Question 29: Part A:What is the downward force exerted by the atmosphere on a fo

Question

Question 29:

Part A:What is the downward force exerted by the atmosphere on a football field, whose dimensions are 380 ft by 160 ft ?

Part B: An inflated basketball has a gauge pressure of 9.1 lb/in2 ... What is the actual pressure inside the ball?

Part C: he water tank in the figure(Figure 1) is open to the atmosphere and has two holes in it, one 0.80 mand one 3.6 m above the floor on which the tank rests.. If the two streams of water strike the floor in the same place, what is the depth of water in the tank?

Explanation / Answer

A) 380ft = 115.824 m

and 160ft = 48.768 m

then area = 115.824 x 48.768 = 5648.504832 m^2.

atmospheric pressure = 100000Pa.

force = pressure x area = 564850483.2 N

B) Actual pressure = atmospheric pressure + gauge pressure = 14.7 psi + 9.1psi = 23.8 psi.

Now convert to Pa where 14.7psi = 1.1013x10^5 Pa

So. 23.8psi*1.013x10^5Pa/14.7psi = 1.64 x 10^5 Pa

C) First, determine how long it takes the water to reach the floor.

d = .5 * a * t^2
t^2 = 2 d / a
t = sqrt(2d/a)

t1 = sqrt(2 * .8m / 9.81m/s^2)
t1 = .40s
t2 = sqrt(2 * 3.6m / 9.81m/s^2)
t2 = .86s

Now we can determine the ratio of horizontal speeds at exit. If the spot on the floor is "d" from the tank then the velocities are:

v1 = d / .40s
v2 = d / .86s
v1:v2 = 2.15

Now by Bernoulli, if the tank is open to the atmosphere, then v = sqrt(2gh)
v^2 = 2gh
h = v^2/2g or
h = k * v^2
depth1 = k (d/.40)^2
depth1 = 6.25 kd
depth2 = k (d/.86)^2
depth2 = 1.35 kd

depth1 / depth2 = 4.63
depth1 = 4.63 * depth2

Also, because of the heights, they are separated by 3.6 - .8m or 2.8m.
depth1 - depth2 = 2.8m

This gives 2 equations with 2 unknowns. Combine to solve:
depth1 = 2.8m + depth2
(4.63 * depth2) = 2.8m + depth2
3.63 * depth2 = 2.8m
depth2 = 0.77m
depth1 = 3.57m

Depth of water = depth1 + .8m
Depth of water = 3.57m + .8m
Depth of water = 4.37m

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