Search results

a quicker way \int \frac{dx}{x\sqrt{1-x^2}}\\ =\int \frac{dx}{x^2\sqrt{(\frac{1}{x})^2-1}} use reverse chain rule/let u=1/x, as this is a standard integral

1. factorise. 2. let 1=6

i don't know long division

electricity/waves is physics hsc physics is pretty bad

yeah

i can't quite follow your working, so i'll just provide a solution Let\ the\ frustrum\ be\ the\ region\\ 0\leq kx\leq y;\ a\leq x \leq b\ \ rotated\ about\ the\ x-axis\\ so\ V=\pi \int_{a}^{b}y^2.dx\\ = \pi \int_{a}^{b}k^2x^2.dx\\ = [\pi k^2b^3 - \pi k^2a^3]/3\\ = (b-a)/3*(\pi k^2a^2+\pi k^2b^2...

um just use a rotation of a line, eg. y = kx, between values a and b find the expression for this volume, and the expressions for B1 and B2 in terms of the other variables, sub in. provide your working so far if you are still stuck

have you drawn a diagram of the situation?

Re: This made me sick What 9000?!

99.00+

timothy, you've basically got it all right x<-3, -1 < x < 1 or x > 3, but as x=4 lies on the particle's motion, it must satisfy x > 3

how do you know that it's wrong? what is the correct answer?

..has already been posted (twice) in the thread

yep, that working's correct

let u=1/x ....

you know the frequency of the light, so you know the energy per photon (E=hf) so you know the energy being given to each electron. this energy is going to be equal to the work function for the zinc plate (if youre given it) + the kinetic energy of the electron, i.e. hf = hf0 + 1/2 mev2, and you...

yeah probably would need to

if b^2 - 4ac < 0, (and a > 0) the quadratic is always positive for m^2 + m + 4, the discriminant is 1 - 16 = -15 < 0 so always positive

if m<-1, then m^2 > -m -> m^2 + m > 0 so m^2 + m + 4 > 0 if -1 < m < 0, then 0 < m^2 < 1 so m^2 + m + 4 > 3 and if m > 0, m^2 > 0 m^2 + m > 0, as m^2 + m + 4 is greater than zero for each case it is always postive works well for seeing it intuitively, but as for actually proving you're best to...