problem15_80

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

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15.80: a) The tension is the difference between the diver’s weight and the buoyant force, N. 392 )) s m 80 . 9 )( m 0800 . 0 )( m kg (1000 kg 120 ( ) ( 2 3 3 water = - = - = g V ρ m F b) The increase in tension will be the weight of the cable between the diver and the point at x , minus the buoyant force. This increase in tension is then ( 29 x π g Ax ρ x ) s m 80 . 9 )( ) m 10 00 . 1 ( ) m kg (1000 m kg 10 . 1 ( ) ( 2 2 2 3 - × - = - μ x ) m N 70 . 7 ( = The tension as a function of x is then . ) m N 70 . 7 ( ) N 392 ( ) ( x x F + = c) Denote the tension as , ) ( 0 ax F x F + = where N 392 0 = F and m. N 70 . 7 = a Then, the speed of
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Unformatted text preview: transverse waves as a function of x is μ ax F v dt dx ) ( + = = and the time t needed for a wave to reach the surface is found from ∫ ∫ ∫ + = = = . dx ax F μ dt dx dx dt t Let the length of the cable be L, so ( 29 2 2 F aL F a μ ax F a μ ax F dx μ t L L-+ = + = + = ∫ s. 98 . 3 ) N 392 ) m )(100 m N (7.70 N 392 ( m N 70 . 7 m kg 10 . 1 2 =-+ =...
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  • Force, Buoyant Force, kg, ax, dx. dx dt

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