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Vinil7 [7]
3 years ago
15

Which of the following types of technologies has best helped scientists to study very high-energy objects in outer space, such a

s exploding supernovae, black holes, and pulsars?
 A. 

radio telescopes

 B. 

space stations

 C. 

X-ray telescopes

 D. 

lunar modules


​
Physics
2 answers:
azamat3 years ago
5 0

X-Ray telescopes are the one which helped the scientist to study very high energy objects in outer space such as exploding supernovae, black holes and pulsars efficiently.

Answer: C

Explanation:

The X-rays can penetrate through larger distance and also through the mirrors of the satellite at grazing angle with least diffraction.

The X-rays are emitted in the form of high energy photons during explosions of supernovae, black holes and pulsars.

All kinds of explosions in the space release light and heat. From the heat energy X-rays will also be released during these explosions. So these X-rays emitted during explosions can be detected by a space telescope with X-ray detector mostly photo detector.

As soft X-rays can easily penetrate the mirror of the satellite containing the telescope, soft x-rays can be detected in space.

The X-ray telescopes could be used once the telescope is placed in space because most of the x-rays will be absorbed in the earth's atmosphere if the telescope is used in earth.

So by satellites, the telescope is made to reach the uppermost layer of atmosphere or in space such that x-rays could be detected.

erica [24]3 years ago
3 0

Very high-energy objects and events spit out very high-energy photons, so the instrument you need in order to detect them is the       X-ray telescope. <em>(C)  </em>

Inconveniently, X-ray telescopes only work when they're up in orbit, because X-rays get seriously soaked up in Earth's atmosphere, and most of them never make it down to the surface ... (lucky for us !) .

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A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
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<h2>Given :</h2>

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Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

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  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

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_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

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Now, in 10 seconds the number of electrons passing through it is :

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  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

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