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vodka [1.7K]
2 years ago
15

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

on) of 17.5 m and a chord (the length parallel to the flow direction) of 3 m. The airplane is flying at standard sea level with a velocity of 200 m/s. If the flow is considered to be completely laminar, calculate the boundary layer thickness at the trailing edge and the total skin friction drag. Assume that the wing is approximated by a flat plate. Assume incompressible flow.
Physics
1 answer:
Sidana [21]2 years ago
7 0

Solution :

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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A hollow metal sphere has 7 cm and 11 cm inner and outer radii, respectively. The surface charge density on the inside surface i
Alexus [3.1K]

Answer:

1)  E = 0 , 2) zero, 3)    E = - 2,162 10⁴ N/C , 4) directed towards the center of the sphere , 5) E = 1.412 10⁴ N / C , 6)direction coming out of the sphere

Explanation:

The electric field is a vector quantity, therefore we can calculate the field due to each charge distribution and add vector  

          E = E₁ + E₂

To calculate each field we can use Gauss's law, which states that the flow is equal to the charge  by the Gaussian surface divided by ε₀

For this case, let's take a sphere as a Gaussian surface

          Ф = ∫E dA = q_{int} /ε₀

 The area of ​​a sphere is

        A = 4π r²

         E = 1 / 4πε₀   q_{int} / r²

1) r = 4 cm

This radius is smaller than the radius of the sphere, therefore the charge inside is zero and therefore the field is zero

            E = 0

2) there is no field

3) r = 8 cm

Let's calculate each field, for the inner surface

This radius is larger than the internal radius, so the field is

           σ = q_{int} / A

The area of ​​the sphere is

          V = 4 π R_in²

         Rho = q_{int} / 4π R_in²

          q_{int} = ρ 4π R_in²

         E₁ = 1 /ε₀ ρ r_in² / r²

For the outer surface

This radius is smaller so there is no load inside the Gaussian surface and therefore the field is zero

          E₂ = 0

Total E

         E = E₁ + 0

          E = 1 /ε₀  ρ₁ R_in² / r²

Let's calculate

           E = 1 /8,854 10⁻¹²   250 10⁻⁹ (7/8)²

           E = - 2,162 10⁴ N / C

4) as the electric field is negative, it is directed towards the center of the sphere

5) r = 12 cm

In this case the two surfaces contribute to the electric field,

Inner surface

        Q₁ = ρ₁ 4π R_in²

        E₁ = 1 /ε₀  ₁rho1 R_in² / r²

Outer surface

         Q₂ = ρ₂ 4π R_out²

          E₂ = 1 /ε₀  ρ₂ R_out² / r²

The total field is

          E = E₁ + E₂

          E = 1 /ε₀  | ρ | [- R_in² + R_out²] / r²

Let's calculate

          E = 1 /8,854 10⁻¹² 250 10⁻⁹ [- 7² + 11²] / 12²

          E = 1.412 10⁴ N / C

6) As the field is positive, it is directed radially with direction coming out of the sphere

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3 years ago
The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field
frosja888 [35]

Answer:

N = 3032  turns

Explanation:

The magnetic field produced by a solenoid is described by

        B = μ₀ n I

Where is the permittivity in a vacuum with a value of 4π 10⁻⁷ N /A²,  n  is the turn density and I the current

Let's apply this equation to the problem, the turn density is the number of turns per unit length, in this case it is the same magnet length

       L = 8 cm = 0.08 m

Let's calculate

      B = μ₀ N/L   I

      N = B L / μ₀ I

      N = 0.10 0.08 / (4π 10⁻⁷  2.1)

      N = 3,032 103 turns

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Is being social media famous worth real-life relationships?
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Answer:

No

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Social media fame isnt worth it, because youre not really able to connect with people. Real life relationships with friends, family, and everything else is way more important. having people notice you or be attracted to you through the internet is nothing compared to real life social interactions. Dont leave people just to be online, youre going to regret it.

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Write<br>any<br>to increase<br>magnitude of current in dynamo​
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Answer:

Explanation:

It can be increased by: increasing the rate of rotation. Increasing the strength of the magnetic field. Increasing the number of turns on the coil.

Hope this helps

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3 years ago
Which of the following is true about wedges?
timofeeve [1]

Answer:

wedges are a type of inclined plane.

Explanation:

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