<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: 1.22 m
Explanation:
The equation of motion in this situation is:
(1)
Where:
is the final height of the ball
is the initial height of the ball
is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)
is the time at which the ball lands
is the acceleration due gravity
So, with these conditions the equation is rewritten as:
(2)
(3)
Finally:
Answer:
They sh0uld g0 t0 the reactor and then, see what the issue is...
Explanation:
then, see if they can fix the problem, im sorry if its wr0ng.
Try this option, the answers are marked with colour.
Answer:
Answer:
118.4 N
Explanation:
weight of chair, mg = 95 N
Push, F = 39 N
Ф = 37 ° below x axis
Let n be the normal force.
So, by using the diagram and resolve the components of Force F.
n = mg + F SinФ
n = 95 + 39 Sin 37°
n = 95 + 39 x 0.6
n = 118.4 N
Explanation: