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Sidana [21]
3 years ago
14

An aircraft increases its speed by 2% in straight and level flight. If the total lift remains constant determine the revised CL

as a percentage of its original value to three significant figure
Engineering
1 answer:
damaskus [11]3 years ago
7 0

Answer:

96.1%

Explanation:

We know that lift force

F_L=\dfrac{1}{2}C_L\rho AV^2

                                                                    ------------(1)

Where C_L is the lift force coefficient .

          ρ is the density of fluid.

         A is the area.

        V is the velocity.

Now when speed is increased by 2 % and all other parameter remains constant except C_L .

Let;s take new value of lift force coefficient is C_L' .

F_L=\dfrac{1}{2}C_L'\rho A(1.02V)^2

                                                                         -----------(2)

Now from equation 1 and 2

C_L\times V^2=C_L'\times1.0404 V^2

⇒C_L'=0.961C_L

So we can say that revised value of  lift force coefficient is 96.1% of original value.

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Answer:

IDK

Explanation:

same thing is happening to me

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Three communications channels in parallel have independent failure modes of 0.1 failure per hour. These components must share a
ozzi

Answer:

the MTTF of the transceiver is 50.17

Explanation:

Given the data in the question;

failure modes = 0.1 failure per hour

system reliability = 0.85

mission time =  5 hours

Now, we know that the reliability equation for this situation is;

R(t) = [ 1 - ( 1 - e^{-0.1t )³] e^{-t/MTTF

so we substitute

R(5) = [ 1 - ( 1 - e^{-0.1(5) )³] e^{-5/MTTF = 0.85

[ 1 - ( 1 - e^{-0.5 )³] e^{-5/MTTF = 0.85

[ 1 - ( 0.393469 )³] e^{-5/MTTF = 0.85

[ 1 - 0.06091 ] e^{-5/MTTF = 0.85

0.9391 e^{-5/MTTF = 0.85  

e^{-5/MTTF = 0.85 / 0.9391

e^{-5/MTTF = 0.90512

MTTF = 5 / -ln( 0.90512 )

MTTF = 50.17

Therefore, the MTTF of the transceiver is 50.17

7 0
3 years ago
4. Calculate the square root of potato using the sqrt() function. Print out the value of potato again to verify that the value o
N76 [4]

Answer:

potato<-100

print(potato)

sqrt(potato)

potato<-potato*2

print(potato)

Explanation:

The complete question is as follows

Create a variable called potato whose value corresponds to the number of potatoes you’ve eaten in the last week. Or something equally ridiculous. Print out the value of potato.

Calculate the square root of potato using the sqrt() function. Print out the value of potato again to verify that the value of potato hasn’t changed.

Reassign the value of potato to potato * 2.

Print out the new value of potato to verify that it has changed

The question was answered using R programming language.

At line 1, I assumed that I ate 100 potatoes in the previous week.

So, potato = 100

At line 2, the value of potato is printed as 100.

At line 3, the square root of potato is calculated using sqrt function: Square for of 100 = 10

At line 4,the initial value of potato is doubled and stored in potato variable. 100 * 2 = 200

At line 5, the new value of potato is printed: 200.

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