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d1i1m1o1n [39]
3 years ago
12

The formula length x width x height is used to calculate the volume of a type of

Physics
1 answer:
hram777 [196]3 years ago
6 0

Answer:

D

Explanation:

egwfefeewfwvebefhbdhahwuw

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In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
Which of the following is true of our current knowledge of electrons?
frosja888 [35]
They have a negative charge and rotate around the nucleus
8 0
3 years ago
Intermolecular forces hold
blagie [28]
I believe Intermolecular forces hold, <span>molecules, ions, and atoms? But I would see if that doesn't sound familiar check it with a site or something?</span>
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3 years ago
Which of the following statements about the energy of a pendulum is true?
salantis [7]
I think the correct answer is C.
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3 years ago
A series RLC circuit with R = 15.0 ohms, C = 4.70 uF, and L = 25.0mH is connected to an AC voltage source with V(t) = 75.0 Vrms
Brilliant_brown [7]

(a) The rms current in the circuit is 2.58 A.

(b) The rms voltage of Vab is 38.7 V, Vbc is 158.83 V, Vcd is 222.93 V, Vbd is 64.11 V, and Vad is 75 V.

(c) The average rate at which energy is dissipated in each of the 3 circuit elements is 193.23 W.

<h3>Capacitive reactance of the circuit</h3>

The capacitive reactance of the circuit is calculated as follows;

X_c = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 550 \times 4.7 \times 10^{-6}} \\\\X_c = 61.56 \ ohms

<h3>Inductive reactance of the circuit</h3>

Xl = ωL

Xl = 2πfL

Xl = 2π x 550 x 25 x 10⁻³

Xl = 86.41 ohms

<h3>Impedance of the circuit</h3>

Z = \sqrt{R^2 + (X_L - X_C)^2} \\\\Z = \sqrt{15^2 + (86.41 -61.56)^2 } \\\\Z = 29.03 \ ohms

<h3>rms current in the circuits</h3>

I_{rms} = \frac{V_{rms}}{Z} \\\\I_{rms} = \frac{75}{29.03} \\\\I_{rms} = 2.58 \ A

<h3>rms voltage in resistor (Vab)</h3>

V_{ab} = I_{rms} R\\\\V_{ab} = 2.58 \times 15\\\\V_{ab} = 38.7 \ V

<h3>rms voltage in capacitor (Vbc)</h3>

V_{bc} = I_{rms} X_c\\\\V_{bc} = 2.58 \times 61.56\\\\V_{bc} = 158.83 \ V

<h3>rms voltage in inductor (Vcd)</h3>

V_{cd} = I_{rms} X_l\\\\V_{cd} = 2.58 \times 86.41\\\\V_{cd} = 222.93\ V

<h3>rms voltage in capacitor and inductor (Vbd)</h3>

V_{bd} = I_{rms} \times (X_l - X_c)\\\\V_{bd} = 2.58 \times (86.41 - 61.56)\\\\V_{bd} = 64.11 \ V

<h3>rms voltage in resistor, capacitor and inductor (Vad)</h3>

V_{ad} = I_{rms} \times \sqrt{R^2 + (X_l- X_c)^2} \\\\V_{ad} = 2.58 \times Z \\\\V_{ad} = 2.58 \times 29.03\\\\V_{ad} = 75 \ V

<h3>Average rate of energy dissipation in the 3 circuit element</h3>

P = I_{rms}^2 Z\\\\P = (2.58)^2 \times 29.03\\\\P = 193.23 \ W

Learn more about RLC series circuit here: brainly.com/question/15595203

6 0
2 years ago
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