Like my comments above say, "atomic bomb" is usually used to refer to fairly primitive designs that don't make much, or any, use of fusion fuel.

It almost entirely comes from the weapon itself.

>how much hydrogen is actually used in a hydrogen bomb

That's not going to be public knowledge, and a lot of those back-of-the-envelope sort of calculations end up making unfounded assumptions, so i's not a surprise that answers differ so wildly.

In terms of the relevant reaction Q-values (the amount of energy released by one single reaction), fission does on the order of 10 times better than fusion.

But if you divide the Q-value by the mass of the fuel particles, fusion does better than fission.

People often get that confusion, because they've heard that fusion releases more energy per unit fuel mass, and are then surprised to find out that fission releases much more energy per reaction.

And then in a weapon, the crucially important role that fusion fuel plays is that it produces fast neutrons that can induce more fission in an already supercritically-multiplying system of fission reactions. Each additional neutron produced by fission therefore has the chance to cause many more fission reactions, each of which comes with on the order of ten times more energy than the initial fusion reaction.

So thermonuclear weapons are able to much much more yield for a given total fuel mass.

According to your definitions then, all nuclear weapons would be called "nuclear" (great, this is right), but then all chemical explosives could be called "atomic" (highly confusing, given that the term "atomic bomb" usually refers to a subset of nuclear weapons).

Amount of energy released by a single reaction.

Yes, that's what I was saying. There are weapons that are loosely referred to as "pure fusion", but they don't actually purely consist of fusion fuel. Some amount of fission fuel is needed to ignite the thermonuclear burn of the fusion fuel.

Yes, but the same could be said of many "true thermonuclear" weapons too. The main contribution of the fusion fuel to the yield is due to the high-energy fusion neutrons inducing more fission, rather than the fusion reaction Q-values, which are on the order of 10 times lower than the relevant fission Q-values.

All nuclear weapons use fission in some way. Even "pure fusion" weapons are not really pure fusion.

Reaching a state where a thermonuclear burn can take place, outside of stars and certain scientific research devices, requires fission.

Modern designs are all thermonuclear (fully thermonuclear, not just boosted fission). But not all countries have modern stockpiles. So we still need words to refer to old designs and differentiate them from modern ones.

Yes, I agree that boosted weapons are technically "thermonuclear" too, just with a much smaller amount of fusion fuel. I don't think the common terminology is good, but this is what it is.

"Atomic bomb" and "hydrogen bomb" are not really modern terminology, but they're both subsets of nuclear weapons.

But generally speaking, "atomic bomb" refers to nuclear weapon designs which are either fission-only, or boosted fission.

And then "hydrogen bomb" (also known as thermonuclear) refers to weapons which derive a significant amount of their yield from fusion reactions, in addition to fission.

>Moreover, is hydrogen bomb more powerful and destructive than a nuclear bomb?

"Hydrogen bombs" are a type of nuclear weapon. And generally higher yields can be achieved with them than with "atomic bombs" (fission-only or boosted fission), for a given mass of fuel.

Light has no mass, so all of its energy is kinetic.

That is a quadratic function of v, not exponential.

The pressure on the surface of the window is equal to the sum of the static pressure and the dynamic pressure far upstream. And the dynamic pressure is proportional to v^(2), where v is the speed of the wind.

https://en.wikipedia.org/wiki/Nuclear_cross_section

You can think of it as the probability that a reaction occurs.

Lower Coulomb barrier, which means that the cross section begins to increase at lower energies, which means that the convolution of the cross section with a Maxwellian distribution function is higher at lower temperatures.

It could certainly work, it just hasn't been done yet at large scale.

It means that the Coulomb barrier is a little bit lower. It's unrelated to the stability tritium, it's just possible to make this reaction occur at a reasonable rate at lower temperatures.

The cross section for that reaction is very low over all energies. DT and DD have the most favorable Maxwellian-averaged cross sections as a function of temperature, meaning that they “turn on” at the lowest temperature, and therefore, it’s easiest to create the required conditions in a reactor.

What you're looking for is the Poynting vector.

It's pictured here for a DC circuit. The red arrows are the electric field, the green are the magnetic field, and the blue are the Poynting vector (the direction of power flow).

In the AC case, the red and green arrows change direction sinusoidally, but the blue vectors always point in the same direction; from the power source to the load.

It's not because they necessarily "interact more readily", it's just that the kinematics is more favorable when they do interact.

If you want to slow something down, you want to take as much kinetic energy away from it as possible with each collision, and simple kinematics shows that the optimal way to do that is for the neutron to collide with a nucleus of roughly the same mass, so ideally a proton.

That's why hydrogen, and hydrogen-containing compounds are very good neutron moderators. The lighter the nucleus the better.

There's not really any meaningful sense in calling a neutron star a giant nucleus. A nucleus is bound by the residual strong force, and the heaviest nuclei have radii on the order of tens of femtometers or so.

A neutron star is bound by gravity, and has a radius of around a few kilometers.

>What happens when an object enters Earth's atmosphere traveling faster than its terminal velocity?

Drag slows it down toward its terminal velocity.

>As a side question, what would happen if something enters another planets atmosphere, say, Jupiter, faster than it's relative terminal velocity?

Same thing, just with its terminal velocity on Jupiter.

Liquids clump up like that because of surface tension. For gases, there is not really any surface tension that would cause that.