ch01-p030 - 30. To solve the problem, we note that the...

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0.2 2.00 s 3 g 1kg 60 s 4.00(5.00) 3.00 g/s 0.101g/s 0.101 s 1000 g 1 min 6.05 10 kg/min. t dm dt = ªº = =− ¬¼ × 30. To solve the problem, we note that the first derivative of the function with respect to time gives the rate. Setting the rate to zero gives the time at which an extreme value of the variable mass occurs; here that extreme value is a maximum. (a) Differentiating 0.8 ( ) 5.00 3.00 20.00 mt t t + with respect to t gives 0.2 4.00 3.00. dm t dt The water mass is the greatest when
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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