In a hydraulic machine, the area of small piston is 50 cm^2 and that of the large piston is 5 m^2. If an effort of 20 N is applied on the small piston, how much load can be lifted on the large piston? Calculate.
*I want the Process of this and the answer is down below me.*
[Answer : 2 × 10^4 N]
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In a typical hydraulic press, a force of 20 N is exerted on small piston of area 0.050 m2. What is force exerted by large piston on load if it has an area of 0.50 m2?
answer:- 200N
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The areas of the pistons in a hydraulic machine are 5cm2 and 625cm2. What force on the smaller piston will support a load of 1250N on the larger piston?
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10
F1=1250NA1=625cm2
F2=xnA2=5cm2
Since pressure is constant.
A1F1=A2F2⇒6251250=5x⇒x=10N
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