EXAM!!!!!!!!!!!!

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EXAM!!!!!!!!!!!!

Post  Suzy on Thu May 21, 2009 8:39 pm

hey Smile was doing the 2005 past paper paper one and got stuck at these questions;

8. A potential difference of 5000V is applied between two mental plates. The plates are 0.10m apart. A charge of +2.0mC is released from
rest at the positively charged plate as shown. The kinetic energy of the charge just before it hits the negative plate is
A 4.0 x 10^-7 J
B 2.0 x 10^-4 J
C 5.0 J
D 10 J....................(Answer's D)
E 500 J

and

16. A liquid and a solid have the same refractive index. What happens to the speed and the wavelength of light passing from the liquid into
the solid?
Speed....................Wavelength
A stays the same........stays the same....................(Answer's A)
B decreases...............decreases
C decreases...............increases
D increases................increases
E increases................decreases

I dunno how they get the answers though Shocked

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Re: EXAM!!!!!!!!!!!!

Post  Admin on Fri May 22, 2009 10:11 am

Suzy,
2005 Q8
Using W = QV; where Q = 2mC and V = 5000 you get
W = 2x10^-3 x 5000
W = 10J, this is the work done, remember work done will be converted to kinetic energy hence the answer D. Smile
Q16
Rememeber n2 / n1 = v1 / v2 = w1 /w2 and frequency never changes:
where n = refractive index, v = speed and w = wavelength (can't find lamba sorry)
Since "A liquid and a solid have the same refractive index."
then n1 = n2 so v1 = v2 and w1 = w2
so answer a is correct.
Hope this helps, how did maths go yesterday?

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Re: EXAM!!!!!!!!!!!!

Post  Suzy on Fri May 22, 2009 5:41 pm

Awright! I get it Cool
thanks
I always forget about W=QV thats really annoying

Maths was horrific.
Felt like crying, but was in too much shock to.
Paper 1 was okay - Paper 2 was impossible, not like past papers at all pale
Dreading August

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Re: EXAM!!!!!!!!!!!!

Post  ANDYQ on Fri May 22, 2009 11:19 pm

seats goin to be numbered this year admin(I cant remember if they were last year) lol!

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Re: EXAM!!!!!!!!!!!!

Post  Admin on Sat May 23, 2009 2:19 pm

I think thy were it came in either last year or the year before but I was not around last year so not too sure.

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Re: EXAM!!!!!!!!!!!!

Post  ANDYQ on Sat May 23, 2009 4:33 pm

paper 2005 q22 ai

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Re: EXAM!!!!!!!!!!!!

Post  Admin on Sat May 23, 2009 9:05 pm

22.a.i.
Andy tension is the force exerted by a cable, towbar, rope etc.
In the diagram Vertical is adjacent to to 21 degrees,
Horizontal is opposite
so vertical component is 4.5 x 10 ^ 3 cos 21 degrees.
I know that most examples are the other way around and
that this does bug you please try not to worry about it just
look for the opposite and adjacent in the question and
you will be fine!

If you need help with the rest let me know although Sunday is the earliest I can get back to you.

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Re: EXAM!!!!!!!!!!!!

Post  ANDYQ on Sat May 23, 2009 9:24 pm

cheers need help with q 27 bii) please think its a potential divider q and am still clueless about them

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Re: EXAM!!!!!!!!!!!!

Post  Admin on Sat May 23, 2009 11:25 pm

Short Answer 25 kilo ohms,
explanation Vo = V2 - V1 x gain
where P - Q = V2 - V1, Vo = 2V and gain = ( 1000000 / 500000 ) also known as 2
so P - Q = 1 V that is answer to bi.
the voltage at Q = Vs x 20k / (100k + 20k) and from Question Vs = 12V
Voltage at Q = 2V
so P - Q = 1; P - 2 = 1
therefore voltage at P = 3 V
again using voltage divider equation VP = Vs x Rth / ( Rth + 75k)
we get 3 = 12 Rth / (Rth + 75k)
3 x (Rth + 75k) = 12Rth
3Rth + 225k = 12Rth
225k = 9Rth
25k = Rth
Hope that helps

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Re: EXAM!!!!!!!!!!!!

Post  Suzy on Sun May 24, 2009 4:45 pm

hi Smile
what exactly does reverse-biased and forward-biased mean???

and see when we shine UV light onto the gold leaf, why doesn't it fall if it's positively charged? I don't get why it's just the one with the electrons that falls???

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Re: EXAM!!!!!!!!!!!!

Post  Admin on Sun May 24, 2009 5:37 pm

Forward biased allows current to flow so the arrowhead of the diode is pointing to the negative terminal of the battery in circuit diagrams.
Reversed biased does not allow current to flow and the diode points to the positive terminal.

The UV light removes electrons so it will remove the negative charge making a negatively charged gold leaf neutral.
Remember the leaf is standing due to the repulsion between the excess electrons. So when we remove the electrons it falls.

Now it the leaf is positively charged then the shortage of electrons creates a positive to positive repulsion. Since the UV light will remove electrons it makes the leaf even more positively charged there is still a positive to positive repulsion the leaf will not fall.

Does this help?

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loell

Post  Scoobasteve on Sun May 24, 2009 11:17 pm

BRICKING IT

sir please be in tomorrow, amma follow you about .

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Re: EXAM!!!!!!!!!!!!

Post  Admin on Mon May 25, 2009 10:25 am

Monday is a holiday, so you can contact me on the Forum on Monday.

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Re: EXAM!!!!!!!!!!!!

Post  S.McGrath on Mon May 25, 2009 1:59 pm

sir have u got any free periods tuesday morning before the exam? for last minute stuff Laughing

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Re: EXAM!!!!!!!!!!!!

Post  ANDYQ on Mon May 25, 2009 3:21 pm

tommorrow is ma last exam and the FREEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEDOM lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol! lol!

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