Question

В заливе Кара-Богаз-Гол на глубине 30 м давление воды составляет 360 кПа. Определите по этим данным плотность воды в заливе. Помогите решить пожалуйста :с

Answer 1

P=p*g*h
p=P/g*h=360 000/10*30=1200 кг/м3

Answer 2

Calculation of Water Density in Kara-Bogaz-Gol Bay

To determine the density of water in Kara-Bogaz-Gol Bay, we can use the given information about the water pressure at a depth of 30 meters. The pressure of the water can be related to its density using the hydrostatic pressure equation.

The hydrostatic pressure equation is given by:

P = ρgh

Where: - P is the pressure of the water (in pascals or Pa) - ρ is the density of the water (in kilograms per cubic meter or kg/m³) - g is the acceleration due to gravity (approximately 9.8 m/s²) - h is the depth of the water (in meters)

In this case, the pressure of the water is given as 360 kPa (kiloPascals) at a depth of 30 meters. We need to convert this pressure to pascals before using it in the equation.

1 kPa = 1000 Pa

Converting the pressure to pascals:

360 kPa * 1000 = 360,000 Pa

Now, we can rearrange the hydrostatic pressure equation to solve for the density of the water:

ρ = P / (gh)

Substituting the given values:

ρ = 360,000 Pa / (9.8 m/s² * 30 m)

Calculating the density:

ρ ≈ 1234.69 kg/m³

Therefore, the approximate density of water in Kara-Bogaz-Gol Bay, based on the given information, is 1234.69 kg/m³.

Please note that this calculation assumes that the temperature and salinity of the water remain constant.