Question

the pressure exerted by a person's lungs can be measured by having the person blow as hard as possible into one side of a manometer produces a 70-cm difference in the heights of the columns of water in the manometer arms, what is the lung pressure?

Answer 1

The lung pressure exerted by blowing into a manometer with a 70-cm difference in water column height is calculated as 6867 Pa using the density of water and the acceleration due to gravity.

Step-by-step explanation:

When a person blows into one side of a manometer and produces a 70-cm difference in the heights of the columns of water in the manometer arms, the pressure exerted by the lungs can be calculated using the height of the water column and the density of the fluid in the manometer. The pressure (P) exerted by the lungs is equal to the height difference (Δh) multiplied by the density (ρ) of the fluid (water in this case), and the acceleration due to gravity (g), which is P = ρ×g×Δh. Given that the density of water is 1000 kg/m³ and the acceleration due to gravity is approximately 9.81 m/s², the lung pressure can be calculated as follows:

P = 1000 kg/m³ × 9.81 m/s² × 0.70 m = 6867 Pa

This is the gauge pressure, which is the pressure relative to atmospheric pressure. If we wish to express this in other units such as mm Hg (torr) or in atmospheres, we can convert from pascals using the appropriate conversion factors.

Answer 2

`p=6860\ Pa`

Step-by-step explanation:

Given that:

• Pressure difference created in the water level of a manometer,
  `h=70\ cm`

We have:

density of water,
`\rho=1000\ kg.m^(-3)`

As we know that pressure in terms of liquid column is given as:

`p=\rho.g.h`

`p=1000* 9.8* 0.7`

`p=6860\ Pa` is the pressure created by blowing through the lungs.