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denpristay [2]
3 years ago
11

What is 3600Hz has in rpm?​

Physics
1 answer:
Setler [38]3 years ago
8 0

Answer:

Explanation:

N=Rotor Speed in Revolution per minute(rpm)

for P=4 and N=3600, f comes out to be 120 Hz.

So frequency of voltage produced is 120 Hz. But this is not practical. Generally 4-Pole generator has N=1500rpm(for 50 Hz) or 1800rpm for 60 Hz. Two pole generator can have N=3600rpm(f=60Hz).

The most practical situation is generator having N=3600Hz with 2 Poles.

Hope It will be helpful!!!

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The answer is b. The crest.

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Joanna wants to determine the speed of sound in xenon. When she plays a tone with a frequency of 440 Hz, the resulting sound wav
iVinArrow [24]

The speed of the sound in the xenon is 178 m/s. And the right option is b 178 m/s

<h3 /><h3>What is speed?</h3>

Speed can be defined as the ratio of the total distance traveled by a body to the total time taken.

To calculate the speed of the sound in the xenon, we use the formula below.

Formula:

  • v = λf............. Equation 1

Where:

  • v = Speed of the sound in xenon
  • f = Frequency
  • λ = Wavelength.

From the question,

Given:

  • f = 440 Hz
  • λ = 40.4 cm = 0.404 m

Substitute the values above into equation 1

  • v = 440(0.404)
  • v = 177.76 m/s.
  • v ≈ 178 m/s

Hence, The speed of the sound in the xenon is 178 m/s. And the right option is b 178 m/s

Learn more about speed here: brainly.com/question/4931057

7 0
2 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
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