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Mademuasel [1]
4 years ago
11

During a total lunar eclipse, the moon

Physics
2 answers:
Anit [1.1K]4 years ago
8 0

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.

lidiya [134]4 years ago
5 0
The answer would be A) Moves into Earth's umbra. Hope this helps!!
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A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that
FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

t=\sqrt{\dfrac{2s}{g}}

t=\sqrt{\dfrac{2\times 32}{9.8}}

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

v=\dfrac{d}{t}

v=\dfrac{11\ m}{2.55\ s}

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

3 0
3 years ago
Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object be?
Serjik [45]

Answer:

I belive it would be ture

Explanation:

It's been a while since I learned this but I think that is right.

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