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Mademuasel [1]
3 years ago
11

During a total lunar eclipse, the moon

Physics
2 answers:
Anit [1.1K]3 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]3 years ago
5 0
The answer would be A) Moves into Earth's umbra. Hope this helps!!
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Describe three important ways we use the electromagnetic spectrum in our everyday lives.
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Answer:

We use X-rays to help the injured, Radiowaves to communicate or entertain, and Visible light to see.

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3 years ago
What is the wavelength (in air) of a 13.0 khz vlf radio wave?
bonufazy [111]
Since,
Speed = Frequency * WaveLength

=> WaveLength = Speed / Frequency --- (A)

Frequency = 13.0 kHz.
As the radio waves are electromagnetic waves, their speed is equals to the speed of light. Therefore,
Speed = C = 3 *  10^{8}  m/ s^{2}

Plug in the values in equation(A):

A => WaveLength = \frac{3 * 10^{8}}{13*10^{3}}

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-i
5 0
3 years ago
A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The
uysha [10]

Answer:

f = 692 N

Explanation:

given data:

f =800N

a =1.2 m s^{2}

m= 90 kg

from newton's second law

net force F_{net} =\sum F = F_1 +F_2 +..... = ma

therefore we have from above equationF_{net} = F -f = ma

ma =F - f

putting all value to get force of friction

1.2*90 = 800 - f

f = 692 N

8 0
3 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7
Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I=\frac{P_{avg}}{A}

I=\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}

I=7.296\times 10^{-6} W/m^2

Amplitude of electric field at receiver end

E_{max}=\sqrt{2I\mu _0c}

Amplitude of induced emf

=E_{max}d

=\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75

=17.591\times 10^{-2}=0.1759 v

7 0
3 years ago
What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s
stiv31 [10]

Answer:2800000j

Explanation:

For us to know the kinetic energy of the vehicle,

Where m is the mass

And v is the velocity

Then, K.E=1/2mv^2

While, K.E=1/2×3500×40^2

Therefore, our answer will now be

K.E=2800000j

5 0
3 years ago
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