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crimeas [40]
2 years ago
9

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.
Physics
1 answer:
scoundrel [369]2 years ago
3 0

Answer:

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

P3=\alpha gh

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

P=\frac{F}{A} \\F=PA\\F=(244.3kPa)(2m^2)=488.6KN

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Answer:

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Explanation:

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Newton's 2nd law is a vector equation, so we can decompose the forces along perpendicular axis in order to convert it in two algebraic equations.

We can choose one axis as parallel to the horizontal surface (we call it x-axis, being the positive direction the one of  the movement of the blocks due to the horizontal force applied to the 6.0 kg block), and the other, perpendicular to it, so it is vertical (we call y-axis, being the upward direction the positive one).

Taking into account the forces acting  on both masses, we can write both equations as follows:

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If we choose to the mass of 6.0 kg, in the horizontal direction, there are two forces acting on it, in opposite directions: the  horizontal applied force of 45 N, and the tension in the string that join both masses.

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F = m₁*a = 9.0 kg* 3 m/s² = 27 N

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