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Vesnalui [34]
2 years ago
9

Consider a glider flying at 400 meters altitude, when suddenly all its static ports become blocked by volcanic ash. The pressure

at 400 meters altitude is 96,610 Pascals. The glider then descends to near sea-level and flies at a true airspeed of 25 m/sec.What is the airspeed indicated by the pitot-tube driven airspeed indicator in m/s?
Physics
1 answer:
Furkat [3]2 years ago
5 0

Answer:

the airspeed indicated by the pitot-tube driven airspeed indicator is 91.23m/s

Explanation:

Pitot tube

U = \sqrt{\frac{(p_t - p_s)2}{d} }

U = velocity(m/s)

p_t= stagnation pressure (pa)

p_s= static pressure (pa)

d = fluid density(kg/m³)

p_t = p_a_t_m + \frac{1}{2} dv^2

v = true velocity

= 101325 + 1/2(1.225)(25)²

p_t = 101,707.8125pa

p_s = 96,610pa

d = 1.225kg/m³

U = \sqrt{\frac{2(101,707.8125 - 96,610)}{1.225} } \\\\U = 91.23m/s

the airspeed indicated by the pitot-tube driven airspeed indicator is 91.23m/s

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If we reduce the distance in such a way that r₁₂f = r₁₂₀/3, replacing in (1), we get:

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The result is reasonable, as the gravitational force is inversely proportional to the square of the distance between masses, which is consequence of our 3-D space.

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If we apply the same considerations to the same mass on the surface of a planet with a mass equal to one-sixth the mass of Earth, and a radius one third that of Earth, we can apply the same equation as above:

Fgp = G*m*(me/6) / (re/3)² = m*ap

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As the mass is the same, we conclude that the gravitational force exerted by the unknown planet on the mass (which we call weight) is 3/2 times the one experimented on Earth's  surface.

So, the right answer is D. 3/2W.

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