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Brilliant_brown [7]
2 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]2 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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Nastasia [14]

Answer:

Example A will appear green, while Example B will appear greenish-blue.

Explanation:

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In example B, green and blue are being reflected so the object will appear a mix of green and blue. This color is cyan (greenish blue).

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2 years ago
What type of wave passes through the spring in the frog toy? Why?
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3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

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lina2011 [118]

Answer:

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