@0celo7 It looks fine to me. Adding other things like sun dried tomatoes would make it different but not necessarily better.
At the end of the day the point is to enjoy eating it.
Sun dried tomatoes seem fashionable in the cooking world these days, but I have to say I find them a bit chewy. I'm not sure the texture would work ina meal like yours.
@DanielSank Well, correct me if I'm wrong, but for an object rolling without slipping the formula is $\overrightarrow{v} = - \overrightarrow{\omega}R$, right?
Let me tell you something though: if you don't learn to disconnect from your work, drink a cup of tea, listen to music, and then go to sleep knowing you can think more tomorrow, you're going to have a bad time.
@SirCumference If $\hat{z}$ points up from the ground, then that makes no sense.
Have we got a What is an elementary particle? question? We get so many related questions that I think a canonical q/a would be good if one doesn't already exist.
That would be addressed in the q/a. I would interpret the question as what are particles in QFT? since in non-relativistic QM we just assume their properties.
So we'd say particles are the objects described by Fock states
My confusion about quantum theory (beyond my obvious unfamiliarity with its terminology) is basically twofold:
I lack an adequate understanding of how the mathematics of quantum theory is supposed to correspond to phenomena in the physical world
I still have an incomplete picture in my mind of...
The title isn't what you're after, but the content is close.
Also, if we tell someone that particles are the things that we represent with Fock states, then we'd better explain why those things behave particle-y.
@JohnRennie From inequality of Clausius, $\Delta S \ge \dfrac{q}{T}$
For an isolated system, $q=0$
Then $\Delta S \ge 0$
But my book states, that for spontaneous process within an isolated system $\Delta S >0$
Why no equality, JR?
If anyone knows the answer, please let me know.
@SirCumference The thing is that the point of contact doesn't slip, so $v_{cm}- R\omega = 0 \implies v_{cm}= R\omega$. to understand this, you can split the motion of the rigid body into two: Pure rotational and pure translational. FOr the latter, every point on the rigid body has the velocity of com while for the former the velocity of any point is $R\omega$ tangentially.
@Abcd I believe that the gist is (it's been a considerable while since I've done any thermodynamics, so it's very possible that I'm missing something or other): isolated system $\implies$ no exchange of energy and it seems reasonable that this means change in enthalpy ($H$) is $0$, while 'spontaneous process' is defined as the change of Gibbs Free energy being negative, which gives $\Delta G = \Delta H - T\Delta S < 0$ for constant temperature, giving $\Delta S>0$
> A pressureless perfect fluid can be interpreted as a model of a configuration of dust particles that locally move in concert and interact with each other only gravitationally, from which the name is derived.