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problem11_99

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

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11.99: a) From Eq.(11.10), mm 0.66 or m, 10 62 . 6 4 ) m 10 00 . 5 )( Pa 10 20 ( ) m 50 . 1 )( s m 80 . 9 )( kg 50 . 4 ( 2 7 10 2 - × × × = = - l to two figures. b) J. 022 . 0 ) m 10 0500 . 0 )( s m 80 . 9 )( kg 50 . 4 ( 2 2 = × - c) The magnitude F will be vary with distance; the average force is N, 7 . 16 ) cm 0250 . 0 ( 0 = l A Y and so the work done by the applied force is J. 10 35 . 8 ) m 10 0500 . 0 ( N) 7 . 16 ( 3 2 - - × = × d) The wire is initially stretched a distance m 10 62 . 6 4 - × ( the result of part (a)), and so the average elongation during the additional stretching is m 10 12 . 9 4 - × , and the average force the wire exerts is N. 8 . 60 The work done is negative, and equal to
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Unformatted text preview: J. 10 04 . 3 ) m 10 0500 . )( N 8 . 60 ( 2 2--×-= ×-e) See problem 11.82. The change in elastic potential energy is ( 29 J, 10 04 . 3 ) ) m 10 62 . 6 ( ) m 10 62 . 11 (( m 50 . 1 2 ) m 10 00 . 5 )( Pa 10 20 ( 2 2 4 2 4 2 7 10----× = ×-× × × the negative of the result of part (d). (If more figures are kept in the intermediate calculations, the agreement is exact.)...
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