problem11_64

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

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11.64: a) m. 3 . 1 ) m 10 30 . 0 ( 4 ) 10 0 . 9 )( 23 . 0 ( ) ( 2 4 4 0 μ w l l σ w = π × × - = - = - - b) N, 10 1 . 3 m 10 0 . 2 m 10 10 . 0 42 . 0 ) m) 10 0 . 2 ( ( Pa) 10 1 . 2 ( 1 6 2 3 2 2 11 × = × × × × = = = - - - π w w σ AY l l AY F where the Young’s modulus for nickel has been used.
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  • Ay, M., Modulus

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