A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3 kPa, determine the speed of the aircraft.

The speed of the aircraft = 81.2310 m/s

Explanation:

Using Bernoulli equation: (p1/p) + (V1^2/2) + gh1 = (p2/p) + (V2^2/2) + gh2

Since the relationship highlights the relationship between the pressure in the fluid and the velocity of the aircraft at a given height.

Where g = acceleration due to gravity

            p1, p2 = the pressure at point 1 and the pressure at point 2 respectively

            p= the density of the fluid

        The density of air at the height of 3000 m = 0.9093 kg/m^3

            V1, V2 = velocity at point 1 and velocity at point 2 res[ectively

             h1, h2 = height at point 1, height at point 2 respectively

height is constant, which means h1 = h2

The velocity at point 2 is not stated which makes it zero

So from the Bernoulli equation above, we now have:

(p1/ p) + (V1^2/2) = p2/p

Making V1 the subject

V1 = √(2(p2 - p1)) /(p)

 V1 = √(2(3000) / (0.9093)

 V1 = √(2 x 3,299.24117)

V1 = √6,598.4823

V1 =  81.2310 m/s

answer: its c lololo

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge