problem08_p107

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

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8.107: a) For 0 < t the rocket is at rest. For t 0 90 s, Eq. ( 29 40 . 8 is valid, and ( 29 ( 29 ( 29 ( 29 ( 29 . s 120 / 1 1 ln s m 2400 t t v - = At 90 = t s, this speed is 3.33 km/s, and this is also the speed for . s 90 t b) The acceleration is zero for 90 and 0 < t t s. For 90 0 t s, Eq. (8.39) gives, with ( 29 ( 29 s 120 / 1 m/s 20 0 2 s, 120 t dt dm a m - = - = . c) The maximum acceleration occurs at the latest time of firing,
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Unformatted text preview: 90 = t s, at which time the acceleration is, from the result of part (a), ( 29 2 120 / 90 1 m/s 20 m/s 80 2 =-, and so the astronaut is subject to a force of 6.0 kN, about eight times her weight on earth....
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