A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from the bottom of the tank. the water flows out of the small hole into an empty, 0.2 m-diameter container placed below the tank. the student uses a garden hose to keep the water level in the tank at the original height. 1) h1 = 0.50 meters and h2 = 0.03 meters
a. what is the velocity of the water as it leaves the hole near the bottom of the tank? (4 pts)

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aristocles aristocles · Answer 1 · 2018-08-18T02:14:35+02:00

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

[tex]Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0[/tex]

Now by equation of continuity

[tex]A_1v_1 = A_2v_2[/tex]

[tex]\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2[/tex]

from above equation we can say that speed at the top layer is almost negligible.

[tex]v_1 = 0[/tex]

now again by equation of continuity

[tex]\rho g h = \frac{1}{2} \rho v^2[/tex]

[tex] v = \sqrt{2 g h}[/tex]

here we have

[tex] h = h_1 - h_2[/tex]

[tex]h = 0.50 - 0.03 = 0.47m[/tex]

now speed is given by

[tex] v = \sqrt{2* 9.8 * 0.47}[/tex]

[tex]v = 3.03 m/s[/tex]