The magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.
<h3>What is angular momentum.?</h3>
The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.
It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.
The magnitude of the angular momentum of the two-satellite system is best represented as;
L=∑mvr
L=m₁v₁r₁-m₂v₂r₂
Hence, the magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.
To learn more about the angular momentum, refer to the link;
brainly.com/question/15104254
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<span> The boiling point of water at sea level is 100 °C. At higher altitudes, the boiling point of water will be.....
a) higher, because the altitude is greater.
b) lower, because temperatures are lower.
c) the same, because water always boils at 100 °C.
d) higher, because there are fewer water molecules in the air.
==> e) lower, because the atmospheric pressure is lower.
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Water boils at a lower temperature on top of a mountain because there is less air pressure on the molecules.
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I hope this is helpful. </span>
Answer:

Explanation:
Let the charge on the ball bearing is q.
charge on glass bead, Q = 20 nC = 20 x 10^-9 C
Force between them, F = 0.018 N
Distance between them, d = 1 cm = 0.01 m
By use of Coulomb's law in electrostatics

By substituting the values


Thus, the charge on the ball bearing is 
Answer:
the time comes eventually.
Explanation:
ur body just be giving up
Answer:

ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:

Where:


Now:

Where
is the mass of the car,
is the initial velocity of the car,
is the mass of train,
is the final velocity of the car and
is the final velocity of the train.
Replacing data:

Solving for
:

Changed to cm/s, we get:

b) The kinetic energy K is calculated as:
K = 
where M is the mass and V is the velocity.
So, the initial K is:



And the final K is:




Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J