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vazorg [7]
3 years ago
5

What were the key factors that made the wright brothers’ wind tunnel a useful tool for wing design?

Physics
2 answers:
Anna71 [15]3 years ago
7 0

Answer:

Balancing equipment.

Explanation:

The wind tunnel made by Wright's brother was made up of a wooden box with glasses on it to see the inner instruments work.

The key factor that made the Wright's brother wind tunnel a useful tool for wind design is the balancing equipment used in it.  It was designed to measure lift and drag. Old hacksaw blades and bicycle spokes were used to make the balances which measure the forces of lift and drag action on a wing.

IrinaK [193]3 years ago
5 0
The Wright brothers built a wind tunnel<span> and developed model-testing </span>techniques<span> including a </span>balance to more accurately determine the lift and drag of their aircraft. I hope my answer has come to your help. God bless and have a nice day ahead!
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Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

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a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
2 years ago
"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3
avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

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We know that magnetic field due to long wire is given by B=\frac{\mu _0i}{2\pi r}

It is given that magnetic field is zero at some distance so

\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}

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So \frac{3}{x}=\frac{8}{24-x}=6.5454\ m

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Benjamina started her walk from the front door of her ground floor apartment. She walked 6 meters to the corner of the building
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The distance is the total distance she walked which is 16 meters adding the 6 meters to the corner and 10 meters to her friend's apartment. Her displacement is the distance from her original starting point so you set up a triangle with side lengths of 6 and 10 and solve for the hypotenuse which gives you a displacement of 11.66 meters.  
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A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti
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Answer:

v_f = 3.45 m/s

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now we have

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so we will have

10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)

2(9.81) sin10 \times 3.5 = v_f^2

so final speed is given as

v_f = 3.45 m/s

6 0
3 years ago
3. A skater moves across the ice a distance of 12 m before a constant frictional force of 15 N cause him to stop. His initial sp
RSB [31]

Explanation:

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