Recent content by MattRSK

Well if A and B are both n x n then they will always be n x n matrices. So does that mean that it is possible?

Homework Statement Matrix A and B are both square matrices of the same size. Is it possible to compute, AB^2009 If not, why not?The Attempt at a Solution Just had this question in an exam, the exam did not allow calculators and I could not answer this question, I gather it is not possible...

2 seconds on physics forum and I found what I'm after. Thanks very much.

Sounds like a kinematic equation problem to me.

Just an update! I got an A for my assignment, thanks again for your help. Regards Matthew

Hi, I am new here also and struggled with the same decision recently. At the university I go to here in New Zealand the computer engineering course is basically a blend of computer science papers and electrical engineering papers. They don't have core computer engineering papers. I decided...

Hi, Looks like an interesting project. Are the books displayed at the end of your project worth a read for a beginner. Could you suggest a book that would explain what happens with a particle accelerator?

Hi, sorry this doesn't relate to the materials you have but I thought you might be interested anyway. The best idea would be to slow the rate at which the egg stops as minger has described. Acceleration due to gravity is 9.8m/s2

Good points to think about. I will give it a try. Thanks!

Hi My brother has a logitech pure fi anytime alarm clock radio. The problem with it is that when it is cold the speakers do not work properly distorting the sound. This is not very useful as it is cold in the mornings when the alarm goes off (playing a song from the ipod). Now I had an...

Christchurch Cathedral, New Zealand

Thank you to everyone who replied to this thread. This is my first experience with the Physics Forum and its just great! I will endeavor to be helpful to others on the site. You guys have given me a huge helping hand so thanks very much! Cheers Matthew

Firstly thanks for all your replies you are very generous with your time. ln(vt2) - ln(vt1) = -k(t2-t1) Becomes ln v(t) - ln v0 = -kt which then becomes ln v(t) = -kt + ln v0 and I plot the ln v against time. and the last equation becomes 1/v(t) = -kt + 1/v0 and I plot...

Could it be V(t)-1= v-1 -kt

By simplify do you mean saying, t2 would be anytime t and t1 would be the initial voltage ie t1=0 So simplifying the expressions would give: for ln (vt2)/(vt1) = -k[t2-t1] Becomes ln V(t)/V=Kt making that ln V(t)=ln V(-kt) and for k[t2-t1]= [(v(t2)-1)/-1] - [(v(t1)-1)/-1]...