Question

Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The orifice is 52.5 cm above the base of the container. What is the speed of liquid coming out from the orifice?

Answer 1

Answer:50m/s^2

Explanation:

Use formula V=√2gh

Modify the equation as V^2 = 2g(h1-h2)/1-a^2/A^2

Put values in equation

V^2 = 2×10m/s^2×(3m-0.52m)/1-(0.1)^2

So we get V^2=50m^2/s^2