The cross-sectional area of the output piston in a hydraulic device is 3.0 times the area of the input piston. For each 1.5 cm the input piston moves, how far does the output piston move

For each 1.5 cm the input piston moves, then output piston will move 0.5 cm

Explanation:

Let cross-sectional area of the input piston = A

Let cross-sectional area of the output piston = 3A

pressure (P) = Force (F) / Area (A)

F = PA

For a constant gravitational force on the inlet and outlet piston;

P₁A₁ = P₂A₂

But pressure = ρgh

where;

ρ is density of water

g is acceleration due to gravity

h is the distance or height moved by the piston

(ρg)h₁A₁ = (ρg)h₂A₂

h₁A₁ = h₂A₂

Area of output piston = 3 times area of input piston

h₁A₁ = h₂(3A₁)

For each 1.5 cm the input piston moves, then output piston will move;

1.5A₁ =  h₂(3A₁)

1.5 = 3h₂

h₂ = 1.5 / 3

h₂ = 0.5 cm

Thus, for each 1.5 cm the input piston moves, then output piston will move 0.5 cm

not sure 12

explanation: