Part 1
When the solar atmosphere accumulates a lot of magnetic energy
to a point that cannot accumulate more, all that magnetic energy is suddenly released,
and with it, a lot of radiation. So much, that in fact it covers all of the
electromagnetic spectrum; from radio waves to gamma rays. That burst of
radiation is called a solar flare. In a single solar flare the amount of
radiation released is millions of times greater than all the nuclear bombs in
the face if the earth exploding together. Lucky for us, most of the high-energy
radiation dissipates before reaching the Earth, and the radiation that do reach
us, is deflected by the Earth’s magnetic field.
Part 2
1. Not all the radiation
of solar flares that reach the Earth is deflected by its magnetic field; some
of them reach us and charges the upper atmosphere with ionized particles. Those
particles react with the gases in the atmosphere and produce a light; that
light is what we call Auroras borealis or southern nights; One the most beautiful
natural spectacles in earth, who thought Auroras begin their lives as deadly
solar flares.
2. Solar flares
contain a lot of high-energy radiation that is extremely dangerous for our
electronic devices; when they reach the Earth, they can damage sensible
electronics like satellites. A very powerful solar flare could even damage all
the electronic devices on the surface of the Earth.
Answer:
10042.6 ohm
Explanation:
f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm
C = 5 nF = 5 x 10^-9 F
XL = 2 x π x f x L
XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm
Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)
Xc = 3184.7 ohm
Total impedance is Z.
Z^2 = R^2 + (XL - Xc)^2
Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2
Z = 10042.6 ohm
Answer:
Period of brightness variation and luminosity.
Explanation:
The Cepheid variables are used as distance indicators. This requires estimation of periods and (usually) intensity-mean magnitudes in order to establish a period—apparent luminosity relation. It is particularly important for the techniques employed to be as accurate and efficient as possible.
Explanation:
It is given that,
Frequency of the laser light, 
Time, 
(a) Let 
is the wavelength of this light. It can be calculated as :
or
(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :
n = 13440
Hence, this is the required solution.
Answer: Option (D) is the correct answer.
Explanation:
The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.
So, it means more electrons are added to the same energy level.
Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.
