ×--times smaller than the pressure amplitude at sea level (Example 16-1), so pressure amplitude decreases with altitude for constant frequency and displacement amplitude....
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  • Wavelength, 8 m, 3 Pa, 2 Pa, 0.27 295 m

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