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Lesechka [4]
3 years ago
9

air passing over an airplanes wing travels ,and therefore exerts pressure.than air traveling beneath the wing.

Physics
1 answer:
pickupchik [31]3 years ago
5 0
Bernoulli's principle of laminar/lamellar air flow, I think. High flow speed = low pressure, low flow speed = high pressure I think. So, the wings/aerofoils are designed to induce a low pressure on the top side of the wing and a high pressure on the underside of the wing, thus producing an "aerodynamic upthrust" (a static upthrust comes from an object in water via Archimedes) and LIFT. 

Two "particles" of air one going topside and the other underside meet again at the end of their motion across the wing. So, top side has to travel faster than bottom side. So top side has a lower "dynamic pressure" than underside.

And all that for 5 points ????????? (If I'm right, of course ... )
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Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

A(t)=A_oe^{\frac{-bt}{2m}}

or amplitude for unforce oscillation

\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}

or

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Now, provided in the question Amplitude of the driven oscillation

A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}

the value of the maximum amplitude is obtained (k=m\omega_d^2)

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A=\frac{F_{max}}{(b\omega_d}

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b=3b

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A=\frac{1}{(3}0.075

or

A = 0.025 m = 25 mm

6 0
3 years ago
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