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lozanna [386]
2 years ago
11

This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e

lectrical energy is designed to yield 10 kW of electrical power in a 20 mi/h wind. [The density of air is 1.2 kg/m3.]
What length rotor blades does this wind turbine have?
Physics
1 answer:
Volgvan2 years ago
4 0

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

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Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

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We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        \frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}

        \frac{1}{f_2} = 0.06

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the expression for the power of the lenses is

          P = \frac{1}{f}

where the focal length is in meters

           

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A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to
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The frictional force between the block and the horizontal surface is determined by applying Newton's law;

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Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

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Integrating the velocity equation, we will see that the position equation is:

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To get the position equation we just need to integrate the above equation:

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Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

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Now we remember that $u=\cos (\omega t)$

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