To solve this problem it is necessary to take into account the kinematic equations of motion and the change that exists in the volume flow.
By definition the change in speed is given by
[tex]v_f^2-v_i^2 = 2ax[/tex]
Where,
x= distance
[tex]v_f =[/tex]final velocity
[tex]v_i =[/tex]initial velocity
a = acceleration
On the other hand we know that the flow of a fluid is given by
[tex]\dot{V} = Av[/tex]
Where,
A = Area
v = Velocity
PART A )
Applying this equation to the previously given values we have to
[tex]v_f^2-v_i^2 = 2ax[/tex]
[tex]v_f^2-0 = 2*(9.8)(15.6)[/tex]
[tex]v_f^2=305.76[/tex]
[tex]v_f = 17.48[/tex]
Therefore the velocity of the water leaving the hole is 17.48m/s
PART B )
In the case of the hole we take the area of a circle, therefore replacing in the flow equation we have to,
[tex]\dot{V} = \pi r^2 v[/tex]
[tex]r = \sqrt{\frac{\dot{V}}{\pi v}}[/tex]
[tex]r = \sqrt{\frac{3*10^{-3}*\frac{1}{60}}{\pi (17.48)}[/tex]
[tex]r = \sqrt{9.10*10^{-7}}[/tex]
[tex]r = 0.54*10^{-4}[/tex]
The diameter is 2 times the radius, then is [tex]1.91*10^{-3}[/tex]m or 1.91mm
Note: The rate flow was converted from minutes to seconds.
Answer:
The speed at which water leaves the hole is 17.48 m/s.
The diameter of hole is 0.0147 m.
Explanation:
Given data:
Height below the water level is, h = 15.6 m.
Rate of flow is, [tex]Q=3.00 \times 10^{-3} \;\rm m^{3}/min[/tex].
(a)
The speed at which water leaves the hole is,
[tex]v=\sqrt{2gh}[/tex]
Here, g is gravitational acceleration.
[tex]v=\sqrt{2 \times 9.8 \times 15.6}\\v =17.48 \;\rm m/s[/tex]
Thus, speed at which water leaves the hole is 17.48 m/s.
(b)
The rate of flow is,
[tex]Q= Av[/tex]
A is cross-section area of hole.
[tex]Q= (\pi r^{2})v\\3.00 \times 10^{-3}= (\pi r^{2}) \times 17.48\\r=7.39 \times 10^{-3} \;\;\rm m[/tex]
Then diameter is,
[tex]d=2r\\d=2 \times 7.39 \times 10^{-3} \;\;\rm m\\d=0.0147 \;\rm m[/tex]
Thus, diameter of hole is 0.0147 m.
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