Recent content by cookiesandcream

excluded means you can't take both

oh..sorry i thought everyone was talking about the later part... We just derived it for non-uniform circular motion and apparently it's the exactly the same way as doing it for uniform circular motion

I believe what is in the syllabus is that you have to learn to derive the tangential and normal components of acceleration for non-uniform circular motion. I still don't think you need to solve problems when the particle is moving in non-uniform circular motion, and since my teacher insists...

Our teacher gave us a copy of the syllabus, and it says 1989, so I'm pretty sure the syllabus was updated since 1981. It says: 6.3.1 Motion of a particle around a circle and some points about proving angular velocity 6.3.2 Motion of a particle moving with uniform angular velocity around...

if the scholarships are closing on the 30th, does that mean it's due at midnight today, or can we still submit our answers online tomorrow afternoon? I've kept putting off answering the questions since some of them are incredibly annoying and suddenly it's due already :(

I would actually rather be taught proper grammar and spelling english, but my school's teachers insist that it's the content and how you write it that matters, and that they don't deduct marks for horrible spelling.. My english teacher said that apparently by 2020, there won't be anymore...

Tn = 5/2 + (n-1) 5/2 i think?

have u got the answer? i'm not sure if this is right.. i hope it is :) d/dx (y^2) = d/dx (4ax) 2y.(dy/dx) = 4a dy/dx = 4a/2y i.e. dy/dx = 2a/y At P (at^2, 2at) -> dy/dx = 2a/2at = 1/t i.e. gradient of normal will be -t y-y1 = m(x-x1) y - 2at = -t (x - at^2) y - 2at = -tx + at^3 For Q, let y...

Sketch the graph of arg (conjugate of z) = 45 degrees i'm just confused :confused: i thought it'd just be a reflection of arg z except i'm not really sure after thinking about it... thanks!

lol joining in the fun :cool: x^2 + 2xyi - y^2 +2x -2yi + 1 =0 Equating reals: x^2 -y^2 + 2x + 1 = 0 ------------------- [1] Equating imaginary: 2xy - 2y = 0 i.e. 2y(x-1) = 0 y = 0 or x = 1 Sub y = 0 into equation [1] x^2 +2x + 1 = 0 (x...

does (-4,0) have to be considered? I think the pt (4,0) should be excluded from the solution, but i'm probably wrong :)

to integrate of 5/3x, you need f'(x) on the top and f(x) on the bottom, so take 5 out of the picture: 5 S 1/3x dx (S meaning the integral sign) = 5/3 S 3/3x dx(as you need a 3 on top, you borrow a three and then divide by it) = 5/3 ln 3x + c hope that helps edit*lol i think i posted this a...

Re: Best Chemistry Textbooks/Guides? i thought the new 2008 excel chem was pretty :D

I don't think in 2/3 u maths it's just memorising formulas, eg in parametrics you actually have to know how to derive the formula, they never just tell you to sub points into the formula. If you read the last questions even in 2u tests, they're quite challenging, which require more thinking than...

uhh am i the only one here who actually finds 4 unit hard? :( I have never been able to do the entire 4u hsc papers without peeking at the solutions, even the more recent easier ones...and someone has got to have found the complex number questions in cambridge hard..lol maybe i'm just hopeless...