Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown
1 answer:
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A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that , which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
The frequency of the electromagnetic wave is 9.55 × 1014 Hz and it is classified as ultraviolet.
<h3>What is meant by electromagnetic waves?</h3>Electromagnetic waves are forms of energy that are invisible and travel throughout the universe. However, some of the effects of this energy are visible. The light that we see is a component of the electromagnetic spectrum.
Electromagnetic waves, or EM waves, are produced by vibrations between an electric field and a magnetic field. In other words, electromagnetic waves are made up of oscillating magnetic and electric fields.
<h3>How do you calculate the speed of an electromagnetic wave?</h3>
The wavelength and frequency of any periodic wave are used to calculate its speed. v = λf.
In free space, the speed of any electromagnetic wave is equal to the speed of light, c = 3 108 m/s.
The frequency of the electromagnetic wave is 9.55 × 1014 Hz and it is classified as ultraviolet.
To learn more about electromagnetic wave refer to:
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