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astraxan [27]
3 years ago
7

Which is true about Pluto

Physics
2 answers:
andrew-mc [135]3 years ago
8 0

Answer:

D.)it orbits near the Kepler belt

Explanation:

The Kuiper belt is an area similar to the asteroide belt extending from the orbit of Neptune to about 50 AU from the Sun. It mainly consists of icy asteroids and dwarf planets, which are rocky objects big enough to be defined as planet but that do not have enough gravity to clear their orbit from other obejcts.

Pluto was discovered in 1930 - initially it was classified as a planet, although it is much smaller than the other 8 planets of the Solar System. However, it has been recently de-classified to dwarf planet because its gravity is not enough to clear its orbit from other objects (asteroids). Pluto is located inside the Kuiper belt, so option D is correct. Other dwarf planets in the Kuiper belt are for instance Haumea and Makemake.

mel-nik [20]3 years ago
6 0

Answer:

<h2>D.) it orbits near the Kuiper belt.</h2>

Explanation:

You mean the Kuiper Belt, not the Kepler Belt.

The Kuiper Belt refers to the circumstellar disc that is border region of the Solar System, when Pluto it's present. Actually is from the Neptune's orbit to the outer Solar System.

The Kuiper Belt, we can define it as a larger asteroid belt, that contains three dwarf planets: Pluto, Haumea and Makemake. Dwarf planets refers to "planets" that have little mass, but they are not satellites, however they don't domain its region. So, it's like a satellite but we some characteristics of planets.

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Explanation:

Given , for a converging lens

Focal length : f=20\ cm

Height of object : h=1\ cm

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Using lens formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}, we get

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}, where v = image distance from the lens.

On solving aboive equation , we get

\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm

Formula of Magnification : m=\dfrac{v}{u}=\dfrac{h'}{h} , where h' is the height of image.

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\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm

Hence, the height of the image is 2 cm.

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C

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