So the answer is approximately like this? The molecules move randomly, but when they collide, they lose kinetic energy and slow down. Then they start moving randomly again, but collide again and again lose kinetic energy. Therefore groups of particles that have collided with each other will tend to stay close to each other because they keep colliding and losing kinetic energy.
Patashu: very possibly! Thanks for sharing that video. It seems like it's not so much a binding or attractive force as it is an emergent property of having lots of inelastic collisions, as Warp describes.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
If you have a spinning wheel, time passes more slowly for the outer edge of the wheel than its surroundings, as a consequence of relativity.
This is related to the consequence of relativity, that if someone were to travel far from Earth and then come back, he would have aged less than people here (the reason for this having to do with acceleration and changing frames of reference.) There's nothing that would stop this from working at a smaller scale: If you are standing at the end of a football field, and a friend runs to the other end and back, he will have aged less than you (even if it's just by a staggeringly small fraction.) We can make the setup even smaller and have just a spinning wheel: Every point on the edge of the wheel will travel farther and then closer. Therefore the outer edge of a spinning wheel will age slower than its surroundings.
So my question is: How fast would a wheel have to spin in order for its outer edge to age, let's say, 10% slower than its surroundings? (In other words, for every 10 seconds that pass, the edge of the wheel only ages 9 seconds.)
So my question is: How fast would a wheel have to spin in order for its outer edge to age, let's say, 10% slower than its surroundings? (In other words, for every 10 seconds that pass, the edge of the wheel only ages 9 seconds.)
I don't know the answer, but your question prompted me to find the relativistic rocket which just blew my mind.
My only contribution is that you probably need to consider the acceleration (from centripetal force) as well as the velocity. But I could be wrong.
I know when you're working out projectiles at school you're always doing it without factoring air resistance. Does anyone know how to factor it into calculations?
Air resistance is, broadly speaking, proportional to the square of speed and to the surface area of the object in the direction of motion. That squared term is why it costs so much more to get to e.g. 80MPH than it does to get to 60MPH.
As such, any attempt to involve it immediately gets you into some rather tangled calculus. And that's ignoring the drag effects of turbulence behind the object.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
I know when you're working out projectiles at school you're always doing it without factoring air resistance. Does anyone know how to factor it into calculations?
Pages and pages of complex equations being simulated by a supercomputer. And even then it's just an approximation.
So my question is: How fast would a wheel have to spin in order for its outer edge to age, let's say, 10% slower than its surroundings? (In other words, for every 10 seconds that pass, the edge of the wheel only ages 9 seconds.)
In SR, dT²=dt² - dx² - dy² - dz². Changing to the coordinates of a frame rotating with angular velocity w, we have x=r cos(theta - wt), y = r sin(theta - wt), after evaluating the differentials with some algebra, we get to dT² = (1-w²r²)dt² -dr² - r²d(theta)² - dz² - 2wr²dt d(theta). The metric becomes singular for wr larger than the speed of light, but since spacetime is still flat, it's just a coordinate singularity, not a physical one. Bodies at rest in this frame are rotating at angular velocity w in the inertial one. So, setting dr=dz=d(theta)=0. We have dt/dT = 1/sqrt(1-w²r²).
tl;dr It's (maybe surprisingly) the old time dilation equation dt = gamma*dT. For the situation in the statement the inertial frame must measure a clock cycle at the edge 10/9 larger. Thus, dt/dT = 10/9, which implies v ~ 43,6 % of the speed of light.
Mitjitsu wrote:
I know when you're working out projectiles at school you're always doing it without factoring air resistance. Does anyone know how to factor it into calculations?
For a more complete answer, there's a parameter in fluid mechanics called Reynolds number. It measures how strong are the resistance forces compared to the body's inertia. This number is important because its value indicates which approximation to use. For low Reynolds numbers, drag can be considered proportional to the body's velocity. For higher ones, it is proportional to the square of velocity.
For very high Reynolds numbers, turbulence has a considerable effect. At this point, chaos happens. You could try to derive some solutions, but you wouldn't be able to use them for anything useful. According to my aeronautics engineer friends, what they do is make some big restrictions to the pressure, temperature, etc. of the fluid, make experiments, run simulations in supercomputers and try to derive formulas using numerical data. The problem is that those formulas are very complicated and only work for a very specific subset of the problem. With current knowledge, no unified way of dealing with turbulence exists.
If you're talking about carbonated water/sodas suddenly turning to icy slush in your bottle or glass, then it's because excess heat is removed when bubbles form and dissipate, cooling the solution down (which has a lower freezing point than normal water, anyway) below the freezing point and creating an easy slushie. It's effectively nucleate boiling at low temperatures; gas particles have this wonderful tendency to trap heat in a substance, and when they build and expand on their way out, they tend to take a lot of that heat with them. It's also why you'll see some nuclear reactors using very minor, controlled boiling in their core; it creates more heat transfer, thus making the energy transfer from point A to point B much more efficient.
If you're talking about actually carbonating your water, creating your own sodas and such... I have no idea. :|
If you're talking about actually carbonating your water, creating your own sodas and such... I have no idea. :|
Home carbonating devices consist basically of a bottle of compressed CO2 that shoots the gas into the water. The water gets carbonated.
When doing so, especially if the water is cold, ice tends to form for a short period of time (before it quickly melts.)
If you're talking about actually carbonating your water, creating your own sodas and such... I have no idea. :|
Home carbonating devices consist basically of a bottle of compressed CO2 that shoots the gas into the water. The water gets carbonated.
