swansont

For case 3 the mass travels less than 6 meters. Last I checked that was finite, and bounded.

Opinions aren’t necessarily supported by evidence. I think chocolate is better than vanilla. Am I a cult?

You aren’t making it obvious what you already know, given your mistaken and/or vague assertions. No, electrons are in orbitals that are mostly outside of the nucleus, and protons and neutrons are in the nucleus; they do not “make up an atom’s electrons”. Proton number dictates the number of electrons in a neutral atom, because the magnitude of charge on each is equal. Neutrons, being neutral, have no effect on the number of electrons, and have a limited impact on chemistry. (they e.g. affect reaction rates because more massive objects at a given energy move slower)

! Moderator Note Then post it in speculations, and beef up the rigor, because there’s not much here

! Moderator Note You posted this in mathematics. Where’s the math?

The problem is that it has to fit with all of physics, not just cosmology.

You are making claims about chemistry. Like it or not, the bulk effects of an atom interacting with another generally falls under the umbrella of chemistry. And the electron structure of an atom dictates the chemistry

Argon has more electrons than oxygen. Doesn’t mean you can substitute one for the other.

Yes. Nuclear reactors don't undergo nuclear explosions. AFAIK the reactor in question is not the same design as the Chernobyl reactor (which underwent a steam explosion and did not have a containment vessel), so comparisons to it are limited This is not to say that bad results won't happen from bombing a nuclear plant, but "the end of Europe" is hyperbole. Chernobyl didn't end Europe, just as Fukushima didn't end Japan.

! Moderator Note Technical aspects of the discussion on the reactor have been split https://www.scienceforums.net/topic/126806-nuclear-reactor-technical-discussion-split-form-war-games-russia-takes-ukraine/

Money is probably a factor. You need to buy these items and maintain these systems, both of which cost money. In a pressurized water reactor a turbine is in a second loop. You have more opportunity for leaks with each new penetration. Plus the fun of either potentially having a thermal shock if the loop is cold and all of the sudden you fire it up, or if you keep it hot you are wasting some of the generated heat, making the plant less efficient. It's risk/reward, which is skewed by the nature of the beast: you want to make money, so a certain amount of value engineering goes on, cutting back on costs that are deemed unnecessary. Are four layers of redundancy required, or can you get away with three? And if you are having problems with the reactor, perhaps it's best not to rely on the reactor itself. Some reactors can use natural convection and not rely on pumps, but it might not work on a commercial scale plant. Decay heat can be something like 7% of full power, so if you have a 1 GW plant, that's 70 MW you need to remove - that's a lot of water that needs to be moved, and you aren't pulling the energy out by driving a steam turbine. I'm not sure how much "broken by an invading force" gets considered in the design of a plant.

Humans ovulate without the intervention of a male, too.

No. Superluminal speeds are inconsistent with causality: there are situations where you can get a response to a signal before you send the signal. Depending on what you mean by parallel dimensions, they are science fiction or an interpretation of QM. Not actually science itself.

Light moves at the same speed as gravity because photons have no mass. Massless things move at c. Massive things move at speeds slower than c.

! Moderator Note Moved to Science News

Different kinds of lead? Which sidesteps the question about cost. None of this addresses my post. What is your point? I don't think that was it. The difference in chemistry was addressed immediately by exchemist. Instead of asking for a followup, the OP doubled down on dubious claims trying to support the notion. I suspect that's the source of any real antagonism. The title was just foreshadowing.

And how does that change the cost of procuring and processing? In a word, no. Whatever you read you interpreted incorrectly. Fragmentation makes for dust which can be inhaled. Alpha decayers are very destructive internally, but not so much as an external dose. More dangerous/more damaging is NOT the same as more radioactive DU is less radioactive than natural uranium of equal mass (U-235 has a shorter half-life)

He fights to overcome a stutter, which isn’t about energy.

Where? No, that means you have fewer atoms for the same mass, which (all else being the same) means fewer chemical reactions

When your prediction fails it’s a clue that your idea is flawed. This is a rather spectacular failure, befitting the flaw

Dividing by zero is undefined, which is what happens when you try and use a crappy definition, which is what W/g is.

No, since mass=gravity/weight is nonsense, as I have explained in another thread. Nothing valid can be derived from it.

First of all, let me say this: post non-mainstream material (i.e. any pet theory) in speculations You need to clarify what you mean by gravity in this equation. Typically we use variables that are precisely defined. I can't really come up with a meaning that's correct. Weight is a force, meaning W = mg, where g is the gravitational acceleration. So m = W/g, but you have weight in the denominator. So you are claiming the gravity is weight * mass, or m^2*g, and I don't know what that means. And since that's nonsense, everything derived from it is nonsense.

I don't know how you would expect me to do that, when I have told you that it's not, and conditions under which it fails to hold.

c is invariant, which means it's the same in all inertial reference frames. If you are at a fixed location in a gravitational field, you are in an accelerated frame, not an inertial one. You will locally measure c to be the same, but that might not be the case if you measure it in reference to some other frame. The difference, though, might be negligible with regard to the precision of a measurement you are making.