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11 months ago

Which gives the work done by a torque during angular displacement?

Physics

1 answer:

Nutka1998 [239]11 months ago

4 0

Only when the torque is constant the work done is done during angular displacement. So the correct option is d.

Work done by the force is calculated as the dot product of force and angular displacement of the point of application of force. It is equal to the change in rotational kinetic energy of the body.

Work done by a torque can be calculated by taking an analogy from work done by force.

Work done = torque × angular displacement

So work is done by a torque during angular displacement only when torque is constant.

If a torque applied on a body rotates it through an angle ω, the work done by torque is

W = ζ × ω

To know more about the work done by torque refer to the link given below:

brainly.com/question/17083207

#SPJ4

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Answer:

D. move up to another shell that would form

Explanation:

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2 years ago

HELP

Free_Kalibri [48]

Answer:

Explanation:

Although there is absolutely NO regard for significant digits, I can help you with this, nonetheless.

The equation for Potential Energy is PE = mgh. We have everything but the height of the ball. We have to solve for that using a one-dimensional motion equation:

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2 years ago

A car accelerates at a rate of 3 m/s^2. If it's original speed is 8 m/s, how many seconds will it take the car to reach a final

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2 years ago

A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b

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Answer:

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

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2 years ago

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azamat

Answer:

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2 years ago