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DochEvi [55]
3 years ago
9

Difference between hydraulic brake and hydraulic lift

Physics
2 answers:
aliya0001 [1]3 years ago
7 0

Answer:

The hydraulic lift and the hydraulic brake are virtually the same machine. Both of them use some means of applying pressure to a hydraulic fluid that forces a piston or pistons elsewhere to move outward with greater force. There are only two differences between the hydraulic lift and the hydraulic brake. One is that all the push is delivered by one cylinder in a lift and by either two or four cylinders in a brake system. The other difference is that in a hydraulic lift the pump moves the lift piston out in a series of movements while in a brake system the brake pedal moves the brake pistons out in one short movement.

As fluid is forced into the lift cylinder it forces the lift piston upward. The lift cylinder is a lot wider than the pump cylinder. This means that the area of the lift piston is a lot greater than the area of the pump piston. Because Pascal's Principle guarantees that the pressure in a fluid is the same everywhere and the area that pressure is applied over is greater at the lift cylinder, the force applied by the lift cylinder must be greater. The hydraulic lift takes the long low-force motions of the pump piston and converts then into short high-force motions of the lift cylinder.

hope it helps! please mark me brainliest

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CaHeK987 [17]3 years ago
6 0

Answer:

The hydraulic lift and the hydraulic brake are virtually the same machines The other difference is that in a hydraulic lift the pump moves the lift piston out in a series of movements while in a brake system the brake pedal moves the brake pistons out in one short movement.

Explanation:

hope this helps have a good night/day :)

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W-ΔEk = W'..................... Equation 1

Where W' = Work done by friction on the chair, W = Work done on the chair by me, Ek = change in Kinetic energy of the chair as a result of the slide.

From the question,

W = FdcosФ.............. Equation 2

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Where F = Force applied on the chair, d = distance of slide, Ф = angle between the force and the horizontal, m = mass of the chair, v = final velocity of the chair, u = initial velocity of the chair

Substitute equation 2 and equation 3 into equation 1

W' = FdcosФ-1/2m(v²-u²)........................ Equation 4

Given: F = 156, d = 5 m, Ф = 26°, m = 18.8 kg, v = 3.1 m/s, u = 1.3 m/s

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