# Tag Info

1

Your equation for gas permeation mostly applies to hydrogen, which will dissociate into hydrogen atoms before entering the material. For nitrogen (major component of air), the equation has no square roots. Nitrogen permeation is extremely slow; if the box is welded shut and there are no cracks, then this process will be negligible, even over a 10-year ...

2

Apart from the heating due to sound absorption, as per the comment by HolgerFiedler, I don't think you will find a mechanism that can radiate due to polarization effects in the medium. Any EM radiation would be at the frequency of your acoustic waves. With the difference between the speed of sound and the speed of light, that would be very-long-wave ...

0

You wrote the proper Bernoulli equation for this problem, then you went on a cancellation frenzy with the pg terms which was incorrect, after properly neglecting any change in flow velocity. Delta P / (pg) = Delta h is what you should have left in your Bernoulli equation after neglecting the change in velocity. Since the problem is looking only for the ...

2

I was going to make this a comment, but I thought of a phrase that I think intelligent non-physicists should grasp, so it might be worth keeping in your "teaching toolkit". At one level it's wholly a matter of taste and English usage that tells which phrase is correct, but the phrase "rushes in" is probably more evocative of the true physics than "sucked ...

2

I have discussed this problem with someone who solves this type of problems numerically and got the following response: The expression on the right-hand-side of (**) in my question is evaluated at $t_0+dt$ instead of $t_0$. This together with (*) provides two equations for $\mathbf{v}(t_0+dt)$ and $P(t_0+dt)$, \begin{align} ...

2

Orbital insertion is hard enough with ships where the initial stages can be aimed in directions favorable to the final orbit. This design calls for a fixed launch direction, which would be a terrible waste of on-board fuel for all but those satellites whose orbits coincide with the gun's trajectory. Even that can be overcome with fuel. The real issue is ...

0

In osmosis, normally, water will move toward the side of membrane with higher concentration of salt, and this can happen with or without heat (though much faster with higher temp. because of the fast motion of molecules). So my guesses are: 1. This can prevent over-bloating of the rice due to over-absorption of water, and perhaps prevent grain damage. 2. ...

4

One more nasty factor: What is the expansion speed of your propellant. Take the Jules Verne approach and your spacecraft falls far short no matter how much powder you put in the gun because the expansion velocity is too low. Your craft will never exceed the expansion velocity of the propellant. Note, however, that you don't have to use explosives (or ...

0

To begin with ask yourself two questions: Has the device been re-calibrated in the middle of data taking? Is the calibration known to drift over time similar to the length of the data taking? If both of the answers are "no" then you can reasonably assume that the calibration effect is the same on each and every data point. So the mean is off by that ...

68

Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether. In order to achieve orbit, at least two impulses must be applied to the projectile. The first ...

11

I think the heart of the question is whether one could arrange a continuous combustion of propellant along the length of the barrel. In that way the acceleration occurs along the length of the barrel in a more gentle way. Since the expanding gases from the propellant in a shell casing expand and the pressure of the expanding gases declines along the way it ...

31

Anything launched into orbit by such a gun needs to travel at orbital velocity (in fact above orbital velocity) in the lower atmosphere. That's generally undesirable, to put it mildly: there will be really serious heating.

0

Let's say you have got such gun. Next logical step will be to install on the satellite a smaller gun that would shot-back several small shells and so accelerate the satellite, indeed? If this small on-board gun would use really many small shells (size of molecula) then your are getting just a traditional rocket. Apparently it is not much difference for ...

16

Aside from the interior ballistic aspects of these various projects, it was quickly realized that any satellites launched by gun would have to withstand high g-loadings during firing of the gun and the size and mass of the satellite would be greatly constrained by the dimensions of the bore of the gun and the maximum impulse which could be provided by the ...

1

Poiseuille's law will tell you that a pipe of 0.1 m diameter will achieve a flow velocity of around $v=0.1\mathrm{~m/s}$, which is actually more than I expected. For pipes a bit wider than this, the flow will be turbulent (Re>2200), for which you can't apply Poiseuille. For turbulent flow, you can look into the Darcy-Weisbach equation. The interesting part ...

0

Normalization Factor Let us define a generalized Gaussian probability density function (PDF) as: $$f_{s}\left( x \right) = A_{o} \ e^{^{\displaystyle - \frac{ (x - x_{o})^{2} }{ 2 \sigma^{2} } }} \tag{0}$$ where $A_{o}$ is the normalization constant, $x$ is the argument, and $s$ denotes the set of distributions (e.g., particle species), $x_{o}$ is the ...

2

Using Torricelli's law (https://en.wikipedia.org/wiki/Torricelli%27s_law) you get $v=\sqrt{2gh}$ irregardless of how big the opening is. Now you can calculate how much mass is leaving the tank at any time by multiplying the volume that is leaving the tank by its density: $$\Delta m=A\Delta s\cdot\rho$$ with $\Delta m$ the mass leaving the tank at any second, ...

