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elixir [45]
3 years ago
6

How is the q of an rlc parallel resonant circuit calculated?

Physics
1 answer:
notka56 [123]3 years ago
3 0

Answer:

It is calculated by dividing Resistance, R, by Inductive reactance, XL.

Explanation:

Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.

Q factor is the inverse in the resonance series circuit.

Q factor of a resonance parallel circuit,

<h3>Q = R/XL</h3>

R = Resistance

XL = Inductive reactance

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Answer:

D. your brain is processing things on the conscious and unconscious

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Consciousness is the awareness of ourselves and our environment.The two-track mind means that perception, memory, thinking, language, and attitude all operate on two levels conscious and unconscious level.

Glad to help!!

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You have a meteorite sample and you decide to use the uranium-235/lead-207 system to date it. After analysis, you find that it h
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Originally there must been

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Describe how the coriolis effect acts differently based on location on earth
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How does an ocean wave transfer energy across the ocean?
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The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be
Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
  • It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
  • Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
  • The time period of geosynchronous orbit is 24 hours or 1440 minutes.
  • As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
  • If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
  • a1= (T1/T2)⅔×a2

           = (1440/90)⅔×6780

           = 43,090 km

  • Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

#SPJ4

   

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