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tankabanditka [31]
3 years ago
7

What happens to a nucleus undergoing alpha decay?

Physics
1 answer:
Law Incorporation [45]3 years ago
4 0
It looses two protons, it changes from one element to other
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Answer this question thanks
maria [59]

Like they said the leopard seal will overpopulate and kill out all of the other animals. Because it would be at the top of the chain then without the killer whale. Therefore, since it is at the top of the chain it will overpopulate because it has nothing to eat/kill it so then they would eat the animals lower than them<span />
7 0
3 years ago
. When fusion begins in a high mass protostar, which kind of star is formed? blue star. yellow star. red star. 2. To which galac
guapka [62]

The reason that fusion of light elements produces energy to support a star is because of the “mass defect” we discussed when we studied the proton-proton chain. The product of hydrogen fusion (one helium nucleus) has less mass than the four hydrogen nuclei that created it. The extra mass has been converted into energy. Each fusion reaction of light elements in the core of a high mass star always has a mass defect. That is, the product of the reaction has less mass than the reactants. However, when you fuse iron, the product of iron fusion has more mass than the reactants. Therefore, iron fusion does not create energy; instead, iron fusion requires the input of energy.

When iron builds up in the core of a high mass star, there are catastrophic consequences. The process of fusing iron requires the star's core to use energy, which causes the core to cool. This causes the pressure to go down, which speeds up the gravitational collapse of the core. This causes a chain reaction: core collapses, iron fusion rate increases, pressure decreases, core collapses faster, iron fusion rate increases, pressure decreases, core collapses faster, iron fusion rate increases, etc., which causes the star's core to collapse in on itself instantaneously. After the core collapses, it rebounds. A large quantity of neutrinos get created in reactions in the core, and the rebounding core and the newly created neutrinos go flying outward, expelling the outer layers of the star in a gigantic explosion called a supernova (to be precise, a type II or core collapse supernova).

For a brief period of time, the amount of light generated by one star undergoing a supernova explosion is greater than the luminosity of 1 billion stars like the Sun. These explosions are so bright that they are visible at immense distances. If a nearby star were to undergo a supernova explosion, it would be so bright it would be visible during the daytime. In modern history, no supernova has gone off close enough to us to be visible during the daytime. However, both Tycho Brahe and Johannes Kepler observed naked-eye supernovae during their lifetimes. In 1987, a supernova went off about 50,000 parsecs away from us. Below is a ground-based telescope image of the supernova about 2 weeks after the explosion. Note how bright the exploding star (lower right corner) is compared to all of the rest of the objects in the image.

8 0
3 years ago
How is energy involved in physical changes and in chemical changes?
GaryK [48]

During phase changes, energy changes are usually involved. For example, when solid dry ice vaporizes (physical change), carbon dioxide molecules absorb energy. Meanwhile, when liquid water becomes ice energy is released.

hope this helps :)

4 0
3 years ago
Describe how air particles differ in the way they move when air begins to cool.
grin007 [14]
<span>when air cools , the temperatures drops, the molecules move slowly taking up less space , the amount of space the air  takes ups shrinks , or reduce the air pressure 
</span>
7 0
3 years ago
A rock is thrown at an angle of 30 degrees above the horizontal with initial velocity 15m/s what is the displacement when the ro
Delvig [45]

The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by

x=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\cos30^\circ\,t

so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:

y=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\sin30^\circ\,t-\dfrac g2t^2=0

where g=9.8\,\dfrac{\mathrm m}{\mathrm s^2} is the acceleration due to gravity. Then we find that t\approx1.5\,\mathrm s, at which point we find x\approx20\,\mathrm m.

6 0
2 years ago
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