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Pre-Calc Homework Solutions 262

Pre-Calc Homework Solutions 262 - 262 Section 6.4 26 0.16...

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22. continued (c) e 2 kt 5 0.16 2 kt 5 ln 0.16 t 5 2 } ln 0 k .16 } 5 2 } 5700 ln ln 2 0.16 } < 15,070 years The animal died about 15,070 years before A . D . 2000, in 13,070 B . C . 23. Note that the total mass is 66 1 7 5 73 kg. v 5 v 0 e 2 ( k / m ) t v 5 9 e 2 3.9 t /73 (a) s ( t ) 5 E 9 e 2 3.9 t /73 dt 5 2 } 21 1 9 3 0 } e 2 3.9 t /73 1 C Since s (0) 5 0 we have C 5 } 21 1 9 3 0 } and lim t s ( t ) 5 lim t } 21 1 9 3 0 } (1 2 e 2 3.9 t /73 ) 5 } 21 1 9 3 0 } < 168.5 The cyclist will coast about 168.5 meters. (b) 1 5 9 e 2 3.9 t /73 } 3 7 .9 3 t } 5 ln 9 t 5 } 73 3 l .9 n 9 } < 41.13 sec It will take about 41.13 seconds. 24. v 5 v 0 e 2 ( k / m ) t v 5 9 e 2 (59,000/51,000,000) t v 5 9 e 2 59 t /51,000 (a) s ( t ) 5 E 9 e 2 59 t /51,000 dt 5 2 } 459 5 , 9 000 } e 59 t /51,000 1 C Since s (0) 5 0, we have C 5 } 459 5 , 9 000 } and lim t s ( t ) 5 lim t } 459 5 , 9 000 } (1 2 e 2 59 t /51,000 ) 5 } 459 5 , 9 000 } < 7780 m The ship will coast about 7780 m, or 7.78 km. (b) 1 5 9 e 2 59 t /51,000 } 51 5 , 9 0 t 00 } 5 ln 9 t 5 } 51,00 5 0 9 ln 9 } < 1899.3 sec It will take about 31.65 minutes. 25. y 5 y 0 e 2 kt 800 5 1000 e 2 ( k )(10) 0.8 5 e 2 10 k k 5 2 } ln 1 0 0 .8 } At t 5 10 1 14 5 24 h: y 5 1000 e 2 ( 2 ln 0.8/10)24 5 1000 e 2.4 ln 0.8 < 585.4 kg About 585.4 kg will remain. 26. 0.2 5 e 2 0.1 t ln 0.2 5 2 0.1 t t 5 2 10 ln 0.2 < 16.09 yr It will take about 16.09 years. 27. (a) } d d p n } 5 kp } d p p } 5 k dh E } d p p } 5 E k dh ln ) p ) 5 kh 1 C e ln ) p ) 5 e kh 1 C ) p ) 5 e C e kh p 5 Ae kh Initial condition: p 5 p 0 when h 5 0 p 0 5 Ae 0 A 5 p 0 Solution: p 5 p 0 e kh Using the giving altitude-pressure data, we have p 0 5 1013 millibars, so:
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