A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the air is stationary. What is the DIFFERENCE IN PRESSURE between the inside and outside?
[?] Pa

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

The given data in the problem is;

dP is the change in the presure=?

Using Bernoulli's Theorem;

[tex]\rm \rho\frac{V^2_{12}}{2} +P_1= \rho \frac{V^2_{22}}{2} +P_2 \\\\\ P_2-P_1=\rho \frac{v_2^2-v_1^2}{2} \\\\ P_2-P_1= 1.21 \times \frac{17.4^2-0}{2} \\\\ \triangle p=369.36 \ N/m^2[/tex]

Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²

To learn more about the pressure refer to the link;

https://brainly.com/question/356585

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A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the air is stationary. What is the DIFFERENCE IN PRESSURE between the inside and outside? [?] Pa​

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A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the air is stationary. What is the DIFFERENCE IN PRESSURE between the inside and outside?
[?] Pa