# Tag Info

17

Arnold Neumaier's comment about statistical mechanics is correct, but here's how you can prove it using just thermodynamics. Let's imagine two bodies at different temperatures in contact with one another. Let's say that body 1 transfers a small amount of heat $Q$ to body 2. Body 1's entropy changes by $-Q/T_1$, and body 2's entropy changes by $Q/T_2$, so ...

17

Nature has no preferences, and therefore entropy tends to increase. Sounds paradoxical? The point is that each microscopic state (describing the exact position and velocity of each atom in the system) is equally likely. However, what we typically observe is not a micro state, but a course-grained description corresponding to incredibly many micro states. ...

16

To make it fall you need a torque. This torque is provided by the weight force acting on the center of mass of the object and by the offset between the center of mass and the edge of the object. Imagine your domino standing upright then tilt it. You are moving the center of mass. When the center of mass (blue) is on the right of the edge (red) then you have ...

14

From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness = inverse temperature $\beta=1/k_BT$. This changes continuously. If it passes from a positive value through zero to a negative value, the temperature changes from very large positive to infinite (with indefinite sign) to very large negative. ...

10

There was some doubt about Lubos' answer (which I've accepted), so this is just a verification. I copied the method Lubos described and found the potential difference for an ellipsoid with different eccentricities. Sure enough, for an oblate spheroid, if you make the center-equator distance a fraction $e$ larger than the center-pole distance, the ...

10

If you made the most perfect cone possible, so that its tip was a single atom, and stood it on the most perfect surface possible (a perfectly smooth, perfectly hard sheet of atoms), and completely removed all forces other than gravity, it would still topple. This is because those atoms are all jiggling around due to thermal motion. This effect fundamentally ...

7

The forces are never balanced, as there is only ever one force - gravity. The key is to remember Newton's second law: $F = ma$. Force and acceleration are paired, not force and velocity. Knowing just an object's current velocity tells you nothing about what forces are acting on it. There are two ways to see how the velocity goes to zero. Either the initial ...

6

If a plane is flying without any rudder input, then the banner will always fly straight behind the plane, with nose-tail-banner in a straight line, no matter what the speed or direction of the plane and/or wind. The only thing that affects the plane and banner is the flow of air over the control surfaces. How would the banner know that there was wind ...

5

It takes a lengthy proof, but Lyman Spitzer shows in the second chapter of Physical Processes in the Interstellar Medium (the standard text in interstellar matter studies) that the velocity distribution of interstellar gas particles (which is what forms nebulae) is very nearly Maxwellian - the deviation is less than 1%. Other larger systems, probably not so ...

5

Some engineering texts use "moment" and "couple" to talk about forces that tend to rotate an assembly (what physicist mean when they say "torque", but the engineers sometimes have a slightly different meaning for that word). A roughly translation guide is... A "couple" is a pair of opposite forces whose points of action are not co-linear. A couple is ...

5

Using the method of images, you can calculate the force between the ring of charge and the sphere. Assume the sphere is on the z axis with it's center on the point $z$, a radius of $R_s$ and the ring's radius is $R_r$ with a charge density $\lambda$. So $z$ denotes the center of the sphere. To calculate the force, you can replace the sphere with a charged ...

4

Think of the layer of skin on the finger tip which is in contact with the object that has a lower temperature than the body. If the object has a high heat capacity and a high thermal conductivity (like metal), then the skin of the finger will come to an equilibrium temperature that is lower than if it is in contact with an object that has a lower heat ...

4

Strictly speaking there are no reversible processes in Nature; it is an idealization that enables one to get bounds on efficiency of nonequilibrium processes by using techniques of equilibrium thermodynamics only. A reversible process is therefore primarily a theoretical concept for discussing what would happen in a process if dissipation were absent. It ...

4

You gave an expression for the force, whereas Prathyush's answer treats it as the potential. If that expression really is the force, then the answer is different. You don't really need to do any calculation in this case to see what is going on. Clearly $F(0) = 0$, so this is indeed a point of equilibrium (the question should probably read something like ...

