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the 1-mark questions (1 Viewer)

lepton_index

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I reckon this test is easier than most others... Certainly much easier than the 2003 one.
I got everything, except 4c(iii) and 7a(iii). And they are 1 mark questions! (Must have had a brain-dead then).
But anyway, I'm too lazy to think now. So what's the answer for 4c(iii) and 7a(iii) ? (anybody? Please?)
 

Rorix

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yeah, am=ah

7aiii) is let a1=sin^2, a2=cos^2 and i was working out a3 when time ran out:(
 

Giant Lobster

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7 a iii

for case n = 3
let a1 = sin^2 a2 = cos^2 a3 = tan^2

that question has the hardest (for me) in the whole paper, stress wise :p
 

mojako

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this is strange...
for me part (iii) was really easy

but obviously its not for you guys
and the top student in my school couldnt do this part, whilst he did everything else in the whole paper

but instead I couldnt figure out part (i) and skipped it
@Rorix: what I wrote in the other thread was right, but I forgot to include part (ii) as well
 

SeDaTeD

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Yeah, q7a(iii) was the only one I left out. I threw heaps of numbers at it, but wasn't able to get that one.
 

Logix

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ummmmmmmmm........

wasnt 4c(iii) the conics question?

and isnt Am=Ah the circles question?
 

mojako

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SeDaTeD said:
Yeah, q7a(iii) was the only one I left out. I threw heaps of numbers at it, but wasn't able to get that one.
looks like u need some idiotic brain to be able to work out this part :D
 

fantasia

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whatd u guys get for the 5cii) show that cos@=g/Rw^2 or @=0

i got the first bit.. i didnt understand where the @=0 bit comes from.
 

SeDaTeD

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Hmmm, i think my solution was fudgy.
I said when w^2 = g/R, cos@ = 1
therefore, @ = 0.
Don't think that's what they're looking for though.
 

Logix

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me 2, i thought that when @=0, the particle does not rotate, because it will just stay on the same spot.

btw Am=AH is for 5b(iii). i think lepton was asking for 5c(iii) :p
how bout 5a(iii)?
 

mojako

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fantasia said:
whatd u guys get for the 5cii) show that cos@=g/Rw^2 or @=0

i got the first bit.. i didnt understand where the @=0 bit comes from.
before any simplification, you'll have two sin@'s, move one so u have sin@ on LHS and on RHS
when sin@=0, the equation holds
when sin@=0, @=0

well probably that explanation is a bit dodgy because if u make the two sin@'s on the LHS then you're dividing by zero..
the best explanation might be to say that it can be zero, when there's no centripetal force and the marble rotates on the spot..

if @=0, the particle rotates but on the same spot.. just like the whole sphere.. it rotates on the same spot... like earth's rotation and not revolution
 
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SeDaTeD

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Oh crap i read it as the marble was the one moving, not the sphere.
 

Logix

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oh yeh, ur rite. i forgot bout the sin @s, just cancelled them out :(

wat did u's get for the volume question? i wasnt sure wat the limits of integration were
 

mojako

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Logix said:
oh yeh, ur rite. i forgot bout the sin @s, just cancelled them out :(

wat did u's get for the volume question? i wasnt sure wat the limits of integration were
0 and 2
I dont understand why they give two marks there
for me part (i) looks harder :p
 

Giant Lobster

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SeDaTeD said:
Yeah, q7a(iii) was the only one I left out. I threw heaps of numbers at it, but wasn't able to get that one.
lmao
i almost ended up like that

i paid 20 minutes for that 1 mark =/ i damn well deserve it ! ahahaha
 

Logix

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wrong question mojako, thats the other volumes question, i meant the one that u had to do by cylindrical shells
 

mojako

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Logix said:
wrong question mojako, thats the other volumes question, i meant the one that u had to do by cylindrical shells
ooh.. sorryy...
its x=0 and x=2a
also dont forget the radius is (x+2a) because its rotated about x=-2a
am I correct? (hoping so...)
 

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