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rusak2 [61]
2 years ago
8

Two plane mirrors make the angle α = 46.0 ° α=46.0° between them. A ray of light incident on one of the mirrors is reflected and

it hits the second mirror. Find the angle between the ray incident on the first mirror and the ray reflected off of the second mirror.

Physics
1 answer:
user100 [1]2 years ago
6 0

Answer:

β = 88°

Explanation:

Given,

Angle between the mirror, α = 46.0 °

Let 'i' and 'r' be the angle made with mirror by incident and reflected ray respectively.

Now,

α + i + r = 180°

i + r = 180°-46°

i + r = 134°

we know, angle of incidence and angle of reflection are equal.

Let β be the required angle

β + (180°-2i) + (180°-2r) = 180°

β -2(i+r) = -180

β = -180° + 2 x 134°

β = 88°

Hence, the angle between them is equal to 88°.

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At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen
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  2. The power generation potential of the wind turbine at such place is of 2290 kW
  3. The actual electric power generation is 687 kW

Explanation:

  1. The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: \frac{1}{2} V^2 where V is the velocity of the air.
  2. The specific kinetic energy would be: \frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}.
  3. The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
  4. To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: \frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}
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  6. Then, the Power generation potential is: 40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW
  7. The actual electric power generation is calculated using the definition of efficiency:\eta=\frac{E_P}{E_I}}, where η is the efficiency, E_P is the energy actually produced and, E_I is the energy input. Then solving for the energy produced: E_P=\eta*E_I=0.30*2290kW=687kW
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