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 v1=1.79 m/s. At a lower point, the diameter of the stream has decreased by a factor of 0.769. What is the vertical distance h between these two points?

Answers

GPT 3.5
To solve this problem, we can use the principle of conservation of mass and the equation of continuity for the flow of an incompressible fluid:

A1 * v1 = A2 * v2

Where:
A1 = cross-sectional area of the stream at the upper point
A2 = cross-sectional area of the stream at the lower point
v1 = velocity of the stream at the upper point (1.79 m/s)
v2 = velocity of the stream at the lower point
We are given that the diameter of the stream decreases by a factor of 0.769, which means that the cross-sectional area decreases by the square of this factor:

(A2/A1) = (0.769)^2
A2 = (0.769)^2 * A1

Since the cross-sectional area of the stream is proportional to the square of the diameter, we can write:

(A2/A1) = (d2/d1)^2
0.769 = d2/d1
d2 = 0.769 * d1

Now, we can express the cross-sectional areas in terms of the diameters:

A1 = π*(d1/2)^2
A2 = π*(d2/2)^2
A2 = π*(0.769 * d1/2)^2
A2 = π*0.769^2*(d1/2)^2
A2 = 0.591*(π*(d1/2)^2)
A2 = 0.591*A1

Substitute these values into the equation of continuity:

A1 * v1 = A2 * v2
A1 * 1.79 = 0.591*A1 * v2
v2 = 1.79/0.591
v2 = 3.03 m/s

The velocity of the stream at the lower point is 3.03 m/s.

Now, we can apply the principle of conservation of mechanical energy to determine the vertical distance h between these two points:

ΔE = ΔKE + ΔPE = 0
0.5*m*v1^2 + m*g*h1 = 0.5*m*v2^2 + m*g*h2
0.5*1*v1^2 + 0 = 0.5*1*v2^2 + 1*9.8*h
0.5*1*1.79^2 = 0.5*1*3.03^2 + 9.8*h
0.5*3.2041 = 0.5*9.1809 + 9.8*h
1.60205 = 4.59045 + 9.8*h
1.60205 - 4.59045 = 9.8*h
-2.9884 = 9.8*h
h = -2.9884/9.8
h = -0.305 m

Therefore, the vertical distance h between these two points is approximately 0.305 m.

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