Question

A force of 3600 N is exerted on a piston that has an area of 0.040
`m^(2)`. What force is excepted on a second piston that has an area of 0.20
`m^(2)`?

Use
`(F_(1) )/(A_(1) ) = (F_(1) )/(A_(2) )`

A. 18,000 N
B. 7200 N
C. 180,000 N
D. 90,000 N

Answer 1

A.

Step-by-step explanation:

18,000N

Answer 2

A. 18,000 N

Step-by-step explanation:

By using Pascal's principle, we can say that the pressure on first piston is equal to the pressure on the second piston. Therefore we have:

`p_1 =p_2`

`(F_1)/(A_1)=(F_2)/(A_2)`

where:

`F_1 = 3600 N` is the force on the first piston

`A=0.040 m^2` is the area of the first piston

`F_2 = ?` is the force on the second piston

`A=0.20 m^2` is the area of the second piston

Re-arranging the equation, we find:

`F_2 = A_2 (F_1)/(A_1)=(0.20 m^2)(3600 N)/(0.040 m^2)=18,000 N`