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Sagot :
Answer:
0.022 m²
Explanation:
By Pascal's law, we have the following relationship
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]Where F1 is the effort force, A1 is the effort piston area, F2 is the resistance force and A2 is the resistance piston area.
Since the resistance force is the weight of a patient, we get:
F2 = 82kg (9.8 m/s²) = 803.6 N
Now, we can replace F1 = 14.4 N, F2 = 803.6N, and A2 = 1.2 m²
[tex]\frac{14.4N}{A_1}=\frac{803.6N}{1.2m^2}[/tex]Solving for A1, we get:
[tex]\begin{gathered} 14.4(1.2)=803.6(A_1) \\ 17.28=803.6A_1 \\ \frac{17.28}{803.6}=A_1 \\ 0.022=A_1 \end{gathered}[/tex]Therefore, the area of the effort piston area should be 0.022 m²
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