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Murrr4er [49]
3 years ago
5

If an astronaut has a mass of 16 Kg on Earth, what would be his mass on the moon and on the space station

Physics
1 answer:
nlexa [21]3 years ago
3 0

Answer:

The astronaut's mass is 16 kg.

Explanation:

Mass can be defined as a measure of the amount of matter an object or a body comprises of. The standard unit of measurement of the mass of an object or a body is kilograms.

Irrespective of the location of an object or a body at a given moment in time, the mass (amount of matter that they're made up of) is constant. This ultimately implies that, whether you're in the moon, space, earth or any other place, your mass remains the same (constant).

Therefore, if an astronaut has a mass of 16 Kg on Earth, his mass on the moon and on the space station would remain the same, as his original mass of 16 Kg because mass is indestructible.

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The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.

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Does the efficiency of a motor depend on mass?
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Practically yes

  • Efficiency=Output/input

So

If mass is more output may come less so it affects the efficiency practically

But thepritically it doesn't

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2 years ago
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Which of the following could classify as a non-essential need according to the hierarchy of needs? A. healthy food B. human cont
sergejj [24]
D. According to the hierarchy of needs the body and mind must be taken care of first and foremost.

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3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

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P = A*I

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r_1 = 18m

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\text{Power} \propto r^2

For both panels we would have to

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Therefore the correct option is option C.9

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