A, C, and D all happen at different stages
of a total lunar eclipse.
I'll describe the stages of the eclipse, but before I do, I just need
to clarify: The Earth doesn't have an umbra or a penumbra, but
its shadow does.
-- the eclipse begins when the first edge of the moon
moves into the penumbra of Earth's shadow; ( C )
this part of the moon grows steadily.
-- After a while, the first edge of the moon begins to move
into the umbra of Earth's shadow ( A ), and gets very dark.
-- The total phase of the eclipse begins when the ENTIRE
moon is in the umbra of Earth's shadow.
Then everything happens in reverse.
-- Eventually, the leading edge of the moon moves out
of the shadow's umbra, into the penumbra. This part
steadily grows.
-- After a while, none of the moon is in the umbra, and
the whole thing is in the penumbra. The moon is
fully illuminated, but not quite as bright as it should be.
-- Soon, the leading edge of the moon leaves the penumbra
of Earth's shadow, and gets brighter. This portion of the moon
steadily grows, until ...
-- the moon completely leaves the penumbra, all of it is as bright
as it's supposed to be. The eclipse is completely over. ( B )
==> The whole process lasts several hours.
==> Everybody on the night side of the Earth sees the same thing
at the same time. It doesn't matter WHERE you are on the night
side ... if you can see the moon in the sky, you see the present
phase of the eclipse.
==> The lunar eclipse can only happen at the Full Moon. In fact, the
mid-point of the total phase is the exact moment of Full Moon.
Answer:
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
Explanation:
The expression for conservation of the angular momentum (L) is

Where
initial moment of inertia and angular velocity
is the final moment of inertia and angular velocity
The expression of moment of inertia of the satellite (a solid sphere) is

Where
is the satellite mass
r is the radus of the sphere
Substititute 1900kg for m and 4.6m for r

The final moment of inertia of the satellite about the centre of mass

Where
is the antenna's mass and
I is the length of the antenna

So, the Final rotation rate of the satellite is:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.
The distance traveled by the wood after the bullet emerges is 0.16 m.
The given parameters;
- <em>mass of the bullet, m = 23 g = 0.023 g</em>
- <em>speed of the bullet, u = 230 m/s</em>
- <em>mass of the wood, m = 2 kg</em>
- <em>final speed of the bullet, v = 170 m/s</em>
- <em>coefficient of friction, μ = 0.15</em>
The final velocity of the wood after the bullet hits is calculated as follows;

The acceleration of the wood is calculated as follows;

The distance traveled by the wood after the bullet emerges is calculated as follows;

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.
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Answer:
The student hears the wave that is transmitted by the desk
Explanation:
Mechanical waves need a material medium to be able to be transmitted, in the case of sound waves, one of the most common media is air, but it is also transmitted in other media in this case, stationery is transmitted.
The student hears the wave that is transmitted by the desk
The speed of the wave is proportional to the density of the material, so the wave that the student hears arrives much faster through the desk than through the air