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Leno4ka [110]
3 years ago
14

You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength

regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the:____________
Physics
1 answer:
GREYUIT [131]3 years ago
7 0

Answer:

Period of the signal.

Explanation:

So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.

When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.

Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.

NB: not the amplitude but the period.

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Select the correct answer from each drop-down menu.
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A concave lens can only form a. A. real image. B. reversed image. C. virtual image. D. magnified image.
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A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.
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What number and type of hybrid orbital(s) form(s) when one p and one s atomic orbital mix?
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Two equivalent hybridized orbitals will form from the mixing of one s-orbital and one p-orbital, that is (sp) orbital.

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3 0
2 years ago
Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8
umka2103 [35]

Answer:

a) F=2.048\times 10^{-7}\ N

b) a=0.1138\ m.s^{-2}

Explanation:

Given:

  • mass of raindrops, m=1.8\times 10^{-6}\ kg
  • charge on the raindrops, q=+21\times 10^{-12}\ C
  • horizontal distance between the raindrops, r=0.0044\ m

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

we have:

\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}

<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

a=\frac{F}{m}

a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}

a=0.1138\ m.s^{-2}

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