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Ksivusya
3 months ago
8

A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an a

rea of 10 cm2 , what is the cross-sectional area of the larger piston
Physics
1 answer:
kicyunya [3.2K]3 months ago
5 0

Answer:

The cross-sectional area of the larger piston is 392cm ^{2}[/tex]

Explanation:

To find the solution, we apply the following equation:

Pascal's principle: F=P*A   Formula (1)

F=Force applied to the piston

P: Pressure

A= Area of the piston

Nomenclature:

Fp= Force on the primary piston= 500N

W= weight of the car =m*g=2000kg*9.8m/s2= 19600N

Fs= Force on the secondary piston= W = 19600N

Ap= Primary piston area=10cm^{2} =10*10^{-4}m^{2}

As= Area of the secondary piston=?

Pressure applied on one side is distributed to all liquid molecules since liquids are incompressible.

From equation (1)

P=F/A

Pp=Ps

\frac{Fp}{Ap} = \frac{Fs}{As}

As= \frac{Fs*Ap}{Fp}

As=\frac{19600*10*10^{-4} }{500}

As=0.0392m^{2} =0.0392*10^{4}cm^{2}

As=392cm ^{2}

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