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4 years ago

In a series RCL circuit the generator is set to a frequency that is not the resonant frequency.

Physics

1 answer:

Pani-rosa [81]4 years ago

8 0

Answer: 624 Hz

Explanation:

If the ratio of the inductive reactance to the capacitive reactance, is 6.72, this means that it must be satified the following expression:

ωL / 1/ωC = 6.72

ω2 LC = 6.72 (1)

Now, at resonance, the inductive reactance and the capacitive reactance are equal each other in magnitude, as follows:

ωo L = 1/ωoC → ωo2 = 1/LC

So, as we know the resonance frequency, we can replace LC in (1) as follows:

ω2 / ωo2 = 6.72

Converting the angular frequencies to frequencies, we have:

4π2 f2 / 4π2 fo2 = 6.72

Simplifying and solving for f, we have:

f = 240 Hz . √6.72 = 624 Hz

As the circuit is inductive, f must be larger than the resonance frequency.

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All atoms of the element potassium have 19 protons. One of the most stable types of potassium atoms has the mass number 39. How

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The formular for determinatio of mass number is this:
Mass number = number of proton + number of neutron
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Mass number = 39
Proton number = 19
Number of proton = ?
Number of neutron = mass number - number of proton
Number of neutron= 39 - 19 = 20
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4 years ago

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Name two things radio waves have in common with visible light

kondaur [170]

Radio waves and visible light, as we perceive it, are both part of the EM spectrum. The only difference between them in terms of our perception is that we don't have receptors to 'see' radio waves. That's where radio receivers come into the picture*, to convert radio waves into sound impulses that we can hear.

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3 years ago

The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.

Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

The angular velocity of the telescope is approximately 0.199 rad/s

Second question:

The telescope should accelerates away by approximately F = 0.0005·m

Third question:

The pulling force between the Earth and the satellite

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

The angular velocity of the telescope, ω ≈ 0.199 rad/s

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$ , therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ 0.0005·m (away from the Sun)

Third part:

Other forces include;

The force of attraction between the Earth and the telescope which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817

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Answer:

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Explanation:

Brainlist Tho??

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I think the answer is .579 m

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