Recent content by ValeForce46

If we say that a proton has a kinetic energy of ##50## GeV, can we say that each of the three quarks that compose it have roughly a mean energy of ##\approx\frac{50}{3}=17## GeV? If not, what can we say about the energy of each individual quark inside a baryon with a known energy?

My professor and the book I'm reading (Particles and Nuclei: An Introduction to the Physical Concepts by Povh et al.) says that "The emitted nucleons are primarily neutrons since they are not hindered by the Coulomb threshold" which means that a neutron has a separation energy lower than a...

Well, thank you anyway for the replies. I suppose, if my questions have no answers, you might as well delete this thread.

Well, according to the book I am reading (from which I took the pictures I posted), "Introduction to the Structure of Matter: A Course in Modern Physics" by John J. Brehm and William J. Mullins, there is a classical interpretation, at least for the "normal" version (no spin is involved yet) of...

If it can't be moved, yes. I'll re-post on classical physics. Sorry.

Sorry, I shouldn't have posted here. It was my first impulse, because this is just the introduction to the Zeeman effect and later the book use quantum physics to explain it.

I'll put pictures from the book as I think they are relevant to understand the problem: I have trouble understanding the case where the observer watches the source in a direction perpendicular to the magnetic field. The electron will rotate around B axis, so the observer will only see a linear...

oops... I did the integral wrong:doh: ##ΔS_{Universe}=ΔS_{System}=m*c*\ln (\frac{T_f}{T_0})+n*c_v*\ln (\frac{T_f}{T_0})=0.02 \frac{J}{K}## Hope this is right ...

Yeah, you made me realize that. Thanks a lot. However, for completeness, the variation of entropy of the universe is the variation of entropy of the system, because the walls of the container are adiabatic so there's no variation of entropy of the environment: ##\Delta S_{Universe}=\Delta...

##K=6.13 J ##. So you're silently (not that much :oldbiggrin:) saying that my result ##T_f=300.013 K## is right? I even doubted my equation.

The kinetic energy of the pendulum ##K=\frac{1}{2}\cdot m\cdot v^2## will turn into heat (entirely). So both the air and the block of iron will change their temperature. To find ##n## (moles of the gas) I can use the ideal gas law: ##n=\frac{pV}{RT}=0.9 mol## Do I have the following equation...

This is how I solved part a) : ##Q_1=C\cdot (T_1-T_i)## This quantity is negative because object 1 loses heat. (positive for the machine) ##Q_2=C\cdot (T_2-T_i)## This one is positive because the object 2 absorbs heat.(negative for the machine) Then the exchanged heat FOR THE MACHINE is...

...Yes, actually there's no reason to change, in this problem. Thanks a lot :oldsmile:

For the second part (the tube is completely closed before it is on the vertical), do I have to repeat the reasoning I did before? Now, when the air under the mercury is compressed, the air above the mercury expands. Boyle's Law: ##p_{atm}*A*l_1=p_1*A*(l_1-x)## for the air under the mercury...

Is this right? ##ρ_{Hg}*A*l_2*g+p_{atm}*A=p_f*A## The first term is the weight of mercury. (On the homework statement I did a mistake ##ρ_{Hg}=13.6*ρ_{H2O}##) If yes, then ##l_1'=0.3 m##.