A piston of cross-sectional area 2.5 times 10 | vedclass

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(B) According to Pascal's Law, the pressure applied at one end of a hydraulic system is transmitted equally throughout the fluid.Therefore, the pressu

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$2.94$

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(B) According to Pascal's Law, the pressure applied at one end of a hydraulic system is transmitted equally throughout the fluid.Therefore, the pressure $P_1$ at the first piston equals the pressure $P_2$ at the second piston:$P_1 = P_2$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$Given:$F_1 = 250 	ext{ N}$$A_1 = 2.5 	imes 10^{-2} 	ext{ m}^2$$F_2 = m 	imes g = 3000 	ext{ kg} 	imes 9.8 	ext{ m/s}^2 = 29400 	ext{ N}$Now, substitute the values into the formula to find $A_2$:$A_2 = \frac{F_2 	imes A_1}{F_1}$$A_2 = \frac{29400 	imes 2.5 	imes 10^{-2}}{250}$$A_2 = \frac{29400 	imes 0.025}{250}$$A_2 = 117.6 	imes 0.025 = 2.94 	ext{ m}^2$Thus, the cross-sectional area of the other piston is $2.94 	ext{ m}^2$.

$A$ piston of cross-sectional area $2.5 \times 10^{-2} \text{ m}^2$ is used in a hydraulic lift to exert a force of $250 \text{ N}$ on water. The cross-sectional area of the other piston which supports a car of mass $3000 \text{ kg}$ is $(g = 9.8 \text{ m/s}^2)$ (in $\text{ m}^2$)

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