A constricted horizontal tube of radius r1 = 4.00cm tapers to a tube of radius r2 = 2.50cm. If water flows at a speed 3.50m/s in the larger tube, (A) find its speed in the smaller tube. (b) Find the water pressure difference ΔP=ΔP1-ΔP2 in kPa and in atm.
I have no clue.
The Attempt at a Solution
Succeeded in doing part A.
A1V1 = A2V2
V2 = (A1V1)/A2 = (pi(4)^2(3.50)) / (pi(2.50)^2) = 8.96 m/s