Question

a car is 3000 kg is to lifted by a hydraulic lift the area of cross section of piston carrying car is 15 cm ^2 and the area of cross section smaller is 10cm^2 calculate the force needed to be applied on the smaller pistons to lift the car.​

Answer 1

Answer:

Pressure on the piston P=AF

Force F=m×a

              =3000×9.8

              =29400 N

Area of cross section A=425×10−4sqm

Therefore the pressure P=425×10−43000×9.8=6.92×105Pa.

Solve any question of 

Answer 2

Pressure on the piston P=AF

Force F=m×a

=3000×9.8

=29400 N

Area of cross-section A=425×10−4sqm

Therefore the pressure P=425×10−43000×9.8=6.92×105Pa.