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Genrish500 [490]
4 years ago
10

What happens when a star's core runs out of hydrogen and why does this occur?

Physics
2 answers:
UNO [17]4 years ago
7 0

Answer:

Explanation:

Sun is made up of gases mainly helium and hydrogen. The energy getting from sun is due to the fusion reaction between the hydrogen molecules in the sun. After a long time, when all the hydrogen is converted into helium, the fusion reaction stops, only very few hydrogen molecules fuse which are in the interior core of the sun.

The sun starts contracting due to the gravitational pull and the pressure the temperature increases. As the temperature reaches to a certain maximum value, the fusion of helium starts and then the sun again gives us energy in different forms.

Ad libitum [116K]4 years ago
5 0
When the core runs out of hydrogen to fuse, then it starts fusing the helium. This happens because a star cannot create more hydrogen.
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A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio
LuckyWell [14K]

Answer:

U = √Rg/sin2θ

Explanation:

Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.

Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.

According to the range formula,

R = U²sin2θ/g

Cross multiplying we have;

Rg = U²sin2θ

Dividing both sides by sin2θ, we have;

U² = Rg/sin2θ

Taking the square root of both sides we have;

√U² = √Rg/sin2θ

U = √Rg/sin2θ

Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ

3 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
3 years ago
Hurryyyyyyyy
padilas [110]
The answer would be C. Gamma Rays and High Frequency EM waves travel at the speed of light and are transverse waves.
7 0
3 years ago
Read 2 more answers
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
Which statement best describes an atom?
SpyIntel [72]
The correct answer is going to be  <span>C, because in the nucleus of an atom there are protons and electrons; which can't move, and are surrounded by electrons on the electrical cloud</span>
5 0
3 years ago
Read 2 more answers
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