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3 years ago

Should aircraft wings have infinite stiffness?

2 answers:

5 0

Answer: 2.1.3 The evolution of aircraft wing structures-form follows function Torsional stiffness is fundamental to all aeroelastic approaches unity, 0 will..

HOPE IT HELPS GIVE ME BRAINLIST PLEASE :)

Colt1911 [192]3 years ago

3 0

Answer:

No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.

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Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

zvonat [6]

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

$\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}$

= 0.4167 = $1.0396e^{-0.4888*Fo}$

therefore Fo = 3.8264

Now determine how long the plate should be left in the furnace

Fo = $(\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )$

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

8 0

3 years ago

Answer my question I will mark brainliest

Iteru [2.4K]

Answer:

150

Explanation:

Mark me Brainliest

6 0

3 years ago

Read 2 more answers

Need help solving math problem using integration

notka56 [123]

Ummm did you try to add or subtract and multiply or divide that can get your answer

8 0

2 years ago

Much of the workd went to bed hungry

Marysya12 [62]

The workers went to bed hungry probably because they are hard workers and so didn’t want to eat because they didn’t want to take break┌(;￣◇￣)┘

7 0

3 years ago

Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin

Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$ ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4}$

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$ ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

$=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0

3 years ago