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Luba_88 [7]
2 years ago
15

A 1200 kg car is travelling qt 20m\s determine the work that must be done on the car by the frictional force of the brakes to st

op the car
Physics
1 answer:
Anestetic [448]2 years ago
4 0

Answer:

240000 Joules

Explanation:

Applying,

The work that must be done by the frictional force of the brake to stop the car must be equal to the kinetic energy of the car.

W = mv²/2.................. Equation 1

Where W = work done by the frictional force of the brake, m = mass of the car, v = velocity of the car.

From the question,

Given: m = 1200 kg, v = 20 m/s

Substitute these values into equation 1

W = (1200×20²)/2

W = 240000 Joules

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to find

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f=\frac{v}{4L}\\ f=\frac{335m/s}{4(0.029m)} \\ f=2887.93Hz

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3 years ago
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5 0
2 years ago
A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free
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Consider that the bar magnet has a magnetic field that is acting around it, which will imply that there is a change in the magnetic flux through the loop whenever it moves towards the conducting loop. This could be described as an induction of the electromotive Force in the circuit from Faraday's law.

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Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

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