I really need the answer to this question please

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

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

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

Franklin needs to ship a box with FedEx. In order to calculate his shipping costs, he needs to measure the mass of the package.

balandron [24]

Answer:

meter

Explanation:

please mark as brainlist answer

8 0

3 years ago

Read 2 more answers

Calculate the average power required to lift a 750 N object a vertical distance of 10 meters in 4.0 seconds.

Nina [5.8K]

Answer:

500

Explanation:

5 0

3 years ago

Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a

timurjin [86]

Answer:

Acceleration of the second particle at that moment is given as

$a_2 = 2.12 m/s^2$

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

$F = 0 = m_1a_1 + m_2a_2$

now we have

$m_1 = 0.200 kg$

$m_2 = 0.250 kg$

$a_1 = 2.65 m/s^2$

now we have

$0.200(2.65) + 0.250a_2 = 0$

so we have

$a_2 = 2.12 m/s^2$

8 0

3 years ago

The wheels of an automobile are locked as it slides to a stop from an initial speed of 30.0 m/s. If the coefficient of kinetic f

Amiraneli [1.4K]

Answer:

x = 76.5 m

Explanation:

Let's use Newton's second law at the point of contact between the wheel and the floor.

fr = m a

fr = miy N

N-W = 0

N = W

μ mg = m a

a = miu g

a = 0.600 9.8

a = 5.88 m / s²

Having the acceleration we can use the kinematic relationships to find the distance

$v_{f}$ ² = v₀² + 2 a x

$v_{f}$ = 0

x = -v₀² / 2 a

Acceleration opposes the movement by which negative

x = - 30²/2 (-5.88)

x = 76.5 m

8 0

3 years ago