Bender

This is something I've been wondering about as well: can potential gravitational energy also be seen as mass, and if so, what is the reference? Moving them arbitrarily close together, the potential energy can become any arbitrarily large negative figure.

While the original question is adequately answered, I like to add an interesting point (at least, I think it is interesting): the gun used the same amount of energy to fire the bullet, there is no difference in the combustion process, but to someone standing besides the rail, the bullet now has no kinetic energy. Where did the energy go?

Bernoulli's equation is what you need: if a fluid flows at a higher speed, the pressure is lower. See Strange's link above for a good explanation.

I feel your pain (at least I did last time I went to the dentist who successfully anaesthetised half my face, but somehow not my tooth.)

I guess he means that the universe is flat. I have no clue what is asked, though.

Yes, but I thought Michell deserved mentioning because of how sad his story is.

No oxytocine, no love, no babies. Why would an individual only interested in his own survival procreate? Seems like a huge liability, hindering mobility and camouflage. Sure, one such individual could function in a social group by rationally weighing his actions, but there wouldn't be a social group if all the individuals were that way. Even the most solitary of (sexual) animals have hormones to make them attracted to others for at least short periods of time.

The orbits where known long before Kepler, by measuring the position of the planets in the sky with ever increasing accuracy and fitting them to a model (either geocentric or heliocentric models were used). I think Kepler was the first to speak about elliptical orbits, as his first law describes (at least, he was the one to become known for this, as happened often throughout history of science, he may not actually have been the first). In fact, it was John Michell who built the setup, but he died when he was finished and Cavendish executed the experiments. Those were very impressive measurements and it took a long time to improve them; probably the most impressive experiment predating the digital age. Even now our best estimate of G is barely more accurate.

How are Schrödinger and Dirac equations in conflict with Newton's second law?

The satellite would rotate in the same direction as the prop without countermeasures, like what MacSwell said. Anyway, there would be some motion between fan and satellite, so some energy could be generated. It is not like it is a very practical idea anyway.

Why does the 2nd law assume that it is not reversible? Part of the rocket energy is converted to entropy (which you cannot recover), the other part is converted to kinetic and potential energy of the satellite (which you can recover partially).

The wind turbine will generate less electric energy than the satellite has kinetic and potential energy at the start. In practice, it will probably be a lot less.

I think it does apply if you write it as [math]F=\frac{dp}{dt}[/math]

Sure. Humans could too (btw, I retract my previous statement about slowing down in lower orbit and instead point to Janus' clear and more correct explanation)

F and v are vectors in that equation. You are applying 5 N at one side of the cylinder and -5 N at the other, for a resulting force of 0 N, as evidenced by the fact that (ignoring the fall) the cylinder remains at the same spot. Each part of the cylinder has a linear momentum mv, but adding them all together in a vector sum, the resulting linear momentum is 0. In fact, for a small particle, the linear and angular momentum are interchangeable. Just multiply your equation with the radius r and you get [math]Frt=mvr[/math] split up v [math]Frt=mr^2 \omega[/math] And rewrite with torque M and moment of inertia I [math]Mt=I \omega[/math] Or where did you think the formula's for calculating angular momentum or moment of inertia come from, if not by deriving them from the formula's for linear motion? I suggest at least reading the wikipedia article if you are genuinely interested in the subject. You will never convince anyone if you don't even know the basics. Of course, the last equation is more practical for rotating objects because integrating all the little parts doesn't yield 0. (in case you care, it is because in vector-notation all the vectors of angular momentum are now pointing in the same direction, orthogonal to the original force) Also, the correct abbreviation for seconds is s, not sec. Edit: in case you are unfamiliar with summing vectors, look at this image: a) shows what happens if all vectors (eg forces) act in the same direction. b) shows how to add vectors with different direction: you can draw them end to tail, and the resulting vector is going from the start to the end. c) is the situation we are discussing: the end of the last vector arrives at the start of the first, so the resulting vector is 0. If this is not clear, you are welcome to ask more detail.

No, Janus is right: for a lower orbit, you have to slow down. Your speed will be lower, but the distance you need to travel is also smaller, since it is a smaller orbit. Net result is that you go around the Earth faster. Simply speeding up or slowing down will result in an elliptical orbit without course corrections. Just curious: Why would there be a difference when AI's want to build a megastructure vs humans wanting to do the same? Why would they put something in orbit at the wrong side of the earth and then move it to the other side?

[math]2\pi=\tau[/math] obviously makes the most sense as it appears most often.