4

No. You are wrong. Even if we exclude drag, the force of gravity does not suddenly drop to zero. The force acts to continuously accelerate the ball downwards. Because it starts off with a velocity upwards this will be reduced because of this force. At some point it will be zero as the velocity reverses sign, but there is no equilibrium as you can prove by ...

4

Other than User58220's answer, I'm reading a lot of nonsense here. When you fly an airplane (I and many other people on this site do), when you are cruising in the air, you center the rudder. The plane has no awareness of the movement of the air mass over the ground (wind). The plane has a vertical stabilizer (tail) which causes it to point into its ...

3

This is well outside my area of expertise but I believe ou can apply Maxwell–Boltzmann statistics, at least loosely to clusters of galaxies, clusters of stars and in many cases the gas molecules in nebulae. For clusters this is known as the Virial Theroem and Andrew described it quite well in this question Stellar Viscosity in Galaxies. For the nebular ...

3

Since at higher altitudes, the air pressure is lower, the boiling point of water decreases, since it's easier for the energy insde the water to get free. When A liquid starts to boil, you reach a critical point where the liquid loses a lot of heat, much more than when not boiling, thus requiring much more energy for the same increase in temperature, and ...

3

Take a hydrogen gas in a magnetic field. The nuclei can be aligned with the field, low energy, or against it, high energy. At low temperature most of the nuclei are aligned with the field and no matter how much I heat the gas I can never make the population of the higher energy state exceed the lower energy state. All I can do is make them almost equal, as ...

3

I look at two models of a "fat earth": a spherically symmetric interior with an aspherical surface layer in hydrostatic equilibrium. This analysis generalizes from the constant density assumed in other answers and thereby exhibits the sensitivity of the flattening to the surface density. I compare the result to those of various other answers. To estimate ...

3

In this answer, I will present a framework to use, and then I will frame the prior answers within that framework. Let me sum up the values we have here. I'll use the same notation (as best as possible) as everyone else and Wikipedia for an oblate spheroid where $a$ is the large, equatorial, radius. Mark1, method in the question, $2 (a-b) = 21.6 km$ ...

3

You may demonstrate numerical friction by applying a Runge-Kutta method (or even the Euler method if RK is too advanced) to a conservative system (such as Sun - Jupiter - Saturn) and notice the dissipative effect of stochastic perturbances (aka discretization errors): Ultimately, the planets will fall into the sun, though this takes many revolutions. ...

3

Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with $n$ degrees of freedom and Lagrangian of the form: $$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$ where $T$ is quadratic in ...

3

The banner is basically a wind vane while the aircraft can resist torque by using its rudder. Even without using the rudder the torque on the craft due to angled wind speed will be lower than the torque on the banner due to the difference in masses. Both @user58220 and @Gregsan give valid responses but with relevance to the question I think it is fair to ...

3

To answer your question, you should first understand when is a system most stable. Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero. We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$ Secondly, it should be able to maintain ...

2

Here I would like to numerically check the theoretical prediction of a factor $\frac{2}{5}$ in difference from Mark Eichenlaub's original monopole argument. In practice, this means calculating the difference in the gravitational potential between the North pole and the Equator, and divide by the corresponding difference in monopole potentials. For numerical ...

2

You have 3 obvious weak points in the system. Where the (longer) arm is attached to the existing bracket Where the bracket attaches to the pole Where the pole is attached to the ground / base. Notation I'll use: $W$ = weight of monitor $L$ = maximum arm length $=> M = WL$ = moment at point of pole attachement $x$ = distance between attachment ...

2

The direction that the domino falls is determined by the location its center of mass. It falls to the left or right depending on whether the center of mass is to the left or right of the bottom-most edge. If the center of mass is precisely above the edge, then it balances on that edge in an unstable equilibrium.

2

Chemical equilibrium is a subset of thermodynamic equilibrium. So there are no examples of systems in thermodynamic equilibrium that are not also in chemical equilibrium. Here's an example of a hypothetical system in chemical equilibrium but not in thermodynamic equilibrium. A system containing two phases, one a condensed phase containing two equilibrating ...

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