Recent content by porcupine6789

Somebody correct me if I'm wrong but 1atm is 760mm of Hg. So when you invert the tube the pressure exerted by the air and Hg is the same so therefore the Hg will not fall. And I don't think the pressure of the air would change.

Can someone explain these to me?

Ok, so I've been working with a set up similar to this one this summer and I can kind of tell you how we used each thing. First off why do you need the Wollaston between the lamp and the specimen? We just used the half wave plate to control the amount of power that was entering the objective (an...

Ok, so say I have a system at a global temperature, I add an energy source (laser in this case) and it heats the system in a localized area. This leads to a maximum temperature peak in the system. Then you can get out a change in temperature from this maximum to the global. OK, now my...

The half wave plate does not polarize the light into a single direction. This is why the prisim is needed, in order to polarize the light before it enters a spatial light modulator. It can also be used to reduce the power of the laser so you don't burn out anything in the path. But from my...

They are the same thing. a and b are arbitrary letters to denote which two points you are looking at.

If I understand this correctly the separation of colors is due to frequency changes caused by the refraction going on inside the prism. As for when light is traveling inside the prism it is still traveling at speed c. It is getting bent and absorbed from the atoms and molecules inside the...

Usually weight is defined as mass times gravity. So they are not the same thing. As for on the Earth the weight would change a little bit. But not enough for you to really feel. I think the gravity is stronger around the equator than the poles, or it could be the other way around, I forget...

No I mean the one will be positive and the other will be negative.

Well if there is nothing touching the two spheres together then they will remain with their respective charges.

If I understand your question correctly, the sphere closer to the rod would be negatively charged while the other one would become positively charged. When the spheres were touching the electrons would build up on the sphere closer to the rod because of opposite charges while the other one...

So, you have the parallel section of the circuit. Treat that as it's own individual circuit. Add the three resitors, then you plug that back into the circuit. So, you end up with the battery, the Req and the 20 and 5 ohm resitors in series. Then find the current and you can find the...

Sorry I read the original problem wrong. Treat the parallel section of the circuit as its own serperate circuit. Collapse the three resistors into a single resistance. Then try proceding from there.

remember when adding resistors in parallel you add the inverse. (1/R)+(1/R)=1/Req.

Yeah, should be. Except the units of EMF is in volts. Resitance is ohms.