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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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Parallel rays of monochromatic light with wavelength 583 nm illuminate two identical slits and produce an interference pattern o
romanna [79]

Answer:

  I = 2.18 10⁻⁴ W / m²

Explanation:

The two-slit interference pattern is described by the expression for constructive interference.

             d sin θ = m λ

If we also want to know the distribution of intensities we must perform the su of the electric field of the two waves, and find the intensity as the square of the velvet field, obtaining the expression

              I = I_max cos² ((π d /λ L) y)

where d is the separation of the slits, λ  the wavelength, L the distance to the screen e and the separation of the interference line with respect to the central maximum

 

let's reduce the magnitudes to the SI system

λ  = 583 nm = 583 10⁻⁹ m

L = 75.0 cm = 75.0 10⁻² m

d = 0.640 mm = 0.640 10⁻³ m

y = 0.900 mm = 0.900 10⁻³ m

let's calculate the intensity of this line

        I = 5 10⁻⁴ cos² ((π 0.640 10⁻³ /583 10⁻⁹ 0.75 10⁻²) 0.900 10⁻³)

        I = 5 10⁻⁴ cos2 (413.84)

         I = 5 10⁻⁴ 0.435

        I = 2.18 10⁻⁴ W / m²

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3 years ago
While you are studying for an upcoming physics exam, a lightning storm is brewing outside your window. Suddenly, you see a tree
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3 years ago
A vector quantity is the magnitude of a given quantity? True Or False​
Anettt [7]

Answer:

false

Explanation:

A vector quantity is the magnitude of a given quantity? True Or False

6 0
4 years ago
A carpet is to be installed in a room of length 9.72 meters and width 17.30 of the room retaining the proper number of significa
Usimov [2.4K]

Answer:

168 m^2, 380 m^2

Explanation:

length of the room, l = 9.72 m

width of the room, b = 17.30 m

Area of teh rectangle is given by

A = length x width

So, A = 9.72 x 17.30 = 168.156 m^2

the significant digits should be 3 in the final answer

So, A = 168 m^2

Now length = 72 m

width = 17.39 feet = 5.3 m

Area, A = 72 x 5.3 = 381.6 m^2

There should be two significant digits in the answer so, by rounding off

A = 380 m^2

3 0
3 years ago
Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis
Veronika [31]

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

3 0
4 years ago
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