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The mass of a fundamental particle turns out to be quite an elusive concept, because massless particles act as a source of gravity and they carry momentum. What then is special about mass? Where mass comes in is in explaining the relationship between the total energy of a particle and its momentum. For any particle we have the expression for the total ...


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When you are given acceleration, density, area and time, you can indeed find an expression for mass in terms of these. Here is how you go about it: Make a table of the units that occur in each, and their exponents: L M T a 1 -2 D -3 1 A 2 t 1 As you can see, you need to use D (density) as the only one that contains mass. But ...


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The particle data group has a thorough article on neutrino oscillation experiments. Oscillations are the way to measure mass differences between neutrinos, and in table 13.7 experimental values are given for delta(m^2) of the three neutrino species. These put a lower limit to the mass values. From the definition it is obvious that the limits are of the ...


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The mass of each of the three mass types of neutrinos is unknown (although a lower limit has been placed, as Anna V's answer indicates), chiefly because they exist in three weak classifications as well as the three mass classifications, and the combinations can not be simultaneously isolated and identified with certainty. The combined mass of all three mass ...


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A tensile design for a space elevator cable requires there to be a counterweight some distance beyond the point of geostationary orbit. The counterweight would be hurled with centrifugal force and would keep the entire cable taut. The cable must not only support a payload, but also must be a tether for the counterweight, and provide for centripetal force ...


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This new paper addresses exactly this question, albeit with simulations. Here's a partial breakdown of the distribution of matter in the Universe, summarized from the above paper: Dark matter makes up about 26% of the critical energy density budget of the Universe, while "baryonic matter" (which is jargon for "visible matter" and includes all baryons as ...


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Massless particles can carry and do carry confined charges (gluon in QCD) and they may carry charges under spontaneously broken generators (photon is transforming e.g. under the generators of $SU(2)$ associated with the W-bosons) but they cannot carry charges under unconfined $U(1)$ force like electromagnetism. The reason may be explained in different ways. ...


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We integrate $x dm$ to say "for every piece of mass $dm$, sum up the amount of mass times the coordinate $x$ of that mass". It makes sense to integrate over $m$, because every piece of mass has an $x$-coordinate. Likewise, in kinematics, it makes sense to integrate over $t$ because for every $t$ we have a velocity $v(t)$. However, calculus doesn't care ...


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The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry. Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum ...


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A classical "shell" of charge of radius $R$ will appear to have a field, outside the shell, corresponding to the charge all being at the center of the shell. If the charge on the shell is $q$ then a charge $dq$ being pulled in from infinity will cost an energy $k_e~q~dq / R$; without loss of generality this is the same if $dq$ is spread over an infinitely ...


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There are 4 general contributors to the mass density of our co-moving patch of what may be a larger universe: (1) Visible baryonic matter (including clouds of baryonic matter which may be visible only as shadows blocking galaxies). NASA estimates 4.6% of all matter is baryonic (http://map.gsfc.nasa.gov/universe/uni_matter.html). (2) Dark matter (not ...


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You can find an excellent description of what a topological insulator is in this brief presentation from the Yazdani Group at Princeton: Topological Insulators. To answer your question on the meaning of negative effective mass: The effective mass is actually determined by the behavior of the energy levels $E({\bf k})$ as functions of the crystal wave ...


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You would have to apply a force upwards to stop the body. Stopping the gravity would stop the acceleration but not the speed that it already has. A good place to start to check the effects of g forces in a human body is this wiki Changing the mass won't stop the fall either. You cannot make the mass zero, you can cut legs and arms but I think you'll be ...


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You can actually (in principle) do an experiment for one of those. If you had a huge, dense, thin sheet of matter in empty space you could cut a person sized hole through it then attach a cylinder to the hole and put a flat bottom on the far away end. So on one side it looks like a flat tower coming up from a big plane and on the other side it looks like ...


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Well, you can think of this as a car pulling a wagon with $a = 0.2 \text{ m/s}^2$. Naturally, the car's motor needs to generate enough force to accelerate both the car itself and the wagon it is pulling. This is true of the situation in this question. The steam engine needs to generate enough force to accelerate both itself and the two wagons at $a = 0.2 ...


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You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." ...


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When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously. When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much ...


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regarding the first half: Density is Mass/Volume (not the other way around), so you need a Volume.. You could achieve that by a sufficient power of the area or find yourself a length-scale from acceleration and time regarding the second half: that's the whole point of dimensional analysis, a mass-like quantity is a mass-like quantity and must come from ...


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I can see that this question has been downvoted but I think it still deserves a proper answer. First, it is the Chandrasekhar (one word) limit, named after the Indian-American astrophysicist Subrahmanyan Chandrasekhar. Second, the Chandrasekhar limit does not mean that an object cannot be more massive than 1.4 times the mass of the Sun. There are plenty of ...


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If you consider that gravity is weak compared to the electromagnetic force because $G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2} $ and $k_e \approx 8,987 \times10^9 N m^2 C^{-2} $ it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, ...


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If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.


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I don't see the paradox you are describing and i have some issues with your text: "it will always fall with an acceleration of 9.8m/s": If there is air resistance, it will reach a terminal velocity at which the object is no longer accelerating. Let's assume no air resistance to make things easy. "If I throw it very high up in the sky, it falls with a ...


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While your object is in motion its acceleration is $g$, or -9.81 m/sec$^2$ (we'll take the upwards direction to be positive). This constant acceleration is why the velocity decreases from its initial value of $+v$ when you throw it to $-v$ when it lands. So far so good. But I think the force you are talking about is the force required to stop the object ...


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By your logic, there is no gravitational force on the ball while you are holding it, because it is not accelerating. And if there is no gravitational force you should be able to throw the ball as high as you want, right? The truth, though, is that there are two forces on the ball - one from your hand and one from gravity - which are in balance. Keeping that ...



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