Recent content by Stevecgz

We're also very interested in seeing the work. Your work. See the sticky thread "FAQ: Why hasn't anybody answered my question?"

Thanks for the replies and the link Fred. I'll be working on it and see what I can come up with.

To clairify from my 2nd post, I know that I want mil spec dimensions and tolerancing for the rail sections, but I do not know what those dimensions and tolerances are because the mil specs are given in GD&T and I don't understand GD&T.

No, I do not.

No, I don't have access to ASME Y14.5. I thought my school library would have it but a quick search revelaed nothing. I'm trying to tolerance the whole part as BD&T, not GD&T, so for most of the part I already know the tolerances I want and how to apply them. My issue is specifically...

Now that I've completed my first CAE course I'm attempting to design a part (to actually have built) using solidworks. I'm want to build this part to mil spec, problem is mil specs are given in GD&T and I only know BD&T. I've been trying to convert the GD&T to BD&T so I can dimension and...

Thanks for all the replies. I'm starting to see there might be more to linear algebra than just solving equations. The articles about google are quite interesting. Definently a legitimate application, yet as concrete as the basic concept seems to be it's still quite difficult to concieive on...

So I'm about 10 weeks into my first course on linear algebra. Things are going well I sopose, I'm finding it much easier than diff eq or calc, but I'm also finding it to be terribly boring. Unlike many other math courses, this linear algabra course (and text) has yet to present any...

I think the variable W and W as the symbol for watt caused confusion. So the correct answer would be: Energy = Charge*Potential = 260C*6V = 1560J Right?

Ok, I see where I went wrong. I was looking at the variable W as if it had the unit of watts, which in power, not energy. I now see that W is energy is joules, and the formula you wrote does work. Thanks Number2Pencil, I think I can solve the problem now.

It seems the units don't work out for that formula. Did you mean V = W/I? If so I think it doesn't help for this problem since I don't know the current and can't find the current from the given values.

Homework Statement Determine the energy required to move 260C through 6V. Homework Equations q = Cv U = (1/2)*C*v^2 The Attempt at a Solution C = q/v = 260C/6V = 130/3 F U = (1/2)(130/3 F)(6V)^2 = 780J I applied the voltage/charge relationship and energy equation for a...

+1 Unless your instructor has stressed conceptual understanding and definitions on previous exams, your time will be better spent doing as many practice problems as possible, as opposed to merely studying equations or rereading material. When doing practice problems, attempt them without...

Right :smile:

I think dextercioby missed a term. Start by putting in the form y' + y*(1/x) = -5x^3 Next, your integrating factor is p = e^[integral(1/x)dx] = e^(lnx) = x Continue...