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Flauer [41]
3 years ago
13

What happens in the core of a star that causes it to shine?? pls make it short

Physics
2 answers:
vodka [1.7K]3 years ago
8 0
Nuclear fusion, which releases energy.
We use the same process in the "hydrogen" bomb.
Oksi-84 [34.3K]3 years ago
3 0
The procedure that makes stars sparkle is the same one that we have outfit for utilization as a weapon: atomic combination.

Stars begin as gigantic districts of gas, for the most part hydrogen. This gas will begin to contract, and it warms up. As it gets, its temperature rises. At the point when the warmth gets sufficiently high, it causes the individual hydrogen iotas to impact and consolidate into helium with the arrival of vitality. This is called fusion.

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What mass has a rest energy of 100J?
alexandr1967 [171]

Answer:

option a is correct

Explanation:

<h2>I hope it's help you ❣️❣️</h2>
6 0
3 years ago
Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium
ivolga24 [154]

Answer:

A

Explanation:

Iron and gadlinium are both very easily made into magnetic substances.  Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

5 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brak
Fudgin [204]

Answer:

V = 11.83 m/s

Explanation:

Given the following data;

Mass = 2000 kg

Force = 10000N

Distance = 14 m

To find the final velocity of the car;

First of all, we would determine the acceleration of the car;

Acceleration = force/mass

Acceleration = 10000/2000

Acceleration = 5 m/s²

Next, we would use the third equation of motion to find the final velocity;

V^{2} = U^{2} + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

V² = 0² + 2*5*14

V² = 0 + 140

V = √140

V = 11.83 m/s

5 0
3 years ago
A sprinter moving at 10 m/s slows down at a rate of 1.4 m/s2. how fast is the runner moving after 4 seconds?
AURORKA [14]
V=u+at
v=10+(-1.4 (4))
=10-5.6
=4.4m/s
3 0
3 years ago
Read 2 more answers
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