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9

Yes, there are the quantum numbers Charm, Strangeness, Topness and Bottomness, which are conserved by strong and electromagnetic interactions, but not by weak interactions. Upness and Downness are simply the Isospin, which is also preserved for strong interactions, when the quark masses can be neglected, which is usually a very good approximation as ...


6

When a radioactive element decays, part of its mass is converted to energy - no obvious need for antimatter anywhere. Instead, the energy is released because the binding energy of the sum of the fragments might be higher than that of the parent nucleus. However, to fully convert matter to energy you do need the antiparticle. Otherwise, you run into ...


5

The website to which you linked doesn't seem to understand the purpose and results of Prof. Schwab's experiment. In fact, it didn't really describe the experiment at all. It just rehashed a lot of quantum mumbo-jumbo to make it look as though some power of "mind" causes quantum effects, rather than the more mundane cause-and-effect of having to use tools ...


4

First, let us look at the current rate at which the moon slows down. I have a few different sources, and they don't all give me the same answer. First, there is this claim that Earth slows down at a rate of about 0.005 seconds per year per year. A year has approximately $365.25 \cdot 24 \cdot 3600 = 3.15\cdot 10^7 \mathrm{sec}$, so 0.005 seconds change ...


4

Energy is never created nor destroyed, and to say "X is converted into energy" is just meaningless. We don't convert things distinct from energy into energy, all we ever do is convert one form of energy into another. The badly posed question from your book probably intends to ask why we cannot convert the mass energy that any chunk of matter contains as per ...


3

How does the kinetic energy of a ballerina increase? Conservation of angular momentum: $$L_1=L_2 \implies I_1\omega_1=I_2\omega_2\quad\quad (1)$$ Pulling in your arms reduces moment of inertia $I$, since the same mass is now distributed over a volume closer to the spin centre, $I=\sum mr^2$. As you say, reducing $I$, so $I_2<I_1$, implies ...


3

This decay (occurring via the strong interaction) violates the charge conjugation since $J^{PC}(\pi^0) = 0^{-+}, J^{PC}(\rho^0) = 1^{--}, J^{PC}(\eta'^0) = 0^{-+}$. The charge conjugation transforms a particle in its anti-particle. In the case of the 3 particles involved in this decay, they are all their own anti-particle, and the effect of the charge ...


3

No, the law of conservation of momentum definitely applies to photons. The modification is that the momentum of a lightlike particle is constrained to be equal to its energy divided by $c$. Therefore, for photons, where $E = h f$, then the momentum is equal to $p = \frac{h f}{c}$


3

You're right that the reaction fails to conserve baryon number. The change in strangeness is a strike against the reaction, but not a fatal one; after all, the strange $K$ mesons decay into various mixtures of zero-strangeness mesons, charged leptons, and neutrinos. The thing to notice is that only the charged weak current, mediated by the $W$ boson, ...


2

The thrust coming from a rocket engine is exerted on the engine bell, and it is directed along its axis of symmetry. It's not completely clear how you're modelling your ship but it is probably more realistic to apply the force to the "thruster fire" block, whatever that is. It's important to note, though, that if applying the force to the engine bell and ...


2

If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ...


2

Even if the original dust cloud only had a relatively small angular velocity (which it might have had for all sorts of reasons), the process of collapsing would have amplified it. That is, the collapse process preserves the angular momentum, but it translates to a much larger rotational speed in the newly-collapsed system. Think of what happens to a spinning ...


2

The Moon rotates around the Earth slower than the rotation of the Earth itself. That's why, from a fix point on the Earth, the Moon appears to be moving. The Moon creates the tide on Earth. So the tide "follows" the Moon. However as the Earth rotates faster than the Moon it will tend to carry the tide with itself "forward". The Moon pulls the tide toward ...


1

HyperLuminal asked: "Does that mean that electrons are infinitely stable?" Think about Dirac's model of an electron, which includes left and right handed contributions. Now add the (Nobel-worthy) Brout-Englert-Higgs idea, that the left-handed bit interacts with a condensate of weak hypercharge, while the right-handed bit does not. This suggests a ...


1

As the Earth isn't a closed system with regard to other objects in the Solar System, including atomic particles, its momentum is affected by collisions. But if the collisions were inelastic (if the atomic particles were absorbed into elements of the crust and the atmosphere), the momentum of the atomic particles and the momentum of the Earth involved in the ...


1

Explain why the mass of a tree cannot be converted directly into energy. That's a tricky one, because it could turn out that it is possible to turn matter alone into energy. Floris hinted at this with radioactive decay, but there are potentially other methods such as melting hadrons in a quark-gluon plasma (QGP), see for example this report. The interesting ...


1

The ping pong ball would lose a tiny amount of kinetic energy to the truck. The truck ends up with a momentum of just under twice what the ping pong ball had. However, energy is 1/2 m*v^2 = 1/2(m*v)^2/m. Since the truck is much more massive than the ping pong ball, it carries much less energy for a given momentum. The end result is that the small amount of ...


1

Yes. I'm not sure you can call this "inelastic" collision as that implies that some kinetic energy is lost (due to heat from the collision for instance). If that's wrong feel free to correct me. Start with writing out the conservation of momentum for the situation where two objects merge after colliding. $$p_1+p_2=p_3$$ $$m_1v_1+m_2v_2=m_3v_3$$ You know ...



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