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2 years ago

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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 obey

ing Bernoulli's equation.

1 answer:

Sidana [21]2 years ago

7 0

This is an incomplete question, here is a complete question.

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, $1.11\times 10^3Pa$

Step-by-step explanation :

As we are given:

Speed = 150 km/h = 41.66 m/s

Density = $\rho=1.28kg/m^3$

Area = A = 7.00 m × 12.0 m

Formula used :

$\Delta P=\frac{1}{2}\times \rho \times v^2$

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

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

$\Delta P=1.11\times 10^3Pa$

Thus, the pressure difference will be, $1.11\times 10^3Pa$

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