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myrzilka [38]
3 years ago
5

Based on discoveries to date, which of the following conclusions is justified?a) Most stars have one or more terrestrial planets

orbiting within their habitable zones.b) Planets are common, but planets as small as Earth are extremely rare.c) Planetary systems are common and planets similar in size to Earth are also common.d) Although planetary systems are common, few resemble ours with terrestrial planets near the Sun and jovian planets far from the Sun.
Physics
1 answer:
KIM [24]3 years ago
6 0

Based on discoveries to date, the conclusion as “Planetary systems are common and planets similar in size to Earth are also common” is justified.

Answer: Option C

<u>Explanation: </u>

Some studies show that on average, each star has at least single planet. This means that most stars, such as the Solar System, possess planets (otherwise exoplanets). It is known that small planets (more or less Earthly or slightly larger) are more common than giant planets. The mediocrity principles state that planet like Earth should be universal in the universe, while the rare earth hypothesis says they are extremely rare.

Size is often considered an important factor, because planets the size of the Earth are probably more terrestrial and can hold the earth's atmosphere. The planetary system is a series of gravitational celestial objects orbiting a star or galaxy. Generally, planetary systems describe systems with one or more planets, although such systems may also consist of bodies such as dwarf planets, asteroids and the like.

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Answer:

Explanation:

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horizontal circle at a rate of 2.33 revolutions per second

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\Delta A = \frac{1}{2} \Delta \theta R^2

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E = \frac{\Delta  \phi}{\Delta  t}

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E=\frac{\Delta (BA) }{\Delta  t}

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B is the magnetic field strength

substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

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E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V

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Answer:

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Explanation:

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Answer:

Please see attachment

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