Water enters a leaky cylindrical tank (D = 1 ft) at a rate of 8 ft3/min. Water leaks out of the tank at a rate of 17% of the flow into the tank. At what rate will water rise in the tank (answer in ft/min)?

Water would rise in the tank at a rate of 8.45 ft/min

Explanation:

Diameter of leaky cylindrical tank (D) = 1 ft

Base area = πD^2/4 = 3.142×1^2/4 = 0.7855 ft^2

Volumetric flow rate at which water enters the tank = 8 ft^3/min

Volumetric flow rate at which water leaks out = 0.17 × 8 = 1.36 ft^3/min

Volumetric flow rate at which water rises = 8 - 1.36 = 6.64 ft^3/min

Rate at which water would rise in ft/min = volumetric flow rate at which water rises ÷ base area of the cylinder = 6.64 ft^3/min ÷ 0.7855 ft^2 = 8.45 ft/min

answer:

a

explanation: