Recent content by Karmaslap

Dave answered your questions, I just gave an analogy to understand a relay.

Anything conductive can be used as a ground reference. "Ground" just means the place you define to be 0 volt potential. It does not mean actual earth, or somewhere current dissapears to, it's just convention.

Something can only have voltage with respect to something else. Hooking up a single wire is meaningless, unless you are defining that wire as ground or "0" volts. When we say, for instance, 9 volts, we mean 9 volts potential with respect to our choice of ground. A single wire can't charge...

You need to connect both terminals of the cap to the anode and cathode of the battery to charge it. The voltage at a point has to be in respect to something. I think that if you hook up your battery to Earth ground and the capacitor to Earth ground, and then run a wire between the positive...

You could make a tesla coil, and the long arc that comes off of it is a plasma. Much cooler.

Nor would iron powder oxidize inside of an oil.

A relay could also be solid state, a transistor is essentially the same thing. OP- think of it like a dam full of water. By pulling a lever, you use a small amount of force to open the floodgates and release the water, which generates a LOT more force. Your small force activated a larger one...

If increasing your number of turns is no good, can you decrease the diameter of the air gap? That will also give you a stronger field. Materials with high magnetic permeability won't multiply your field- they will "guide" it to some extent so you can more easily measure it. Any ferromagnetic...

Velocity is max when KE is maxed. Force and Acceleration are both max when the PE is max. You should have known when force and velocity were maxed if you know what KE and PE are.

You should state your question better. I don't know what you're asking. At equilibrium Δx = 0 and at max amplitude Δx is greatest. The force the spring exerts at equilibrium is zero, and is max at amplitude. K never changes.

Δv = v1-v0 = (π,0) - (0,-π) = (-π,-π). This is wrong. Δv= (π,π). By basic arithmetic, your pies are positive. 0 - - π = π, and π - 0 = π

Jumping onto his thread for a question I had, will AA batteries output more current than AAA will?

Assuming you're using a solenoid shape which is longer than it is wide (and not all that wide) you can estimate the strength of the field in the center of the solenoid with B = μNI/L. As you can see you will have a very, very small field with micro or nano amps even with a high number of turns...

I'll draw up a diagram to help explain why I am interested. (Also, a solenoid-based accelerator would be a coilgun, not a railgun. This is a linear motor: This would use the same principle as a coilgun, except that I don't need a huge discharge from a capacitor into the coil for rapid...

Hi, thanks for looking at it. I should edit out the sawtooth looking image and replace it for one with a flat top, which i would prefer. I wrote that above it... But huge wall of text means notes get skipped over. The second figure is exactly how a normally-shaped field would be. Thinking of...