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Jet001 [13]
3 years ago
12

Help, anyone?? please:/

Physics
1 answer:
NeTakaya3 years ago
5 0
A,E,C,B,D is the order
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Name TWO WEAKNESSES of the model pictured below
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Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare,  and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

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2 years ago
The amount of energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin is called its
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Specific heat.  The definition of specific heat is the amount of energy required to raise the temperature of 1g of a substance by 1K or 1°C.
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3 years ago
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You drive at 195 miles down the 101 freeway at 65 miles per hour. How long did it take? time =
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According to my calculations, if you drive an average of 65 miles per hour, with 0 minutes of stop-time, you should reach your destination in 3 hours and 0 minutes.

Explanation:

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A 2.0-kilogram cantaloupe rolling north at 4.0 meters per second collides head on with a 1.0-kilogram orange rolling south at 8.
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I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
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2 years ago
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