A fireman's water hose has a diameter of about 4.0 inches. The water is flowing through it at about 14.0 m/s. The nozzle's diameter drops to about 0.80 inches. If there is no friction, what would the exit velocity of the water be?
2 answers
I got it 350 m/s
area proportional to diameter^2
V2 A2 = V1A1
V2 / V1 = A1/A2 = 4^2/.8^2 = 16 / .64
V2 = 14 * 16/.64 = 350, agree
V2 A2 = V1A1
V2 / V1 = A1/A2 = 4^2/.8^2 = 16 / .64
V2 = 14 * 16/.64 = 350, agree
1 answer