Recent content by harelo

What electives to take? (IC design related) Hello everyone, I'm entering my 4th year in Biomedical and Electrical engineering, where I have a few electives to do, however I am somewhat puzzled as to the options. From all my studies I've found that I really enjoy IC circuit design (digital...

Math classes are VERY different form Math in Engineering. What I have found from all my math classes (Calculus I, II, Multivariable, Linear Algebra I, II, Infinite Series, to name a few) is that they teach these from an almost esoteric point, with lots of beefy mathematical theory and...

Yeah we are planning on using your normal run of the mill Keg. We have a tap, the keg, and the cider. We're just confused as to how it all goes together.

I'd figure a thread in a forum with my fellow engineers would be a good place for a drinking question :P So my buddy's b-day is coming by soon, and we want to surprise him by giving him a keg of our home brewed cider, he loves that stuff However, I've been trying to find out how we can...

I like PSpice, the student version is free. The circuit response simulator works very well.

I have been thinking of becoming a Biomedical Engineer, this is an option course taken from a Electrical, Mechanical, or Material engineering stand point. I really like this option because it offers me the opportunity to engineer for the most complicated system in the world, the human body...

yeah just tried with made up values and I got the same answer, thanks a lot for the help :)

Thank you :) would you mind sharing what your answer was with me tho? I'd like to know if you had another approach

Will do \sqrt(\frac{M}{m})=\frac{x}{R-x} I multiply the B radius to both sides (R-x)\sqrt(\frac{M}{m})=x I expand (R-x) out of the brackets R\sqrt(\frac{M}{m})-x\sqrt(\frac{M}{m})=x Move the expanded x to the other side R\sqrt(\frac{M}{m})=x+x\sqrt(\frac{M}{m})...

Not sure if my edit appeared on time or not, but: After further factoring, I ended up with this: \frac{R\sqrt(\frac{M}{m})}{1+\sqrt(\frac{M}{m})}=x Am I right or am I completely out to lunch?

Yeah that makes perfect sense, so let's recap: but by having: \frac{GMB}{x^2}=\frac{GmB}{(R-x)^2} I understand that you divide out the gravitational constant and the test mass, leaving us with \frac{M}{x^2}=\frac{m}{(R-x)^2} and by cross multiplication, you get...

M being the mass of planet A and m being the test of Planet B I'm assuming? and where did the negative that was there go to? Sorry I just want to understand the problem before I do it blindlyG01, I went under the assumption that the forces should be equal in magnitude but different in...

[SOLVED] Gravitational field between two planets. Homework Statement Planet B and Planet A are separated by a distance of R. Homework Equations At what distance will the net gravitational field be 0, express R in terms of Planet A. HINT: Use Gravitational Force. The Attempt at a Solution...

OHHHHHH how can I be so dumb! that speed I Calculated is V'! wow I'm dumb! So now it should be: m_{1}v_{1} = (m_{1}+m_{2})V^{'} v_{1} = ((m_{1}+m_{2})V^{'}) / m_{1} v_{1} = ((3300)(4.72))/1100 v_{1} = 14.16 m/s I think this should be right, is it?

well I know that Momentum is conserved, however kinetic energy isn't (inelastic collisions, do'h) and the acceleration would be the the coefficient of friction times acceleration due to gravity (like shown above). So, m_{1}v_{1} = (m_{1}+m_{2})V^{'} hmmmm... Where can this get me?