1

For a particular setup, the equations may get very simple: the tank should be massless ($M=0$) and the hole is all the way at the bottom of the container. Then $h$ is proportional to the mass of the water in the container: $m=m_0 h/h_0$, with $h_0$ the initial height. It's straightforward to derive that the force generated by the water jet is $F=2\rho g h ... 0 Just visualize the upper beaker as a tube connected to the siphon tube(both tubes having different area of cross section) 1 In this case, it is allowed. As a thought experiment, you can replace the top bucket with the siphon hose by an S-shaped tube that has a gradual change in diameter, like this: _ / \ |www| | | \www| | | \ww| | | \w| | | \|__/ | | | With this type of analysis, it is more ... 2 To answer your specific question: absolutely none. The Millenium run is a "dark matter-only" simulation. In this sort of simulation gas physics is taken to play a negligible role. All the gas (and stars, indeed all "baryonic matter" as it's called in the jargon) is removed and replaced with additional dark matter. The extra dark matter is added just to keep ... 0 Blood pressure is measured in mm-Hg. If we take a typical blood-pressure 120-80, that can be adjusted to height above or below the heart provided we make a very bad assumption that the human body is a bag of water. In reality, the human body is nowhere near that simple. 120-80 mm-Hg where Hg is a density of 13.59 g/cc and blood is about 1.06, then every ... 1 If the mouth of the bottle is small (e.g., a wine cork with drilled hole for a drinking straw), water indeed won't flow out. Would you also wonder why the water doesn't stay in an upside-down bucket (the limit case for a bottle with a very wide mouth)? What happens is that air enters through the mouth with a volume equal to the volume of water that leaves ... 6 I will try to answer as many questions as I can. I won't presume to give you complete exhaustive answers, but maybe they will be nonetheless useful to you. What variables determine the range of temperatures over which matter is liquid? My understanding of thermodynamics is that matter changes from a solid to a gas when some thermal vibrations create an ... 2 Liquid: molecules form bonds with neighboring molecules for most of the time, but there are enough energy for the bonds to break momentarily and be formed again with another molecule "These explanations seem hand-wavy to me." What is the level of your knowledge of physics? Are you aware of the quantum mechanical nature of atoms and molecules, at a ... 3 The paper by Yusuf Billah and Robert Scanlan (cited in the wikipedia article on the Tacoma Narrows Bridge 1940) distinguishes between resonance as a response to a driving force and what the authors call "self-excitation" or "negative damping." They demonstrate that the Karman Vortex Street (which occurs at the trailing edge of the deck) was not the cause of ... 0 This can be concluded by reviewing Gibbs equation for upstream and downstream stagnation conditions. $$T_0ds_0=dh_0-\frac 1{\rho_0}dP_0$$ Because across the shock wave is an adiabatic process,$dh_0=0$Then Gibbs equation becomes $$ds_0=-\frac 1{\rho_0T_0}dP_0=-\frac {R}{P_0}dP_0$$ We know entropy increases. This leads to conclusion that the stagnant ... 1 This is hard to answer quantitively. A train in open air pushes air out of the way, and sucks air in behind it. Some air is dragged along with the train. Mostly air moves near the train. In a New York subway, you can feel a strong breeze before the train arrives. The tunnel diameter isn't that much larger than the train, so the train acts something like ... 2 You can calculate the air drag along your arms using the formula for air drag,$F = \frac{1}{2} \rho u^2 c_D A$, takin$u$the speed at which they'll effectively move relative to air, and$A$their projected area on the plane perpendicular to motion.$\rho$is the density of air, that's why it's so much less efficient than in water.$c_D$will be close to ... 3 The situation where the equation holds is when the fluid is imcompressible. In your example where you have a fluid flowing in a tube under gravity, you can imagine two situations. One is where there is no dissipative interaction between the fluid and the tube, and one where there is. In the situation with no dissipation, then the fluid should accelerate ... 3 The "buoyant force" does not arise from a principle, but is what remains of the gravity force when you subtract from it an average hydrostatic pressure. Let's assume that you have some reference density$\rho_0$. Then if$\rho=\rho_0$everywhere and the fluid is at rest,$p = \rho_0 g z$. Now let's call this baseline pressure$p_0(z)$, without loss of ... 3 We're talking about the Coanda Effect, right? Then I think this article could provide some useful insights. Quoting from the article: When the fluid flows over the heated curved surface in proximity of the curved surface as the temperature of the curved surface increases, dynamic viscosity of the fluid at the vicinity of the wall is increasing with ... 0 if the negative pressure is due to high velocity, there is possibility of cavitation to occur but if the negative pressure is due to height, it will just try to suck water, may be form other side or from the same side. 2 In general, the term cannot be neglected. It's entirely possible that the term has to stick around when advection is significant. But, since you are asking how it goes away, I'm guessing you are in a situation where it does, in fact, go away. Like any other order of magnitude analysis, you have to non-dimensionalize your equation. This means you have to ... 