<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.
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Answer:
<h2>Kilometer (km) and micrometer (um) respectively</h2>
Explanation:
<h3>One thousand meters is equal to one kilometer represented as km. </h3>
and
<h3>One thousandth of a meter mean 1/1000 m which implies one thousands part of a meter which is equal to micro meter and represented as um.</h3>
Answer:
12.25 meters
Explanation:
s=1/2(v+u)t
s= displacement
v= final velocity
u= initial velocity
t= time
7m/s+0m/s divide by 2= 3.5 m/s velocity Times 3.5 seconds= 12.25 meters
<span>The best and most correct answer among the choices provided by the question is the second choice. The temperature of the substance is directly proportional to the average kinetic energy of the molecules. </span><span>I hope my answer has come to your help. God bless and have a nice day ahead!</span>