Recent content by chopnhack

Step 1: Reduce RHS into singletons F = {AB->C, C->A, BC->D, ACD->B, D->E, D->G, BE->C, CG->B, CG->D, CE->G} Step 2: Reduce LHS redundant attributes ACD->B has closure of A: A, C: C,A and D: D,E,G since A is in the closure of C we can remove A from ACD->B making it CD->B no other LHS could be...

Input: I=1:5; J=transpose(1:5); V=1./(I+J-1) output as copied from Matlab: V = 5×5 1.0000 0.5000 0.3333 0.2500 0.2000 0.5000 0.3333 0.2500 0.2000 0.1667 0.3333 0.2500 0.2000 0.1667 0.1429 0.2500 0.2000 0.1667 0.1429 0.1250...

Thank you gents! I=1:5; J=transpose(1:5); V=1./(I+J-1) I think I got it - by transposing one matrix, MATLAB implicitly extends zeros to create a 5x5 matrix before beginning the operation for V=. Then each value starting at 1,1 fills in the matrix, 1 then 0.5 across the first row, etc. Thanks!

My first attempt was: V=zeros(5,5) a=1; i=1:5; j=1:5; V(i:j)=a./(i+j-1) I figured to create a 5x5 with zeros and then to return and replace those values with updated values derived from the Hilbert equation as we move through i and j. This failed with an error of : Unable to perform assignment...

I have done just that. Hopefully I can get a clarification. It's very difficult to learn magnetism while correcting typos! Thank you!

Yes, but the professor has been known to make mistakes.

sure

Homework Statement What is the force on an electron with a velocity v = (2i – 3j) Mms-1 in a magnetic field B = (0.8i + 0.6j – 0.4k) T Homework Equations F=qvBsinθ The Attempt at a Solution using cross product I got: (1.92i + 1.28j + 5.77k)10^-13 N which is the listed correct answer. My...

Thanks mate! BTW - if you haven't seen the finale... jumping the shark abit... Still a fan though :-) Hi CNH, I think I see the error now. The 5I1 should really be just I1 times the 1 ohm resistor. That makes it all work. Thanks gents!

Homework Statement In the circuit shown, find (a) the current in each resistors (b)The power dissipated in each resistor. Homework Equations V=IR, P=VI, KVL and KCL The Attempt at a Solution See attached. I was going to post this in the main section as its really a question about whether you...

It's a bit tricky. So when calculating capacitance, if we are offered the total area of both plates, we would need to divide by 2? Here are two sample questions that I have answered correctly according to the answer guide: A 15-pF capacitor has a potential difference of 1.50 V between its...

Can anyone explain why? Is it possibly because charge migrates to one side with the other side simply being polarized, but not conducting? I seem to recall that there is no flow through the capacitor, rather a push and pull if you will, on either side.

Hello all, I have a simple question about capacitors. We are learning about them in class and in some questions, I seem to come up with different answers than the solution sets, always by 2. I apparently am using the area of each plate in a capacitor to derive my results. My question is when...

I was so happy to have found the proper way of handling it, I didn't recall the formula! Thanks all.

yes, sadly I recall it now...