All categories

2 years ago

Which layer of the sun is responsible for producing the light shown in the picture above?

Physics

2 answers:

jasenka [17]2 years ago

7 0

It is photosphere, so your answer would be C. I took the quiz and this was the answer.

The photosphere basically radiates light into space, that's why that's the answer! Have a great day!

Eva8 [605]2 years ago

3 0

Answer:

C. Photosphere

Explanation:

The lights shown in the figure comes from the outermost layer of the Sun. This layer is called photosphere.

This is the layer from where the light of the Sun is radiated, before travelling through space and reaching us.

The photosphere is the coldest layer of the Sun: its surface temperature is between 4500 and 6000 K. Its width is approximately 100 km.

A characteristic of the photosphere is the presence of the sunspots, which appear as darker spots, and are regions of lower temperature caused by a concentration of magnetic flux.

You might be interested in

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

2 years ago

Electricity from the local power company costs 7.22 cents/kWh. A homeowner wishes to pay a maximum of 25 cents per month for the

mestny [16]

You haven't said how much power the stereo uses. It matters !

Whatever that number is, the maximum hours per month is

(3460) divided by (the # of watts the stereo uses when it's playing) .

7 0

3 years ago

Read 2 more answers

a force of 50 newtons pulls a rope attached to a 150 newton sled across a horizontal surface at a constant velocity of 5 meters

lakkis [162]

Answer:

I don't know the answer I hope you find it tho good luck##

3 0

2 years ago

Why are different constellations<br> of stars seen during different<br> seasons?

slamgirl [31]

Actually, they're not. There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around. And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night.

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ? Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ? Now they're straight in the
direction of the sun. So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.

5 0

3 years ago

Read 2 more answers

A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?

Anon25 [30]

Answer:

57N

Explanation:

$F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }$

6 0

2 years ago

Read 2 more answers