Recent content by irishbob

Homework Statement A 1500-kg roller coaster car starts from rest at a height H=23.0m above the bottom of a 15.0-m-diameter loop. If friction is negligible, determine the downward force of the rails on the car when the upside-down car is at the top of the loop.Homework Equations Conservation of...

Well it's electroSTATIC, so the charge isn't moving. Electrostatic force is exerted by stationary particles. To answer your question more broadly: Imagine a particle is placed into an electric field, and held in equilibrium by some force. Stop time, and remove the force holding it in...

Thanks! :D

Ah. Well there's the B-S law B=(μ_0*I)/(4pi)*∫(d*(l x r(hat))/r^2 and Ampere's law: ∫(enclosed) B·dl=μ_0 I(enclosed) and for Ampere's it has to be highly symmetrical with constant field for us to use it easily. I still don't know which to use. It seems like it would be...

Homework Statement Determine the magnetic field (in terms of I, a, and d) at the origin due to the current loop shown in Figure P30.14. (Use mu_0 for µ0, pi for π, a, d, and I as necessary.) http://www.webassign.net/pse/pse6_p30-14.gif Homework Equations B=mu_0(I)/(2pi*r) for straight...

Cool, just wanted to make sure. I understand this stuff pretty well conceptually, but I'm awful at interpreting the questions.

Homework Statement Two solid spheres, both of radius 5 cm, carry identical total charges of 2 microcoulombs. Sphere A is a good conductor. Sphere B is an insulator, and its charge is distributed uniformly throughout its volume. How do the magnitudes of the electric fields they separately create...

Thanks! That's exactly what I needed. :D

Homework Statement The Sun is lower in the sky during the winter than it is during the summer. How does this change affect the flux of sunlight hitting a given area on the surface of Earth? How does this change affect the weather? (I take it I'm supposed to consider the sun to have a...

I think I figured it out (webassign says I was right both in the problem, and in "practice another version"): 100 C Steam -> 100 C water: Q=-mLv=-m(540.2cal/g) 100 C Water -> 50 C Water: Q1=m(1 Cal/gC)(100-50)C Total Energy lost from condensation: 50m+540.2m=590.2m 20 C water and Cu -> 50 C...

I'll give it a shot that way, but then is the "50g copper calorimeter" unnecessary then? Why doesn't that affect anything? So, x is the quantity of steam that condenses: Q(water)=(0.25kg+x)(4186 J/(kgC)(50C-20C) Q(steam)=-x(2.26x10^6)J/kg Right? How can I relate that? Hold the phone! I think...

Forgive me I've had limited access to the internet until today. So would a better approach be to start with the energy lost as the steam condenses in terms of m, and then use substitution? That wouldn't be viable because that would leave me with 3 unknowns wouldn't it? I understand why...

Homework Statement A 50.0g copper calorimeter contains 250 g of water at 20 C. how much steam must be condensed into the water if the final temperature of the system is to reach 50 C? Homework Equations Q=-mLv where Q is heat energy, m is mass, Lv is latent vaporization constant of water...

That was the problem! it was 3 a.m. my time when I was doing this, and for some reason I thought it would be a good idea to use the radius of the sun for R. Thanks!

what do I use for R though? It's still not coming up right...