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Zarrin [17]
3 years ago
15

Can anyone help me? Physics

Physics
1 answer:
olga2289 [7]3 years ago
5 0
I cannot read please show closer
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What is the EM used for?
Phantasy [73]

Answer:

Applications of electromagnetic waves

Explanation:

"Hot" objects in space emit UV radiation as well. X-ray: A dentist uses X-rays to image your teeth, and airport security uses them to see through your bag. Hot gases in the Universe also emit X-rays. Gamma ray: Doctors use gamma-ray imaging to see inside your body.

3 0
3 years ago
lightning and thunder both occur at the same time why do you see the Lightning before you hear the thunder
denis23 [38]

Because the light from it travels to you about 874 thousand times
as fast as the sound does, so the hearing part falls behind the seeing
part.
5 0
3 years ago
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Solar System Model Comparison and Contrast Chart
kotykmax [81]
What are you trying to here?
8 0
4 years ago
As a pendulum swings back and forth_____________.
Nookie1986 [14]

Answer:e

Explanation:

As the Pendulum swings back and forth its kinetic energy is converted into potential and Potential into kinetic .

At highest point Pendulum is momentarily at rest and thus possess all the energy in the form of Potential Energy which is converted into kinetic energy as the pendulum moves downward.

The kinetic energy is maximum at mean position i.e. at lowest point.

Thus all of the given options are true

8 0
4 years ago
Sherlock Holmes examines a clue by holding his magnifying glass (with
Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

M=-\frac{q}{p}=-\frac{-30}{10}=3

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0
4 years ago
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