Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of 3.00 x 10^5 Pa?

Explanation:

Given:

gauge pressure desired,

As we know that the layers of the liquid column exert pressure due to their weight which is given as:

where:

pressure due to liquid column

density of the liquid for water

acceleration due to gravity

height of the liquid column

Similarly:

Iknow the first part it is the average star i am not 100% sure

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j