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 100 kpa , and the depth of the water will be 14.1 m . the pressure of the air outside the tank, which is elevated above the ground, will be 92.0 kpa . part apart complete find the net downward force on the tank's flat bottom, of area 2.00 m^2 , exerted by the water and air inside the tank and the air outside the tank. assume that the density of water is 1.00 g/cm^3.

488.6KN

Explanation:

Hello!

the first step to solve this problem we must find the pressure exerted at the bottom of the tank (P) which is the sum of the external air pressure (P1 = 92kPa), the pressure inside the tank (P2 = 100kPa) and the pressure due to the weight of the water (P3), taking into account the above we have the following equation

P=P1+P2+P3

to find the pressure at the bottom of the tank due to the weight of the water we use the following equation

where

α=density=1  g/cm^3=1000kg/M^3

H=height=14.1m

g=gravity=3.71m/s^2

solving

P3=(1000)(14.1)(3.71)=52311Pa=52.3kPa

P=P1+P2+P3

P=100kPa+92kPa+52.3kPa=244.3kPa

finally to solve the problem we remember that the pressure is the force exerted on the area

would be false hope this out

the answer to this is proletariats.

hope this : )