Recent content by cubeleg

The liquid He4 is boson gas, so at 2.17K they condensate in a new phase called superfluid. Landau model, or two liquid model, states that the normal and the superfluid phases coexist, but below this temperature the superfluid density is bigger. This idea is the same that Landau uses to explain...

A calculator won't give to you what you need. What you need is to study more on the subjet or ask to your supervisor, because indeed it is dangerous what you are doing, in the best case your battery will have a leakage of a powerful acid. This current cannot be supplied by normal methods, you...

Well at least, you won't break your insert :). But I think that your wires will be completely cooked, even if you evacuate the oxygen the coating will be burnt, at least it happened to me with manganine wires trying to do something similar. I advice to try to rewire the sample holder, with...

I think that the answer is yes, the coating will be degraded. But I wonder why do you have to warm all your insert at that high temperature? I don't know what kind of system are you using, but He3 cryostats use to be delicated, warming them at such a temperature could damage it permanetly. For...

I have to agree with Vanadium50. About the josephson effect, although this a bit out of the topic, of course the quantum of flux is involved, but where is radius of the hydrogen atom? and how this is related to this topic? it is interesting because you are dealing with a macrospic system and you...

Ok, thanks for the answer. I have done like that and seems that 6 correspond with 1 pole, 12 with 2 poles, beyond that I haven't obtained any correlations.

In filters usually you can find two different specifications: number of poles and dB/octave. I know that the poles gives you the transfer functions using this equation for a low pass filter H(f)=1/\sqrt{1+(f/f_{C})^{2P}} Where P is the number of poles of the filter. But how can I translate...

I suppose that you mean voltage V in the output, well I would say that when de load resitance is zero, there is no voltage, and the current is the Norton current. As well if the load in infinite, open output, the voltage is the thevenin voltage. But I don't understand very well your question...

It doesn't matter very much, just calculate the voltage in the output open for the thevenin equivalent and the current in the output shorcircuited. To solve the circuit you have to have to select a zero voltage node, but you can put it where you want, I like to put in the minus of the output...

I have to agree with you that the coincidence is an artifact introduced in the "model". About the physical sense of the capacitance is exactly what I try to say in my first post. Regarding the LC, in my opinion, although the analogy is not very useful, the bohr model obtain those number...

I have to say that this coincidence in the energies is strange. Anyway to consider a single hydrogen nucleus as a conductor sphere is a very rough approximation, isn't it? Another point is that assuming your arguments as valid would imply that to introduce a second electron in the hydrogen...

Looking at the definition of capacitance, the ratio between the charge and de potencial in a system of charges, I would say that in some sense your guess is right. If I understand correctly, the capacitance gives you the charge that you have to introduce in a conductor to increase the voltage...

I add my two cents. From wikpedia I have bolded which I think is important for the discussion. In my opinion the division of physics in applied and fundamental is arbitrary, all physicists are interested in explaining some physical event, if it is useful or not is another thing. I think that...

Although almost everything is answered, I would like to remark few points. Actually, ferromagnetism (also antiferromagnetism, as Cr, and ferrimagnetism, as magnetite) are colective properties. The intrinsic origin of all this properties is the exchange, in brief, the energy of a system of...

In a superconductor there is no losses of electric energy, so once you provide into a coil a energy it is trapped and accumulated in current. Of course there are losses, due to several factors, so the current won't flow forever. But in a perfect superconductor the current will stay forever...