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goblinko [34]
3 years ago
8

Consider a rectangular wing mounted in a low-speed subsonic wind tunnel. The wing model completely spans the test-section, so th

at the flow "sees" essentially an infinite wing. The wing has a NACA 23012 airfoil section and a chord of 0.23 m, where the lift on the entire wing is measured as 200 N by the wind tunnel force balance. Determine the angle of attack if the wing span, airflow pressure, temperature, and velocity are 2 m, 1 atm, 303 K, and 42 m/s, respectively. Refer to the Appendix graphs given below for standard values. (Round the final answer to the nearest whole number.)
Physics
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

Answer for the question :

"Consider a rectangular wing mounted in a low-speed subsonic wind tunnel. The wing model completely spans the test-section, so that the flow "sees" essentially an infinite wing. The wing has a NACA 23012 airfoil section and a chord of 0.23 m, where the lift on the entire wing is measured as 200 N by the wind tunnel force balance. Determine the angle of attack if the wing span, airflow pressure, temperature, and velocity are 2 m, 1 atm, 303 K, and 42 m/s, respectively. Refer to the Appendix graphs given below for standard values" is in the attachment.

Explanation:

Download pdf
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A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e
ddd [48]

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J

The energy extracted by the turbine every second is 2649600 Joules

8 0
3 years ago
A single-turn circular loop of wire of radius 65 mm lies in a plane perpendicular to a spatially uniform magnetic field. During
BabaBlast [244]

Answer:

Explanation:

change in flux = no of turns x area of loop x change in magnetic field

= 1 x π 65² x 10⁻⁶ x ( 650 - 350 ) x 10⁻³

= 3.9 x 10⁻³ weber .

rate of change of flux  = change of flux / time

= 3.9 x 10⁻³ / .10

= 39 x 10⁻³ V

= 39 mV .

Since the magnetic flux is directed outside page and it is increasing , induced current will be clockwise so that magnetic field is produced in opposite direction to reduce it , as per Lenz's law.

5 0
3 years ago
The Earth and Moon are separated by about 4.0 10^6 m. Suppose Mars is [{MathJax fullWidth='false' 2.9 \times 10^{11} }] m from E
Shalnov [3]

Answer:

a) α=7.9x10^-4 rad

b) θ=1.12x10^-4 rad

c) The Earth and the Moon cannot be seen without a telescope.

Explanation:

In this exercise we will use the concepts of angular resolution, which depends on both the wavelength of the rays and the diameter of the eye or lens on the meter. Its unit of measure is the radian. The attached image shows the solution step by step.

4 0
3 years ago
Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers
Dmitrij [34]

Answer:

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

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True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then th
nlexa [21]
I think that this is false but I am not sure
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