Recent content by dchau503

It gets cold inside because of how heat goes from hot to cold. Because the thermal conductivity of air is lower than that of wood, I assume that the heat loss to the outside occurs at a slower pace when I open the windows than when I keep the (wooden) windows shut though. What am I missing?

How about if we just make this problem simpler: i.e. we can ignore radiation, there's just one rectangular slab of styrofoam with a hole in it. If you're just armed with convection and conduction, how would you calculate the heat loss if 10 W is applied equally throughout one side of the slab...

This is mainly just a theoretical question: say you have a rectangular box made of styrofoam (one of those ice cooler things) and it is levitating in air at room temperature conditions. It also has one hole at the top of the container so that heat can get out of it. Inside the box is a lightbulb...

Yes that makes sense. I just needed to remind myself of partial fraction decomposition in order to get to that answer.

Homework Statement My book tried to simplify \frac{1}{4-u^2} into \frac{1/4}{2-u} + \frac{1/4}{2+u} . What algebra trick did it use? Homework Equations \frac{1}{4-u^2} \ \ \frac{1/4}{2-u} + \frac{1/4}{2+u} The Attempt at a Solution \frac{1}{4-u^2} \ = \...

I got that u = \frac{2cx^4-2}{1+cx^4} . I still don't see why the form was chosen.

Homework Statement x \frac{du}{dx} \ = \ (u-x)^3 + u solve for u(x) and use u(1) \ = \ 10 to solve for u without a constant. Homework Equations The given hint is to let v=u-x The Attempt at a Solution This equation is not separable and the book wants me to make it separable...

So then any term as the constant would make sense? E.g. the book could've used \frac{1}{8} \ln(c) \text{or}\ 3\ln(c) as the last term in the integrated equation? I'm just a little confused how they got \frac{1}{4} \ln(c) .

Homework Statement The givenequation is this: \frac{1}{4} \frac {du}{(2-u)} \ + \ \frac{1}{4} \frac{du}{(2+u)} \ = \frac{dx}{x} My book says that when integrated, the above equation becomes \frac{-1}{4} \ln (2-u) \ + \ \frac{1}{4} \ln (2+u) \ = \ln (x) + \frac{1}{4} \ln (c) I...

Thanks for the answer. I finally understand the problem and it's solution, even with the typo in the problem. :( Crash-course study books frequently have mistakes in them, lol.

Your advice put me on the right track, but I still don't have the complete answer: I did 14.5 cm^3 x (thickness) = 1X10^(-4) cm^3 From the equation, I solved that thickness is 6.9 x 10^(-6) cm^3. I got the right significant figures, but not the right units. How do I convert that answer...

There's no area measurement listed in the original question. But that would make it a lot easier, if the problem did list it. Edit: actually, the problem did say the "area of 14.5 cm^3"...but that's weird because a cubed unit signifies volume, not area.

Homework Statement A piece of gold leaf (density 19.3 g/cm^3) weighing 1.93 mg can be beaten into a transparent film covering an area of 14.5 cm^3. what is the volume of 1.93 mg of gold? What is the thickness of the transparent film in angstroms? I only need help with the second question...