A car is lifted by a hydraulic jack that consists of two pistons. The diameter of the larger piston is 2 m and that of the smaller piston is 50 cm. If the force applied on the smaller piston is 240 N, find the weight of the car.
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Answered by
5
Answer:
240/(0.25^2) = wt./ 1^2
wt = 240/0.0625
= 3840 N
Answered by
5
ANSWER:2M=200CM
RADIUS OF LARGE PISTON=200/2=100CM
AREA(1) =πr^2=π×50^2=2500πCM^2
DIAMETER OF SMALLER PISTON= 50CM
RADIUS=50/2=25
AREA(2)= πR^2
=625πF
=W×A(2)/A(1)
=F=W×625π/2500π
=F=W/20
SO MINIMUM FORCE IS W/20
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