problem44_32

# University Physics with Modern Physics with MasteringPhysics™ (12th Edition)

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44.32: a) Using the definition of z from Example 44.9 we have that . λ λ λ ) λ λ ( 1 1 0 0 0 s s z = - + = + Now we use Eq. 44.13 to obtain . 1 1 1 1 1 β β - + = - + = - + = + c v c v v c v c z b) Solving the above equation for β we obtain . 3846 . 0 1 5 . 1 1 5 . 1 1 ) 1 ( 1 ) 1 ( 2 2 2 2 = + - = + + - + = z z β Thus, . s m 10 15 . 1 3846 . 0 8 × = =
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