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Ne4ueva [31]
3 years ago
7

The drawing shows a side view of a swimming pool. The pressure at the surface of the water is atmospheric pressure. The pressure

at the bottom of the pool is greater because of the weight of water above it. As a storm approaches, the atmospheric pressure drops. What happens to the pressure at the bottom of the pool? A The pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases. B Nothing happens to the pressure at the bottom of the pool. C The pressure at the bottom of the pool increases. D The pressure at the bottom of the pool decreases, but not as much as the atmospheric pressure decreases.
Physics
2 answers:
Nikitich [7]3 years ago
7 0

Answer:

A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

Explanation:

Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is

    P =P_{atm} + ρ g y

Where P_{atm} is the atmospheric pressure, ρ  the water density, and 'y' the depth measured from the surface.

Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth

Now we can examine the claims

A) To true. State agreement or with the equation above

B) False. Pressure changes with atmospheric pressure

C) False. It's the opposite

D) False. They are directly proportional

miss Akunina [59]3 years ago
3 0

A The pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

<h3>Further explanation </h3>

Hydrostatic pressure is pressure caused by the weight of a liquid.

The weight of a liquid is affected by the force of gravity.

If a liquid is placed in a container, the higher the liquid content in the container, the heavier the liquid content is and the greater the liquid pressure at the bottom of the container.

The hydrostatic pressure of a liquid can be formulated:

\large{\boxed{\bold {P_h ~ = ~ \rho.g.h}}

Ph = hydrostatic pressure (N / m², Pa)

ρ = density of liquid (kg / m³)

g = acceleration due to gravity (m / s²)

h = height / depth of liquid surface (m)

If the container is open, then the atmospheric pressure (P₀) can be entered into the equation.

\large{\boxed {\bold {P_h ~ = ~ P_o ~ + ~ \rho.g.h}}

The magnitude of P₀ is usually equal to = 1.01.10⁵ Pascal (Pa = N / m²) = 1 atm

For example submarines  :

The deeper a submarine is, the greater the hydrostatic pressure it experiences, so that the hull / wall of the submarine is made thick to withstand that pressure.

As a storm approaches, the atmospheric pressure drops

Because the hydrostatic pressure is proportional to the atmospheric pressure, the pressure at the bottom of the pool will also be reduced

<h3>Learn more </h3>

Isaac Newton's investigations of gravity

brainly.com/question/1747622

Gravitational force

brainly.com/question/7955425

increase the gravitational force between two objects

brainly.com/question/2306824

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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
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1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

given information:

q_{1} =  3 nC = 3 x 10^{-9} C

q_{2} =  2 nC = 2 x 10^{-9} C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

potential energy of the charge, F

F = k \frac{q_{e}q}{r}

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k = constant (8.99 x 10^{9} Nm^{2} /C^{2})

electron charge, q_{e} = - 1.6 x 10^{-19} C

since it is measured at the midpoint,

r = \frac{0.5}{2}

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F = F_{1}+ F_{2}

  = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

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the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

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R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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