Physics - SoM Arrived!

[paulsterio]

Oh whoops, you misunderstood me, i meant the absolute max that I could have gotten for physics was 88 (i dropped 2 marks on a MCQ) and for chem absolute max I could have gotten was 70 (dropped the IR MC and a mark on the first extended response).

LOL! i was expecting a 70 in chem as well, cause i got 2 MC wrong, but i cant believe i lost so many marks for explanations, like freakin fractional distillation!! annoyed =.=

and btw guys, im expecting that some people would have got 90/90, theres always a few every year, mind you i only lost half a mark on core, damnn further electronics!!!

[taiga]

Ahahah yeah 20% got that one right.

and you are not in that 20%, hence you are looking at an ATAR of <80

[thushan]

damn you

[paulsterio]

and you are not in that 20%, hence you are looking at an ATAR of <80

but youre assuming that everyone else who did vce got everything thushan got correct, correct as well

[thushan]

Got 86/90. Somewhat happy with it, but I guess careless mistakes and a lack of studying didn't help.

In AOS1Q20 got 2.5/3, got 0.5/2 for ElecQ9 (I am aware why I lost the marks but I still maintain that my ans is also correct, but who cares), and lost the springboard MCQ.

[paulsterio]

Got 86/90. Somewhat happy with it, but I guess careless mistakes and a lack of studying didn't help.

In AOS1Q20 got 2.5/3, got 0.5/2 for ElecQ9 (I am aware why I lost the marks but I still maintain that my ans is also correct, but who cares), and lost the springboard MCQ.

im guessing you drew sharp points rather than curved points for question 9?

but hmm, congrats on the score

[thushan]

other way, i drew curved points

[luffy]

other way, i drew curved points

I just had a look at the physics exam again.

Just out of curiosity, whats your reasoning behind the curved points?

[thushan]

P(diode)
= V(diode)I(diode)
= V(diode)I(circuit)
= V(diode) x V(r1 + r2)/(R(r1) + R(r2)) --> where V(r1+r2) is the total voltage drop across resistors R1 and R2

Given V(diode) is constant and V(r1+r2) is constant as well, we get

P(diode) = a/(b+R(r1)) where a and b are constants

So P and R(r1) do not have a linear relationship -> it's a hyperbola, so as R(r1) increases and decreases, P follows a hyperbolic characteristic, increasing and decreasing in that kind of fashion.

Showed my physics teacher (assessor) who agreed with my reasoning. Doesn't matter now though, I don't mind.

[luffy]

P(diode)
= V(diode)I(diode)
= V(diode)I(circuit)
= V(diode) x V(r1 + r2)/(R(r1) + R(r2)) --> where V(r1+r2) is the total voltage drop across resistors R1 and R2

Given V(diode) is constant and V(r1+r2) is constant as well, we get

P(diode) = a/(b+R(r1)) where a and b are constants

So P and R(r1) do not have a linear relationship -> it's a hyperbola, so as R(r1) increases and decreases, P follows a hyperbolic characteristic, increasing and decreasing in that kind of fashion.

Showed my physics teacher (assessor) who agreed with my reasoning. Doesn't matter now though, I don't mind.

Ok, now you're getting me worried. I remember having curved, hyperbolic lines in my exam! Were the lines supposed to be straight? I had distinct sharp points at the 1,3,5 seconds though.... Do you reckon they will pay me the mark? T_T Wait a minute, isn't the "curved" answer "more" correct than the others? 

*sigh* Even more eager to see my statements now...

Also, Thushan, you forgot to mention R(LED) and V(LED) when substituting for I(Circuit). However, as R(LED) approaches 0 and V(LED) is constant, it won't make a difference to your answer.... As if VCAA assessors didn't see this when creating their own worked solutions...

[pi]

lol, further amazement of myself getting an A+, I drew curved lines too

Out of curiosity thushan, what is your definition of 'lack of studying'?

[thushan]

Also, Thushan, you forgot to mention R(LED) and V(LED) when substituting for I(Circuit).

Yeah, that was implied. V(LED) is constant and R(LED) is irrelevant in the calculation as I am effectively considering the two resistors r1 and r2 as one resistor and measuring the current and voltage through them.

[paulsterio]

other way, i drew curved points

oh right, i drew sharp points for that and got full marks

Ok, now you're getting me worried. I remember having curved, hyperbolic lines in my exam! Were the lines supposed to be straight? I had distinct sharp points at the 1,3,5 seconds though.... Do you reckon they will pay me the mark? T_T Wait a minute, isn't the "curved" answer "more" correct than the others? 

hmm, well, technically yes, the graph of the current versus time graph, will be a hyperbola, hence so will the power versus time
but youve also got to consider that the brightness against power graph is not linear either
in fact, here it is http://www.flickr.com/photos/therontrowbridge/5322251604/
that function to me looks quite hyperbolic, so using composite functions, yes, the brightness versus time should be a linear graph

but hmm, sharp points are technically right so you should get a mark i guess

also to back up my point, its a composite of a positive and negative hyperbola, which means you should end up with a line of opposite gradient (:

[thushan]

Ah, that's fair enough. I ended up getting 0.5/2 for that question.

[paulsterio]

Ah, that's fair enough. I ended up getting 0.5/2 for that question.

i guess that 0.5 does show that even the examiners arent quite clear where to award marks