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?

Answer 1

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.