Recent content by |<ings

this has probably been said already: Heat flows from objects of higher temperature to objects of lower temperature until both (or all if there are more than two objects) are at the same temperature (i.e. thermal equilibrium). If the space around the ball is colder (at a lower temperature) than...

Much of my confusion at times came from wrongfully thinking that E varied according to dA, thereby necessitating the integral. I thought: if E varies (with respect to r), then it would also vary with respect to A; it's clear now that E can be a function of r, but not vary over the surface at a...

Well, TBH, we did only examples involving simple shapes (spheres, infinitely long cylinders, lines, planes). When using spheres, the Gaussian surface is a sphere (concentric). For cylinders and lines the surface we used was a cylinder of course. And for the plane we used, a cylinder of...

TNX, I think I get it now. I would rep you if I could. ____________________________________________ TNX to the others too. Feel free to add more insight if it will help.

What you are saying is that if the component of E onto A (iow comp_{A} E) is constant for all the segments (patches) of area on the closed Gaussian surface, then the integral form "reduces" to the second form. Is this what you are saying? BTW, vector A is the vector normal to the area.

I have difficulty when trying to decide which form of Gauss's law I should use for a problem. Please, tell me when I should use the integral form: \int E \cdot dA = \frac{Q_{in}}{\epsilon_{0}} and when it's appropriate to use the other form: E \cdot A = \frac{Q_{in}}{\epsilon_{0}}

Help?

Hello?

What...? I always thought T = -mgcos(theta) (i.e. the radial component of the Weight force) in a pendulum... not T = mgcos(theta) + mac. T and the radial component of the Weight force cancel out. The only force acting on the system is the tangential (real word?) component of the Weight...

Once, I inadvertently made two unrelated mistakes on a lengthy math problem and got the correct answer. The second mistake (must have) canceled out the first mistake, once and for all proving that two wrongs do make a right. Well... sometimes they do.

Also note that if energy is conserved then 1/2 kA2 is the total energy of the system (K + U). So erok81 the original equation you supplied for KE is actually, as the other posters have pointed out, the equation for total energy of the system, . I think that's where your confusion arised/arose?

Miss Penguins read bigstar's post and keep in mind that just because a value for a quantity is not given (in this case it is given, just not explicitly), you can't assume that the quantity's value is 0. BTW, I don't have to be a mentor to give help, do I? (I'm new)

MissPenguins, please note that if the equation is valid, the vmax would be given by setting x equal to 0. Also, I'm fairly sure that bigstar is correct: you forgot to subtract x2 from A2.