Answer: A <u>Nebula </u>is left behind. A spectacular explosion in which a star ejects most of its mass in a violently expanding cloud of debris.
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Answer:
They both have the same angular speed.
Explanation:
The mathematical formula for angular speed is:

where
is angular speed,
is a constant, and
is the period (the time it takes the marry-go-round to complete a lap).
What we can see from the formula is that, since the
does not change its value, the angular speed depends only on the period T.
In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.
Thus, since the period for both is the same, the angular speed given by
will also be the same
Answer:
The work is -67.76 J
Explanation:
The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.
This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.
In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.
So, the loss of kinetic energy is 
You know:
- mass=m=0.22 kg
- Initial velocity of the ball:

Final velocity of the ball: 
Replacing:
= -67.76 J
Friction work is always negative because friction is always against displacement.
<u><em>The work is -67.76 J</em></u>
Impulse = Ft = (m)(delta v)
delta v = change in velocity = velocity final - velocity initial.
= -22m/s - +18m/s = -40m/s.
mdeltav = (0.40kg)(-40m/s) = -16kgm/s or -16Ns.
The 3rd Quadrant is the lower left one on a graph,
where ' x ' and ' y ' are both negative.
Your picture is completely unrelated to this question.