problem15_80

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: transverse waves as a function of x is ax F v dt dx ) ( + = = and the time t needed for a wave to reach the surface is found from + = = = . dx ax F dt dx dx dt t Let the length of the cable be L, so ( 29 2 2 F aL F a ax F a ax F dx 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 =-+ =...
View Full Document

Ask a homework question - tutors are online