Answer :
The compressed air must exert a force of 508.47N on the small piston in order to lift the car weighing 13300 N and the pressure exerted is 16.14x10^4Pa.
Given circular cross-section having a radius (r1) = 3.17cm = 3.17x10-2m
The pressure is transmitted by a liquid to a second piston of radius (r2) = 16.2cm = 16.2x10-2m
Force exerted on car (F2) = 13300N
Here pascals principle is applied.
We know that Pressure (P) = force/area
P1 = P2 then F1/A1 = F2/A2
Area of the circle = π*r²
F1/π*(3.17x10-2)^2 = 13300/π*(16.2x10-2)^2
F1 = 508.47N
(b) The air pressure will produce a force of that magnitude P =F/A
P = F1/A1 = 508.47/ π*(3.17x10-2)^2 = 16.14x10^4Pa
(c) As a result, the work is equal in both pistons. The little piston displaces a small volume of liquid, which is dispersed across the large piston in a thin layer.
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complete question: In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radius of r1 = 3.17 cm. This pressure is transmitted by an incompressible liquid to a second piston of radius 16.2 cm. (a) What force must the compressed air exert on the small piston in order to lift a car weighing 13,300 N? Neglect the weights of the pistons. (b) What air pressure will produce a force of that magnitude? (c) Show that the work done by the input and output pistons is the same.
