<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>
Answer:
The index of refraction of the liquid is 1.35.
Explanation:
It is given that,
Critical angle for a certain air-liquid surface, 
Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1
Using Snell's law for air liquid interface as :




So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.
I think it would be WEIGHT and ROUGHNESS OF SURFACE.
Nuclear Fission s a power source with a very low environmental impact.
Answer:
The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions
Explanation:
Given data
= 0
= 5 
Time taken = 7 sec
(1). The angular acceleration is given by



We know that from the equation of motion


-------- (1)
(2). The angular acceleration is given by


- 0.357 
We know that from the equation of motion


= 35.01 rev ------- (2)
Total no of revolution made by the machine is

17.5 + 35.01
52.51 rev
Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51 rev