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oee [108]
3 years ago
15

Gravitational pull between 2 objects is determined by which of the following?

Physics
2 answers:
sp2606 [1]3 years ago
8 0

Answer:

Both distance between and mass of the 2 objects

Explanation:

The gravitational pull between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 is the mass of the first object

m2 is the mass of the second object

r is the distance between the two objects

From the formula, we see that the gravitational pull depends both on the distance between the two objects (r) and on their masses (m1, m2), so the correct answer is

Both distance between and mass of the 2 objects


vampirchik [111]3 years ago
3 0
Both distance between and mass of the 2 objects.
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A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
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Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

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A 0.25 kg baseball is in contact with a bat for 0.0075 s and exerts a force or Tour
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Explanation:

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If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th
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Answer:

1470kgm²

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The formula for expressing the moment of inertial is expressed as;

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Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

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