Question

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the station so she accelerates at 1.50 m/s2. What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall? Give your answer relative to an observer who is space walking and therefore does not accelerate with the space station due to the push.

Answer 1

Answer:

$a = 5.83 \times 10^{-4} m/s^2$

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = $1.50 m/s^2$

now we know that

$F = ma$

$F = 70(1.50) = 105 N$

now for the space station will be same as above force

$F = ma$

$105 = 1.8 \times 10^5 (a)$

$a = \frac{105}{1.8 \times 10^5}$

$a = 5.83 \times 10^{-4} m/s^2$

Answer 2

The magnitude of the acceleration of the space station is mathematically given as

$a = 5.83 * 10^{-4} m/s^2$

What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall?

Question Parameter(s):

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg.

The station pushes off one wall of the station so she accelerates at 1.50 m/s2.

Generally, the equation for the  Force is mathematically given as

F = ma

Therefore

F = 70(1.50)

F= 105 N

In conclusion, the space station force

F = ma

105 = 1.8 * 10^5 a

a = 5.83 * 10^{-4} m/s^2

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