# Tag Info

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No. There is nothing wrong with perturbation theory, or with theories with known, restricted accuracy. The point of theory is to explain the results of observation from as simple an initial theoretical standpoint as possible. Therefore: Since experiment always has a finite uncertainty, one can only ask that theory match the experimental value within its ...

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This is what I think the first bit of the calculation does. Suppose you start with a spherical eye with a hole in it (e.g. the pupil in the human eye): The radius of the eye is $ER$ and the radius of the hole is $AR$, and with the length $DA$ these form a right angled triangle. Pythagoras' theorem tells us: $$DA^2 + AR^2 = ER^2$$ so: $$DA = ... 5 Here's my quantitative attempt at 4. and 1.: The Coandă effect here is the tendency of the airflow to adhere to the surface of the ball. This means that near the surface of the ball, the streamlines are curved with a radius of curvature approximately equal to the radius of the ball R; this curvature results in a pressure gradient just as it does in ... 4 Let me give the naivest possible estimate, so that people have something to criticize. Assuming that the most of the jet interacts with the ball and is deflected at a substantial angle then the force on the ball is roughly the momentum flow through the pen. In your units this is \rho_{air} Q^2/(\pi d^2). Saying the force to levitate a ball is 1\times ... 4 I can't see how a negatively charged electron can stay in "orbit" around a positively charged nucleus. Even if the electron actually orbits the nucleus, wouldn't that orbit eventually decay? Yes. What you've given is a proof that the classical, planetary model of the atom fails. I can't reconcile the rapidly moving electrons required by the ... 4 The article looks indeed wrong. In fact, there are two mistakes. First, you're right that the acceptance-rejection method has to be applied to \rho(r), and not to m(r). To understand how this idea works, suppose we want to generate a one-dimensional normalized distribution function p(y). Now, let's assume we can rewrite this distribution function in ... 3 Mathematics is just a systematic way of stating facts about the world. It is only useful inasmuch as it is internally self-consistent. The latter fact means that there is nothing to "assume" about mathematics. It is a relationship between axioms and conclusions that enables one to succinctly summarize many observations. Something like Galileo's ... 3 If you are interested just in the direct irradiance you can neglect the emmision and scattering terms in the Radiative Transfer Equation RTE wich can in this case be simplified to allow only for absorbtion and is known under the name Beer-Bougert-Lambert's law of absorption$$ \cos\theta_{0}\,\frac{\partial}{\partial p}S^{i} = -\frac{\kappa^i}{g}\,S^{i} $$... 3 Ok, I'm still not sure on what level you want to do this, but I will start you off with some basics. The most important factor is probably the solar elevation angle, \theta. As described on the wiki-page it can be calculated using this formula:$$ \sin\theta=\cos h\cos\delta\cos\Phi+\sin\delta\cos\Phi $$where h is the hour angle, \delta is the solar ... 2 The treatment of electrons as waves has combined with spherical harmonics (below image) to form the foundation for a modern understanding of how electrons "orbit." Tweaks to the spherical harmonic differential equations yields the Schrodinger equation, which yields the accepted models of electron orbital structures: The only element for which the ... 2 I'm a first year physics student, so my answer might not be satisfactory - but I hope it will give some insight to the problem. 1) from what I know we need to consider: Drag - which I will address Turbulence - which I know next to nothing about, and therefore I will ignore with hope someone will be able to expand. we need the drag force to be equal to ... 2 I suspect you have misunderstood what is meant by acceleration in the context of circular motion. Acceleration is the rate of change of velocity, but remember that velocity is a vector so it has both direction and magnitude. You are probably thinking that acceleration means a change in the magnitude of the velocity, e.g. speeding up from 1 m/s to 2 m/s, but ... 1 The CDF, F(x), is related to the PDF, f(x), via the relation:$$F(x) = \int_{-\infty}^xdx'\,f(x')$$In the case of radial distributions, your lower limit is obviously 0 and not -\infty. Thus, your CDF is m(r) and the PDF is 4\pi(r')^2\rho(r') (technically should be \rho(r') with the 4\pi(r')^2 coming from dx\to dr and spatial isotropy, but ... 1 There will always be solutions that can't be analytical. For example, any model of more than two bodies without any special constraints, cannot be solved analytically. From the gravitational interactions between three planets to three particles interacting (electromagnetically or otherwise) in quantum theory. To have mathematically analytical solutions, ... 1 Short answer: that depends on your definition of sound theory. For instance, it is possible to find peer-reviewed papers considering such possibilities. The idea that antimatter can be gravitationally repulsed from ordinary matter is definitely not the most popular one. Nevertheless, some people do try to apply it in astrophysical context. Let us have a ... 1 Presumably, the analytical solution is using $$P(x,y,z) = \lim_{T\to \infty} \frac{1}{T} \int_{-T/2}^{T/2} P(x,y,z,t)\, dT$$ Note the limit that takes T to infinity. If the solution is periodic with period T, then this is precisely equivalent to writing P(x,y,z) = \frac{1}{T} \int_{-T/2}^{T/2} P(x,y,z,t)\, dT ... 1 I'm surprised the "real" schematic doesn't include a transformer turns ratio (actually, a model for the high voltage transformer, which has a primary winding with N_p turns connected to the power oscillatory circuit and a secondary winding with N_s turns (with N_s > N_p) connected to the discharge gap). Maybe everything is referenced to its ... 1 I'm no expert, but ... MIT's "Magnetic Circuits and Transformers" discusses eddy current losses in chapter V.2. An approximate formula for a magnetic sine wave at frequency f and peak amplitude B_{max} (in Tesla, mks units throughout) is:$$ P_e = k_e f^2 t^2 B_{max}^2 V  where $k_e = \pi^2/(6 \rho)$ theoretically (but in practice is often ...

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There are many possible examples of this, and you may need to be more specific in what you want. Here are two that immediately come to mind: 1) A bead in a harmonic trap (or a bending cantilever) that is undergoing thermal kicks from Brownian motion. The strength of these fluctuations depends on temperature; if the temperature of the system changes over ...

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