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Alborosie
3 years ago
4

An empty office chair is at rest on a floor. Consider the following forces: 1. A downward force of gravity. 2. An upward force e

xerted by the floor. 3. A net downward force exerted by the air. Which of the forces is (are) acting on the office chair?
Physics
1 answer:
mario62 [17]3 years ago
3 0

Answer:

Explanation:

When a chair is at rest on the floor , the earth will exert a force of gravity  on it which is also known as its weight. This force acts on the chair in downward direction towards the centre of the earth.

The chair also experiences  another force called reaction force which is exerted by the floor on which it rest . This force acts on the chair in upward direction. This force is equal to the weight of the chair  in case chair is at rest or it is moving in uniform motion.

Due to these two equal and opposite  forces, acting on the chair, it remains at rest, as net force becomes zero.

A net downward force exerted by air , which is the reaction force of buoyant force acting on chair in upward direction acts on the floor on which chair rests . This force does not act on the chair.

Hence out of three forces given , only two forces act on the chair.

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Answer:

4 J

Explanation:

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Now, the net force will be;

F_net = 3 - 1

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4 years ago
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A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.
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Answer:

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Explanation:

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<h3>What is a frictional force?</h3>

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Given: Frictional forces are considered to be negligible.

A few seconds after Rock X is released from rest, Rock Y is also released. Consider moving downward to be constructive. Both rocks are launched from rest, thus their initial velocities are equal to zero (u = 0).

We know that the velocity is given by,

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Rock X is released first, followed by Rock Y, and because Rock X travels faster than Rock Y, it travels farther. This is due to the fact that whenever Rock X has already acquired some velocity (t).

Its velocity (v) is higher, increasing the spacing (s). In conclusion, because Rock X falls faster than Rock Y,'s' rises each time Rock X hits the ground.

The complete question is given below:-

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the same location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward.

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