Starting today August 7th, 2024, in order to post in the Married Couples, Courting Couples, or Singles forums, you will not be allowed to post if you have your Marital status designated as private. Announcements will be made in the respective forums as well but please note that if yours is currently listed as Private, you will need to submit a ticket in the Support Area to have yours changed.
Well, I've never seen such complex diagram of the thing. I've seen some simplified diagrams though. I don't know what most of the elements are. I suppose that some of them have to be protecting Z-diodes. But that out of my competence anyway.It's a MOSFET transistor. N-channel, to be precise. In a simple integrated circuit.
Well, I've never seen such complex diagram of the thing. I've seen some simplified diagrams though. I don't know what most of the elements are. I suppose that some of them have to be protecting Z-diodes. But that out of my competence anyway.
It's a MOSFET transistor. N-channel, to be precise. In a simple integrated circuit.
I'm not a physicist, so you can stump me if you like to.Oh! I've got a real question for a physicist. And this is not just a "try to stump the physicist" question...this is a real question:
It's a guess, but I think it has to do with the fact a charged particle in magnetic field will emit photons. Another possibility is that charged particle with high energy can ionize the matter it goes through. The particles from a high energy collision has enough energy to ionize a lot of atoms.The sensor arrays at CERN (or Fermi, or SSC, whatever). How do you induce an electrical current and/or electric potential from sub-electronic particles?
I don't think that the size has anything to do with it. The energy matters. Singe energetic (fast) electron can knock other electrons from a lot of atoms.What I can't figure out is, the particles are so much smaller than even a single electron. How do you produce a flow of multiple electrons out of that? Does a boson produce a magnetic field or something?
I don't think the current produced by a single particle is what they measure. So, the capacitance does not matter.For that matter, the whole incident happens in the span of attoseconds. How do you capture that??? The smallest capacitance we can get in the circuitry is on the order of femtofarads. It seems like the whole reaction would get lost in the noise. Do you just massively over-sample and stagger the phase over attosecond-long increments or something?
Measuring the curvature of the trails and the trails of the cascade of secondary particles allows measuring the charge, mass, energy of the particle in question.
I really doubt they detect quarks directly, as when you have the energy to make a quark escape a hadron then you have the energy to make new hadron. Thus you detect most of the particles by the way they decay.Well that explains how you can interpret the signatures. You see one trail curve twice as much as another, then one's a top quark and the other's a bottom. You see a trail that does not mathematically match anything you've seen before, then you've discovered a new particle. Cool....
But I thought they were smashing H+ ions (a.k.a. protons) in a vacuum? How do you sense a curve if you don't sense anything at all until you collide with something?
But I thought they were smashing H+ ions (a.k.a. protons) in a vacuum? How do you sense a curve if you don't sense anything at all until you collide with something?
Totally legit.
They're not quarks, they sound more like electrons and positrons (opposite charge, same mass, so equal and opposite spirals). We can't isolate quarks yetWell that explains how you can interpret the signatures. You see one trail curve twice as much as another, then one's a top quark and the other's a bottom. You see a trail that does not mathematically match anything you've seen before, then you've discovered a new particle. Cool....
Because we know what we should see if the Higgs boson (say) exists. A major tool is not in what it does, but in what its decay products do - if the Higgs decays into a shower of pions and kaons (I have no idea if it can or not), we can trace back the identified paths of paions and kaons to a single point - the Higgs.But I thought they were smashing H+ ions (a.k.a. protons) in a vacuum? How do you sense a curve if you don't sense anything at all until you collide with something?
We use multipliers to amplify what each particle does. Take a look at this video - you can see macroscopic effects of single atoms decaying:Oh! I've got a real question for a physicist. And this is not just a "try to stump the physicist" question...this is a real question:
The sensor arrays at CERN (or Fermi, or SSC, whatever). How do you induce an electrical current and/or electric potential from sub-electronic particles? You know, you're trying to track down the signature of a Higgs Boson (or strange quark...whatever). Somehow you capture these signatures in the form of an electric current or something. You amplify it, run that through ADC, digitize it, display it on a computer. What I can't figure out is, the particles are so much smaller than even a single electron. How do you produce a flow of multiple electrons out of that? Does a boson produce a magnetic field or something?
For that matter, the whole incident happens in the span of attoseconds. How do you capture that??? The smallest capacitance we can get in the circuitry is on the order of femtofarads. It seems like the whole reaction would get lost in the noise. Do you just massively over-sample and stagger the phase over attosecond-long increments or something?
"A scientific man ought to have no wishes, no affections, -- a mere heart of stone."- Charles Darwin
If you mean starting from the ground going up and then coming down then you have to specify the climb rate. At one g the drop rate is 10 metres per second squared.How high must one jump (in a vacuum obviously) so that you stay in the air for 1 second. Please give your answer in feet or meters please. Ok, Europeans can answer in metres, but furlongs are right out.
If you mean starting from the ground going up and then coming down then you have to specify the climb rate. At one g the drop rate is 10 metres per second squared.
If it takes you 0.5 seconds to climb then the total distance (start to finish) should be 5 metres thus the height should equal 2.5 metres if you exclude inertia. otherwise you have to calculate the deceleration time before reaching a speed of zero before gravity causes you to reverse direction.
Someone do the maths here because I am too full of Easter cholesterol and wine to think straight
Correct answer. Unfortunately, I can't give you full credit. You took it upon yourself to set gravity to 10m/s*s rather than the universal 9.8. Fittingly, I'm awarding you 98/100.If you mean starting from the ground going up and then coming down then you have to specify the climb rate. At one g the drop rate is 10 metres per second squared.
If it takes you 0.5 seconds to climb then the total distance (start to finish) should be 5 metres thus the height should equal 2.5 metres if you exclude inertia. otherwise you have to calculate the deceleration time before reaching a speed of zero before gravity causes you to reverse direction.
Someone do the maths here because I am too full of Easter cholesterol and wine to think straight
Correct answer. Unfortunately, I can't give you full credit. You took it upon yourself to set gravity to 10m/s*s rather than the universal 9.8. Fittingly, I'm awarding you 98/100.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?