Ridley_Himself
Ridley_Himself t1_jd5po6a wrote
Reply to comment by El_Sephiroth in What the hell is the actual difference between an isotope and a nuclide?? by amypinecone
Bar is a unit of pressure, but it is still derived from the Greek root barus, meaning heavy. So we get a few words from the root such as barometer, isobar, baryon (“heavy particle“), barium (due to the high density of barium minerals).
Ridley_Himself t1_jd16gt8 wrote
Reply to comment by drhunny in What the hell is the actual difference between an isotope and a nuclide?? by amypinecone
I had thought it came from them having about the same atomic weight iso=same, bar=weight/heavy.
But then baryon has the same root.
Ridley_Himself t1_jci18bb wrote
Reply to Energy can not dissappear or be created, only change form, right? Earth is blasted by the sun 24/7, where does the excess energy go, because I guess it doesn't stay here or we'd cook by mr_greenmash
Heat can be transferred in four basic ways: conduction, convection, advection, and radiation. While the first three require a medium, radiation can travel through a vacuum.
All objects will emit thermal radiation at an intensity and range of wavelengths dependent on their temperature. Earth’s thermal radiation is infrared.
Ridley_Himself t1_jce1e0z wrote
Reply to comment by Hiddencamper in Why were the control rods in the reactor featured in the HBO series 'Chernobyl' (2019) tipped with graphite? by Figorama
So, if I'm reading this correctly, the design flaw is that, if you need to insert the control rods to kill a reaction, the graphite tips have to move past the portion of the fuel with the highest rate of reaction?
Ridley_Himself t1_jcc4gt0 wrote
Reply to Was T. rex's skull bulletproof? by aesthetic_rex
If soldiers are coming in with the intent of killing T-rex, they would likely have equipment for it, and gun designs were developed for the express purpose of hunting large animals such as elephants.
Funny enough, I just stumbled across a rifle cartridge called a .577 Tyrannosaur.
Ridley_Himself t1_jbyvn95 wrote
Reply to comment by Oheligud in As they still have a neutral charge, can antineutrons replace neutrons in a regular atom? by Oheligud
The issue here is that protons, neutrons, and their respective antiparticles are not elementary particles; they are made of quarks and antiquarks respectively bound together by gluons. A proton contains two up quarks and one down quark. An antineutron contains two down antiquarks and one up antiquark. A quark from the proton and an antiquark from the antineutron would annihilate and produce mesons from the remaining (anti)quarks, which would quickly decay.
Ridley_Himself t1_jbu8224 wrote
Yes. Radioactive decay can also be expressed by a decay constant, λ, which is effectively an instantaneous decay rate (relative to the amount present).
λ=ln(2)/t1/2.
It gets a bit more complicated with Y-90 since it is itself a decay product of Sr-90, so it will be produced at the same time it decays.
Ridley_Himself t1_jb7xorj wrote
Reply to If I took a cup full of the stuff that exists at the exact center of the Earth, and cooled it, what would I have? by [deleted]
So, it's not just a matter (no pun intended) of density but one of certain chemical affinities.
There is something called Goldschmidt classification. It's rather outdated but can describe the behaviors of elements in very broad strokes.
Essentially, it classifies elements based on which "phase" they preferentially enter. Lithophiles prefer a silicate rock phase, chalcophiles prefer a sulfide phase, siderophiles prefer a metallic phase, and atmophiles prefer a gaseous phase. In this scheme gold is considered a siderophile and would enter a metallic phase. The iron-rich metallic phase in Earth was denser than the silicate rock, so the former sank to form the core while the latter floated to form the crust and mantle. So thinking in these terms, there would be gold in Earth's core, but it wouldn't necessarily separate from the nickle-iron alloy of the core, especially since it probably only exists in trace concentrations. Of course, you'd have to determine experimentally how gold would behave under the pressure and temperature conditions present in the core.
Goldschmidt classification is nowhere near absolute, which is why we can have gold in the crust.
Ridley_Himself t1_jb5weu9 wrote
Reply to What happens at the end of a subduction zone? When the entire plate subducts? by kittens0423
If the last of a plate is subducted, the result is a boundary between the overriding plate and whatever was on the other side of the subducted plate. For instance, the west coast of North America was once a subduction zone between the North American Plate and the Farallon Plate. Eventually, the Farallon Plate was almost completely subducted and North America met the spreading center between the Farallon and Pacific Plates, forming the current boundary, which includes the San Andreas Fault.
If, instead of a spreading center reaching the trench, the plate includes continental crust, then a continental collision results, as in the case of India colliding with Eurasia.
Ridley_Himself t1_jdeglys wrote
Reply to comment by CrustalTrudger in With the extraordinary amount of precipitation that has fallen on California, would that weight have any effect on the tectonic plates/fault lines and could it cause a major earthquake? by barfly2780
Loads aside, I’d heard of pore fluid pressure from heavy precipitation as a possible mechanism for making earthquakes more likely.
I might have expected increased loading to increase earthquakes in areas of normal faulting. All the locations you mentioned on that subject are at convergent boundaries.