Suppose that a tank containing a certain liquid has an outlet near the bottom. Let be the height of the liquid surface above the outlet at time t. Torricelli's principle states that the outflow velocity v at the outlet is equal to the velocity of a particle falling freely (with no drag) from the height h.(adsbygoogle = window.adsbygoogle || []).push({});

(a)

Show that v=[tex]\sqrt{2gh}[/tex], where g is the acceleration due to gravity.

(b)

By equating the rate of outflow to the rate of change of liquid in the tank, show that h(t) satisfies the equation

A(h)(dh/dt)=-α*a*[tex]\sqrt{2gh}[/tex]

where A(h) is the area of the cross section of the tank at height h and a is the area of the outlet. The constant α is a contraction coefficient that accounts for the observed fact that the cross section of the (smooth) outflow stream is smaller than a. The value of α for water is about 0.6.

(c)

Consider a water tank in the form of a right circular cylinder that is 3 m high above the outlet. The radius of the tank is 1 m and the radius of the circular outlet is 0.1 m. If the tank is initially full of water, determine how long it takes to drain the tank down to the level of the outlet.

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I know that for part (a) you can use conservation of energy, mgh=.5mv^2 or bernoulli's principle.

part (b) i just get lost. I'm pretty bad at related rates.

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# Torricelli's principle

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