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:jaw:

ΔGPE = -GMm/r1 + GMm/r2 = -GMm[r2 - r1]/r1r2 Let r2 = 20000km = 2*107m and r1 = 1*107m. Then GMm = (1*106*2*1014)/1*107 = 2*1013 Now the change in energy moving from 20000km to 80000km is given by: ΔGPE = GMm[8*107 - 2*107]/1.6*1015 Using GMm = 2*1013, this comes out to be equal to: 750000 J =...

v2 = u2 + 2as 2as = v2 - u2 {by subtracting u2 from both sides of the equation} a = (v2 - u2)/2s {dividing both sides by 2s} It doesn't really get any simpler than that mate.

We don't need to really know anything about the hydronium ion. But here you go: http://en.wikipedia.org/wiki/Hydronium

d/dx (3cos35x) = -15sin5x.3cos25x = -45sin5x.cos25x, we can verify this by using the substitution u = cos5x to find the integral, and it turns out to be correct.

1 MOL O2 molecules = 2 MOL O atoms. When you calculate the moles of CO2, this is equal to the moles of oxygen molecules, and is half the moles of oxygen atoms present. This is correct because one mole is defined as containing Avogadro's number of elementary particles. Thus: 1 MOL O2 molecules =...

Exactly.

Yeah Na2S is the correct answer as all group 1 compounds are soluble in water.

What? Are you unsure as to why Na2S is the correct answer?

http://www.boardofstudies.nsw.edu.au/manuals/calculators_hsc.html http://www.boardofstudies.nsw.edu.au/manuals/calculators_hsc_gen_maths.html It seems that you can't use the HP 39G for 2 unit mathematics, however it is permitted for general maths and SC examinations.

Lmao, are you trying to say that the syllabus states that you only have to be able to apply simpson's rule and the trapezoidal rule for functions using only up to 5 function values? You must be mistaken, lolz. I would check the syllabus but I can't really be bothered.

3x - y = 5 => m1 = 3 2x - 4y + 1 = 0 => m2 = 1/2 ∴ tan Θ = |3-(1/2)|/|1+(3/2)| = 1 Θ = tan-11 = 45o Does that help?

tan Θ = |m1-m2|/|1+m1m2|, where Θ is the angle between two lines with gradients m1 and m2.

(x-h)2 + (y-k)2 = r2 (1) is the general equation of the circle. Substitute the first point into the general form (1): (7-h)2 + (2-k)2 = r2 => 49 - 14h + h2 + 4 - 4k + k2 = r2 => 53 - 14h - 4k + h2 + k2 = r2 (2) And the second point: (2-h)2 + (3-k)2 = r2 4 - 4h + h2 + 9 - 6k + k2 = r2 => 13 - 4h...

H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H20(l) As you can see, the mole ratio of H2SO4(aq) to NaOH(aq) is 1:2.

:uhhuh:

*2Na(s) + 2H2O(l) ----> 2NaOH(aq) + H2(g) :D Almost there independantz!

Yeah, we get around the same amount.

No problem.

P(x) = 5x4 - 11x3 + 16x2 - 11x + 5 = 5x4 + 5 - 11x3 - 11x + 16x2 = 0 [Divide through by x2] 5x2 + 5/x2 - 11x - 11/x + 16 = 0 5(x2 + 1/x2) - 11(x + 1/x) + 16 = 0 5(x + 1/x)2 - 10 - 11(x + 1/x) + 16 = 0 5(x + 1/x)2 - 11(x + 1/x) + 6 = 0 Let z = x + 1/x; 5z2 - 11z + 6 = 0 (5z - 6)(z - 1) = 0 z =...