If a piston exerts an external pressure of 75.0×10³ N/m² and moves a distance of 2.00 m, what is the magnitude of the work performed by the piston? Given: Cross-sectional area of the piston = 5.00×10² mm².

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Quizplus · Accepted Answer

The work done by the piston can be calculated using the formula $W = F \cdot d$, where $F$ is the force exerted by the piston and $d$ is the distance moved. The force exerted by the piston can be found by multiplying the external pressure by the cross-sectional area of the piston. First, convert the area from mm$^2$ to m$^2$ (1 mm$^2$ = 1×10$^{-6}$ m$^2$), so the area is 5.00×10$^{-4}$ m$^2$. Then, $F = 75.0 \times 10^3 \, \text{N/m}^2 \times 5.00 \times 10^{-4} \, \text{m}^2 = 37.5 \, \text{N}$. Finally, $W = 37.5 \, \text{N} \times 2.00 \, \text{m} = 75.0 \, \text{J}$, which is 7.50×10$^{1}$ J.

If a Piston Exerts an External Pressure of 75