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Professional ApplicationCalculate the increase in velocity of a 4000 -kg space probe that expels 3500 $\mathrm{kg}$ of its mass at an exhaust velocity of $2.00 \times 10^{3} \mathrm{m} / \mathrm{s}$ . You may assume the gravitational force is negligible at the probe's location.
$\Delta v = 4.16 \times 10 ^ { 3 } \mathrm { ms } ^ { - 1 }$
Physics 101 Mechanics
Chapter 8
Linear Momentum and Collisions
Moment, Impulse, and Collisions
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we have the change in velocity. It was the velocity minus V not It was ve Zahn and I over m f now ve is given Toby two times. Change that three meters per second much. But by dawn off the initial mass is 4000. You know, Graham, divided by the final mass is 4000. My study 500 que program, and that comes out to be for one 58 0.8 meeting for a second.
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