A 1500 kg is being lifted at a constant velocity by a hydralic jack attached to a flat plate. Underneath the plate is a pipe with a top radius of 24 cm which is determined to experienxe a preashre of 81235 N/m^2. The other end of the pipe has a radius of 2.00 cm. How much force must be exterted at this end?

The force that must be exerted at the other end is 102.08 N/m₂

Explanation:

Using Pascal's formula; P₁ = P₂

P₁ = F₁/A₁ = m*g/(πr²)

m = 1500 kg; g = 9.8 m/s², r = 24 cm = 0.24 m

P₁ = (1500 * 9.8)/(22/7 * 0.24 * 0.24)

P₁ = 81235 N/m²

P₂ = F₂/A₂

A₂ = πr² where π = 22/7; r = 2.00 cm = 0.02m

A₂ = 22/7 * 0.02 * 0.02 = 0.00126 m²

P₂  = F₂ / 0.00126 m₂

substituting in the formula  P₁ = P₂

81235 N/m² =  F₂ / 0.00126 m²

F₂ = 81235 N/m² * 0.00126 m²

F₂ = 102.08 N/m₂

Therefore, the force that must be exerted at the other end is 102.08 N/m₂

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