A hydraulic lift raises a 4000 kg automobile when a 500N force is applied to the smaller piston. If the smaller piston has an ar
1 answer:
Answer:
The cross-sectional area of the larger piston is 800 cm².
Explanation:
Given;
mass of the automobile, m = 4000 kg
force applied on the small piston, F₁ = 500 N
area of the smaller piston, A₁ = 10 cm²
load lifted by the larger piston, F₂ = 4000 x 10 = 40,000 N
Pressure experienced by each piston is given as;
Where;
A₂ is the cross-sectional area of the larger piston
Therefore, the cross-sectional area of the larger piston is 800 cm².
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