What is the pressure difference Δp=pinside−poutside? Use 1.28 kg/m3 for the density of air. Treat the air as an ideal fluid obeying Bernoulli's equation.

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A hurricane wind blows across a 7.00 m × 12.0 m flat roof at a speed of 150 km/h.

What is the pressure difference Δp = p(inside)-p(outside)? Use 1.28 kg/m³ for the density of air. Treat the air as an ideal fluid obeying Bernoulli's equation.

Answer : The pressure difference will be, [tex]1.11\times 10^3Pa[/tex]

Step-by-step explanation :

As we are given:

Speed = 150 km/h = 41.66 m/s

Density = [tex]\rho=1.28kg/m^3[/tex]

Area = A = 7.00 m × 12.0 m

Formula used :

[tex]\Delta P=\frac{1}{2}\times \rho \times v^2[/tex]

Now put all the given values in this formula, we get:

[tex]\Delta P=\frac{1}{2}\times (1.28kg/m^3)\times (41.66m/s)^2[/tex]

[tex]\Delta P=1.11\times 10^3Pa[/tex]

Thus, the pressure difference will be, [tex]1.11\times 10^3Pa[/tex]

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