During a windstorm, a 25 m/s wind blows across the flat roof of a small home. Find the difference in pressure between the air inside the home and the air just above the roof, assuming the doors and windows of the house are closed. (The density of air is 1.28 kg/m³)• ( 400 Pa)


Given:

Speed of the air inside the home = v₁ = 25 m s⁻¹ 
Speed of the air outside the home = v₂ = 0 m s⁻¹
Density of air = ⍴ = 1.28 kg m⁻³

To Find:

Pressure difference = P₁ - P₂ = ? 
[where P₁ = pressure outside the home and P₂ = pressure inside the home]

Solution:


The pressure difference P₁ - P₂ can be calculated by the relation:

P₁ - P₂ = `\frac {1}{2}` ⍴ (v₂ - v₁) 

by putting values

P₁ - P₂ = `\frac {1}{2}` 1.28 kg m⁻² ((25 m s⁻¹)² - (0 m s⁻¹)²) 



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.