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A rocket is fired in deep space, where gravity is negligible. In the first second it ejects $\frac{1}{160}$ of its mass as exhaust gas and has an accolcration of 15.0 $\mathrm{m} / \mathrm{s}^{2} .$ What is the spoced of tho cxhaust gas relative to the rocket?

$$2.4 \mathrm{km} / \mathrm{s}$$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

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in this question, the rocket is moving by injecting some mass off gas. We know that in the course off one second that rocket objected 1/1 60 off its mass and by doing that, it achieved an acceleration off 15 m per second squared. And with that we have to determine what is the velocity of the gas that is being expelled. To do that, we have to use this equation which relates the achieved acceleration which the velocity of the gas that is being expelled, the initial mass off the rocket to the variation in the mass off the rocket and the time during which that mass was expelled. So using that equation, we have the following they achieved acceleration is 15 m per second squared. Now we have minus the velocity off the gas which we don't know divided by the initial mass off the rocket which I will call em are Then we multiply it by the variation in the mass off the rocket. The variation in the mass off the rocket is equals to minus the mask off gas that was expelled. So we have minus the mask off the rocket divided by 160. And finally we divide by Delta T, which is one Then we get the following 15 is equals to we have a minus sign and another minus sign so they will cancel out. So we have the velocity of the gas divided by the mass off the rocket. Now, anything divided by one is just that thing. That minus sign isn't there anymore. So we have the mass off the rocket divided by 160. As you can see, the mass of the rocket will be canceled. Then we get the following 15 is equals to the velocity of the gas divided by 160. The reform The velocity of the gas is 15 times 160 which results in 2000 and 400 m per second or 2.4 kg meters per second. Any off those two is a correct solution to this question. So this is the answer

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