Asked by Bassel
If a person blowing into one side of an open tube manometer produces 80cm defference btw the heights of the columns of water in the manometer what is the gauge pressure of the lungs
Answers
Answered by
Bosnian
ρ [ kg / m³ ] = density
g [ m / s² ] = acceleration of gravity
h [ m ] = height
p [ Pa ] = pressure
Static pressure = ρ g h
The difference between the pressure in the equilibrium state of the manometer and the pressure after blowing::
Δ p = p1 - p2 = ρ g h₁ - ρ g h₂ = ρ g ( h₁ - h₂ ) = ρ g Δh [ Pa ]
In this case:
ρ = 1000 kg / m³
g = 9.81 m / s²
Δh = 80 cm = 0.8 m
Δ p = ρ g Δh = 1000 ∙ 9.81 ∙ 0.8 = 7848 Pa
g [ m / s² ] = acceleration of gravity
h [ m ] = height
p [ Pa ] = pressure
Static pressure = ρ g h
The difference between the pressure in the equilibrium state of the manometer and the pressure after blowing::
Δ p = p1 - p2 = ρ g h₁ - ρ g h₂ = ρ g ( h₁ - h₂ ) = ρ g Δh [ Pa ]
In this case:
ρ = 1000 kg / m³
g = 9.81 m / s²
Δh = 80 cm = 0.8 m
Δ p = ρ g Δh = 1000 ∙ 9.81 ∙ 0.8 = 7848 Pa
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