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3 years ago

8. What are the seven kinds of transverse waves found on the electromagnetic

Physics

1 answer:

Vinvika [58]3 years ago

4 0

Answer:

Radio Waves, Microwaves, Infrared Waves, Visible Light, Ultraviolet, X-rays and Gamma Rays.

If you need help with anything else in physics hit me up. :D

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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

Find the velocity. 10 points. Will give brainliest!

BARSIC [14]

Answer:

6.060606...

Explanation:

To figure out velocity, you divide the distance by the time it takes to travel that same distance, then you add your direction to it. So the distance would be 1000m and the time would be 2 minutes and 45 seconds and if you convert the minutes into fractions you would get 165 seconds than you would divide 1000m by 165 seconds and you would get 6.060606... seconds as her average velocity

4 0

3 years ago

I need help PLEASE HEEEEEEEEELPP

lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B (8 - 4) m/s / 1 s = 4 m / s^2

From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period

6 0

2 years ago

Sound waves produced by a source pass a point five times every second. Which of the following choices correctly describes the pe

Ugo [173]

The period is 1/5 second, and the frequency is 5 Hz.

Explanation:

Let's start by reviewing some definitions about waves:

- The period of a wave is the time taken for the wave to make one complete cycle. It is indicated with T and it is measured in seconds (s)

- The frequency of a wave is the number of complete cycles made by the wave in one second. It is indicated with f and it is measured in Hertz (Hz). It is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where f is the frequency and T is the period.

In this problem, we have a wave that passes a given point five time per second. This means that the number of oscillations per second is five, and so its frequency is:

$f=\frac{5 cycles}{1 sec}=5 Hz$