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

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explanation:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j