2 years ago

A successful marriage is based on:

Physics

2 answers:

kaheart [24]2 years ago

5 0

Answer:

i made another account to make u brainiest

Explanation:

melisa1 [442]2 years ago

4 0

Answer:

a and d

Explanation:

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A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi

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Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

$= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

= 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

$=\frac{1}{2} \times 1.225 \times 200^2$

$=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

= 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

= 270 N

Therefore the net drag = 270 x 2

= 540 N

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2 years ago