problem18_45

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

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18.45: From Table (18.2), the speed is (1.60) v s, and so 2 2 2 ) 60 . 1 ( 3 3 s v M RT m kT v = = = (see Exercise 18.48), and so the temperature is . ) m s K 10 385 . 4 ( K) mol J 3145 . 8 ( 3(1.60) mol) kg 10 0 . 28 ( ) 60 . 1 ( 3 2 2 2 4 2 2 3 2 2 v v R Mv T × = × = = - - K 987 s) m 1500 )( m s K 10 385 . 4
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Unformatted text preview: 2 2 2 4 = ⋅ ×-K 438 s) m 1000 )( m s K 10 385 . 4 ( b) 2 2 2 4 = ⋅ ×-K. 110 s) m 500 )( m s K 10 385 . 4 ( c) 2 2 2 4 = ⋅ ×-...
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