Question

H=0,15 м (рбензин= 710кг/м3)(рмұнай= 800 кг/м3)(рсу= 1000 кг/м3)(рмай= 900 кг/м3)P P=?​

Answer 1

Ответ:

точно не помню но вроде формула такая p=pgh тоесть. g=9,8=10H/кг на земле

1. 710кг/м3×10×0,15=1065H

2.1200H

3. 1500H

4.1350H

Answer 2

It seems like you're providing some values for density (рбензин, рмұнай, рсу, рмай) and a height (H), and you're looking for pressure (P). To calculate the pressure at a certain height in a fluid column, you can use the hydrostatic pressure formula:

$P = \rho \cdot g \cdot H$

where:

• $P$ is the pressure at the specified height,
• $\rho$ is the density of the fluid at that height,
• $g$ is the acceleration due to gravity,
• $H$ is the height.

Given values:

• Height ($H$) = 0.15 m
• Density of benzene ($\rho_{\text{benzene}}$) = 710 kg/m³
• Density of petroleum ($\rho_{\text{petroleum}}$) = 800 kg/m³
• Density of water ($\rho_{\text{water}}$) = 1000 kg/m³
• Density of crude oil ($\rho_{\text{crude oil}}$) = 900 kg/m³

Assuming $g$ is approximately $9.81 \, \text{m/s}^2$, which is the acceleration due to gravity on the surface of the Earth, we can calculate the pressure for each fluid at the given height using the hydrostatic pressure formula:

1. For benzene: $P_{\text{benzene}} = \rho_{\text{benzene}} \times g \times H$ $P_{\text{benzene}} = 710 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.15 \, \text{m}$

2. For petroleum: $P_{\text{petroleum}} = \rho_{\text{petroleum}} \times g \times H$ $P_{\text{petroleum}} = 800 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.15 \, \text{m}$

3. For water: $P_{\text{water}} = \rho_{\text{water}} \times g \times H$ $P_{\text{water}} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.15 \, \text{m}$

4. For crude oil: $P_{\text{crude oil}} = \rho_{\text{crude oil}} \times g \times H$ $P_{\text{crude oil}} = 900 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.15 \, \text{m}$

Now, let's calculate these values.