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Sagot :
Assuming that the pressure inside the house is at atmospheric pressure, the force on the window is approximately 1378 Newtons.
Given the data in the question;
- Area of the window; [tex]A = 2.5m^2[/tex]
- Velocity of wind; [tex]V = 30m/s[/tex]
Force on the window; [tex]F = \ ?[/tex]
Force
Force is an influence that can cause the change of motion of a particle or object. Also, force normal to the area of a surface is equal to pressure multiply by the area of that surface.
[tex]F = P * A[/tex]
Where P is pressure and A is area of the surface.
Also, we know that;
[tex]Pressure \ P = \frac{D_{air}V^2}{2}[/tex]
Where [tex]D_{air}[/tex] is density of air, approximately( [tex]1.225 kg/m^3[/tex] ) and V is velocity.
Hence
[tex]F = P * A\\\\F = \frac{D_{air}V^2}{2} * A[/tex]
We substitute our values into the equation above
[tex]F = \frac{D_{air}V^2}{2} * A\\\\F = \frac{1.225kg/m^3*(30m/s)^2}{2} * 2.5m^2\\\\F = 551.25 kg/ms^2 * 2.5m^2\\\\F = 1378.13 kgm/s^2\\\\F = 1378N[/tex]
Therefore, Assuming that the pressure inside the house is at atmospheric pressure, the force on the window is approximately 1378 Newtons.
Learn more about force on the surface of an area: https://brainly.com/question/1130378
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