A hydraulic lift in a garage can lift a 1600 kg vehicle when a 400 N force is applied to the smaller piston of area 10 cm2. The area of the larger piston is:

392cm²

Explanation:

The solution to this question was gotten using the law of paschal

F1/A1 = F2/A2

Such that

F2 = f1 x A2/A1

F2 is the force that is lifting the hydraulic lift = 1600kg

= 1600x9.8g

F1 = 400N

A1 = 10cm³

A2 = area of the bigger piston

1600x9.8 = 400 x A2/10cm²

15680 = 400A2/10

15680 = 40A2

A2 = 15680/40

A2 = 392

Therefore the area of the piston is equal to 392cm²

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

this would be about 43%