Question

The speed of air above the wings of an aeroplane is 120 m/s and below the wings it is 90 m/s. If density of air is 1.3 kg/m³ the pressure difference is .......... . (Neglect the thickness of wings.) (A) 156 Pa (B) 39 Pa (C) 4095 Pa (D) 6300 Pa

Answer 1

Answer:

option c is correct

Explanation:

∆p = 1/2 * density* ∆v^2

= 1/2 *1.3 *(120^2 -90^2) = 4095 Pa

Answer 2

The pressure difference is 4095 Pa.

Explanation:

The speed air above the wings of aeroplane, v = 120 m/s.

The speed air below the wings of aeroplane, V = 90 m/s.

The density of the air, d = 1.3 kg/m^3

To calculate the pressure difference use  Bernoulli's principle as,

$P_1 +\frac{1}{2}d v^2 = P_2 +\frac{1}{2}d V^2$

$\Delta P=\frac{1}{2}d( v^2-V^2)$

substitute the given values, we get

$\Delta P=\frac{1}{2}\times1.3kg/m^3[(120m/s)^2 -(90m/s)^2]$

$\Delta P=\frac{1}{2}\times1.3kg/m^3(6300)$

$\Delta P=4095Pa$

Thus, the pressure difference is 4095 Pa.

#Learn More: Bernoulli's principle.

https://brainly.in/question/1457838