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Marizza181 [45]
3 years ago
5

Which layer of the Sun appears pinkish-red during a solar eclipse?

Physics
2 answers:
notka56 [123]3 years ago
8 0

Answer:

the answet is the 3 layer of the sun

Explanation:

isaw a solar eclipse and leanerd is

iragen [17]3 years ago
4 0

Answer:

Chromosphere

Explanation:

It is the second of the three main layers in the Sun's atmosphere and is roughly 3,000 to 5,000 kilmoeters deep.It's rosy red color is only apparent during eclipses.The chromosphere sits just above the photosphere and below the solar transtion region.

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Why are sunlight and gravity not considered matter?
Nady [450]

Sunlight and gravity are not considered matter because sunlight does not have  a rest mass that is characteristic of matter while gravity is just a vector component of matter.

<u>Explanation:</u>

Sunlight has energy and momentum that makes it similar to matter but it cannot be considered matter when you consider all the factors that are characteristic of matter. Light is an electromagnetic wave that is made up of photons and  the protons do not have rest mass.

Gravity is  a vector component of matter that has both magnitude as well as direction and the gravity increases with increase in mass of an object.

6 0
3 years ago
Read 2 more answers
Why can videos be streamed from one computer to another with excellent quality?
AleksAgata [21]

Answer:

(d) they are transmitted using digital signals

Explanation:

i just did the quiz an got it right

3 0
2 years ago
Read 2 more answers
Mr. Smith used 606 kWh for the month. If the service charge was $61.37 (see back page of the bill), what is the approximate char
Murrr4er [49]

Answer:

.10/KWh

Explanation:

divide 606 by 61.37 and you get .1012...

7 0
3 years ago
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g&#10;            y- 0 = 10.0²/2 9.8&#10;            y - 0 = 5.10 m&#10;            &#10;The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system&#10;             y₂ = 5.1 + 44&#10;             y₂ = 49.1 m&#10;Let's use the other equation to find the time&#10;              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from
Sliva [168]

Answer:

spacing between the slits is 405.32043 ×10^{-9}  m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4I_{0} × cos²∅/2

so

I/4I_{0}  = cos²∅/2

here I/4I_{0} is 85 % = 0.85

so

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 ×cos^{-1} 0.921954

∅  = 45.56°

∅  = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅  = 2π d sinθ / wavelength

so

d = ∅× wavelength /  ( 2π  sinθ )

put all value

d = 0.795 × 610×10^{-9} / ( 2π  sin2.95 )

d = 405.32043 ×10^{-9}  m

spacing between the slits is 405.32043 ×10^{-9}  m

7 0
3 years ago
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