<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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60 Miles per hour 60 times 2 is 120
To solve this problem we must keep in mind the concepts related to angular kinematic equations. For which the angular velocity is defined as
Where
Final angular velocity
Initial angular velocity
Angular acceleration
t= time
In this case we do not have a final angular velocity, then
Re-arrange for 
Therefore the mangitude of the angular aceleration is 5449.1rad/s²
The answer would be letter choice B
Answer:
The correct answers are It is the resistance of an object to changes in its motion, and It is a force
