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2 years ago

13

The small piston of a hydraulic lift has a crosssectional area of 2.87 cm2 and the large piston 314 cm2 . What force must be app

lied to the small piston for the lift to raise a load of 2.4 kN

1 answer:

vredina [299]2 years ago

4 0

Answer:

P1 = P2 pressure is uniform

F1 / A1 = F2 / A2

F1 = F2 (A1 / A2) = 2,400 N * (2.87 / 314) = 21.9 N

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The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

$g = \frac{F}{m}$

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And $F =\frac{GMm}{r^{2} }$

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

$g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11} \times 1}{(6.6 \times 10^{6}) ^{2} }$

$g = 0.977 \times 10^1= 9.77\ N$

So, the gravitational field strength is approximately equal to 10 N.

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Answer:

Explanation:

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