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Nikitich [7]
3 years ago
7

How can astronomers infer wich elements are found in a star

Physics
2 answers:
alina1380 [7]3 years ago
8 0

Just recently, astronomers discovered a distant solar system, 127 light years away with up to seven planets orbiting a Sun-like star called HD 10180.

Like the very first exoplanet 51-Pegusus discovered in 1995, this new system was found using the science of spectroscopy.

In fact, most of the roughly 500 planets so far found orbiting other stars, were detected by the same method.

Spectroscopy — the use of light from a distant object to work out the object is made of — could be the single-most powerful tool astronomers use, says Professor Fred Watson from the Australian Astronomical Observatory.

"You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours.

"It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson.

When heated or when electrically charged, certain chemicals emit radiation at very specific colours or wavelengths called emission lines.

There are also absorption lines that appear as dark marks dividing the spectrum at specific wavelengths.

Absorption lines are created when light from something hot like a star passes through a cooler gas, cancelling out the emission lines the chemicals in the gas would normally create.

When you look at the spectrum of a star, for example, you can see absorption lines because the star's outer atmosphere is cooler than the central part, explains Watson.

Alchen [17]3 years ago
5 0
By previous elements found in other stars, accurate guesses and lots of studying
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2.966\times 10^{-11}\ N

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Plug in the given values and solve for 'F'. This gives,

F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})

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A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th
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(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

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(a) To solve the question, we  apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

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Sum of initial momentum = Sum of final momentum

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The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

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