Pre-Calc Homework Solutions 261

Pre-Calc Homework Solutions 261 - Section 6.4 3 k 261 17....

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17. At time t 5 } 3 k } , the amount remaining is y 0 e 2 kt 5 y 0 e 2 k (3/ k ) 5 y 0 e 2 3 < 0.0499 y 0 . This is less than 5% of the original amount, which means that over 95% has decayed already. 18. T 2 T s 5 ( T 0 2 T s ) e 2 kt 35 2 65 5 ( T 0 2 65) e 2 ( k )(10) 50 2 65 5 ( T 0 2 65) e 2 ( k )(20) Dividing the first equation by the second, we have: 2 5 e 10 k k 5 } 1 1 0 } ln 2 Substituting back into the first equation, we have: 2 30 5 ( T 0 2 65) e 2 [(ln 2)/10](10) 2 30 5 ( T 0 2 65) 1 } 1 2 } 2 2 60 5 T 0 2 65 5 5 T 0 The beam’s initial temperature is 5 8 F. 19. (a) First, we find the value of k . T 2 T s 5 ( T 0 2 T s ) e 2 kt 60 2 20 5 (90 2 20) e 2 ( k )(10) } 4 7 } 5 e 2 10 k k 52} 1 1 0 } ln } 4 7 } When the soup cools to 35 8 , we have: 35 2 20 5 (90 2 20) e [(1/10) ln (4/7)] t 15 5 70 e [(1/10) ln (4/7)] t ln } 1 3 4 } 5 1 } 1 1 0 } ln } 4 7 } 2 t t 5 < 27.53 min It takes a total of about 27.53 minutes, which is an additional 17.53 minutes after the first 10 minutes. (b)
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