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A. Calculate the (absolute) pressure at the bottom of a neighborhood swimming pool 30.0 m by 8.9 m whose uniform depth is 2.0 m . Remember that the pressure at the top of the pool is atmospheric pressure, which (in SI metric units) is 1.013×105 Pa.
B. Now find the (total) force exerted on the bottom of this swimming pool.
C. What is the pressure against the side of this pool (somewhere very near the bottom)?

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boffeemadrid

Answer:

120920 Pa

32285640 N

120920 Pa

Explanation:

[tex]\rho[/tex] = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth = 2 m

Length = 30 m

Breadth = 8.9 m

[tex]P_0[/tex] = Atmospheric pressure = [tex]1.013\times 10^5\ Pa[/tex]

Absolute pressure is given by

[tex]P=P_0+\rho gh\\\Rightarrow P=1.013\times 10^5+1000\times 9.81\times 2\\\Rightarrow P=120920\ Pa[/tex]

The absolute pressure is 120920 Pa

Force is given by

[tex]F=PA\\\Rightarrow F=120920\times 30\times 8.9\\\Rightarrow F=32285640\ N[/tex]

Force exerted on the bottom of this swimming pool is 32285640 N

Pressure is uniformly spread across the pool. So, pressure on the side of the pool is 120920 Pa

Eduard22sly

A. The absolute pressure at the bottom of the neighbourhood swimming pool is 120900 Pa

B. The total force exerted on the bottom of the swimming pool is 32280300 N

C. The pressure against the side of the pool somewhere very near the bottom is 120900 Pa

A. How to determine the absolute pressure

  • Height (h) = 2 m
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Atmospheric pressure (P₀) = 1.013×10⁵ Pa
  • Density of water (d) = 1000 Kg/m³
  • Absolute pressure (P) =?

The absolute pressure can be obtained as follow:

P = P₀ + dgh

P = 1.013×10⁵ + (1000 × 9.8 × 2)

P = 1.013×10⁵ + 19600

P = 120900 Pa

B. How to determine the force

  • Absolute pressure (P) = 120900 Pa = 120900 N/m²
  • Dimension = 30.0 m by 8.9 m
  • Area (A) = 30 × 8.9 = 267 m²
  • Force (F) =?

P = F / A

Cross multiply

F = P × A

F = 120900 × 267

F = 32280300 N

C. How to determine the pressure near the button

  • Absolute pressure (P) = 120900 Pa
  • Pressure near the button =?

The pressure in a liquid is uniformly distributed.

Thus, the pressure near the button will have the same value as that of the absolute pressure (i.e 120900 Pa)

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