# Tag Info

1

You know Newton's second law: $$\vec F = m \vec a~.$$ Now you multiply both sides by "$\vec r \times~$", then you get $$\vec r \times \vec F = \vec r \times m \vec a = \frac{d}{dt} (\vec r \times \vec p)~.\tag{1}$$ (If you carry out the time derivative, the first term from the product rule is $\vec v\times\vec v$, which of course is $0$.) This is very ...

3

Your figure of 156.4 grams is a mass, not a weight. If you used a set of masses and a pan balance, then you are comparing it with known masses in a nearly uniform gravitational field. Since gravity is the same on both, it does not need to be known. If you used a digital balance, then that device already converted from the forces it detected to a mass. It ...

0

In fact, $\delta Q/T$ is the entropy of $\delta Q$, and if $\delta Q$ is considered as the heat energy in transfer, it follows that $\delta Q/T$ is the entropy in transfer. For the inequality \begin{align}\oint dS \ge \oint \frac{\delta Q}{T}.\end{align} Whether reversible or not does not need to be considered for $\delta Q/T$ in that the entropy production ...

0

One should never write $d S_S= \delta Q_{E\to S}/T_S,$ unless it is a reversible process in the system, as well, because the temperature $T$ is whatever temperature the heat is supplied and that is the temperature of the environment (heat reservoir), presumably behaving independently of the amount of heat it supplies. The reservoir is always assumed to ...

0

Heat is defined as a transfer from the environment to the system. No subscripts are needed for that purpose. I think you would have to be very clear about the specific situation for the other processes that you mention before there can be any discussion about them. Friction where? Turbulence implies a non-equilibrium condition...

3

Well, shock waves are kind of inappropriately named because they do not oscillate, they are actually a discontinuity. A shock wave is the final stage of a nonlinearly steepening wave that has reached a balance between steepening and energy dissipation (i.e., irreversible energy transformation). An contrasting example would be water waves, where there is ...

2

The definition of a physical concept can be a differential form but can’t be the difference of functions. $\Delta S=S_{final}-S_{initial}$ is an equation but not the definition of entropy. Thermodynamics itself now can hardly explain “what is the entropy really" , the reason please see bellow. 1.Clausius’ definition \begin{align}dS=\left(\frac{\delta ...

1

A conservative force is such that the work needed to go from A to B is path-independent (see for example On conservative forces). If you choose a complicated path which for example goes from A to B several times, you will do positive work in certain traits (meaning that the point you are considering "uses" its energy) and negative work in other traits ...

1

1. No, there is no "shift in meaning". "Accuracy", "precision", and "trueness" is a technical term for measurement not physics. And there is no such thing as a "measurement community" because measurement occurs everywhere. As such, "accuracy", "precision", and "trueness" are heavily overloaded technical terms used in varying fields like maths, computer ...

1

Work done is defined as the dot product of force and displacement. ... should it not be the product of force and the time Were these formulae (for work and energy) actually derived based on some physical understanding or are they just constructs to understand forces better? Neither of the two. Most formulas and definitions have an historical ...

2

The path a free positive test charge would follow if acted upon no other force but the force due to the field itself. This is wrong. (Did you actually have a book that said, this? If so, what was the book? This would be a serious error.) A charge in free space will have an acceleration parallel to the field, but the acceleration is not typically in the ...

4

If the particle moves from the point $x$ to $x+dx$, and assume $dx\gt 0$ for simplicity, then its potential energy increases by $$dU = \frac{dU}{dx}dx$$ Well, it increases if $dU$ is positive and decreases if $dU$ is negative. So far I have only used the definition of the derivative – pure mathematics. However, the total energy is conserved. The sum of ...

Top 50 recent answers are included