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A single-stage rocket is fired from rest from a deep-space platform, where gravity is negligible. If the rocket burns its fuel in 50.0 $\mathrm{s}$ and the relative speed of the exhaust gas is $v_{\mathrm{ex}}=2100 \mathrm{m} / \mathrm{s}$ , what must the mass ratio $m_{0} / m$ be for a foral speed $v$ of 8.00 $\mathrm{km} / \mathrm{s}$ (about equal to the orbital speed of an earth satellite)?

$$

\frac{m_0}{m}=45.1

$$

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Cornell University

Numerade Educator

University of Washington

University of Winnipeg

in this question, this rockets fired from a dips place platform where gravity is negligible and achieves some velocity after burning all of its field. We know that the velocity of the gas that is expelled from the rocket is 2000 and 100 m per second. And we also know that the final velocity off that rocket after burning all of its fuel is eight kilometers per second, which is very fast then. Given this information, what is the ratio between the initial mass off the rocket on the final mass off the rocket? To answer this question, we have to use this equation which tells us what is the final velocity as a function off the initial velocity, the exhaustion, velocity, the initial mass and the final mass. By using that equation, we get the following. The final velocity is equals to the initial velocity which is zero plus the velocity off the gas times. The longer it, um off the initial mass divided by the final mass. So the final velocity V is equals to the velocity of the gas times the luxury item off the initial mass divided by the final mass since the initial velocity is just zero now plugging in the values that we have we got the following The final velocity is eight kilometers per second. The velocity of the gas is 2000 and 100 m per second, which is the same as 2.100 kilometers per second. All you have to do is divide by 1000 to convert from meters per second, two kilometers per second. Then we have 2.100 times the logarithms off the ratio in between the initial mass and the final mass. Now we sent this value to the other side dividing so that we get eight divided by 2.100 Is the cost of the luxury item off the ratio that we are interested in now exponentially ating both sides off this equation we get the following e to this power is equals to e to the logarithms off the initial mass divided by the final mass. It happens that e to the luxury TEM councils out. So e council's elaborate, um, and your left. If the following e to this power is equals to the ratio that we are interested in, then that ratio am zero divided by M is equals to eat 28 divided by 2.100 which is approximately 45.1. And this is the answer to this question.

Brazilian Center for Research in Physics