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jeka57 [31]
3 years ago
15

You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.

Will the period be the same when the buggy is used on the moon? Explain.
Physics
1 answer:
Blizzard [7]3 years ago
3 0

Answer:

Remains same

Explanation:

T = Time period of oscillation

m = mass

k = spring constant

Time period of oscillation is given as

T = 2\pi \sqrt{\frac{m}{k} }

we know that as we move from earth to moon, the value of spring constant "k"  and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

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Answer:

C) one-half as great

Explanation:

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In this case, the sphere starts from rest, so v_0=0. Replacing the given values and solving for g':

g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}

The acceleration due to gravity near Earth's surface is g=9.8\frac{m}{s^2}. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

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Answer:

0.15kg/m³

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