You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m/s^2. The pressure at the surface of the water will be 150 kPa, and the depth of the water will be 13.6 m. The pressure of the air outside the tank, which is elevated above the ground, will be 93.0 kPa.

A) Find the net downward force on the tank's flat bottom, of area 2.15 m^2, exerted by the water and air inside the tank and the air outside the tank.

Question

Kayla Padilla

Physics

13 October, 08:51

You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m/s^2. The pressure at the surface of the water will be 150 kPa, and the depth of the water will be 13.6 m. The pressure of the air outside the tank, which is elevated above the ground, will be 93.0 kPa.

A) Find the net downward force on the tank's flat bottom, of area 2.15 m^2, exerted by the water and air inside the tank and the air outside the tank.

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Answers (1)

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Elsie · Answer 1

630.93 kN of force.

Explanation:

Pressure inside the tank is 150 kPa

The acceleration due to gravity on Mars g is 3.71 m/s^2.

The depth of water h is 13.6 m.

Pressure due to air outside tank is 93 kPa

The density of water p is 1000 kg/m^3

Pressure of the water on the tank bottom will be equal to pgh

Pressure of water = pgh

= 1000 x 3.71 x 13.6 = 50456 Pa

= 50.456 kPa.

Total pressure at the bottom of the tank will be pressure within tank and pressure due to water and pressure outside tank.

Pt = (150 + 50.456 + 93) = 293.456 kPa

Force at the bottom of the tank will be pressure times area of tank bottom.

F = Pt x A

F = 293.456 x 2.15 m^2 = 630.93 kN