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irga5000 [103]
2 years ago
15

What were some of the challenges to safety resulting from such radical airframe designs as highly swept wings, high wing loading

, increasing speed, and long-duration flights are high altitude
Engineering
1 answer:
Andrew [12]2 years ago
8 0
How to take off and land, stopping considerations (stopping distance), control system capability over a large speed range and flutter, and structural integrity for the wing platform and speed range.
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 d. The company uses role-based access control and her user account hasn't been migrated into the correct group(s) yet

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Since Deidre is accessing her e-mail there appears to be nothing wrong with her account or password. Since her role is new, most likely the problem is associated with her new role.

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how are the construction steps for bisecting an angle similar to the steps you would have taken to construct and bisect a line s
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does the question entail anything else?

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Which of the following is a valid reason why a scientist might reject a scientific theory
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C I think hope it helps :)
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3 years ago
A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio
natima [27]

Answer:

1) A=282.6 mm

2)a_{max}=60.35\ m/s^2

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

V=\omega \sqrt{A^2-x^2}

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

3.5=\omega \sqrt{A^2-0.15^2}    ------3

2.5=\omega \sqrt{A^2-0.225^2}   --------4

Now by dividing equation 3 by 4

\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}

1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

3.5=\omega \sqrt{A^2-0.15^2}

3.5=\omega \sqrt{0.2826^2-0.15^2}

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

a_{max}=\omega ^2A

a_{max}=14.61 ^2\times 0.2826\ m/s^2

a_{max}=60.35\ m/s^2

Time period T

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{14.609}

T=0.42 sec

8 0
4 years ago
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