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lozanna [386]
3 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:
Volgvan3 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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3 years ago
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot
Rzqust [24]

Answer:

595391.482946 m/s

3.21875\times 10^{6}

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = 1.6\times 10^{-19}\ C

t = Time taken = 1 second

m = Mass of proton = 1.67\times 10^{-27}\ kg

Velocity of proton is given by

v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s

The speed of the proton is 595391.482946 m/s

Current is given by

I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C

Number of protons is

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons

The number of protons is 3.21875\times 10^{6}

3 0
3 years ago
Light is shone on a diffraction grating
Pani-rosa [81]

Answer:

    λ = 482.05 nm

Explanation:

The diffraction phenomenon and the diffraction grating is described by the expression

         d sin θ = m λ

where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction

in this case they indicate the distance between slits, the angle and the order of diffraction

         λ = \frac{d sin \theta }{m}d sin θ / m

let's calculate

         λ = 1.00 10⁻⁶ sin 74.6 / 2

         λ = 4.82048 10⁻⁷ m

Let's reduce to nm

         λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)

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3 years ago
Difference between on pitch and frequency​
shepuryov [24]

Answer:

 A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. I hope I got it correct !!

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3 years ago
Please help Math Phys
Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land.  Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

6 0
3 years ago
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