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A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0.54 m below the surface of the water. The water surface drops very slowly and its speed is approximately zero. Keep 2 decimal places in all answers. (a) Find the speed v (in m/s) at which water squirts out of the hole.

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cjmejiab

Speed can be found through the application of concepts related to potential energy and kinetic energy, for which you have

[tex]KE = PE[/tex]

[tex]\frac{1}{2}mv^2 = mgh[/tex]

Where,

m = mass

v = Velocity

g = Gravitational acceleration

h = Height

Re-arrange to find the velocity we have,

[tex]v^2 = 2gh[/tex]

[tex]v = \sqrt{2gh}[/tex]

[tex]v = \sqrt{2(9.8)(0.54)}[/tex]

[tex]v = 3.253m/s[/tex]

Therefore the speed at which water squirts out of the hole is .3253m/s

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