/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 For a safe re-entry into the Ear... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 11

For a safe re-entry into the Earth's atmosphere, the pilots of a space capsule must reduce their speed from \(2.6 \times 10^{4} \mathrm{m} / \mathrm{s}\) to \(1.1 \times 10^{4} \mathrm{m} / \mathrm{s} .\) The rocket engine produces a backward force on the capsule of \(1.8 \times 10^{5} \mathrm{N}\) The mass of the capsule is 3800 kg. For how long must they fire their engine? [Hint: Ignore the change in mass of the capsule due to the expulsion of exhaust gases.]

Short Answer

Expert verified
Answer: The pilots must fire their engine for approximately 316.6 seconds.

Step by step solution

01

List down given information and formula

Initial speed \((v_i) = 2.6 \times 10^4 \ m/s\) Final speed \((v_f) = 1.1 \times 10^4 \ m/s\) Backward force \((F) = 1.8 \times 10^5 \ N\) Mass of the capsule \((m) = 3800 \ kg\) Impulse-momentum theorem: \(F \times t = m(v_f - v_i)\) Where, \(t\) is the time the engine must be fired for We need to find \(t\).
02

Rearrange the impulse-momentum theorem formula to find \(t\)

We will rearrange the impulse-momentum theorem formula to isolate \(t\): \(t = \dfrac{m(v_f - v_i)}{F}\) Now, plug in the given values.
03

Calculate the time \(t\)

\(t = \dfrac{3800 \ kg (1.1 \times 10^4 \ m/s - 2.6 \times 10^4 \ m/s)}{1.8 \times 10^5 \ N}\) \(t = \dfrac{3800 \ kg (-1.5 \times 10^4 \ m/s)}{1.8 \times 10^5 \ N}\) \(t = -\dfrac{5.7 \times 10^7 kg \cdot m/s}{1.8 \times 10^5 N}\) \(t = -\dfrac{5.7}{1.8} \times 10^2 s\) \(t = -(3.166 \dots) \times 10^2 s\) Now, since the time cannot be negative, we will revert the sign of \(t\) as the backward force acts in the opposite direction to the initial velocity. \(t = 3.166 \times 10^2 s\) Therefore, the pilots must fire their engine for approximately \(316.6\) seconds (rounded to one decimal place) to safely re-enter Earth's atmosphere.

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