A force of 3600 N is exerted on a piston that has an area of 0.040 [tex]m^{2}[/tex]. What force is excepted on a second piston that has an area of 0.20[tex]m^{2}[/tex]?
Use [tex]\frac{F_{1} }{A_{1} } = \frac{F_{1} }{A_{2} }[/tex]

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

Respuesta :

Answer:

A. 18,000 N

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:

[tex]p_1 =p_2[/tex]

[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

where:

[tex]F_1 = 3600 N[/tex] is the force on the first piston

[tex]A=0.040 m^2[/tex] is the area of the first piston

[tex]F_2 = ?[/tex] is the force on the second piston

[tex]A=0.20 m^2[/tex] is the area of the second piston

Re-arranging the equation, we find:

[tex]F_2 = A_2 \frac{F_1}{A_1}=(0.20 m^2)\frac{3600 N}{0.040 m^2}=18,000 N[/tex]

Answer:

A.

Explanation:

18,000N