# Tag Info

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At a vertex of the diagram the usual conservation rules apply. Lepton number, charge and so on...

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Okay, so the basic idea is: an object in motion tends to stay moving at the speed that it's moving. When we apply this to rotational dynamics we have an interesting effect: an objects speed goes linearly with the radius it is from the center it rotates around. So if something is rotating with a period T, it must go a distance $2 \pi R$ in time $T$, so its ...

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Felt recoil is partly a matter of momentum, partly a matter of force. When a bullet with mass m leaves a gun with a velocity v, the gun must have an equal-but-opposed momentum MV, where M is the mass of the gun and V is the recoil velocity, or $$mv + MV = 0$$. If there are two possible gun sizes, $M_1$ and $M_2$, each will have a recoil velocity $V_1$ and ...

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Newtons 3rd Law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. Momentum is product of mass and velocity. The heavier gun has more mass, so, for the same momentum, it must have less "backwards" velocity, so less felt recoil.

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The larger firearm has more mass, and therefore more inertia for the recoil momentum of the bullet to overcome. Also, small firearms may be more difficult to secure a good grip on.

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There are at least 3 answers to your question depending on what you mean by "world"? 1 - if you mean "universe," the answer is no, because there would not be "anything" that could walk, or a place to walk on. 2 - if you mean "earth," the answer is no, because there would be no earth. 3 - if you mean "surface," the answer is no, because a horizontal force ...

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Newton precisely expressed it in the law of inertia. A body (in this case a normal person with shoos trying to walk on frictionless surface) maintains its state of motion unless unbalanced external force acts up on it. This means, the person on the frictionless surface maintains his state of rest, if he was at rest, or moves with the velocity that he was ...

