Consider light energy that is momentarily absorbed in glass and then re-emitted. Compared to the absorbed light, the frequency o
1 answer:
The frequency of the re-emitted light is identical to that of the absorbed light.
To find the answer, we need to know more about the frequency of light.
<h3> Why the re-emitted light has the same frequency?</h3>- The wavelength of the light that is momentarily absorbed in glass and then re-emitted is the same, which explains why the re-emitted light has the same frequency as the absorbed light and the frequency of the absorbed light is the same.
- An electromagnetic wave's energy is inversely related to its frequency.
- The relationship between the wave's wavelength and frequency depends on the speed of light:
, c is the speed of light.
- Despite not having mass, light still has energy, and that energy is conserved.
- As a result, in order for there to be energy conservation, the energy of the light that is received and reemitted must be equal.
Thus, we can conclude that, the re-emitted light's frequency matches the absorbed light's frequency.
Learn more about frequency here:
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Answer:
The funda mental frequency of the original tube is 182Hz.
Explanation:
See the attachment for the calculation steps.
In order to calculate the fundamental frequency of the original closed tube we need to find the length of the tube which is equal to the sum of the lengths of the two new tubes.
For closed tubes
f = nv/4L (n = 1, 3, 5,...n)
f = nv/2L (n = 1, 2, 3,...n)
The details of calculation can be found below in the attachment.
Answer:
Explanation:
Given that,
Mass of a crate is 22 kg
It moved up along the 15 degrees incline without tipping.
We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.
It means that the horizontal component of force is given by :
So, the horizontal component of force is 208.25 N.
Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
----
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Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
--- Coulomb's law
Substitute 2 for r1
The new force (F2) is
Substitute 4 for r2
Substitute
The new force is 1/4 of the previous force.