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Oxana [17]
3 years ago
11

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 st

ation 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.
Physics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

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

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F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

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3 years ago
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6 0
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Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi
Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

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Where,

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From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

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