Ask question

Ask question

All categories

4 years ago

8

The electric field of a sinusoidal electromagnetic wave obeysthe equation

1 answer:

Firdavs [7]4 years ago

5 0

Answer with Explanation:

We are given that

$E=-(375V/m)sin(5.97\times 10^{15}(rad/s)t+(1.99\times 1067(rad/m)x)$

a.General equation of electric field wave

$E=E_0sin(\omega t+kx)$

Where $E_0$ =Amplitude of electric field wave

By comparing

$\omega=5.97\times 10^{15}rad/s$

$k=1.99\times 10^7rad/m$

a.Amplitude of electric field wave= $E_0=375V/m$

b.Amplitude of magnetic field wave, $B_0=\frac{E_0}{c}$

Where $c=3\times 10^8 m/s$

Amplitude of magnetic field wave= $B_0=\frac{375}{3\times 10^8}=125\times 10^{-8} T$

c.Frequency of wave, $f=\frac{\omega}{2\pi}=\frac{5.97\times 10^{15}}{2\pi}=0.95\times 10^{15}Hz$

d.Wavelength, $\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{1.99\times 10^7}=3.16\times 10^{-7} m$

e.Period of wave, $T=\frac{1}{f}=\frac{1}{0.95\times 10^{15}}=1.05\times 10^{-15} s$

f.Speed of wave, $v=f\lambda=0.95\times 10^{15}\times 3.16\times 10^{-7}=3.00\times 10^8 m/s$

You might be interested in

Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!

krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

$L = 10 Log\frac{I}{I_0}$

$20 = 10 Log{I}{10^{-12}}$

$I_1 = 10^{-10}W/m^2$

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

$L = 10 Log\frac{I}{I_0}$

$50 = 10Log\frac{I}{10^{-12}}$

$I_2 = 10^{-7}W/m^2$

now we know that

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}$

$r_2 = 47 cm$

so now the distance from friend must be 47 cm

8 0

3 years ago

I think these two are simple questions.. but I need help asap..... TT

frosja888 [35]

1) The average velocity is 56 m/min

2) The average velocity is -83 m/min

Explanation:

1)

The average velocity of an object is given by

$v=\frac{d}{t}$

where

d is the displacement (change in position)

t is the time elapsed

In the graph in the problem, the displacement corresponds to the distance, therefore to the change in the y-variable ( $\Delta y$ ), while the time elapsed is the change in the x-variable ( $\Delta x$ ), so the average velocity can be written as

$v=\frac{\Delta y}{\Delta x}$

At point A, we have:

$y_A = 5 m\\x_A = 0.1 min$

At point B, we have:

$y_E = 55 m\\x_A = 1 min$

So, we have

$\Delta y= 55 -5 = 50 m\\\Delta x = 1.0-0.1 = 0.9 min$

So the average velocity is

$v=\frac{50 m}{0.9 min}=56 m/min$

2)

In this part instead, we have the following:

At point F, we have:

$y_F = 55 m\\x_A = 1.3 min$

At point H, we have:

$y_H = 30 m\\x_A = 1.6 min$

So, we have

$\Delta y= 30 -55 = -25 m\\\Delta x = 1.6-1.3 = 0.3 min$

So the average velocity is

$v=\frac{-25 m}{0.3 min}=-83 m/min$

Learn more about velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

4 0

3 years ago

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.201 times the speed of light and shines a powerful las

bagirrra123 [75]

Answer:

The wavelength of observed light on earth is 568.5 nm

Explanation:

Given that,

Velocity of spaceship $v= 0.201c$

Wavelength of laser $\lambda= 697\ nm$

We need to calculate the wavelength of observed light on earth

Using formula of wavelength

$\lambda_{0}=\lambda_{e}\times\sqrt{\dfrac{1-\dfrac{v}{c}}{1+\dfrac{v}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-\dfrac{0.201 c}{c}}{1+\dfrac{0.201c}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-0.201}{1+0.201}}$

$\lambda=5.685\times10^{-7}\ m$

$\lambda=568.5\times10^{-9}\ m$

$\lambda=568.5\ nm$

Hence, The wavelength of observed light on earth is 568.5 nm

8 0

4 years ago

Sasha is interviewing people for a study he is doing. Before he starts the interview, Sasha explains the purpose of the study an

Alex73 [517]

D.

Sasha is informing participants about the study and making sure they wish to participate.

5 0

3 years ago

a hunter on level ground fires a bullet perfectly level with the ground. at the same time the bullet leaves the gun, he drops an

padilas [110]

Mark brainliest and I will help you

5 0

3 years ago