A tank with a square base of area 1.0 ; m^{2} | vedclass

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(D) The base area of the tank is $A = 1.0 \; m^{2}$.The area of the hinged door is $a = 20 \; cm^{2} = 20 	imes 10^{-4} \; m^{2}$.The density of wate

Vedclass · Accepted Answer

$54.88$

Vedclass · Answer

(D) The base area of the tank is $A = 1.0 \; m^{2}$.The area of the hinged door is $a = 20 \; cm^{2} = 20 	imes 10^{-4} \; m^{2}$.The density of water is $ho_{1} = 10^{3} \; kg/m^{3}$.The density of acid is $ho_{2} = 1.7 	imes 10^{3} \; kg/m^{3}$.The height of both liquid columns is $h = 4.0 \; m$.The pressure exerted by the water column at the bottom is $P_{1} = h ho_{1} g = 4 	imes 10^{3} 	imes 9.8 = 3.92 	imes 10^{4} \; Pa$.The pressure exerted by the acid column at the bottom is $P_{2} = h ho_{2} g = 4 	imes 1.7 	imes 10^{3} 	imes 9.8 = 6.664 	imes 10^{4} \; Pa$.The pressure difference between the two columns is $\Delta P = P_{2} - P_{1} = (6.664 - 3.92) 	imes 10^{4} = 2.744 	imes 10^{4} \; Pa$.The force required to keep the door closed is $F = \Delta P 	imes a = 2.744 	imes 10^{4} 	imes 20 	imes 10^{-4} = 54.88 \; N$.

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