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2

The concept of inertia is indeed useful in two ways. I think your notion of it as a technical promotion of the everyday word "sloth" (without the baggage given it by the Roman Catholic translation of the "deadly sin" Ἀκηδία) as extremely close to the mark. In physics the notion of "inertia" has two, very alike uses: The first is practical, through a weak ...


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It boils down to a matter of convention. Nothing stops you from choosing the annihilation operator to be $a^\dagger$. Still, in quantum field theory, you decompose e.g. a scalar quantity in plane waves $$ \Phi(x) = \int \frac{d^4 k}{4 \pi} \left( a(k) e^{ik\cdot x} + a^*(k) e^{-ik\cdot x} \right)$$ where obviousely the second part is the c.c. of the first ...


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I'll give this a shot though I'm uncertain if this will clarify or cloud the issue. electrons are people, OK current is the number of people, No, that's not the correct analogy. Current is a flow and a people (electron) current is a flow of people (electrons) and the amount of current is the number (Coulombs) of people (electrons) passing a ...


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Since your question is kind of pictorial, i will try to stick with this and avoid any deeper mathematical / physical explanations. electrons are people ok with that current is the number of people no, that's a mistake. Actually current is is this case the number of people crossing a line on the street per time. Like Steeven already mentioned in ...


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But what is potential in this example? In your people-in-the-street analogy, the electric potential would be a shop with sale in one end of the street, or maybe an accident or something else interesting. People tend to move towards the interesting end (less potential) and away from the other and less interesting end (higher potential). Just like a stone ...


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In mathematics, complex numbers and rotation are intimately related. After rotation $\theta$ in the complex plane, the number $1$ becomes $e^{i\theta}=\cos\theta + i\sin\theta$. Although $e^{i(2n\pi)} = 1 = \cos(0) + i\sin(0)$, you actually moved by $\theta$ and not by $0$. If you rotated something by $\pi$, you could say the angle is now $-\pi$ - but you ...


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What you can say is that you have a distribution of the results you are going to get (be it a discrete or continuous random variable), and when you calculate the average of a large sample, you are adding the random variables and multiplying by a constant. The addition of random variables translates into a convolution of the probability density functions, ...


3

In general, we can say nothing about finite $n$, but most of the time, we can safely assume some "niceness" of the distributions in question. If, for example, we assume a finite Variance $\sigma^2$ (a quite common feature), we could use Chebyshev's inequality for a rough error estimation of the form $$P(|\bar{X_n} - µ| > \alpha) \leq ...


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This is the exact reason why we do statistical hypothesis confidence testing. In essence, the confidence interval we get from this test is a quantitative measure of "how far we've converged". For example, consider an experiment to test whether a coin is imbalanced or not. Our null hypothesis is that it is not: in symbols, our hypothesis is ${\rm Pr}(H) = ...


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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 ...


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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 ...


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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 ...


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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 ...


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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...


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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 ...



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