Plants use visible light for photosynthesis. Visible light ranges from low blue to far-red light and is described as the wavelengths between 380 nm and 750 nm. The region between 400 nm and 700 nm is what plants primarily use to drive photosynthesis and is typically referred to as Photosynthetically Active Radiation (PAR). Plant biologists quantify PAR using the number of photons in the 400-700 nm range received by a surface for a specified amount of time, or the Photosynthetic Photon Flux Density (PPFD) in the units μmol/s.
During photosynthesis, plants take in carbon dioxide (CO2) and water (H2O) from the air and soil. Within the plant cell, the water is oxidized, meaning it loses electrons, while the carbon dioxide is reduced, meaning it gains electrons. This transforms the water into oxygen and the carbon dioxide into glucose. The plant then releases the oxygen back into the air, and stores energy within the glucose molecules.
Well I can’t tell you the exact gene would need to be changed, there are plenty of animals able to deal with the salinity of sea water. Basically the kidney has a certain ability to concentrate salt. The greater the ability the saltier the animal’s urine is and the saltier water can be consumed.
If I am remembering correctly the threshold for salinity of water that can be consumed is up to a level just below that of the threshold of the kidney’s ability to concentrate salt.
So we would not need “super kidneys” just normal ones similar to those of one of many animals in the sea. But we might have a shortened kidney lifespan because of it.
<span style="color:#323232;">May be useful:
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So that’s why, Sync tries to load this as an image, but there’s no image link…
But apparently the link still works and is invisible in Sync.
Well I do get a code block so I'm not sure if that's Kbin-specific markdown but I also get a bunch of HTML inside it too. I can't view the raw comment like I could with RES, but I would suspect that the raw Markdown characters should be sent with the post, for the website to support in its style-specific way, or something. This seems more like someone's client or app inappropriately inserting display style information into the post? But I could be totally wrong, I'm not an expert
Being big is advantageous as long as the animal is able to find enough food to sustain itself. Food was plentiful at the time, so dinosaurs grew quite large.
In modern times, most mega fauna is gone because Humans hunted them to extinction.
I think it would ultimately depend on a use case for that metric, otherwise you're putting the cart before the horse. There are many measurements and calculations you could come up with, but no obvious (to me, anyway) interpretation of "most discontinuous": something is either in one piece or not. If you needed a metric like this for a practical purpose, your specific needs would be a starting point for designing one. If it's more of a shower thought, you sort of have "too much freedom" to be able to define anything that's necessarily meaningful.
Simple examples would be just "number of 'discrete parts'", "minimal area needed to span all territories" and things like that. Maybe you're more interested in "total distance from all satellites to wherever the capital is" or something, in a different context. The point is they'd all tell you radically different things, so it's important to know which one to ask for.
You could argue that something like Hawaii and Alaska's distance from the rest of the US makes the US score highly.
You could argue that any number of island nations score highly because after all, most of e.g. the US is in one part.
You could argue e.g. Norway's territories near both poles make it pretty high-scoring too.
You could argue that for whatever reason, distribution of area and population matter, and so on.
Unhealthy, processed food is cheaper to produce, cheaper to buy and more appealing to the consumer. Couple that with a society which is trending towards a more sedentary lifestyle and obesity rates climb. The issue is we’re seeing the results of this about 20 years too late to do anything to reverse the effects.
Should mention that I’m in the UK, but the story is a similar one; albeit currently slightly less extreme.
Newton’s 3rd law applies. Both vehicles experienced an equal force of impact because it is implied there were no other relevant external forces, but their individual accelerations will be different.
With even a small increase in uncertainty in electron position, electricity would start to behave differently. Everything electronic, which depends on electron flow through very tiny conductors, would become unreliable as the electron flow would be unpredictable. Even basic light bulbs probably wouldn’t work.
Chemistry is the exchange of electrons between atoms. All molecular bonds happen through the exchange of electrons from one atom to another… so an increase in uncertainty would result in the bonds breaking down. Molecules would break apart, every material you think of as solid would disintegrate into its base atoms.
So, your rolling ball would cease to be a cohesive ball, and the surface it is rolling on would also cease to be.
This post on physics.stackexchange.com claims that hot accretion disks can reach 10^12^ K, hot enough for fusion to occur but not at rates that would make fusion a significant source of radiation from the disk.
I suppose the elements formed by fusion of hydrogen are pretty fixed. It’s mostly helium. Only the rate depends on the pressure and temperature of the fuel.
So I don’t know whether fusion happens in an accretion disk. But if it does, the elements formed are mostly helium, and a few other light elements at trace amounts.
It’s basic math. You can do the gravitational calculations yourself. Basically any sphere of uniform density is going to exert gravity uniformly. So if you’re in the center the pull from the mass on any direction will be counteracted by the pull in the opposite direction. It’s one of the basic introduction to physics calculus examples.
So to your question about what the zone of negligible gravity would be, you can define negligible gravity, and then figure out how large that zone would be based on the material on the outside of the shell.
Basically the further you get from the exact center of the sphere, you’re going to have more gravity from the closer edge pulling you, and less gravity from the further edge offsetting that. So there’ll be a gradient of increasing gravity as you get further away from the center
This feels very close to answering the question in a way my brain can interpret it. So, going outward makes complete sense to me but the area at the center, the way I under your answer is, yes, the area or zone will increase proportional to its mass?
This may be asking too much, but, have any idea the size of that low gravity zone of earth bs our Sun?
I can’t answer that question for you. Because you’re using a relative term. Only the exact center will have no gravity. Anything outside the exact center will have some gravity. So you have to define what negligible it means.
So once you define low gravity. You can do the math to figure out the size of that zone of low gravity.
I highly recommend doing the math anyway. Follow along with a YouTube example or a written example on gravitational attraction of a sphere. It’s really good calculus. Then you can you know put it into octave and get the exact answer for yourself. Just plugging in numbers for the relative density of the Earth and the mass and the sizes. These will be approximate of course. Because nothing is perfectly uniformly dense so it’s just a rule of thumb anyway
Depends what you’re trying to do. If you want to balance something so it never moves you can only use the exact center. If you want something to stay relatively in the center for a period of hours then you’re going to have a much larger area. If you’re okay with minutes it’s going to be much much larger area. If it needs to be stable for years in the area is smaller. Gravity is going to apply a force of acceleration and on an object, and if there’s nothing resisting that acceleration things will just fall off the center. You know imagine trying to balance something on top of a cone.
My original question stemmed from thinking about the possible different area sizes of low gravity within different size stars - and if that area was gradient.
Not quite. If by “edge” you mean the surface of the earth, then the force of gravity from the closer edge will always exactly offset the gravity from the farther edge. So if the earth were hollow, then you would experience zero gravity at any point in the hollow portion.
Of course, the earth is not hollow. And any mass under you (i.e. closer to the center than you are) will not be offset, and all of it will pull you towards the center. As you move further away from the center, more of the earth’s mass will be closer to the center than you and therefore the force of gravity will increase.
So if the original poster wanted to have a 10km sphere in the center of the earth of zero gravity (earth gravity at least), then all they have to do is hollow out 10km and they are good to go?
Yes you can, as long as you are inside a perfectly spherical shell.
The net gravitational force on a point mass inside a spherical shell of mass is identically zero! Physically, this is a very important result because any spherically symmetric mass distribution outside the position of the test mass m can be build up as a series of such shells. This proves that the force from any spherically symmetric mass distribution on a mass inside its radius is zero.
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