Physics, 23.12.2020 02:10 ejohnstonee111

A(n) 70.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 41.4 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.916 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m/s in the direction away from the shuttle. How long will it take for her to reach the shuttle

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Answer from: sonthatshannah

4.4 min

Explanation:

To solve this, we use the law of conservation of momentum

0 = -m(a).v(a) + m(c).v(c), where

m(a) = mass of the astronaut

m(c) = mass of the camera

v(a) = velocity of the astronaut

v(c) = velocity of the camera

Making v(a) subject of the formula, we have

v(a) = m(c).v(c)/m(a)

Now, we make one last assumption, that the astronaut is moving at constant speed and time, thus the time needed will be

t = s/v -> since s = vt

Now, we already have our v from above, so we substitute

t = s/[m(c).v(c)/m(a)]

t = s.m(a)/m(c).v(c)

Applying the values, we have

t = (41.4 * 70.1) / (0.916 * 12)

t = 2902.14 / 10.992

t = 264 seconds or 4.4 minutes

Answer from: Quest

Well the magnitude of its acceleration related to that car starts with the oil change like when you have to change the oil in your car you have to make sure that your precious car is starting really propley and very well

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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A(n) 70.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself...

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