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Alex17521 [72]
2 years ago
12

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

verage distance from the sun was about 5km/s. What was the orbital speed of this material when it was 40,000 AU from the sun (before it fell inward with the collapse of the nebula)? Use the law of conservation of angular momentum. I have the angular momentum = m x v x r. Is this correct?
Physics
1 answer:
attashe74 [19]2 years ago
3 0
<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
A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.
Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

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NemiM [27]

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4 0
2 years ago
A rock is rolling down a hill. At position 1, it’s velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, it’s velo
mr Goodwill [35]

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation =change in velocity/change in time

a= {velocity final-velocity initial}/(change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity =2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a= (44-2)/12  = 42/12

3.5 m/s²

Best Wishes!

5 0
2 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
2 years ago
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