1$V_1 A_1$is basically the volumetric flowrate. So in a time$\Delta t$, a particle moving with flowrate$V_1 A_1$, sweeps out a volume$V_1 A_1 \Delta t$. 3 You have to define what a soliton is. The most accepted definition in field theory is that a soliton is a stable, localized and finite energy/energy density solution of the equations of motions of the theory. A vortex ring is localized in space, it has finite energy and definitely is the solution of some equation of motion. Then if it is stable, it can ... 1 I guess your question is how to increase evaporation amount because the time is the droplet traveling time. To increase evaporation amount without changing air condition, followings can be considered. reduce droplet size: the evaporation rate is proportion to$\frac {1}{D^2}$, where D is droplet diameter. increase droplet speed: this can increase ... 0 If the propane in the tanks are under pressure,then on connecting them, both tanks will be full of propane as it is a gas. Had it been water, or in a liquid form, then both the tanks will have the same level no matter how you arrange them. 20 Yes, as you can see in this video. As you can see, the droplet will hit the surface, partially coalesce (merge) with the bulk, re-emerge as a smaller droplet, bounce 1-3 times, partially coalesce again, re-emerge again as an even smaller droplet and so on. This process is known as coalescence cascade. You can find another video here. Eventually, the ... 4 if you watch extreme slow motion film of a single water droplet falling into other water, usually a completely flat horizontal surface, you see many circular ripples being created, most of the droplet being absorbed by the body of water, but a tiny ( about 5% of original drop ) droplet being formed, and returning back, vertically upwards. How much of this ... 3 The role of viscous forces with respect to turbulence is analogue to the one of diffusion for mixtures: it sets a scale below which it smoothes out gradients. So there are no vortices smaller than this. When this scale is much smaller than the scale of the experiment, tubulence can happen. Turbulence is not linked with a peculiarity of the molecules. It can ... 1 Quite simply, a viscous flow is a flow where viscosity is important, while an inviscid flow is a flow where viscosity is not important. Gases and liquids alike are considered fluids and any fluid has a viscosity. So a gas bubble surely has a viscosity, albeit relatively low compared to some liquids; liquids are generally more viscous by a factor of 1000. ... 1 In fluid mechanics, a gas (like air) is considered a fluid and operates under the mathematical formulae developed in this field. I believe the very essence of viscosity is displayed in the deformation characteristics of a fluid with regard to shear stresses. For example, the shear stress for a Newtonian fluid with respect to a flat plate is given by Newton's ... 2 From your question, I guess that you don't really visualize what the flow geometry is. It's true that there's one unclear bit in the exercise: I guess it should read "The ﬂow in the boundary layer must match with the constant inviscid bulk velocity$\vec{U} = (U, 0)". The overall geometry is a flow meeeting a plate which is aligned with it (0 incidence). ... 0 I wasn't able to figure out what you did, so here is my analysis, without the resistance. Let: Q = Total volume flow rateQ_a$= Volume flow rate into converging pipe$Q_b$= Volume flow rate into diverging pipe$p_1a$= static pressure just after entrance to a$p_2a$= static pressure just before exit from a$p_1b$= static pressure just after ... 1 I assume that by 'Stokes regime' you mean the drag force a object travelling through a viscous fluid experiences, in laminar flow conditions. For a perfectly spherical object and assuming flow of the fluid around the object is laminar, then acc. Stokes' law: $$F_d=6\pi \mu Rv$$ Where$\mu$is the dynamic viscosity of the fluid,$v$the object's speed and ... 1$\mathbf{u\cdot n}=0$is indeed the answer for an inviscid fluid, and$\mathbf{u = 0}$for a viscous one. 1 As you mentioned, a phonon is a quantized sound wave. Let us ask then what is the difference between a quantum and classical sound wave in a crystal? At room temperature, for most solids, the specific heat is pretty close to$3Nk$. This is the classical Dulong-Petit answer. However, at low temperature, the specific heat deviates away from the Dulong-Petit ... 1 Light, passing through a transparent medium, is scattered when it interacts with that medium's spatial & temporal variations producing the medium's refractive index. This is called Brillouin scattering and can be described as interaction of a photon with a phonon representing the medium's compressional deformation. Although mostly studied in the context ... 1 In a first approximation, and if you have a laminar flow in the pipes, the flux will be$Q=\frac{-\kappa A}{\mu} \frac{(p_b - p_a)}{L}$, from Darcy's law, see notations there. So increasing the area$A$will increase the flux in proportion. Conversely, as you can see from the equation, the pressure drop is inversely proportional to$A\$, if you impose flux. ...

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