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4 years ago

Approximately what core temperature is required before hydrogen fusion can begin in a star?

Physics

1 answer:

valkas [14]4 years ago

8 0

Answer:

The temperature required is near about 3 million kelvin

Explanation:

The brilliance of the star results from the nuclear reaction that take place in the core of the star and radiate a huge amount of thermal energy resulting from the fusion of hydrogen into helium.

For this reaction to take place, the temperature of the star's core must be near about 3 million kelvin.

The hydrogen atoms collide and starts and the energy from the collision results in the heating of the gas cloud. As the temperature comes to near about $1.5\times 10^{7 {\circ}C$ , the nuclear fusion reaction takes place in the core of the gas cloud.

The huge amount of thermal energy from the nuclear reaction gives the gas cloud a brilliance resulting in a protostar.

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A bullet with a mass ????b=13.5 g is fired into a block of wood at velocity ????b=249. The block is attached to a spring that ha

larisa86 [58]

Answer:

$M = 0.436 kg$

Explanation:

As per energy conservation we can say that energy stored in the spring at the position of maximum compression must be equal to the kinetic energy of bullet and block system

so here we have

$\frac{1}{2}(m + M)v^2 = \frac{1}{2} kx^2$

here we know that

k = 205 N/m

x = 35 cm

$\frac{1}{2}(m + M)v^2 = \frac{1}{2}(205)(0.35)^2$

now by momentum conservation we know that

$mv_o = (m + M)v$

$v = \frac{m}{m + M} v_o$

now plug in all values in it

$v = \frac{0.0135}{0.0135 + M}(249)$

now from above equation

$\frac{1}{2}(0.0135 + M)( \frac{0.0135}{0.0135 + M}(249))^2 = 12.56$

$\frac{5.65}{0.0135 + M} = 12.56$

by solving above equation we have

$M = 0.436 kg$

3 0

4 years ago

It is the obligation of researchers to review and comment on the research of other researchers. True or False.

romanna [79]

It is false that it is the obligation of researchers to review and comment on the research of other researchers. It is not their obligations - they don't have to do it, although they can if they want to and if they are allowed by the author him or herself. However, they are not bound by law or something like that to do this, it's just due to their kindness or genuine interest that they do this.

4 0

3 years ago

Read 2 more answers

How will the electrostatic force be changed if the distance between electric charges is reduced one half of the original distanc

GuDViN [60]

I think the answer to your question is A

8 0

4 years ago

A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo

Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A) Energy

B) MRUV

Energy:

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0 ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2 = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) ---> $V2 = \sqrt{(2*g*(H1-H2)}$

Replaced values: V2 = 14,0 m/s

MRUV:

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0

4 years ago

Read 2 more answers

How fast was the storm that hit Galveston and how men died?

vodomira [7]

Answer:

The storm was a category 4 hurricane that struck Galveston, Texas, on September 8, 1900, bringing winds of 130 miles (210 km) per hour and high tides that overwhelmed the low-lying coastal city, demolishing buildings and claiming more than 8,000 lives.

00p- now I can actually answer :)

Hope that I helped you a little :0

3 0

3 years ago