A thin stream of water flows smoothly from a faucet and falls straight down. At

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 v 1.79 m/s. At a lower point, the diameter of the stream has decreased by a factor of 0.913. What is the vertical distance h between these two points? Number cm Previous Give Up & View Solution Check Answer e Next Exit Hint You can think of the thinning of the stream as an effect of mass conservation. If no water is getting trapped anywhere in the stream, then the volume flow rate of the water must be the same at all points in the stream Volume flow rate (VFR) is simply the product of the flow velocity v and the cross-sectional area A of the stream: VFR=vA= constant This relation is usually referred to as the Continuity Equation. Since the velocity of the stream increases due to ravity, the cross-sectional area must decrease in order for the volume flow rate to be contant. How exactly does e stream's velocity change as it falls under the influence of gravity?

Explanation / Answer

the continuity equation is,

A1 v1 = A2 v2

A1 v1 = [0.913]2 A1 v2

hence, the speed v2 is,

v2 = v1/ [0.913]2 = 1.2 v1

from the bernoulli equation, we have

1/2 *density* v1^2 + density*gh = 1/2 density*v2^2

0.5*1.79^2 + 9.81*h = 0.5*(1.2*1.79)^2

hence, the vertical distance is,

h = 0.07186 m

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