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New answers tagged mass

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If You know density $rho_r$ at some temperature $T_r$, there is a following formula for density: $rho=rho_r[1+b(T-T_r)]$, where $rho$ is the density at temperature $T$ and $b$ is called coefficient of cubical expansion, evaluated at reference temperature and density ($rho_r$ and $T_r$). It is valid for liquids.

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I think you're confusion is with notation since unfortunately, two notation often used to denote projected spinors. One notation is to write: $$\psi \equiv \left( \begin{array}{c} \psi _L \\ \psi _R \end{array} \right)$$ In this notation $\psi _L$ and $\psi _R$ are two component Weyl spinors. However, a second notation ...

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In principle one has to calculate the pole of correlation functions involving gauge invariant operators like $\text{Tr}F_{\mu\nu}F^{\mu\nu}$. The problem is that due to asymptotic freedom, QCD is not solvable perturbatively at low energies. This is why nonperturbative techniques like lattice QCD are used to calculate such spectra. A key achievement in this ...

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I haven't checked the arithmetic, but the method to calculate the acceleration is right. The easiest way (to my mind) to find the tension (T) in the string is to apply F=ma to the smaller mass block. Hope fully if you do that, you will also find that (T-6.4) will also accelerate the larger mass at the required rate as well.

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The tension in the string has nothing to do with gravity. Just consider the forces acting in the horizontal direction for each mass separatedly.

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As the blocks are moving horizontally, the gravity and normal forces cancel each other. So your logic about finding the tension is faulty. You should use Newtons second law on each separate block. To find the tension you could suffice with the last one. If there is friction you could add the friction force to the equation, and think about what this would ...

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Your error is that one pound-force is not a pound. A pound-force is a force while a pound is a mass. A pound-force has Imperial units of $$1\,{\rm lbf}=1\,{\rm slug\cdot\,\frac{ft}{s^2}}$$ where slug is the Imperial unit of mass. One pound is equal to 0.45359237 kilograms, thus 150 lbs = 68 kg as you expect when doing the conversions.

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There's a thing called a "slug". " It is a mass that accelerates by 1 ft/s2 when a force of one pound-force (lbF) is exerted on it." (wikipedia). Sometimes you'll see reference to a "pound-mass" to indicate a mass which weighs one pound at sea level (on Earth, thank you! :-) ). The problem is that pounds and kilograms have been used colloquially since ...

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The first equation is a very close approximation since m (satellite's mass) << M (Earth's mass) so m can be ignored. The second equation is the mathematically correct one.

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My textbook uses $R$ for activity as corresponding to the decay rate.

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Electricity does have mass, yes. Indeed, one of Einstein's 1905 papers, "On the Electrodynamics of Moving Bodies" specifically demonstrates this. A moving magnet becomes more massive due to its increase of energy, and this additional inertia causes its electric field to increase in strength as well. Hence E = mc^2. If you wished, with sufficiently accurate ...

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This is a fundamentally pointless question because negative mass doesn't exist (or so we think!), but I'll answer anyway because the answer is so unexpected. Suppose we take our two massive bodies: Then the gravitational force between them is repulsive because: $$F = \frac{G m_1 m_2}{r^2}$$ and $m_1$ and $m_2$ have different signs. But let's ...

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You are quite right. Energy is not conserved between the reference frames. That is the biggest mystery. It is certainly going to change one's concept of understanding of energy. I m here giving a simple example where there is total failure of 'law of conservation of energy' Take an example of spaceship in space. suppose you start the spaceship and ...

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Non-conservation of charge in Majorana terms The Dirac mass term is $m\bar\psi \psi$ where one field-factor $\bar\psi$ is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not. So one may assign a fermion number $1$ to $\psi$ which means that $\bar\psi$ automatically carries $-1$ and in the product, the ...

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Mass so small that it cannot be observed, measured, tested, is unknown, but believed to exist, is not mass. it is pre-mass. It is alpha. The substance and evidence is confirmed only by our strong, unwavering belief that all things visible are made of things not visible.

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Someone posted the energy conservation way (if you have spring constant and distance stretched)so here is the Newtons laws way: If you know the mass of the object and the force on the object and the distance the has the force applied on it then you get the velocity as it leaves the slingshot (given newtons laws: \begin{align}F=ma \\\ F/m=a\end{align} ...

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I believe if you're modelling the slingshot as a spring, you can consider the potential energy stored within it when it's loaded is fully transferred to the kinetic energy of the projectile (neglecting the friction during the short acceleration etc.), and then it gets down to the ballistic of a freefalling object with air friction (see the drag coefficient ...