Let's calculate the angular momentum in the two extremes (assuming a thin cylinder and small balls with respect to cylinder radius R): [math]L_1=4.5 R^2 \omega_1[/math] [math]L_2=2.5 R^2 \omega_{2,cylinder}+2 (R+l_{rope})^2 \omega_{2,balls}[/math] The kinetic energy: [math]E_1=\frac{4.5 R^2 \omega_1^2}{2}[/math] [math]E_2=\frac{2.5 R^2 \omega_{2,cylinder}^2}{2}+\frac{2 (R+l_{rope})^2 \omega_{2,balls}^2}{2}[/math] Since angular momentum and energy have to be conserved, we get two equations from which [math]\omega_{2,cylinder}[/math] and [math]\omega_{2,balls}[/math] can easily be calculated: [math]4.5 R^2 \omega_1=2.5 R^2 \omega_{2,cylinder}+2 (R+l_{rope})^2 \omega_{2,balls}[/math] [math]\frac{4.5 R^2 \omega_1^2}{2}=\frac{2.5 R^2 \omega_{2,cylinder}^2}{2}+\frac{2 (R+l_{rope})^2 \omega_{2,balls}^2}{2}[/math] Now let's look at the linear momentum: [math]p_{cylinder}=mv=m \cdot 0=0[/math] [math]p_{balls}=m_{ball}v_1+m_{ball}v_2=m_{ball}v_1-m_{ball}v_1=0[/math] As you see, there is no linear momentum. Could you clarify how you would make the calculations with linear momentum if there is no linear momentum? note: earlier, I assumed a quasi-static system where the balls are released slowly and remain at the same orientation with respect to the central mass, in that case there is indeed a loss in kinetic energy. There are no significant losses in the system in the video.

Assuming a conventional silicon transistor, and assuming the US "trillion" so 10 trillion is 10^13 I guess such a transistor can get down to at best 10 nm, and then it takes a signal at least 10^-16 s to get through at light speed. At least one electron still needs to jump from one atom to the next, so that might take longer. You can't make a transistor too small because tunnelling will start to kick in. I would say: perhaps. 10^13 electrons flowing in 100 nm² is a current of 10^13 x 10^-19 or about 0.01 µA/nm² or 10^4 A/mm². 0.7 V over each transistor means 10^10 W/mm^2 The density of silicon is 2x10^-6 kg/mm². Converting that to energy by multiplying with c² gives about 10^11 J. In short, the transistor would dissipate the same energy every 10 seconds or so than could be generated by converting its entire mass to energy. I would say: not likely With future technology: who knows?

I discovered this in the Dutch translation of a physics handbook (Giancoli), which is translated, but not adapted to the difference in grid.

On what basis do you make that claim? I see no violation of conservation of angular momentum or conservation of energy. It would really help if you attempted to do some math. Perhaps that way it would be easier to point out your misconceptions, at least if you are here to learn.

If the trap fails, the stench is unbearable. I don't see why an S-trap or a bottle trap would fail sooner than a P-trap. But, like I said, I understand the regulations for the systems I mentioned. Last year I renovated all my outdoor water drainage, according to regulation, but I cannot find a single rule for indoor systems. There are rules for not causing discomfort for your neighbours, including stench, and I think that should suffice.

Catching the load with the second flywheel is a nonelastic collision, conserving momentum but not necessarily kinetic energy. The difference in energy has to be absorbed somehow by the collision (possibly in a spring so you can recuperate it?). Why do you want to do this?

How is private plumbing any business of Minnesota? Does anyone know why they would regulate something like that? I have several bottle traps, which I installed myself. I had to replace a bottle trap installed by the previous owner, but the main reason was that the pipe leaving the trap went uphill, trapping water (and dirt) in that pipe. The bath tub and shower came with their own variations of bottle traps (installed by a plumber). I like that, because that way the hair gets stuck in the trap rather than somewhere down the pipe, as my bottle traps are all very easy to clean without having to disassemble any pipes. In modern showers there is simply no room for a P-trap. Btw, I don't live in Minnesota. I understand the need to regulate the electric circuit, because of the health risk, and the fresh water circuit, to prevent contamination. I even understand the need to regulate outdoor draining pipes for environmental purposes or to be able to connect to the sewer. But if home owners choose to mess up their indoor drainage, who cares?

Some examples: - the nucleus of an atom has less mass than the protons and neutrons separately - a molecule has less mass than the atoms it consists of separately (at least for stable molecules) - a charged battery has more mass than a depleted one The last two differences are, as far as I know, not measurable, which makes it easy to forget it is there. That is also the reason that a nuclear reaction is so much more powerful than chemical reactions, because there the difference is measurable.