Question

Помогите решить задачи по географии 1.Атмосферний тиск біля земної поверхні становить 770 мм рт. ст. Визначте яким буде тиск в шахті глибина якої 900м 2.Висота м. Кременчук над рівнем моря =190м Обчисліть нормальний атмосферний тиск нашого міста.

Answer 1

Ответ:

Объяснение:

При спусканні на глибину на кожні 10м тиск підвищується на 1мм.рт.ст.

900м : 10 м = 9 0мм.рт.ст. на стільки  підвищився тиск

770мм.рт.ст. + 90мм.рт.ст. = 860мм.рт.ст. таким буде тиск в шахті глибина якої 900м

2)

760 мм.рт.ст. - нормальний атмосферний тиск над рівнем моря

відмітка рівня моря - 0 м

100м – 1 мм.рт.ст.

190 м - х мм.рт.ст.

х = 190 : 100 = 1,9 мм.рт.ст. різниця в тискові

760 - 1,9 = 758,1 мм.рт.ст.нормальний атмосферний тиск  м.Кременчук

Answer 2

Task 1: Атмосферний тиск біля земної поверхні становить 770 мм рт. ст. Визначте яким буде тиск в шахті глибина якої 900м.

To determine the atmospheric pressure at a depth of 900m in a mine, we can use the concept of hydrostatic pressure. The hydrostatic pressure is the pressure exerted by a fluid at a given depth and is directly proportional to the density of the fluid and the depth.

The density of air decreases with increasing altitude or depth. At sea level, the density of air is approximately 1.225 kg/m³. We can assume that the density of air remains constant up to a certain height and then decreases linearly with increasing altitude or depth.

To calculate the atmospheric pressure at a depth of 900m, we need to consider the change in pressure due to the weight of the air column above the given depth.

The formula to calculate the pressure at a given depth is:

P = P₀ + ρgh

Where: - P is the pressure at the given depth - P₀ is the pressure at the surface (given as 770 mmHg) - ρ is the density of air - g is the acceleration due to gravity (approximately 9.8 m/s²) - h is the depth (given as 900m)

Let's calculate the atmospheric pressure at a depth of 900m using the given information.

First, we need to convert the pressure at the surface from mmHg to pascals (Pa). 1 mmHg is approximately equal to 133.322 Pa.

P₀ = 770 mmHg * 133.322 Pa/mmHg

Now, we can calculate the atmospheric pressure at a depth of 900m using the formula:

P = P₀ + ρgh

Please note that we need the density of air at the given depth to complete the calculation. Unfortunately, the density of air is not provided in the question. Therefore, we cannot provide an exact answer without this information.

However, we can provide a general approach to solving the problem. We can assume that the density of air decreases linearly with increasing depth. In this case, we can estimate the density of air at a depth of 900m using the density at the surface and the given height of Mount Kremenchuk.

Task 2: Висота м. Кременчук над рівнем моря = 190м. Обчисліть нормальний атмосферний тиск нашого міста.

To calculate the normal atmospheric pressure at a given location, we can use the International Standard Atmosphere (ISA) model. According to this model, the atmospheric pressure decreases with increasing altitude.

The normal atmospheric pressure at sea level is defined as 1013.25 hPa (hectopascals) or 1013.25 millibars (mb). As the altitude increases, the atmospheric pressure decreases at a rate of approximately 1 hPa per 8.43 meters.

Given that the height of Kremenchuk above sea level is 190m, we can calculate the normal atmospheric pressure at this location.

First, we need to determine the change in altitude from sea level to Kremenchuk:

Δh = 190m

Next, we can calculate the change in atmospheric pressure using the rate of decrease:

ΔP = Δh / 8.43m/hPa

Finally, we can calculate the normal atmospheric pressure at Kremenchuk by subtracting the change in pressure from the sea-level pressure:

P = 1013.25 hPa - ΔP

Please note that this calculation provides an estimate of the normal atmospheric pressure at Kremenchuk based on the given information. The actual atmospheric pressure may vary due to local weather conditions and other factors.