When doing so, especially if the water is cold, ice tends to form for a short period of time (before it quickly melts.)
Oh, wait. Yeah. It's because you're taking high pressure gas and causing it to reduce in pressure. When a substance undergoes a rapid pressure change, temperature changes as well; compressing something increases temperature while reducing the pressure causes temperature to drop. It's the same reason a can of compressed air used for cleaning out computers and electronics or a can of spray paint will start getting really cold if you just hold the release for a long time.
If you drop the temperature of cold water, it wants to freeze, and so you'll get some ice crystals to form, but when the chemical reaction takes place to create carbonic acid inside the solution, the freezing point drops back below the temperature of the water, and your ice melts.
Snooker rule:
"When the nominated colour is potted, the player will be awarded the correct number of points. The colour is then taken out of the pocket by the referee and placed on its original spot. If that spot is covered by another ball, the ball is placed on the highest available spot. If there is no available spot, it is placed as close to its own spot as possible in a direct line between that spot and the top cushion, without touching another ball."
So, how close is "as close as possible without touching another ball" in terms of physics?
Snooker rule:
"When the nominated colour is potted, the player will be awarded the correct number of points. The colour is then taken out of the pocket by the referee and placed on its original spot. If that spot is covered by another ball, the ball is placed on the highest available spot. If there is no available spot, it is placed as close to its own spot as possible in a direct line between that spot and the top cushion, without touching another ball."
So, how close is "as close as possible without touching another ball" in terms of physics?
In a literal sense it would be one atom. However, snooker is not a sport where the rules tend to be strictly applied. In a disputed situation the referee will usually act more like a mediator, than an enforcer. Its not a complicated or subjective rule IMO. You just place the ball as near as possible to the black spot while still being in line with the brown spot.
In a literal sense it would be one atom. However, snooker is not a sport where the rules tend to be strictly applied. In a disputed situation the referee will usually act more like a mediator, than an enforcer. Its not a complicated or subjective rule IMO. You just place the ball as near as possible to the black spot while still being in line with the brown spot.
I think you are missing my point. I was hoping for the conversation to go towards the theoretical discussion on what it really means for two objects to "touch" each other in physics, and how close you can theoretically get before reaching this definition.
As a matter of practice, I think it's reasonable to say that two objects are touching when you can only get their center of masses closer to each other by causing one or both of them to deform under the force you're applying.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
But how close can you theoretically get without touching?
Speaking of snooker... I find some instances of spin rather incredible. For example, there are cases where the player makes a long pot by making the white ball travel the entire length of the table, with a strong backspin, and when the white ball hits the nominated ball, the backspin kicks in and makes it come back all the way to almost its starting point (without colliding with any walls in between or anything like that.)
How can the white ball travel the entire length of the table, all the way with a backspin strong enough that after it hits the other ball, the backspin is still able to make it traverse the entire length of the table again, in the other direction? One would think that contact with the table would eliminate the backspin completely (the only effect the backspin would have being that it slows down the ball during the first trip.)
The cueball is skidding/sliding on its first pass; contact with the target ball slows the cueball down enough that it's able to grip the table surface and let its spin actually move it. The sliding removes negligible energy from the ball (just some heat loss due to friction, but the ball is quite smooth), so it still has most of its spin when it starts making the return trip.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Both poles of a magnet will attract iron. Given that magnets can attract and repel each other (depending on the orientation), one could easily think that there would exist a material that both poles of a magnet would repel. But as far as I know, there is no such material. Why is this so?
The school-explanation I was given was this:
Imagine that the magnet (and any magnetic material) contains millions of very tiny mini-magnets, each having their own north- and south pole. In a permanent magnet, all these mini-magnets are aligned in the same direction, all the tiny fields add up to a stronger magnetic field that's measurable outside the magnet.
If you snap a magnet in half, you get two parts, each consisting of half the amount of mini-magnets, but they're still aligned, adding up to two new magnetic fields. Each of the parts is a full magnet, just with half the strength. You don't get a part with a north pole and a part with a south pole, but two whole magnets with two poles each.
In unmagnetized (but magnetic) material, like iron, these mini-magnets are ordered randomly. Their fields go in random directions and cancel each other out.
Now what happens when you put iron near a magnet? The magnetic field enters the iron, aligning its mini-magnets in the direction of the magnetic field. Now the iron is a magnet of its own, and it's attracted the same way two magnets would.
Remove the magnet, and the iron's mini-magnets reposition themselves randomly, losing the magnetic properties of the iron. (though I'm sure you've heard of ways to keep the magnetism inside the iron for a while longer, right?)
So hold any magnetic material near a magnet, it'll align with the field and will always attract.
Of course I've been lying.
There actually are some very rare materials that align themselves against the magnetic field and get repelled.
Why would they do that? At this point, let me mumble something about quantum mechanics and refer you to the more advanced literature. :p
I don't know the physical representation of those "mini-magnets", nor the proper term, nor the actual physical and quantum physical effects that are working on them. I tried reading up on it once and got a headache, so I'll stick to the simplified model.
The effect is known as diamagnetism. Essentially, it is energetically favorable for electrons in a diamagnetic material to have their magnetic fields aligned opposite an applied magnetic field.
It isn't all that rare, either. Water is diamagnetic and in a field of about 16 Teslas, you would levitate!
Will you be dryer if you walk or run in the rain, and what effects would the intensity and the angle of the rain falling have on this theory? Logic to me would suggest if the rain falling perfectly downwards then it would make no difference if you were to run or walk due to being hit by an increased volume of rain when running.
