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liberstina [14]
3 years ago
13

The table shows chronological events in the life of our Sun, a medium-sized star. Place the missing events into the correct rows

of the table.

Physics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

The Sun is a medium sized star present at the center of the solar system. It is primarily formed of Hydrogen and Helium. Like other stars, Sun was also born in a Nebula, and moved on to become a Main sequence star. It has a lifespan of about 10 billion years. At the end of its life when all the hydrogen fused into higher elements, the Sun would eventually throw out majority of its mass and the core would survive as a white dwarf.

The correct chronological order of events is:

1. A nebula located in the Milky Way galaxy begins pulling nearby hydrogen atoms in its orbit.

2. The Nebula shrinks in its volume due to gravity, becoming denser and hotter. But, it's not hot enough for nuclear fusion.

3. The temperature in the core of the Nebula reaches 14 million Kelvin.

4. Hydrogen atoms begin shedding their electrons and colliding with one another.

5. The Sun enters the main sequence stage. The energy created as a result of  its radiation begins nurturing life on planet such as Earth.

6. The Sun uses up all the hydrogen in its core.

7. The Sun expands greatly and cools . It is larger and redder.

8. The Sun completely runs out of hydrogen to fuse. Its outer layers are pushed away, and a cloud of ionized gas surrounds its core.

9. The Sun is a white dwarf with a dim glow.

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The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
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Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

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