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Brilliant_brown [7]
3 years ago
11

A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w

hich can be treated as a series resistor in the circuit.Part A: What is the resonant angular frequency?Part B: If the source emf amplitude is Emax = 69 V what is the current amplitude at the resonant angular frequency?
Physics
1 answer:
son4ous [18]3 years ago
7 0

Answer:

5773.50269 Hz

23 A

Explanation:

L = Inductance = 6 mH

C = Capacitance = 5 μF

R = Resistance = 3 Ω

\epsilon = Maximum emf = 69 V

Resonant angular frequency is given by

\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz

The resonant angular frequency is 5773.50269 Hz

Current is given by

I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A

The current amplitude at the resonant angular frequency is 23 A

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Answer:

The current would stop

Explanation:

Electric currents are interesting because they carry little to no momentum. As soon as you remove a power source, the whole current halts.

3 0
3 years ago
A computational model predicts the maximum potential energy a roller coaster car can have given its mass and its speed at the lo
andreyandreev [35.5K]

Answer:

The prediction for its maximum potential energy is 109,375 J

Explanation:

Given;

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speed of the coaster car at the lowest point, v = 25 m/s

The coaster car will have maximum kinetic energy at the lowest point and based on law of conservation of mechanical energy, the maximum kinetic energy of the coaster car at the lowest point will be equal to maximum potential energy at the highest point.

K.E_{max} = P.E_{max}

K.E_{max} = \frac{1}{2} mv^2\\\\K.E_{max} = \frac{1}{2} (350)(25)^2\\\\K.E_{max} =109,375 \ J

Thus, P.E_{max} = 109,375 \ J

Therefore, the prediction for its maximum potential energy is 109,375 J

3 0
2 years ago
A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r
Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

Explanation:

Given that,

Angular acceleration \alpha=\pi\ rad/s^2

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Angular velocity \omega =2\pi\ rad/s

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}

\omega_{0}^2=\omega^2-2\alpha\theta

Where, \alpha = angular acceleration

\omega = angular velocity

Put the value into the formula

\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi

\omega=\sqrt{2\pi^2}

\omega_{0}=\pi\sqrt{2}\ rad/s

Hence, The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

3 0
3 years ago
3. Waxing means "growing." Waning means
Sonja [21]

Answer:

I think it is the last one.

Explanation:

I am not sure because i am stuck on this one, too.

4 0
3 years ago
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Damm [24]

The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.

C = Q/ΔV

C is the capacitance

Q is the stored charge

ΔV is the potential difference

Rearrange the equation:

ΔV = Q/C

We also know the capacitance of a parallel-plate capacitor is given by:

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C is the capacitance

κ is the capacitor's dielectric constant

ε₀ is the electric constant

A is the area of the plates

d is the plate separation

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ΔV = Qd/(κε₀A)

We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.

5 0
3 years ago
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