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If 1500 lbm of air is pumped into a previously empty 900 ft^3 storage tank?

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Final answer:

The pressure exerted by 50.0 kg of gasoline on the bottom of a tank that measures 0.500 m by 0.900 m is calculated using the formula for pressure and is found to be approximately 1089 Pascals.

Step-by-step explanation:

To calculate the pressure exerted on the bottom of a gas tank due to the weight of the gasoline in it, we need to use the basic pressure formula Pressure (P) = Force (F) / Area (A). The force is the weight of the gasoline, which is the mass times the acceleration due to gravity (F = mg), and the area is the surface area of the bottom of the tank (A = width x length).

Given the mass of gasoline (m) is 50.0 kg and the size of the tank is 0.500 m x 0.900 m, the force exerted by the gasoline is F = mg = 50.0 kg x 9.8 m/s2 (since the acceleration due to gravity, g, is 9.8 m/s2). The bottom surface area of the tank (A) is 0.500 m x 0.900 m = 0.450 m2.

Now, calculating the pressure, P = F / A, we get:

  • F = 50.0 kg x 9.8 m/s2 = 490 N (Newtons)
  • A = 0.500 m x 0.900 m = 0.450 m2
  • P = 490 N / 0.450 m2 ≈ 1089 Pa (Pascals)

Therefore, the final answer in a two line explanation in 300 words: the pressure exerted by 50.0 kg of gasoline on the bottom of the tank is approximately 1089 Pa (Pascals).

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