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The question is not clear: If you are asking about "back and forth" motion i.e. force parallel to the dog's path, you are asking about force along that axis, which is essentially the coefficient of friction of the surface, times dog weight, plus the vector component of the force exerted by the dog along that axis (up to the point of weight times coefficient ...

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This question is a little too big, entire text books have been written to answer it. A standard reference is van Kampen, Stochastic processes in physics and chemistry. Roughly speaking, for a Markov process Master equation -> Kramers-Moyal expansion -> Fokker-Planck equation where the master equation gives the microscopic probabilistic rule for ...

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The speed of sound, which is effectively a measure of how fast particles are moving on average in a gas, is given by: $$C_{s}^{2} = \frac{\partial P}{\partial \rho} \tag{1}$$ where $P$ is the pressure and $\rho$ is the mass density. For an ideal gas, one can use an adiabatic equation of state such that Equation 1 goes to: $$C_{s}^{2} = \frac{ \gamma \ P ... 4 Looking around, the root mean square speed of air at 20 C is about 500 m/s, and given that you have \langle v^2 \rangle \propto \, T so that v_{rms}(T) = \sqrt{\langle v^2\rangle} varies with \sqrt{T} then have$$v_{rms}(15) = v_{rms}(20)\times \frac{\sqrt{15+273}}{\sqrt{20+273}} \approx 496 m/s$$and$$v_{rms}(25) = v_{rms}(20)\times ...

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The definition of $v_{avg}$ for 3 dimension is $$v_{avg}=\int_0^\infty |v|f(v)dv$$ However, it needs to be clear that $$v^2=v_x^2+v_y^2+v_z^2$$ For 3D, the distribution can be derived from 1D distribution but a little different. $$f_{3D}(v) = \left(\frac{m}{2 \pi k T}\right)^{3/2} \exp\left(-\frac{m v^2}{2kT} \right)$$ After similar integration as above, ...

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The one dimensional Maxwell distribution for the $i-$component of the velocity vector is $$f_{1D }(v_i) = \left(\frac{m}{2 \pi k T}\right)^{1/2} \exp\left(-\frac{m v_i^2}{2kT} \right)$$ Let's drop the $i$ and $1D$ subscripts for simplicity. You are looking for the average of the absolute value of $v$, $|v|$. To find $\langle |v| \rangle$, we have to ...

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It is because the only force acting on the molecule is acting perpendicular to the wall, i.e. in the $x$ direction. The force is acting perpendicular to the wall because we assume that there is no friction between particle and wall (they are both perfectly smooth and rigid). In the absence of friction, there can be no force not perpendicular to the wall. ...

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Why only x component changes is : The definition of elastic collision (https://en.wikipedia.org/wiki/Elastic_collision) states exactly that. You can see it as ping pong ball that hits the table. All the speed in the direction of the table will make it bounce away from it : if you hit the ball down, it will come up after the bounce on the table. But the other ...

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If we described the collision as a physics problem describing a particle on a background (which includes the wall), the theory is translationally symmetric with respect to translations in the $y,z$ directions. It's ultimately because the wall looks the same if one shifts it in the $y,z$ directions. (It's true for a flat wall, and similarly it holds locally ...

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Friction can be recognized in a statistical description of a microscopic view, not only in macroscopic view. It is when energy (or any other conserved quantity) that was previously organized becomes disorganized. In this case, when the atoms that were flowing along the surface, after interaction with that surface, lose that flow, and the energy associated ...

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1) Your question states that, until the compressor is switched off, a constant pressure of $5\; atm$ is maintained in the box. This answers the 1st part of your question. (However, perhaps you mean that the compressor delivers a certain amount of gas per second while the 2nd hole is open, and you wish to know the pressure of the gas in the box when ...

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The derivation of ideal gas equation from Hamilton's equations will take the same procedure as what you have seen in Wikipedia. Since you said you haven't understood the way in which the equation is derived I will give you a step by step explanation on it. So we have a system of perfect gas molecules. Of course they are non-interacting. We are going to ...

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Short answer: It can be compressed, due to extreme pressure, but only slightly. The Wikipedia page on the Mariana Trench says that the seawater density is 4.96% greater there than at the surface. Also refer to Hydrostatic pressure - doesn't density vary with depth?. Further explanation (and assuming everything you currently know about water is that ...

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Liquids are compressible but in lots of cases the fact that the volume of water decreases by about fifty parts per million for an increase in pressure of one atmosphere can be ignored. A similar approximation is also often made about the compressibility of solids.

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