Physics, asked by CHAMPSHINSHU, 3 months ago

area of pistons in a hydraulic are 10 cm^2 and 125cm^2 .What force on smaller piston will support a load of 1250 N on larger piston

Answers

Answered by aithimouleendra

0

Answer:

Explanation:

F₁= F₂(A₁/A₂)

F₁ =1250{10/(100)²/125/(100)²}

F₁ =1250(10/125)

F₁ =1250*(2/25)

F₁ =50 *2

=100N

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