Ask question

Ask question

All categories

Nutka1998 [239]

3 years ago

6

Which layer of the sun's atmosphere are you looking at when you look at me in image of the Sun

1 answer:

Hitman42 [59]3 years ago

7 0

The answer is Photosphere. The photosphere is the lowest layer<span> of the </span>solar<span> atmosphere. It is essentially the </span>solar<span> "surface" that </span>we see<span> when </span>we look<span> at the </span>Sun in "white" light. It is <span>like a glowing fog, so at a distance, it </span>looks<span> solid, the same way a cloud looks solid from a distance.</span>

You might be interested in

HELP HELP HELP ME!!!

zzz [600]

I believe the answer is C. Hope this helps!!

7 0

3 years ago

The height of a tree is 4 m.

ryzh [129]

Answer:

1.2

Explanation:

4/100 × 30= 1.2

3 0

3 years ago

A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.30 m/s2. At 20.0 s after bla

lesya [120]

Answer:

Explanation:

v = u +at

u = 0

a = 2.3 m /s²

t = 20 s

v = 2.3 x 20

= 46 m /s

Distance covered under acceleration of 2.3 m/s²

s = ut + 1/2 at²

= 0 + .5 x 2.3 x 20²

= 460 m

After that it moves under free fall ie g acts on it downwards .

v² = u² - 2gh , h is height moved by it under free fall

0 = 46² - 2 x 9.8 h

h = 107.96 m

Total height attained

= 460 + 107.96

= 567.96 m

b ) At its highest point ,it stops so its velocity = 0

c ) rocket's acceleration at its highest point = g = 9.8 downwards .

At highest point , it is undergoing free fall so its acceleration = g

6 0

3 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

4 years ago

Planets that are larger in diameter, farther from the sun, and less dense than smaller planets in the solar system are known as

Sergio [31]

Gas giants lol I'm love this kinda stuff nothing else just this question

6 0

3 years ago