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Otrada [13]
3 years ago
12

Give me four reasons pluto is a cool planet / dwarf planet

Physics
2 answers:
jenyasd209 [6]3 years ago
6 0
Pluto is categorized as a dwarf planet. In 2006, Pluto was categorized with three other objects in the solar system that are about the same small size as Pluto: Ceres, Makemake and Eris. These objects, along with Pluto, are much smaller than the "other" planets. Here is another one Pluto is a dwarf planet that lies in the Kuiper Belt. It's an area full of icy bodies and other dwarf planets at the edge of our solar system. Because Pluto is the biggest known object in this region, some call it "King of the Kuiper Belt."

One thing is certain. Pluto and its neighborhood are very peculiar. If scientists could unravel some of their mysteries, we would know more about how our solar system formed. Hope this helps! Mark brainly pls!
AleksandrR [38]3 years ago
5 0

Answer:

1. Just because it is small doesn't mean it needs to be excluded. If that were the case I would've been out of my friend group a while ago

2. Look at it it's fricking beautiful (see attachment)

3. It is just a great planet I don't think there needs to be any reason given I meannn you agree?

4. There's no other reason needed it's fricking gorgeous and amazing it needs no other reason. It needs a freaking

opening announcement to announce the arrival of the gorgeouness.

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Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

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\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

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Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

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E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
3 years ago
A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
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Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

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 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

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2

2 2 2.00 0.178 m

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That is a reason

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