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Lena [83]
4 years ago
7

Shelly was surprised to learn the dwarf planet Pluto used to be considered a planet. What would have had to be true for Pluto to

remain categorized as a planet?
Pluto would need to have an atmosphere.
Pluto would need to clear the neighborhood of its orbit.
Pluto would need to have a larger moon.
Pluto would need to have water or ice.
Physics
2 answers:
tekilochka [14]4 years ago
7 0
The second answer is the correct response, as Pluto is found in the kuiper belt, and there are objects that are roughly the same size as Pluto.
Inga [223]4 years ago
5 0

Answer: Pluto would need to clear the neighborhood of its orbit.

Explanation:

Pluto was once a planet and was demoted to dwarf planet in year 2006. A celestial object is considered to be a planet when it is round in shape, it has elliptical orbit around a star and it clears it neighborhood.

Pluto lies in Kuiper belt and there were many more bodies similar to Pluto. Hence, it was demoted to a dwarf planet. Had Pluto been still a planet, it would need to clear neighborhood of its orbit.

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A skier reaches the end of a 0.9 km track in 1.5 hours, at what speed does it travel?
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Speed = Distance/Time

S = D/T

D = 0.9Km, T = 1.5hours,

S = 0.9/1.5

S = 0.6 m/s

hope it helps!

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4 years ago
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puteri [66]

Answer:

Mass

Explanation:

They have masses ranging from about 5 to several tens of solar masses.

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4 years ago
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A metal wire 1.50 m long has a circular cross section of radius 0.32 mm and an end-to-end resistance of 90.0 Ohms. The metal wir
elixir [45]

Answer:

So after stretching new resistance will be 0.1823 ohm

Explanation:

We have given initially length of the wire l_1=150m

Radius of the wire r_1=0.32mm=0.32\times 10^{-3}m

Resistance of the wire initially R_1=90ohm

We know that resistance is equal to R=\frac{\rho l}{A} ,here \rho is resistivity, l is length and A is area

From the relation we can say that \frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}

Now length of wire become 6.75 m

Volume will be constant

So A_1l_1=A_2l_2

So \pi \times (0.32)^2\times150=\pi \times r_2^2\times 6.75

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So \frac{90}{R_2}=\frac{150}{6.75}\times \frac{1.508^2}{0.32^2}

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7 0
3 years ago
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blsea [12.9K]

Answer: Gravity causes a change in the mass of objects on earth and the moon.

Explanation:

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3 years ago
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Time-outs are ineffective because
maria [59]
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8 0
4 years ago
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