Question

the area of a pistons in a hydraulic machine are to 5CM square and 625 CM square what force on the smaller Piston will support a load of 1250 Newton on the larger piston

Answer 1

Answer:it's 10N

Explanation:

Refer the attachment

Answer 2

Answer:

The force on the smaller Piston will be equal to 10 Newton to support a load of 1250 Newton on the larger piston.

Explanation:

We have given,

The area of smaller pistons in a hydraulic machine, A₁ = 5cm²

The area of larger piston, A₂ = 625cm²

Force applied on the larger piston due to load, F₂ = 1250N

The force will be applied on the smaller piston is F₁.

Since the pressure (P) will be same on both sides. So the pressure is constant.

From Pascal's law, we can write,

$P=\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\frac{F_1}{5} =\frac{1250}{625}$

$F_1=(2*5)N$

$F_1=10N$

Therefore, the force on the smaller Piston will be equal to 10N.

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