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Physics: Post your doubts here!

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An educated guess from the position of the piston shown in the figure :)
oh ok...jazakAllah..i'll post them ... may Allah bless you..

i still dont get Q:4....how did u know that at those particular positions, that'd be the piston position...? :s
 

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When resolving forces in any direction remember

resolving towards the angle is - COS angle
resolving away from the angle is - SIN angle

This makes every thing easy !
Hope this is helpful
that's exactly how I do it ;)
 
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use fleming's left hand rule ( as current is the cause of motion)
so field directed away from the observer, current will be opposite to the motion electron beam as electrons are -ve!
so the motion is downwards!

refer to the figure below
sorry if i'm wrong!!but i think my concepts still work!!!


Making scheme says D towards the observers but i also had the same answer can any one help
 

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Assalamoalaikum wr wb!!

PAPER 4 doubts...

June 2002
· Q:4 b & c
· Q:6a how should we draw that...? :s

Nov 2002
· Q:3 b(ii)

Nov 2003
· Q:5 b (ii) need explanation, I don’t get it... :s
· Q:4 c Can you show, how is the less ripple represented, please.. :s
· Q:2 b (ii) –ve sign?

June 2004
· Q:4 c reason for the answer...
· Q:6 I’m totally blank abt this...i don’t get how they worked this out..
· Q:8 iii who said we didn’t use them? :s
June 2005
· Q:5b
· Q:6b

Nov 2005
· Q:3 I’m sorta confused with the calculation...part b ...and I don’t understand why they give the ans to C to 2s.f when they’re penalising 2s.f in part b?
· Q:4 c Why speed max when displacement is zero? :s
· Q:5b(ii) why when speed is reduced, deflection is larger? :s
· Q:6 b I don’t understand
·Q:6 c (ii) which way do we do...negative of the upper graph gradient?

June 2006
· Q:2 b (ii) how to do?? :s
· Q:4 c can you show, plz?
· Q:6 a what I understand is the field should be circle..right? I don’t get this :s
· Q:6 c what do I write for this?
· Q:7 Can you answer the complete question, I couldn’t do this one... :(
· Q:8 a For such questions where exactly are we supposed to draw the arrow? Like beside the paper shown..or inside that region? Where do we show....?
Nov 2006
· Q:3 c what to write? :s
· Q:4 b ans. Is 5.99 x 10^24 how does ms say 6.00 :s
· Q:5 a (ii) what’s eddy current?
· Q:6 a (ii) How do the diode works then, normally...isn’t that the way we usually connect? :s
June 2007
· Q:3 d (i) Why is the answer zero? Just to confirm, is it because E=F/Q and F = ma so E is proportional to a...and hence a will be max. when E is max? is it?
· Q:4 c (ii) How’ll the diagram be? :s
· Q:7 b I don’t get why the damping is reduced...i took it as, resistance is increased, more power dissioated, hence more energy losses...so oscillations die off quicker..but it say damping is less..:s confused..
· Q:7 c (iii) can you give some examples? :s
 
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lots of problems. someone help ...
http://www.xtremepapers.com/CIE/Interna ... _qp_11.pdf

mcqs : 8, 15, 18, 19, 27, 33, 38 . just need a short brief description . . . thanks a lot.

27 - The answer is zero because gravitational potential is work done to bring a unit positive charge from infinity
point , therefore according to the equation of GP the radius must change to change the GP but here the GP
is the same therefore zero

hope this is helpful
 
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Assalamoalaikum wr wb!!

Nov 2003
· Q:5 b (ii) need explanation, I don’t get it... :s
· Q:4 c Can you show, how is the less ripple represented, please.. :s
· Q:2 b (ii) –ve sign?

Q5 B ii - I think because Tesla is the magnetic flux density acting normal to the object in this case because they
are parallel if you try resolving MFD normal to the object then it is MFD x cos 90 , which is equal to zero
the answer is change is = 9.81 x 10 ^-5

Q 2 ii - acceleration and displacement are vectors therefore minus is used to show that accelaration is always opposite to the direction of the displacement and W is squared is to ensure that constant is always positive.
 
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Assalamoalaikum wr wb!!

Nov 2006
· Q:3 c what to write? :s
· Q:4 b ans. Is 5.99 x 10^24 how does ms say 6.00 :s
· Q:5 a (ii) what’s eddy current?

What is eddy current this was help full to me

It is the current induced in the conductor when there is a changing magnetic field

Hope This is helpful
 
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Got a doubt in O/N 2010 paper 43, question 3a(i). how do we find the displacement??...i mean there's no value given for "t" in the given equation!!

O/N 2010 paper 43, question 3a(i)

d = -4.0 x cos ( 220 t)
W= 2 x pi x f and W=220
then f = 35
then f = 1/ t
then t = 1/35
Substitute and get the answer i think this is the way
 
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Thanks A lot Bro
JazzakAllah
yup...I do ;) it's easy....say you are taking log of x...
now x is suppose.. 83 +/- 5...
log 83 = 1.92
so Upper Limit (UL) = 83 + 5 = 88
and lower limit (LL) = 83 - 5 = 78
find log of (UL-LL)/2 = (88-78)/2 = 5
log 5 = 0.70 => This is the uncertainty...
so log(83 +/- 5) = 1.92 +/- 0.70
Hope this helped..!
There is a problem here
http://www.xtremepapers.com/community/threads/physics-paper-5-tips.12941/#post-190711
See here!

MODERATOR EDIT: Posts have been removed due to some error made in that...please check the link above to find out how to find uncertainty in log.
 
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too many problems, and since THIS time I am attaching the questions please answer people!
M\J 2010 paper 42 Q5 part b part 2
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf
Capacitance Ratio is 1 : 2 and each capacitor can hold 6 volt max.
3 Volts 6 Volts
Total = 9 Volts maximum

another one now
M\J 2010 42 Q7 part b part 3. I don't get it where does the frequency comes from and how?? and also question 10 of the same paper part b, explain the drawing please!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf
formula is V= Vmaxsin(omega t)
omega= 2pie f
120pie = 2pie f
f = 120pie/2pie = 60Hz

will solve question 10 later and scan it for you

acha moving on O\N2010 Q1 part c part 2. Why don't we place "Re" in denominator of earths radius? why "x2"(the 2 over here means square since I don't know how to do that superscript thingy) please also check Q2 part b part2 of the same paper can somebody please explain me the table? I am unable to comprehend the markscheme and question:(
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_41.pdf
Because we are using graviatational fields in solving the question. We equate g of Earth = g of Moon with formula GM/x^2.
If you need help solving it tell me and Ill take a photo of it for you.
 
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