a. Show that v = √2gh, 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 dh A(h). =-aa√/2gh, dt (34) 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 a 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 a for water is about 0.6. c. Consider a water tank in the form of a right circular cylinder that is 3m high above the outlet. The radius of the tank is 1m, 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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(a) The fundamental kinematic equation of motion is used to create the equation v = √(2gh), where v is the speed at which an object is falling to the ground, h is the height from which it is falling, and g is the acceleration caused by gravity.
When an object is dropped from a height h, gravity will cause it to fall freely, and the equation states that the velocity v will grow linearly with time t:
v = gt
The distance traveled during this time is given by the following equation:
h = gt^2 / 2
Rearranging the equation and solving for t, we get:
t = √(2h/g)
Substituting this value of t back into the first equation, we get:
v = gt = g √(2h/g) = √(2gh)
Since g is the acceleration brought on by gravity and h is the height from which an object is falling, the final expression for the velocity at which it is falling to the ground is v = √(2gh).
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