Question

A thin stream of water flows smoothly from a faucet and falls straight down. At one point, the water is flowing at a speed of 1=1.31 m/s . At a lower point, the diameter of the stream has decreased by a factor of 0.805 . What is the vertical distance ℎ between these two points?

Answer 1

Answer:

Explanation:

as the mass cannot change, the volume per unit time is also constant so the area of the stream times the velocity is constant

If the diameter is 1 where velocity is 1.31 m/s

area is proportional to the square of the diameter    A = (π/4)d²

1²(1.31) = 0.805²(v)

v = 1.31/0.805² = 2.02 m/s

v² = u² + 2as

2.02² = 1.31² + 2(9.81)h

h = 0.120819... m

h = 12.0 cm