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Sagot :
Given:
Piston 1 has an area
[tex]A_1=0.065\text{ m}^2[/tex]and the force applied on this piston is
[tex]F_1=\text{ 400 N}[/tex]The force on piston 2, which is the weight of the car, is
[tex]F_2=6000\text{ N}[/tex]Required: The area of piston 2.
Explanation:
The formula that can be obtained by balancing the pressures is
[tex]\begin{gathered} P_1=P_2 \\ \frac{F_1}{A_1}=\frac{F_2}{A_2} \\ A_2=\frac{F_2A_1}{F_1} \end{gathered}[/tex]On substituting the values, the area of the second piston will be
[tex]\begin{gathered} A_2=\frac{6000\times0.065}{400} \\ =0.975\text{ m}^2 \end{gathered}[/tex]Final Answer: The area of the piston lifting the car is 0.975 meter-squared.
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