A rectangular water tank is 80 , m times 40 , | vedclass

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(D) $1$. Convert all units to meters and hours: Area of the tank base = $80 \, m 	imes 40 \, m = 3200 \, m^2$. Area of pipe opening = $40 \, cm^2 = 4

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$5/8$

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(D) $1$. Convert all units to meters and hours: Area of the tank base = $80 \, m 	imes 40 \, m = 3200 \, m^2$. Area of pipe opening = $40 \, cm^2 = 40 	imes 10^{-4} \, m^2 = 0.004 \, m^2$. Speed of water = $10 \, km/h = 10000 \, m/h$. Time = $0.5 \, h$. $2$. Volume of water entering the tank in $0.5 \, h$: Volume = $	ext{Area of pipe} 	imes 	ext{Speed} 	imes 	ext{Time} = 0.004 \, m^2 	imes 10000 \, m/h 	imes 0.5 \, h = 20 \, m^3$. $3$. Calculate the rise in water level $(h)$: Volume = $	ext{Base Area} 	imes h$ $20 \, m^3 = 3200 \, m^2 	imes h$ $h = 20 / 3200 \, m = 1 / 160 \, m$. $4$. Convert $h$ to $cm$: $h = (1 / 160) 	imes 100 \, cm = 100 / 160 \, cm = 5/8 \, cm$.

$A$ rectangular water tank is $80 \, m \times 40 \, m$. Water flows into it through a pipe with an opening area of $40 \, cm^2$ at a speed of $10 \, km/h$. By how much will the water level in the tank rise in half an hour (in $cm$)?

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