Recent content by [V]

Thats awesome, thanks for your reply.Relating to that topic, I have one more question. Now, if I used a generator (rotating magnetic field) to generate electricity to say, REDUCE an unbalanced solution (that is mostly positive in character) or even found a way to reduce the aerosol spray in air...

I see. What about a solution with a very low concentration of one ion? Would this be possible by blasting gaseous ions through a magnetic field to separate them? I imagine eventually it would stop flowing to their respective container since the coulumbic force is so great. But there has to be a...

Why not? if I had a solution of K+(aq) and Cl(-) and placed it within an electric field, my ions would separate to different sides correct? Why can't I just take a beaker and scoop of the liquid toward one end of the electric field? Wouldn't that contain mostly anions/cations and no longer be...

Whats wrong with the latex??

ty''-(t+1)y'+y=t^2 I know I have to use variation of parameters to solve this. But I am stuck and cannot figure out how to get the homologous equation! y''-(1+\frac{1}{t})y'+\frac{1}{t}*y=t I don't know how to solve this homologous equation in this format. Is it R^2+(1+1/t)R+1/t = 0 ? How...

Can you please give me any physics related formula that appears in this format? \frac{x^2}{4} Its part of a really crazy proof I am trying to do. The variables are irrelevant. Anything that looks like this would be fine! Thanks!

Thank you! :) Soo V = -\int E dr If I derive both sides, I get \frac{dV}{dr}=-E E=\frac{-\Delta V}{\Delta r} With this equation, my other problem seems to work. However, now this presents another problem! Using this relation, I want to find the Capacitance of a parallel plate...

Ahh! Cant be! \int_i^f F dr = (-\frac{KQq}{r_{f}})-(-\frac{KQq}{r_{fi}}) W=-(\frac{KQq}{r_{f}}-\frac{KQq}{r_{fi}}) \Delta U=-W \Delta U=\frac{KQq}{r_{f}}-\frac{KQq}{r_{fi}} U=\frac{KQq}{r} Correct?

I am having a hard time figuring out the signs for some of these formulas: First of all, U=Potential Energy V=\frac{U}{q} \int F dr = \int Eq dr = - Work Since, Work=K=-U Therefore \int Eq dr=+U U=\frac{QqK}{r} Therefore, V=\frac{KQ}{r} BOTH U & V have a positive slope of 1/r...

Up until point 1 it is uniform, then at 100v the slope (E) changes. How does taking the average of the two points give me the correct answer for point 1. It worked, but can you please explain this in detail? Thank you!

How would that work? This is not a uniform E field, I'm sure.

That would be zero correct? I'm not sure how this answers my problem though. Any charge that moves around the equapotential line will have moved perpendicular to an E vector, therefore would have preformed zero work, and ΔV=0 But how would I go about figuring out the E field at any particular...

Homework Statement Determine the magnitude and direction of the electric field at point 1 in the figure. Homework Equations E=\frac{QK}{r^2} V=-\frac{QK}{r} The Attempt at a Solution \int E dr=-V When I take the derivative of both sides of this relation wrt to R, I get...

Homework Statement 4.80×10^23 nitrogen molecules collide with a 16.0 cm^2 wall each second. Assume that the molecules all travel with a speed of 410 m/s and strike the wall head on. Answer in pascals Homework Equations P=mv/A The Attempt at a Solution Ordinarily I would use...

The Solubility Constant for AgI(s) at 25 degrees Celsius is 8.3*10^-17 how do i find \DeltaG(rxn) for [Ag+]=9.1*10^-9 & [I-]=9.1*10^-9 Using the equation: ΔG=ΔGº+RTln(Q) Let K=Our solubility constant: 0=ΔGº+RTlnK ΔGº=-RTlnK Now, ΔG=-RTlnK+RTlnQ ΔG=RTlnQ-RTlnK ΔG=RT(ln(Q/K)) ΔG=-5.67KJ/mol...