Water in a canal 30 dm wide and 12 dm deep is flowing with a velocity of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is required for irrigation?
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Actually this doesn't require use of calculus at all. It is an algebra question.
Solution
Velocity of water in the canal = 1 0 k m / h = 1 0 0 , 0 0 0 d m / h
Hence, Volume of water pumped in half an hour = 2 3 0 ∗ 1 2 ∗ 1 0 0 , 0 0 0 d m 3 = 1 8 , 0 0 0 , 0 0 0 d m 3
Since height of the standing water is given as 8 c m or 0 . 8 d m and V o l u m e = a r e a ∗ h e i g h t :
A r e a = h e i g h t V o l u m e = 0 . 8 d m 1 8 , 0 0 0 , 0 0 0 d m 3 = 2 2 , 5 0 0 , 0 0 0 d m 2 = 2 2 5 , 0 0 0 m 2
can any 1 solve it in details.
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Cross section 3.6 m^2 flowing 10,000 m/hour. That's 36,000 cubic m of water in one hour. In half an hour, that will be 18,000. Divide by 0.08 meters to get 225000. It would have been nice to specify which unit to use, especially since they jump like 3 times but assume meters.