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Gelneren [198K]
3 years ago
15

A 75.0 kg students sits 1.15 m away from a 68.4 kg student. What is the force of gravitational attraction between them?

Physics
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

F=2.589×10⁻⁷ Newtons

Explanation:

The gravitational force of attraction between two bodies is given by

F= Gm₁m₂/r²

Where m₁ and m₂ are the masses of the two bodies, G is the universal gravitational constant while r is the distance between the two bodies.

G=6.67408 × 10⁻¹¹m³kg⁻¹s⁻²

r=1.15m

m₁=75.0kg

m₂=68.4kg

Therefore F= (6.67408×10⁻¹¹m³kg⁻¹s⁻²×75.0kg×68.4kg)/(1.15m)²

F=2.589×10⁻⁷ Newtons

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Answer:

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Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

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with \theta=41.9^{\circ}

- The frictional force, whose magnitude is

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==> This in turn means that all of the horizontal forces are balanced,
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All of the forces on this list must add up to zero. So ...

(sliding friction force) = (pushing force), in the opposite direction.

If nobody pushing the box, then sliding friction force = zero.

Vertical forces:
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All of the forces on this list must add up to zero, so ...

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3 years ago
Read 2 more answers
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