The lasers are red, meaning that they are giving off light in the _____ region of the electromagnetic spectrum

2 answers:

8 0

The lasers are red, meaning that they are giving off light
in the longest wavelength part of the visible region of
the electromagnetic spectrum

Softa [21]4 years ago

4 0

As per the question the color of laser light is given as red.

If we arrange all the electromagnetic waves in the decreasing order of frequency ,then the electromagnetic spectrum contains gamma ray as the first which is followed by all other electromagnetic waves according to their frequency.

The visible light ranges from 400 nm to 700 nm which contains sunlight i.e white colors with it's constituent colors starting from violet to red. The red color is the longest wavelength part of the visible region.

The wavelength of visible light is longer than ultraviolet wave but smaller than infrared radiation. Except the bisible region,the color of radiation is invisible to eye.

As per the question the color of emiited laser radiation is red .Hence it must lie in the visible region of the electromagnetic spectrum.

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Read 2 more answers

Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a dista

luda_lava [24]

To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

$V = \frac{kq}{r}$

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

$V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}$

Replacing the values we have that

$V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}$

$V_1 = 215.12V$

Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

$V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}$