The answer would be ?

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

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago

To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,

bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1 = 800 Hz

harmonic frequency f2 = 1120 Hz

harmonic frequency f3 = 1440 Hz

solution

first we get here fundamental frequency that is express as

2F = f2 - f1 ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength = Speed ÷ Fundamental frequency ................2

here speed of waves in air = 340 m/s

so put here value

Wavelength =340 ÷ 160

Wavelength = 2.125 m

so

Length of the pipe will be

Length of the pipe = 0.25 × wavelength ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

7 0

2 years ago

Phosphorus trichloride is used in the production of chemicals that keep plants from growing in a specific area. This compound is

Sever21 [200]

The Answer Is P should come first.

5 0

2 years ago

Read 2 more answers

A cheetah runs 5 miles in 60 seconds. How fast did the cheetah run in MILES per HOUR? Convert seconds into hours first, then sol

Vinil7 [7]

Answer:

300 miles per hour

Explanation:

Speed is distance per unit time, expressed as s=d/t where t is the time taken, d is distance covered and s is the speed.

Convering s to hrs

To convert seconds to hours, we knkw that 1 hour has 60 minutes and each minute has 60 seconds. Therefore, 1 hour has 60*60=3600 seconds

If 3600s=1 h

60 s=?

By cross multiplication 60s*1 h/3600s=1/60 hours

Given distance as 5 miles and time as 1/60 hours then the speed will be 5 divided by 1/60 hrs which is equivalent to 5*60=300 miles per hour

6 0

2 years ago