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The simplest setup is for small displacements. Suppose the spring rest lengths are $L_1,L_2,L_3$, the mass has mass $m$, the springs have constants $k_1,k_2,k_3$, the angle is 120 degrees between attachments, and the attachment points are set up so that at rest, the springs are all unstretched. The potential becomes ...

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Arguing in classical terms, it is just related with the fact that in addition to the external force (e.g electric field) which is applied on the crystal there is a complicated interaction between the electrons and the crystal (periodic potential), which results in a net force that is dependent on the part of the band that the electron occupies and may be in ...

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This going to be a rather approximate answer because it involves lots of estimated quantities like the current density of matter and the value of the cosmological constant. The second Friedmann equation tells us: $$\frac{\ddot{a}}{a} = \frac{-4\pi G}{3}\left( \rho + \frac{3p}{c^2} \right) + \frac{\Lambda c^2}{3}$$ It's conventional to take $a = 1$ at ...

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Yes it fluctuates but it is a very small fluctuation. Note that unstable particles have a decay rate or width $\Gamma$ that is related to its lifetime $\tau$ by $$\Gamma=\frac{\hbar}{\tau}$$ when you measure the mass/energy of such particles in experiments you always get a Lorentzian or Breit-Wigner distribution like this from which you can measure the ...

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Take a simple quantum mechanical potential that describes an atom. The mass of the atom is fixed. Take hydrogen. The electron is in an orbital around the proton which is a probability distribution of its location in time and space: if you measure it, i.e. interact with it, where you may find it. Correspondingly there exists an energy width to the energy ...

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The so called Copenage Interpretation avoids the question about whether the electron is a particle or a wave. This question is directly not allowed. In fact the wave function is an instrument of the theory with not physical meaning. Acording to CI, the goal of the theory is only to make predictions about the results of an specific experiment. In the case ...

