The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3×
/ 1.72×
or approximately 1.74×
m.
The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:
f = c / λ
where, f = frequency of light
c = speed of light
λ = wavelength of light
Given data = f = 1.72×
Hz
Therefore, λ = 3×
/ 1.72×![10^{15}](https://tex.z-dn.net/?f=10%5E%7B15%7D)
λ = 1.74×
m
The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3×
/ 1.72×
or approximately 1.74×
m.
Learn more about light here;
brainly.com/question/15200315
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What term do you mean? like what he did to the dog is he stopped the dog
Answer:
The magnitude of the electric field at a point equidistant from the lines is ![4.08\times10^{5}\ N/C](https://tex.z-dn.net/?f=4.08%5Ctimes10%5E%7B5%7D%5C%20N%2FC)
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
![\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})](https://tex.z-dn.net/?f=%5Ctheta%3D%5Csin%5E%7B-1%7D%28%5Cdfrac%7B%5Cdfrac%7Bd%7D%7B2%7D%7D%7B2d%7D%29)
![\theta=\sin^{-1}(\dfrac{1}{4})](https://tex.z-dn.net/?f=%5Ctheta%3D%5Csin%5E%7B-1%7D%28%5Cdfrac%7B1%7D%7B4%7D%29)
![\theta=14.47^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D14.47%5E%7B%5Ccirc%7D)
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
![E=\dfrac{2k\lambda}{r}\times2\cos\theat](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B2k%5Clambda%7D%7Br%7D%5Ctimes2%5Ccos%5Ctheat)
Put the value into the formula
![E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B2%5Ctimes9%5Ctimes10%5E%7B9%7D%5Ctimes24.00%5Ctimes2%5Ctimes10%5E%7B-6%7D%5Ccos14.47%7D%7B2.05%7D)
![E=408094.00\ N/C](https://tex.z-dn.net/?f=E%3D408094.00%5C%20N%2FC)
![E=4.08\times10^{5}\ N/C](https://tex.z-dn.net/?f=E%3D4.08%5Ctimes10%5E%7B5%7D%5C%20N%2FC)
Hence, The magnitude of the electric field at a point equidistant from the lines is ![4.08\times10^{5}\ N/C](https://tex.z-dn.net/?f=4.08%5Ctimes10%5E%7B5%7D%5C%20N%2FC)
<span>Ohm's law deals with the relation between
voltage and current in an ideal conductor. It states that: Potential difference
across a conductor is proportional to the current that pass through it. It is
expressed as V=IR.
V = IR
200 = 20R
R = 10 ohms</span>
False, because the ions of the solar wind collide with atoms of oxygen and nitrogen from the Earth's atmosphere. The energy released during these collisions causes an aurora.
hope this helps!