Recent content by DollarBill

Could you be a little more specific? I've checked my signs a few times, but am still having trouble seeing where I messed up.

Homework Statement I'm looking for the maximum power transferred to resistor R. I know how to approach the problem, but not getting the right answer and was wondering if someone could just look over my work. http://img854.imageshack.us/img854/1412/screenshot20110604at115.png Homework...

I posted this in "Materials & Chemical Engineering", but it seems a little more active here. I'm confused with trying to find the slip systems of a simple cubic. I understand the slip plane would be along the plane with the highest atomic density (a close packed plane). But I don't quite get...

That definitely cleared it up -- can't believe I didn't get it before. Thanks!

Considering only the 4 lower-right atoms (one unit cell), I'm somewhat confused as to why the coordinates for the center atom (b) is (1/3, 2/3, 1/2) with respect to (a1, a2, c). c is pretty self explanatory, but I don't quite get the other 2. I think one of my main confusions is where the...

I just finished my sophomore year in Aerospace engineering and am weighing my options for grad school. I definitely want to go grad school, but not sure if I want to pursue a MS or PhD. I know when I finish school, I want to work in lab rather than an office building. I do like research...

Ok, thanks got the answer (3). Yea, I probably should write out the expressions more often

Still not sure if I'm doing it right \int_0^x{\sqrt{t^3+1}dt} {\sqrt{x^3+1}} {\sqrt{2^3+1}}-{\sqrt{0^3+1}} 3-1=2

I was thinking of integration by reverse chain rule :confused:

Homework Statement I'm not sure if I'm doing this right or not: If F(x)=\int_0^x{\sqrt{t^3+1}dt}, then find F'(2) The Attempt at a Solution \int_0^x1/2(t3+1)-1/2*3t2 1/2(x3+1)-1/2*3x2 1/2(23+1)-1/2*3(2)2 Answer=2

Homework Statement http://img401.imageshack.us/img401/46/impulsemw6.jpg Before the force is applied (t<0), the particle moves along the x-axis with a velocity of -6.3 m/s. Find the velocity of the particle after the force stops acting on it. The particle has a mass of 5.2 kgHomework...

1362.771 J I'm not sure why I multiplied it by d again. I guess I forgot to look at the unit and thought they were forces. Thanks.

Homework Statement A 15.8 kg block is dragged over a rough, horizontal surface by a constant force of 90.9 N acting at an angle of angle 28.6 above the horizontal. The block is displaced 22.8 m, and the coefficient of kinetic friction is 0.18. What is the net work done on the block...

Ok now I see what I did. Forgot to square the .53 :blushing: gM=3.8 m/s2

Forgot the parenthesis... gM= G(.11 ME) / (.53 rE)2.