Question

На какой глубине H находился пузырек воздуха если в процессе всплытия на поверхность воды его радиус успел увеличится в 2 раза атмосферное давление 100кПа плотность воды 1000кг/м в 3 температуру воды считать постоянной

Answer 1

PV=C. Следовательно, P'=2P0 => h=10м, т. к. Ro*g*h=10^5

Answer 2

Calculation of the Depth of the Air Bubble

To calculate the depth at which the air bubble was located during its ascent to the water surface, we can use the ideal gas law and the principle of Pascal's law.

According to Pascal's law, the pressure in a fluid (in this case, water) is transmitted equally in all directions. Therefore, the pressure at the depth of the air bubble is equal to the sum of the atmospheric pressure and the pressure due to the column of water above it.

Let's denote the depth of the air bubble as H and the radius of the bubble as r. We are given that the radius of the bubble increased by a factor of 2 during its ascent, and the atmospheric pressure is 100 kPa.

Using the ideal gas law, we can relate the pressure inside the bubble to its radius. The ideal gas law states that the pressure of a gas is inversely proportional to its volume, assuming constant temperature. Since the bubble is in equilibrium with the surrounding water, we can assume that the temperature remains constant.

Let's denote the initial pressure inside the bubble as P1 and the final pressure as P2. The initial radius of the bubble is r, and the final radius is 2r.

According to the ideal gas law, we have:

P1 * V1 = P2 * V2

Since the bubble is spherical, the volume of the bubble can be expressed as:

V = (4/3) * π * r^3

Substituting the values, we have:

P1 * (4/3) * π * r^3 = P2 * (4/3) * π * (2r)^3

Simplifying the equation, we get:

P1 * r^3 = P2 * 8r^3

Dividing both sides by r^3, we have:

P1 = 8P2

Since the pressure inside the bubble is equal to the sum of the atmospheric pressure and the pressure due to the column of water above it, we can write:

P1 = P_atm + ρ * g * H

where P_atm is the atmospheric pressure, ρ is the density of water, g is the acceleration due to gravity, and H is the depth of the air bubble.

Substituting the values, we have:

8P2 = P_atm + ρ * g * H

Since P2 is the atmospheric pressure, we can rewrite the equation as:

8P_atm = P_atm + ρ * g * H

Simplifying the equation, we get:

7P_atm = ρ * g * H

Solving for H, we have:

H = (7P_atm) / (ρ * g)

Now, let's substitute the given values:

- Atmospheric pressure (P_atm) = 100 kPa = 100,000 Pa - Density of water (ρ) = 1000 kg/m^3 - Acceleration due to gravity (g) = 9.8 m/s^2

Calculating the depth H:

H = (7 * 100,000) / (1000 * 9.8) = 7142.86 meters

Therefore, the air bubble was located at a depth of approximately 7142.86 meters during its ascent to the water surface.

Please note that this calculation assumes ideal conditions and neglects factors such as surface tension and the compressibility of the air bubble.