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The basic point has already been mentioned, but I would like to give my version of the answer, and point out a subtlety. The basic equations of fluid dynamics are the conservation of mass, momentum, and energy $$\frac{\partial \rho}{\partial t} - \vec{\nabla}\cdot\vec{\jmath}_\rho \\ \frac{\partial \pi_i}{\partial t} = - \nabla_j\Pi_{ij}, \\ ... 0 Navier-Stokes equations describe fluid in an approximate way which neglects diffusion of molecules altogether. In ordinary case, the velocity field is supposed to be smooth. "Diffusion of momentum" mentioned in connection to the term proportional to \Delta \mathbf u is not really a diffusion in the molecular sense. It is rather kind of metaphor to ... 1 I left this as a comment but I'll expand it here since it provides another viewpoint. Imagine you have a box of gas molecules bouncing around in them. Every molecule is identical so they have the same mass, temperature and pressure. Let's also say this box has a diaphragm in the middle separating the box into two. You now remove the diaphragm and start ... 1 So it's actually a really simple reason, but you're going to have to think a little bit about what's going on. The transport equation states that everything which is a "stuff" can be viewed in this way: "A small box flows downstream; the time rate of change of the stuff inside of the box is equal to the flow of stuff through the boundary of the box, plus ... 5 For a single-component fluid, the conservation of mass follows$$ \left(\begin{array}{c}\text{mass of fluid } \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of fluid } \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{sources or} \\ \text{sinks in }\Delta V\end{array}\right) $$In terms of a ... 2 If there is no friction, then you can not rely on it providing traction. However, other forces must still exist, or you'd fall through the floor! So, there is simple artificial solution to being able to walk in frictionless world. All that is needed is sufficiently contoured surface, something like this kind of knobs covering the surface: #### #### ... 1 Some people have suggested throwing something out--that's not walking. However, a world being frictionless does not mean that it is perfectly flat. You can walk by exploiting this fact--if a foot is in a depression you can develop a horizontal force no greater than the sine of the angle of the walls of the depression times your weight. Note, furthermore, ... 2 It will spin about its axis. In general terms, if you attach a flywheel to a big motor in space and turn it on, the flywheel will spin in one direction with some angular velocity, and the motor and whatever is attached to the motor will spin in the opposite direction with a different angular velocity. This is the basic principle behind a reaction wheel. ... 2 Can it be shown mathematically using principles of mechanics that it is not possible to walk in a friction-less world, or is it only by experiment? It all depends on your definition of walk. If you mean, slide along a horizontal surface thanks to friction, then by definition no, it is not possible. Your critical angle is zero, so the horizontal ... 4 YES walking is possible in a frictionless world with appropriate limb movements. The idea is to twirl the arms and legs to cause propulsion by the Magnus effect. The technique has been developed in a British government ministry https://en.wikipedia.org/wiki/Ministry_of_Silly_Walks Named for its investigator Gustav Magnus, the Magnus effect causes an ... 1 In addition to the answers given already which cover the momentum and traction sides of the problem, I'd also like to toss in magnetism. You could just use magnetic fields to either push or pull you in the direction you want to go. Bonus points for using simple electromagnets that could be switched on and off so you can accelerate/decelerate at will. 4 The purpose of the question, as I read it, is related to an ideal condition (perfectly smooth surface) and a man with regular feet/shoes. In this case, normal daily walking would not be possible, as the horizontal force that moves the person ahead is coming from the friction between the feet and the ground. F = k*N; F : Force (horizontal) k: friction ... 2 Yes. You need ridges (about 2mm tall): _____|~~~~~|______|~~~~~|____ You can walk on that if your tread is the right shape friction or not. Boot tread will interlock with the ridges providing non-friction-based traction and so the ability to walk. 15 Although friction is not one of the four basic forces of nature, it exists because those basic forces exist. Friction is the resistance to motion of two objects held against each other. Friction that allows us to walk depends on gravity to convert our mass to weight which holds our feet against the surface where static friction enables the soles of our ... 5 You cannot walk at all if there is no horizontal component of the force of interaction between you and the ground; by this definition "no friction" is tantamount to "no ability to move horizontally". But to add to Hapa's ANswer: you can move by throwing stuff. How do you do this? Here's one way if you don't have a handy sack of hammers ready in your pocket ... 16 I assume by zero friction you mean no roughness or deformity in the ground. Perfectly smooth. Even so there is a way to walk. As you plunge your foot into the ground you compress it a little based on the atomic theory of matter. This impression allows your foot to be slightly lower than adjacent atoms and can therefore push away from them. On ice this ... 28 If there is no friction, you can still move by conservation of momentum. Take some stuff with you that you don't need. Throw it away in the opposite of the direction you want to go! 0 I agree with you. The existence of microwelds on the surface and the object is essentially what causes our feet/shoes/socks/whatever to push back and forward upon the surface to propel us forward. Without something to back up on (or more accurately, to hook on to), we can't expect to propel forward. Newton's Third Law says every force has an equal and ... 3 The second solution will mathematically satisfy the conservation equations, but corresponds the objects not actually colliding. Or they ghost'' and fly right through each other. :) 3 If you the rate of change of mass is \dot{m} = \mathrm{d}m/\mathrm{d}t and we're talking about passing into the future direction (\mathrm{d}t > 0), then \mathrm{d}m is necessarily negative. Thus, m\mapsto m+\mathrm{d}m represents a decrease of the rocket mass. The opposite convention would require that \dot{m} = -\mathrm{d}m/\mathrm{d}t works ... 0 I think the first law of thermodynamics could be restated as $$\Delta U_S + \Delta U_{\Omega\setminus S} = 0$$ i.e. $$U_S+U_{\Omega\setminus S}=\text{constant}$$ and this would clarify (to me) the relationship between the two concepts. What you describe is a general law of conservation of ... 1 Details have to be filled in, but I think the general idea went along these lines: the variation of the action with respect to the metric g_{\mu\nu} is given by$$ \delta_gS \sim \int T^{\mu\nu}\delta g_{\mu\nu}\,. $$Now specialize to particular variations of g_{\mu\nu}, the diffeomorphisms. For an infinitesimal diffeomorphism of the form x^\mu\to ... 2 First, the chemical reaction that takes place in the cartridge of a gun produces high temperature, high-pressure gas, without the need for external oxygen. The gases then push the gun and the bullet apart. Secondly, the bullet will leave the barrel with a certain amount of momentum, found by multiplying the mass of the bullet by the velocity of the bullet. ... 2 Without mathematics: The divergence operator tells you the net flow into or out of a volume element. Imagine a car park. They count the number of cars that are coming in and the number going out, and a sign says "there are 5 spaces". Not because they count spaces- they count in and out flow. For a steady state (same number of cars in the car park) the ... 2 This is arising from the charge continuity equation. This condition occurs when \partial \rho / \partial t = 0 because the full equation is given by:$$ \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0 $$if there are no sources or sinks. So without charge creation, this holds. Example Application Consequently, the restriction that \nabla ... 3 He is saying that the surface integral over any closed surface must be zero for time-independent current distribution, because otherwise there is a net flux of charge into or out of a volume, and we can't have that going on indefinitely. If \Sigma is a volume with surface \partial\Sigma, we have by Stokes' theorem that$$ \int_{\Sigma} \vec\nabla ...

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Depends on the nature of the collision. If there is a mechanism that takes energy from the system, i.e. a deformation, than energy is lost. You could think of your example as the center of mass of your rod as a point mass that starts rotating on a massless string once is passes the pivot. As usual, energy and momentum are conserved. You could do the ...

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Some people have very long explanations of why the universe can be created out of nothing "because the universe always does this". As a counter argument: There is no proof that the universe always does this. It is only theoretical. If you believe that the whole universe can be created out of nothing, it means you could create energy and matter out of ...

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I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by $$... 7 These types of theories that physicists such as Krauss espouse of a "Universe Coming From Nothing" are quite flawed, as by no means are they talking about nothing! Further, the concepts of particles, mass, and energy are not even well-defined when talking about the universe in general. I wrote a paper on this (excuse the shameless self-promotion), it can be ... 3 The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by:$$D_t = d_t + \omega(t)\times\tag{1} where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body. Since there there are no torques ...

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Indeed. It is due to the law of conservation of angular momentum. The angular momentum of the rotating element within the motor will exactly cancel that of the rest of the motor, thereby giving zero net angular momentum, as with the initial conditions.

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The casing will spin in the opposite direction. That is the principle of reaction wheels.

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The angular momentum of the earth && mouse wheel system does not change. When the earth pulls on the mouse, the mouse pulls on the earth, so no net moment is seen any arbitrary point in the universe.

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In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ...

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