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(II) $\mathrm{A} 3500$ -kg rocket is to be accelerated at 3.0 $\mathrm{g}$ at take-off from the Earth. If the gases can be ejected at a rate of $27 \mathrm{kg} / \mathrm{s},$ what must be their exhaust speed?

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5100 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

University of Sheffield

Lectures

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um nine. Dash 19 8th of the book says that some of the external forces, well, the rocket will be equal to mass times acceleration. Did you do to minus relative velocity? Uh, times, lot rate of change of mass. And so relative velocity is exhaust velocity and the external force medics No forces just, uh, zero minus the force of gravity so minus mg. And that's equal to m a beauty of acceleration minus V x. That's the exhaust velocity times Cmdty, therefore, exhaust velocity. The X is just negative m times uh, a plus. She, um acceleration. Explosion Did gravity over d m E t over dt on. So this gives us this gives us, um minus M, which is minus 3500 kilograms times a policy. This is a S three juice, a three plus one that's four duty four times 9.8 meters per second squared over negative 27 kilograms per second because, uh, you're losing 27 kilograms. Have a second, and so you work that out the kilograms cancel one of the seconds cancel. And where you left with is the unit meters per second and the answer is 5100 meters per second is the exhaust velocity. And, um 01 crucial step by for God here is that exhaust velocity, um, will be This will be a positive here because exhaust velocity is acting in the opposite is acting in the opposite direction, So that's the answer.

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