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If you substitute the decomposition in, you get: $$\partial_t \rho^0 + \partial_t \rho^{E1} + \nabla \cdot (\rho^0 \mathbf{u}^E) + \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0$$ Typically the decomposition used assumes that $\rho^0$ is constant in time and that $\rho^{E1}$ is random in time, such that it's mean value is 0. Therefore, $\partial_t \rho^0 = ... 0 Now for an easy answer: Physics moving (at a constant speed) is physics standing still. From one person looking at another, they will generally disagree about how fast the other is moving and what time it is, but the underlying formulas of physics are unchanged. The space and time part is a really strange idea, but just imagine what the hell this place ... 0 I've heard that a spacecraft could never exceed the speed of light because it's (relativistic) mass quickly approaches infinity and therefore there could never create a big enough rocket to propel it faster and faster. In fact, the spacecraft could never even reach, much less exceed the speed of light. I think that you'll agree that the ... 0 Benford's law is pretty cool. It states that, for many sets of data, a leading digit of n has a probability of$Pr(n) = log_{10}(1+1(n))$Plugging in our n values we find that we can expect low values of n to have a higher probability of being our leading digit. The most (initially) boggling thing is that our$Pr(1) = .301$stays independent of units. If ... 1 The problem is- The rocket is not 'fighting' with any force-field, it is 'fighting' with the very nature of space-time. So unless we have something of zero rest mass 'things' will tend to infinity. And yes the thrust will increase but space-time will distort( following Lorentz transformation, no GR effect here) in such a way that reaching 'c' 'tests' our ... 1 Both Newton and Einstein say that all the laws of physics remain same in a frame at rest with respect to the observer and also in a frame moving with constant velocity. Now Maxwell showed that speed of EM wave in vacuum equals 'c' is a law of physics. But this law was inconsistent with the Galilean transformation, hence Lorentz transformation were ... 1 This question is really about history and what was known to the protagonists in your tale and at what time. As a principle, relativity was embraced every bit as fully by Newton and Galileao as it was by Einstein - it's just that Einstein had a few more experimental results he had to gather into relativistic thinking. As in dgh's answer the whole point of ... 2 Physics occurring in one spacecraft traveling fast (at a uniform speed) is the same as physics in another spacecraft "standing still". In fact, the whole point is that these words - traveling fast and standing still - are relative. All the things you are describing like relativistic mass are only apparent to observers in other reference frames. The person in ... 3 The wave-particle duality thing becomes important when you are dealing in a microscopic scale where quantum mechanics becomes relevant and you have to discard your ordinary notion of particle and wave. So don't expect to relate "particle" or/and "wave" notion that you usually get from picturing a marble or water wave from classical world surrounding you. ... -4 mass=energy(actually mass is a form of energy) so everything who has energy(wave) has mass also if you squeeze all wave in a very little place you get a solid item(like you). if you squeeze a solid(like you again) in a very little place you will get a black hole 1 Electron is accompained by waves, so there still exists electron which has mass. This solves your problem I hope. Look here at what de Broglie says in his Nobel lecture of 1929 (this is an extracted portion): I thus arrived at the following overall concept which guided my studies: for both matter and radiations, light in particular, it is ... 34 I don't really like the whole wave-particle duality business because it obscures the more startling truth about particles: they aren't sometimes waves and sometimes particles, and they also don't transform into waves sometimes before reforming as particles, they are something completely different. It's like the story of the blind men and the elephant: a ... 9 The waves of quantum mechanics are probability waves. The solutions of quantum mechanical equations are the wave functions and the square of the wave function gives the probability of finding the particle at$(x,y,z,t)$. That is why the solutions for the electrons in the field of a nucleus are not orbits, but orbitals, i.e. probability distributions. The ... 7 It seems that you have misundrstood the wave-particle duality. What happens in the double slit experiment is that the electrons impact at the screen as they were particles. But they also interfere, just as waves. So you can see a wave-particle behaviour. But it doesn't say that the electron is destroyed, becomes a wave and then a particle again (as you ... 0 Before considering how rest mass of an object turns into energy, it must be considered what relativistic mass is. Relativistic mass is expressed by equation(1) and equation(2). Mass(M) is made of countless relativistic masses which have kinetic energy and small rest masses as it is expressed by equation(2). vi is the velocity of individual mass(mi). When ... 1 Atmospheric pressure, at some altitude, is largely due to the weight of the atmosphere above that altitude. The atmosphere also virtually behaves like an ideal gas. If evaporation of liquid water increases the concentration of H2O molecules in the air the density decreases and the air becomes buoyant and rises. As it rises it expands and cools. Rising air ... 3 The atmosphere is more or less at equilibrium with regards to being compressed by gravity, so there is no atmospheric heating caused by gravity (otherwise the atmosphere would gradually get closer and closer to the Earth, which clearly is not the case). Heating of the atmosphere is almost entirely due to radiant energy from the sun, along with terrestrial ... 3 The gravitational self-energy of any arbitrarily-shaped object with density function$\rho(\mathbf{r})$and corresponding gravitational potential$V(\mathbf{r})$is $$E=\frac{1}{2}\left\langle V,\rho\right\rangle=\frac{1}{2}\left\langle \nabla^{-2}\rho,\rho\right\rangle=\iiint\nabla^{-2}\rho(\mathbf{r})\rho(\mathbf{r})d\mathbf{r}.$$ Under a scale-parameter ... 0 The charge of a particle is completely independent of its mass. If you had some technique (and there isn't one) to just remove charge but keep the proton exactly the same it would not change its mass. Particle masses arise from a combination of the mass of its constituents and their interactions (the potential energy of particle interactions give it mass ... -1 D=m/v, (with rounded numbers) Sphere for volume= (4/3)(pi)(radius^3) r (cm)=(1.0x10^-13)cm mass (g)= (1.7x10^-24) =(1.7x10^-24)/((4/3)pi(1.0x10^-13)^3) =(1.7x10^-24)/(4.1887x10^39) = 4.0585x10^14 2 It can be seen to follow from a more general statement, namely: "if a parameter in the theory is such that the symmetry gets enhanced when it vanishes, then at every order in perturbation theory the corrections to this parameter will be proportional to its bare value". This is because perturbation theory respects the symmetry of the classical theory. If ... 3 Pretty much no. The problem is that you are not (pardon me :-) ) a rigid body, so you're going to feel a certain amount of force from the wind regardless of what sort of weights you're carrying. What can help is walking with your feet farther apart, which gives you a more stable base to work from, and to learn to turn your body sideways to the wind as ... 1 Suppose such a particle existed. Question is what would happen if it was to enter an electric field? Consider$p$($m = 0$,$q > 0$) entering an electric field$E_i$, on a manifold$M (i,j)$$$F_i = q E_i \; \; \;\text{but} \; \; \; F_i = m a_i$$ It follows that$F_i = 0$since$m = 0$meaning either$q = 0$or$E = 0$, but such is not the case,$F_i$... 1 Amplitude is the maximum displacement at steady state. That is okay for a working definition. So, all you need to do is find out is the maximum value of$|y(t)|$at steady state. There some small algebraic errors in your solution. That said,$y(t) = y_{cf}(t) + y_p(t)$To illustrate my point better, let me use the following form for$y_{cf}(t) ...

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Since this question is about how a photon can travel at light speed and yet have no mass, I will answer by saying that photons having no mass is precisely why they can travel so fast, and without mass, it becomes intangible for anything to make it go slower.

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