Like

Report

A $70-\mathrm{kg}$ astronaut floating in space in a 110$\cdot \mathrm{kg}$ MMU (manned maneuvering unit) experiences an acceleration of 0.029 $\mathrm{m} / \mathrm{s}^{2}$ when he fires one of the MMU's thrusters. (a) If the speed of the escaping $\mathrm{N}_{2}$ gas relative to the astronaut is 490 $\mathrm{m} / \mathrm{s}$ , how much gas is used by the thruster in 5.0 $\mathrm{s} ?$ (b) What is the thrust of the thruster?

(a) The ejected mass is 0.053 $\mathrm{kg}$

(b) $F=5.19 \mathrm{N}$

You must be signed in to discuss.

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

University of Winnipeg

{'transcript': "problem. 8.61. So you have an astronaut who is experiencing some acceleration from his Mm. You injecting some nitrogen gas with some given exhaust philosophy with respect to him, and we know it is acceleration is so first, we'd like to figure out how much of this nitrogen propellant is used in five seconds, and then we would also like to know what the thrust is. So to start off with since five seconds is not very long, we're going to assume that the rate of which the masses changing his constant in the equation the acceleration is equal toe negative. Uh, exhaust cast velocity divided by the mass times the rate at which the mass changes. Now we want to find the radio which the mass changes and then multiply that. Since we're saying this is constant, we could just multiply that by the five seconds over which were interested. And that will tell us how much the mass has changed. Also, I guess I should mention the mass of the astronaut. Plus the mm you is 180 kilograms. They give a and old birth exhaust gas speed. Now we're going to assume also that the mass itself is constant because the rate at which it's losing mass we're going to assume is small compared to the overall amount of mass president to begin with. And so if we put these numbers and we find that it's losing mass at naked zero planes 0106 kilograms per second and so then in five seconds that it applies that change in mass is good 0.5 three kilograms. And so we concede that our assumption works out pretty well. That mess is constant because 53 grams out of 100 80 kilograms is pretty negligible and especially, you know, you have to be out like six digits of President. Be able to worry about that. Now we know that basically, by multiplying this by Ami of N A and then it's really in here, the trust is the negative of the exhaust guests times the rate at which masses is lost. You may already know both of these things now so we can find this is five point 19 rivers negative. I guess"}