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CocoPop last won the day on July 16 2010

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About CocoPop

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    IB Survivor

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    May 2010
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    United Kingdom

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  1. http://www.onemotion.com/flash/sketch-paint/
  2. Just found this online paint tool so decided to use your problem to test it out
  3. You're close, but if you differentiate (1/3)arctan(u/3) you get (1/3) * 1/((u/3)^2 + 1) * (1/3) = 1/(u^2 + 9) - use the chain rule. If you'd like to know where arctan comes from in the integration, you need to substitute u/3 = tan(theta). Can't be bothered to write it out in text, so I've done it on Word and screenshotted:
  4. I can't remember if you look at the dot product in maths HL, but you can write the equation of a line in vector form: r = p + λb Where b is the direction vector. If you have two direction vectors b1 and b2 for two different lines then you know that b1 • b2 = |b1||b2|cos(theta) where theta is the angle between the two.
  5. If you do maths/engineering/physics/chemistry at uni there will be a lot of calculus involving imaginary or complex numbers. Remember that i does not vary (di = 0), so you treat it as a constant in calculus. EDIT: In my experience, when it comes to visualising complex functions, one tends to draw two separate plots - one for magnitude and the other for phase. So for example, if you had the function f(x) = x/(i + x), you'd have: arg(f) = arctan(0) - arctan(1/x) = - arctan(1/x) You can plot that against x and you'll find that the phase varies from -90 degrees (at x = 0) to 0 degrees (as x-> infinity). |f(x)| = x/sqrt(1 + x^2) It's very common to use this method when looking at frequency responses or transfer functions of circuits. For example, if x is the frequency of an AC system, it lets you see the amplitude and phase of the different frequency components in a signal.
  6. CocoPop

    The Oxbridge Guide

    It's true that you'd be hard pressed to find financial aid whilst applying to Oxford, but many colleges have generous bursary schemes that you will be able to apply for when you get there. You'll usually have to write an essay or something of that sort...
  7. From the people I've spoken to, it seems that Physics EE grades tend to be pretty random. My school had a pretty consistent history of brilliant students getting Cs in physics extended essays - essentially the reason why I didn't do it. It seems that teachers are always very happy with the essays yet students come out with low grades, which begs the question, who the hell are these examiners?
  8. CocoPop

    SAT IIs

    This is true, and you can find a full list of schools that have this policy here: http://www.compassprep.com/admissions_req_subjects.aspx There are a few big ones like Yale that allow you to take ACT + Writing as a substitute for SAT I + II.
  9. The American guy at my college did 18 APs and got 5s in all of them. He even did some of them in year 10. Whilst I'm sure he's a very smart guy, I doubt one would be capable of taking the equivalent number of IB subjects, even over a four year span. By contrast, it's not unheard of to hear of people taking ~10-15 APs. EDIT: Also, putting an AP vs IB poll on an IB forum will probably get a fairly biased response!
  10. Purdue also has rolling admissions - good reputation for engineering.
  11. I have an exam during the royal wedding (hence why I'm awake at 5am). My university doesn't care very much for public holidays.
  12. I can only work out a), because i'm doing Dynamics right now. b) F = m(v-u)/r = 0.55(11-0)/(150*10^-3) = 40.3N c) i) The ball's movement is restricted to a circular motion due to the string. This causes a centripetal acceleration with force F=mv^2/r which causes an increase in tension in the string. c) ii) T = mg + mv^2/r = 0.55(9.81 + 11^2/7.5) = 14.3N.
  13. I don't know why I'm taking a break from studying by doing more physics, but here's a quick go. a) Equate maximum kinetic energy (1/2*m*v^2) and maximum potential energy (mgh). Maximum velocity occurs when h=0 (where h is measured from the rest position) and your answer should round up to 11. b) Force equals rate of change of momentum. F = m(v-u)/t. v = final velocity = 11. u = initial velocity = 0. t = 150ms. c i) Centripetal force. c ii) Resolve forces due to weight (mg) and centripetal acceleration (mv^2/r). EDIT - Since the string is vertical this is quite simple. Just add the forces.
  14. You don't actually need to do the derivation to work that out. It should be clear from looking at that expression that the answer is zero. The reason: axb is going to give you a vector perpendicular to a, let's call it vector c. When you take the dot product of vector a with this vector c, the angle between them (say, theta) is 90 degrees. So the cos(theta) term will be zero hence the dot product will also be zero. Haven't done probability in ages, but here's a quick stab at it (looks like an infinite geometric series). You wan that player A gets the first six. So you could have the following combinations: - six - not six, not six, six - not six, not six, not six, not six, six - not six, not six, not six, not six, not six, not six, six And so on. The probability of not getting 6 is 5/6. The probability of getting six is 1/6. So we have that the probability of A getting the first six is: