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If you recognize the identity sin\,x = cos\,(90^{\circ}-x) , then the above is just a re-arrangement, in radians.

Re: HSC 2014 4U Marathon I take it you mean right angle CDB rather than CBD. Let OM = l and BD = 3l Applying Pythagoras we obtain the following: \\BM = \sqrt{r^2 -l^2} OD = 3l-r so considering \Delta CDO, CD = \sqrt{6lr-9l^2} but considering \Delta CDB, and noting that BC = 2BM, (any radius...

Sorry, which substitution has been changed? I stuck with whatever was given in the thread You always consider both the substitution and the limits for any changes to a definite integral anyway.

x = u^2 -1, 2u\,du = dx for \int_0^1 {x\sqrt{x+1}} \,dx Case 1: u = +\sqrt{x+1}. Integral is \int\limits_{1}^{\sqrt{2}} {\left(u^2 -1\right)u\cdot2u\,du} Case 2: u = -\sqrt{x+1}. Integral is \int\limits_{-1}^{-\sqrt{2}} {\left(u^2 -1\right)(-u)\cdot2u\,du} =...

You forget that we reverse the limits due to the substitution being u = -\sqrt{x+1} in the second case. This means that in our integrand the \sqrt{x+1} becomes -u.

Still works. We end up with even powers throughout the integrands, making it a case of evaluating symmetric areas under an even function in u.

Suppose the table can support an infinite weight. Then yes, I suppose so. Balancing is simply dependent on the center of gravity in our book stack, which should be at or behind the table's edge. The greatest horizontal distance beyond the edge occurs when a stack's center of gravity is...

Re: hmmm. Acceleration is m/s^2

Don't know what you mean by 1:1, but this is a comparison of gas moles between the cumulative moles of nitrogen and hydrogen, with that of ammonia (4:2). The increase, therefore, affects the reactant side more than the product side. Compensation for this must come by a decrease on the left and...

Anyone with the paper, feel free to post up your MC answers! :spin:

For Q.13 MC: B is correct. There are 2 ways to justify this: (1) [Paraphrasing the sample answers for Q23 of 2009] Halving the volume will increase the concentration of all reagents proportionally, according to the stoichiometry of the reaction. The left hand side (nitrogen and hydrogen)...

What did people get for the question about halving the volume of the vessel in the Haber process?

aDimitri, Did you get out the circle geo q. by proving the angles add to 180 degrees?

Re: HSC 2014 4U Marathon - Advanced Level 1. 2.