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Katen [24]
3 years ago
12

n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad

ius 5.00 cm. This pressure is transmitted by a liquid to a piston that has a radius of 15.0 cma) What force must the compressed air exert to lift a car weighing 13300 N
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
5 0

Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

F1/A1 = F2/A2

From our question,

The radius of the smaller piston is 5 cm = 0.05 m

The radius of the larger piston is 15 cm = 0.15 m

The force of the larger piston is 13300 N

The force of the smaller piston is unknown = F

A1 = πr² = 3.142 * 0.05² = 0.007855 m²

A2 = πr² = 3.142 * 0.15² = 0.070695 m²

F1/0.007855 = 13300/0.070695

F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

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Mint is a dicot.

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3 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
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Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

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Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

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making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

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Substituting the values;

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velocity at t = 2.50

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Through the next 5 second;

t2 = 5 seconds

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v2 = 6.4 rev/s

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Substituting the values;

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Answer:

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