A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston?​

Explanation:

ratio of the forces is equal to the ratio of the surface areas

which is another way of saying the pressure is constant in the system

2000(9.8) / 500 = A / 10

A = 2000(9.8)(10) / 500 = 392 cm²

e/c, where e is its total energy