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KIM [24]
3 years ago
15

how does the temperature change in the sun's atmosphere differ from the temperature change in the suns interior

Physics
1 answer:
svp [43]3 years ago
6 0
<span> the mechanisms that provide sufficient energy to heat the solar atmosphere. A layer beneath the Sun's surface, acting as a pan of boiling water, is thought to generate a small-scale magnetic field as an energy reserve which, once it emerges from the star, heats the successive layers of the solar atmosphere via networks of mangrove-like magnetic roots and branches.</span>
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A body travels a distance of 20m in the 7th second and 24m in the 9th second. How much distance shall it travel in the 15th seco
DaniilM [7]

Answer:

<u>36 m</u>

Explanation:

We can consider this to be an AP.

Then,

  • a₇ = 20
  • a₉ = 24

<u>Subtract a₇ from a₉.</u>

  • a + 8d - a + 6d = 24 - 20
  • 2d = 4
  • d = 2

  • a + 6(2) = 20
  • a = 8

<u>Finding a₁₅</u>

  • a₁₅ = a + 14d
  • a₁₅ = 8 + 14(2)
  • a₁₅ = 8 + 28
  • a₁₅ = <u>36 m</u>
4 0
2 years ago
Read 2 more answers
What adaptation in frogs enables them to swim​
Flauer [41]

Answer:

Most of the frogs have webbed feet which helps them swim. The thin skin between the toes helps them to push through the water. The frog's feet is webbed ,So they can easily swim.. The skin between toes are flexible soo they can freely push and pull it

Explanation:

6 0
4 years ago
Read 2 more answers
Rub one balloon with wool, fur, or hair and the other balloon with a piece of plastic wrap. Again, suspend the two balloons so t
tester [92]

Answer:

Attract

Explanation:

8 0
3 years ago
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Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

 m_1 = m_2

\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

5 0
3 years ago
Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C

Therefore, the final temperature of the tea is 7.39⁰C.

4 0
3 years ago
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