A force of 20 N acts on a smaller piston of a hydraulic press with an area of ​​10 dm2. How much force does the liquid act on a larger piston of 120 cm2?​

166.67 N

Explanation:

Applying Pascal's principle,

Presure in the smaller piston(P') = Pressure in the bigger piston(P)

But,

Pressure = Force/Area

Pressure in the smaller piston(P') = Force applied to the smaller piston(F')/Area of the smaller piston(A')

Pressure in the bigger piston(P) =  Force applied to the bigger piston(F)/Area of the bigger piston(A)

F'/A' = F/A Equation 1

Make F the subject of the equation

F = F'A/A' Equation 2

From the question,

Given: F' = 20 N, A = 120 cm², A' = 10 dm² = (10×100) = 1000 cm²

Substitute these values into equation 2

F = 20(1000)/120

F = 166.67 N

0.11 m/s and 0.28 m/s should be correct

be subject to a force andchange its motion