Recent content by BrainMan

Oh ok I get it now. The answer should be \frac {20\pi r^2(0.020+.020t)}{R}

So I = \frac{20\pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)]}{R} So like this?

Homework Statement A 5.0-cm-diameter coil has 20 turns and a resistance of 0.50Ω. A magnetic field perpendicular to the coil is B=0.020t+0.010t^2, where B is in tesla and t is in seconds. Find an expression for the induced current I(t) as a function of time. Homework EquationsThe Attempt at a...

Here is the net y-component of the force. \frac{Qqs}{4\pi\epsilon_{0}(r^2+\frac{s^2}{4})^{3/2}}

Homework Statement A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. The dipole is initially oriented so that Q is in the plane bisecting the dipole. Assume that r≫s. A) Immediately after the dipole is released, what...

So I changed my answer to 2KQπ/L^2 and it worked. Thanks!

So I did (Kλ/r) ∫ cosθ dθ and got (Kλ/r) sinθ ] -pi/2 -> pi/2 So my final answer is 2Kλ/r = (2KQ) / (RL) But my homework keeps saying "The correct answer does not depend on: KQ, RL."

Here's what I did: E = ∫dE = ∫(K*dQ*cosθ)/r^2 = ∫(K*λ*dS*cosθ)/r^2) = (Kλ)/r^2 ∫cos(S/r)ds = (Kλ/r) ∫ sin(S/r) ] 0 to L = (Kλ/r)[sin(L/r)-1]

OK I substituted the linear charge density and integrated from 0 to L. Somethings still wrong though.

Homework Statement Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in the figure (Figure 1) . Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a...

I looked ahead and it seems it will be covered in the next chapter. OK so that solves that problem. What I'm wondering about now is the work. My book wasn't very clear about work and the two ways it gave to calculate work were to W = -∫ p dV or to find the area under the curve of the pV diagram.

OK you're right. You get the right answer if you just add MwLv. So basically, the Lava is heating up the water to boiling point and then the rest of the energy is removed when the water evaporates to a gas. The steam then "just rises upward never to be heard from again." Thanks!

Cv is one the things that is confusing me. In my book it has a table for a couple of Cv's for different gases so I'm not sure which to use...

OK I see. So I need to add MwLv because the water is going to evaporate. Then I need to add nCΔT for water in the gas state until it reaches 800 deg C, right? But what should I use for C for water in the gas phase?