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oksian1 [2.3K]
2 years ago
10

What are the wavelengths of electromagnetic wave in free space that have the following frequencies?.

Physics
2 answers:
irina [24]2 years ago
7 0

The wavelengths of the light are 4.3 * 10^-12 m and 0.2 m respectively.

<h3>What is wavelength?</h3>

The term wavelength has to do with the horizontal distance that is covered by a wave. We know that a long wavelength implies that the wave is able to travel a long distance from one point to another.

Given that;

c = λf

c = speed of light

λ = wavelength of ight

f = frequency of light

Thus;

λ = 3 * 10^8/ 7.00 x 10^19

λ = 4.3 * 10^-12 m

λ = 3 * 10^8/1.50 x 10^9

λ = 2 * 10^-1 or 0.2 m

Learn more about wavelength:brainly.com/question/13533093

#SPJ1

Missing parts:

What are the wavelengths of electromagnetic wave in free space that have the following frequencies? (a) 7.00 x 10^19 Hz______ pm (b) 1.50 x 10^9 Hz__________ cm

natulia [17]2 years ago
4 0

~400 nm to ~700 nm are the wavelengths of electromagnetic wave in free space.

What are the wavelengths of electromagnetic wave in free space?

Electromagnetic waves are categorized on the basis of their frequency f, which is equivalent to wavelength λ = c/f. Visible light has a wavelength that ranges from ~400 nm to ~700 nm. Violet light has a wavelength of ~400 nm, and having a frequency of ~7.5*1014 Hz.

So we can conclude that: ~400 nm to ~700 nm are the wavelengths of electromagnetic wave in free space.

Learn more about electromagnetic wave here :brainly.com/question/25847009

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(c) A moving train has a kinetic energy of 8.1 x 10(power of 6)J.
tamaranim1 [39]

the friction force provided by the brakes is 30000 N.

<h3>What is friction force?</h3>

Friction force is the force that opposes the motion between two bodies in contact.

To calculate the average friction force provided by the brakes, we apply the formula below.

Formula:

  • K.E = F'd............. Equation 1

Where:

  • K.E = Kinetic energy of the train
  • F' = Friction force provided by the brakes
  • d = distance

Make F' the subject of the equation

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From the question,

Given:

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Substitute these values into equation 2

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Hence, the friction force provided by the brakes is 30000 N

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