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anonymous
 5 years ago
Iraq is pumping crude oil into the Persian Gulf from an oil storage3 tank (a cylinder, V=πr^2h) that is 80 feet in diameter. The pump will deliver a constant flow of crude oil at 9 cubic feet per second.
A. How fast is the oil level in that tank falling?
B. Hollied Forces are planning an air strike to stop the flow of the oil. If the original oil level was 18 feet above the base and it will be 3 hours before the air strike can be conducted, will the air strike be necessary? Why or why not?
anonymous
 5 years ago
Iraq is pumping crude oil into the Persian Gulf from an oil storage3 tank (a cylinder, V=πr^2h) that is 80 feet in diameter. The pump will deliver a constant flow of crude oil at 9 cubic feet per second. A. How fast is the oil level in that tank falling? B. Hollied Forces are planning an air strike to stop the flow of the oil. If the original oil level was 18 feet above the base and it will be 3 hours before the air strike can be conducted, will the air strike be necessary? Why or why not?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A) Volume = 1600*pi*h, since diameter is 80 then r^2 = 1600 so V is a function of h, V(h) they give us the rate at which the volume is changing or dV/dt = 9 the rate we are looking for is dh/dt or the rate that height is changing to obtain this we will need another rate say dV/dh which is the derivative of our function V(h). dV/dh = 1600*pi dV/dt = (dV/dh)*(dh/dt) **Notice how the dh will cancel Now plug in what we know => 9 = 1600*pi *(dh/dt) solve for dh/dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0B) use the rate dh/dt from part A which is in feet per second and multiply by 3600*3 which is num of sec in 3 hours. This will tell us how much the oil level has dropped, if it is >18 then air strike is not necessary
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