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Daniel [21]
3 years ago
11

Compare between chromosphere, corona and photosphere

Physics
1 answer:
gladu [14]3 years ago
5 0

Answer:

Explained

Explanation:

Photosphere: The lowest layer of the sun is called photo sphere . It is about 300 miles thick from the surface. It is the source of solar flares. It is  marked by bright bubbling granules of plasma.

chromosphere emits a reddish glow as the super heated hydrogen burns off but the red rim can only be seen during total solar eclipse.

The third layer of the sun atmosphere is Corona. It can also only be seen during during a total solar eclipse. Temperature in corona can reach as high as 3.5 million degree fahrenheit.  As the gases cool they become solar winds.

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Tapping into this energy source could provide at least 10 times the energy that can be obtained from the nation’s known coal res
ipn [44]

Tapping means the maximum use of the energy by the people.

<h3>What is energy?</h3>

The energy is the ability to do work.

The energy that can be tapped is solar energy. There are many ways developed to tap the solar energy and make the best use of it by solar plant, water heating, cooking, etc.

Thus, tapping means the maximum use of the energy by the people.

Learn more about tapping.

brainly.com/question/18623323

#SPJ4

6 0
2 years ago
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely i
blsea [12.9K]

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

8 0
3 years ago
Read 2 more answers
What happens when the objects submerged in a fluid at rest?​
Talja [164]
It will act upon a buoyant force on the magnitude of which is equal to weight of the fluid
3 0
2 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00
ira [324]

Answer:

A. 51.42 m.

B. 17.14 s.

Explanation:

Using equations of motion:

vf^2 = vi^2 + 2 * aS

Where,

vf = final velocity

a = acceleration

S = distance to which swan traveled

vi = 0 m/s

6.00^2 = 2 * 0.350S

S = 36/0.7

= 51.42 m.

B.

vf = vi + at

6 = 0 + 0.35t

t = 6/0.35

= 17.14 s.

5 0
3 years ago
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