Recent content by cmcc3119

Cristo, thanks, but please don't jump to conculsions and accuse me of something I have not done. I HAVE NOT CHEATED IN ANY WAY AND DID NOT INTEND TO USE THIS FORUM FOR THAT. If you look at my previous posts you would realize that! I have only ever asked for help with questions that I have...

Not getting in trouble from my uni tutor for asking for help with assignment questions...

Well it is ridiculous that you force people to still hold an account on here. What a friendly place! Is it just so you can get member numbers? Anyway, Cristo, thanks for pointing that out to me have a nice day :)

But it is my account... I opened it, I should be allowed to close it... Any other forum would respect the wishes of its user in a matter such as this, and what I am asking does not impede the forum at all... If for some reason you can't find anyway to delete it, I would rather not be banned...

I would like to close my account on here, can someone please advise me how to do so. Thanks

Homework Statement This was a lab task in which we are asked: Determine the static and dynamic coefficients of friction for a block of wood of mass 82.2grams on a plank of wood. PART A: To determine the static coefficient we found the angle at which point the block just started to...

Yep I figured out the right anwer which is 14.9 I don't know what I meant with that massive negative figure! Thanks for you help anyway.

Two moles of an ideal gas undergo a reversible isothermal expansion from 1.97×10−2 m^3 to 4.82×10−2 m^3 at a temperature of 20.0 C. What is the change in entropy of the gas? WORKING: Using \DeltaS = nRIn(V2/V1) = 2(8.31*ln(4.82*10^-2/1.97*10^-2) ANS = -138.20 J/K Is this...

I understand that that is how the question is written and the sign indicates the point position in the plane. Sorry I worded the question wrong. Basically, z = \sqrt{}3 + i Polar form of this is: 2( cis \pi/6 ) then working out 1/z the answer is 1/2 (cis -\pi/6) I do not know why...

Hi There. I was given this question and the answer: Find the polar forms of 1/z where z = \sqrt{}3 + i and 1/z where z = 4\sqrt{}3 -4i Answers respectively are: 1/2 cis(-\pi/6) 1/8 cis(\pi/6) Can someone please explain to me why it is that the sign of the argument...

Thanks so much for that :)

Hi There, Can someone please tell me where I can find a table/ data that converts radians to surds. I don't know what to call it but, for instance to tan^-1(-1) = -Pi/4 and, cos(-pi/6) = root3/2 ? Thank you :)

4 Questions: (1 + i) / (1 - i) Ans: i (2 + 3i) / (5 - 6i) Ans: (-8+27i)/61 1/i - (3i)/(1-i) Ans: (3-5i)/2 i^123 - 4i^9 - 4^i Ans: -9i Could someone please explain the method (detailed) as to how these answers were obatined? I understand other questions in the same...

Its so hard to put all these number and boxes back into simple ax = b and again rearrange. Thanks cristo for your help!

So how would I know that I have to multiply the left by the inverse of A to find the answer? Is there some basic principle or law that I am unaware of?