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FinnZ [79.3K]
3 years ago
13

What is the energy conversion in a wind turbine?

Physics
1 answer:
Elena L [17]3 years ago
5 0

Answer:

The answer is option C.

Wind energy to kinetic energy to electric energy

Hope this helps

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Which type of electromagnetic waves has highest frequency​
vesna_86 [32]

Answer:

Gamma rays have the highest energies.

Explanation:

HOPE IT WILL HELP ^_^

6 0
3 years ago
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Estimate the force required to bind the two protons in the He nucleus together. (Hint: Model the protons as point charges. Assum
scoray [572]

Answer:

 F = 2.30 10⁴ N

Explanation:

The force required to link two gates must be equal to or greater than the electrostatic force of repulsion, because the protons have equal charges.

                 F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C²

   In this case the proton charge is 1.6 10⁻¹⁹ C and the distance between them is approximately the diameter of the core r = 10⁻¹⁵ m

Let's calculate

              F = 8.99 10⁹ (1.6 10⁻¹⁹)² / (10⁻¹⁵)²

              F = 2.30 10⁴ N

The bond strength must be equal to or greater than this value

3 0
3 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
Give me four reasons pluto is a cool planet / dwarf planet
AleksandrR [38]

Answer:

1. Just because it is small doesn't mean it needs to be excluded. If that were the case I would've been out of my friend group a while ago

2. Look at it it's fricking beautiful (see attachment)

3. It is just a great planet I don't think there needs to be any reason given I meannn you agree?

4. There's no other reason needed it's fricking gorgeous and amazing it needs no other reason. It needs a freaking

opening announcement to announce the arrival of the gorgeouness.

5 0
3 years ago
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Which of the following types of electromagnetic radiation has the shortest frequency? (2 points) Infrared waves Microwaves Ultra
Irina18 [472]

In order from shortest to longest frequency, the electromagnetic waves can be ranked :

1. X-rays

2. Ultraviolet

3. Infrared

4. Microwave

X-rays has the shortest frequency which ranges between 30 petahertz to 30 exahertz. X-rays have the smallest wavelengths at 0.01-10 nanometers. Its energy ranges between 100eV to 100keV.

5 0
3 years ago
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