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aalyn [17]
3 years ago
7

A beam of light has a wavelength of 506 nanometers. What is the frequency of the light? What colors is the light?

Physics
1 answer:
Mila [183]3 years ago
8 0
<span>Beam Light wavelength = 506 nanometer
Frequency of light = F
The color of light = L

To fine the electromagnetic field:
=> S = wavelength x Frequency
Frequency = (3 x 10^8) / (5.06 x 10^ -7)
F = 5.96 x 10^14 Hz

According to the visible light spectrum, wavelength from 495 to 570 is equals to the color of green. Since the given wavelength is 506, thus, the color of light is green.

F = 5.96 x 10^14 Hz
L = green</span>



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A wave has a wavelength of 4. 9 m and a velocity of 9. 8 m/s. The medium through which this wave is traveling is then heated so
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The wavelength of a wave is obtained by taking the ratio of wave speed and frequency.

The wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

What is frequency of a wave?

The number of oscillations completed by a wave in one second is known as the frequency of a wave. It is expressed as the ratio of the velocity of the wave to its wavelength.

Given data-

The wavelength of the wave is, \lambda = 4.9 \;\rm m.

The velocity of the wave is, v = 9.8 m/s.

The mathematical expression for the frequency of the wave is,

f = \dfrac{v}{\lambda}

Solving as,

f = \dfrac{9.9}{4.9}\\\\f =2 \;\rm Hz

Now, with constant frequency and the double magnitude of velocity (v' = 2 × 9.8 = 19.6 m/s). The wavelength of the heated wave is calculated as,

f = \dfrac{v'}{\lambda'}

here,

\lambda' is the wavelength of the heated wave.

Solving as,

2 = \dfrac{19.6}{\lambda'}\\\\\lambda' = \dfrac{19.6}{2}\\\\\lambda' = 9.8 \;\rm m

Thus, we can conclude that the wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

Learn more about the frequency of wave here:

brainly.com/question/1324797

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If a plane is moving at a constant velocity what is happening to the acceleration?
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The plane is not accelerating.

Hope this helps!
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Erik Erikson’s theory of psychosocial development emphasizes that development occurs by overcoming an emotional crisis in each o
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2 years ago
A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi
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Answer:

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

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