208 votes
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Is $\pi^2 \approx g$ a coincidence?

The differential equation for a pendulum is $$\ddot{\phi}(t) = -\frac{g}{l}\cdot\sin{\phi(t)}$$ If you solve this, you will get $$\omega = \sqrt{\frac{g}{l}}$$ or $$T_{1/2}=\pi\sqrt{\frac{l}{g}}$$ $...
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54 votes

Is $\pi^2 \approx g$ a coincidence?

It's annoyingly unclear how far it's a coincidence, but at any rate it isn't completely a coincidence. As you can see in e.g. the Wikipedia article about the metre, a unit almost equal to the metre ...
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49 votes
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How is it possible to differentiate or integrate with respect to discrete time or space?

Let's say space is really a lattice with spacing $\Delta x$. It turns out that this idea has more trouble with experiment than you might think, but we can plow ahead for the purposes of this question. ...
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41 votes

Far away from a charged conductor, the field is like a point charge. Where's the point located?

The answer is that it doesn't matter. The distance at which fields resemble that from a point charge is also the distance at which it does not matter where that point is located within the structure. ...
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40 votes
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Why doesn't physics like infinity (or does it)?

There's no rule against infinity, only a rule against being empirically wrong. Historically, the infinite has hinted we're missing something, but so have plenty of other things too. Let's discuss some ...
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38 votes
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The reasoning behind doing series expansions and approximating functions in physics

The key reason is that we want to understand the behavior of the system in the neighborhood of the state rather than at the state itself. Take the equation of motion for a simple pendulum, for ...
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37 votes
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Is there a rigorous definition of 'much greater than'?

There is a consistent definition, but it involves a couple of arbitrary thresholds, so I doubt you'd consider it rigorous. The construction $X \gg Y$ means that the ratio $\frac{Y}{X}$ is small enough ...
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36 votes
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Far away from a charged conductor, the field is like a point charge. Where's the point located?

Based on some of the back-and-forth I see, I think you're asking the wrong question. I think the question you want to ask is "Given a charge distribution $\rho(\mathbf{r})$, where should I place a ...
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  • 1,026
36 votes
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What is the point of a voltage divider if you can't drive anything with it?

Oh, but you can. You can drive an high impedance input with it...including a buffer, which can then in turn be used to drive whatever you want. The more current you draw the more the voltage will ...
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  • 7,044
34 votes
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Why does a simple pendulum or a spring-mass system show simple harmonic motion only for small amplitudes?

A simple pendulum does not strictly show simple harmonic motion unless you allow some approximations and uncertainties. It approximately behaves as a harmonic oscillator for small amplitudes. An ...
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  • 6,903
33 votes
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What is the true nature of gravity?

The job of physics is to construct models that are able to explain and predict empirical observations. You can never be completely sure that a given model is the "true" description, only ...
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32 votes
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Elliptical Trajectory, or Parabolic?

A parabola and an ellipse are both conic sections, which can be constructed in a plane as all the points where the distances from some reference point (the "focus") and some reference line (the "...
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  • 70.2k
30 votes
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Why is a particle non-relativistic when its kinetic energy is small compared to its rest energy?

When we say a particle is non-relativistic we mean the Lorentz factor $\gamma$ is close to one, where $\gamma$ is given by: $$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$ So saying $\gamma$ is close to ...
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27 votes
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Why is the Newtonian expression for kinetic energy called the "first order" approximation of the relativistic expression?

The way I see it, there are four possible answers. You can pick the one you like the most, because in the end it doesn't matter very​ much. You're right, it's a second order approximation and those ...
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24 votes
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Is the existence of a sole particle in an hypothetical infinite empty space explicitly forbidden by QM?

That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, ...
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  • 106k
24 votes
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Near Earth vs Newtonian gravitational potential

Your equation (2) is the change in potential energy when the object moves vertically by a distance $h$ i.e. when the object moves from $r$ to $r+h$. Let's use equation (1) to calculate this: $$ \...
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23 votes

Why do physicists say that elementary particles are point particles?

Scattering experiments can be used to determine the size of a particle. The results for an extended object are different than that of a point particle. But all of these scattering experiments depend ...
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22 votes

Why is a particle non-relativistic when its kinetic energy is small compared to its rest energy?

'Non-relativistic' means $v\ll c$. That is effectively the same as $\gamma \approx 1$ as $\gamma={1 \over \sqrt{1-v^2/c^2}}$. But also $\gamma={E_{tot}\over E_{rest}} \equiv 1+{E_{kin} \over E_{...
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22 votes
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If charges is quantised, how can we use integrals in electrostatics?

You may also wonder why we use the concept of density of a material despite that material is made of molecules, or atoms, or in general quantized entities. This is the basis of hydrodynamics and solid ...
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22 votes

Why doesn't physics like infinity (or does it)?

Infinity is a shorthand for unbounded. When we say that $\frac 1 x$ goes to infinity as $x$ approaches $0$, what we really mean is that we can make $\frac 1 x$ as large as we like by using a value of $...
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  • 33.3k
21 votes
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Understanding this quote by Feynman

[...] the wave theory was unable to explain things like the Photoelectric Effect and Compton Scattering [...] This is true. As a result, we know that classical electrodynamics is only an ...
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  • 51k
19 votes

Approximating an expression for a potential

There is nothing wrong with your first approximation. You got the leading term $2k/l$ correct. You just did not expand out far enough in powers of $x/l$ to see the $2k x^2/l^3$ term. If you were to ...
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  • 41.3k
19 votes

Is projectile motion an approximation?

Just as the motion of body around the earth is ellipse (1st Kepler law replacing sun by earth), so is the motion of a projectile. Notice that almost everything we deal is an approximation, the earth ...
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  • 461
18 votes

Elliptical Trajectory, or Parabolic?

If gravity is uniform - the force has the same magnitude and direction everywhere, the trajectory is a parabola. This is a very good approximation for trajectories that don't go very far. But in ...
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  • 26.1k
18 votes
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Accuracy of physics laws

Accuracy can mean different things. While the question asks about the statistical accuracy, what immediately comes when talking about the Newton's laws is that they are non-relativistic, i.e., they ...
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  • 37.4k
17 votes
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Should not we apply pseudo-forces all the time as the Earth is a non-inertial frame of reference?

The way to properly answer your question is, for lack of a better word, by enumeration. Consider all the ways the Earth moves, compute the fictitious forces corresponding to them and check that they ...
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16 votes
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Why don't the Navier-Stokes equations simplified for hydrodynamics contain gravitational acceleration?

If you go through the process of non-dimensionalizing the equations, the math becomes more clear. If you start with the momentum equation (ignoring viscous forces because they aren't important for the ...
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  • 16k
15 votes
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Calculating the total time elapsed until two pendulums "stop colliding" gives a divergent result

Your calculation is perfectly correct, under the standard idealizations in mechanics. From a mathematical point of view this isn't that surprising; divergent times are pretty common. For instance, ...
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  • 94.9k
15 votes

Is there an approximation for the Lorentz factor for very large velocities?

This is roughly the simplest you can get it: $$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-v/c} \sqrt{1+v/c}} \approx \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-v/c}}.$$ In other words, if $\Delta ...
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