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ANEK [815]
3 years ago
12

The weight of the North American P-51 Mustang airplane is 10,100 lb and its wing platform area is 233 ft2 . Calculate the wing l

oading (airplane weight per wing area) in both English engineering and SI units. Also, express the wing loading in terms of the nonconsistent (but commonly used) unit kgf/m2 (kilogram-force per square meter).
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer

given,

weight of airplane,W = 10,100 lb

Wing area,A = 233 ft²

now,

Wing Loading Calculation in English engineering

        = \dfrac{W}{A}

        = \dfrac{10,100}{233}

        = 43.35 lb/ft²

Wing Loading in SI unit

        = 43.35\times \dfrac{4.448\ N}{1\ lb}\times (\dfrac{1\ ft}{0.3048})^2

        = 2075.51\ N/m^2

Wing loading in kgf/m²

      = \dfrac{2075.51}{9.81}

      = 211.57 kgf/m²

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So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

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3 years ago
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now its distance from Bambi is given as

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t = \frac{d}{v}

t = \frac{350}{31.3}

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