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Physics: Post your doubts here!

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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
q5 part c
ms says plastic is an insulator / not a conductor / has no free electrons B1
charges do not move (on an insulator) B1
either so no single value for the potential
or charge cannot be considered to be at centre
wats the purpose of this point that charges do not move (on an insulator)
Just defining the term 'insulator'
Thats all, otherwise i agree there is no need
;)
 

XPFMember

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AoA,
Q5
When an electron is moving fast in a magnetic field, it experiences the magnetic force for a shorter duration of time thus is deflected less!
When it is passing slowly (with smaller value of speed), it experiences the magnetic force for a longer duration of time thus is deflected more!

Imagine you have to pass a patch of quicksand;
If you run through you will go through easily and will not be much drenched in the sand
If you walk slowly, you will be going down and down in the sand ;)

^ Just came across my mind, hope it helps :D

Q6(b)
You just have to derive the expression for magnetic flux linkage (Φ) :
We know that
Φ = BAsinθ
For N number of lines,
Φ = NBAsinθ
Done
Small derivation so worth 2 marks only
:):p:)
waalaikumassalam wr wb
haha...nice one :D

ok and for ques 6..i wanna know what's the difference b/w flux and flux linkage..see ms...it says smthng like that...

and jazakAlllahhhhhhhhhhhhh for the help...
many duassss 4 u
 

Jaf

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Wavelength of the waves [from the graph] = 50m
Velocity of the wave = 8m/s
[f]Frequency = Velocity/Wavelength = 8/50 = 0.16 Hz
[a]Amplitude (or max displacement) of the waves [from the graph] = 2m
Maximum velocity= 2(pi)af = 2 x pi x 2 x 0.16 = (16/25)pi
Maximum KE of the object = (1/2)m(v^2) = (1/2) x (2x(10^-3)) x ((16/25)pi)^2 = 0.00404 J. Approximately 4mJ.
 
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tha
Wavelength of the waves [from the graph] = 50m
Velocity of the wave = 8m/s
[f]Frequency = Velocity/Wavelength = 8/50 = 0.16 Hz
[a]Amplitude (or max displacement) of the waves [from the graph] = 2m
Maximum velocity= 2(pi)af = 2 x pi x 2 x 0.16 = (16/25)pi
Maximum KE of the object = (1/2)m(v^2) = (1/2) x (2x(10^-3)) x ((16/25)pi)^2 = 0.00404 J. Approximately 4mJ.
thanx very much...if u may jaf im spending more than 1 hour to solve q1 b in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf, we all know that cosx=adj/hyp....so it must be 25/cos(35) to get R but the mark scheme says 25cos(35)... ill blow pls say how???
 

Jaf

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^I don't get why they're complicating things and finding 'R' in the marking scheme lol. This is how I got the answer (sorry my handwriting is kinda messy :p) :
IMG_1872.JPG
 

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...some more doubts in 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 values do I use? :s
· Q:7
 

Jaf

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thank you but, ALL of this i do know jaf...but i cant understand why 25cos(55)??
Whenever a force is given and you have to resolve it and you want to make a closed triangle make sure the given force is always the hypotenuse of your triangle. So to make the triangle in my diagram, drop a perpendicular from the arrowhead of the 25N on the dotted extension of T. This perpendicular will be the vertical. cos theta is adj/hyp. Adjacent side the dotted extension of T. So cos 55 X hyp = 25cos55.
 
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Whenever a force is given and you have to resolve it and you want to make a closed triangle make sure the given force is always the hypotenuse of your triangle. So to make the triangle in my diagram, drop a perpendicular from the arrowhead of the 25N on the dotted extension of T. This perpendicular will be the vertical. cos theta is adj/hyp. Adjacent side the dotted extension of T. So cos 55 X hyp = 25cos55.
jaf i really dont no how to thank you...and this note about the given force being the hypotinse it really save me thank you
 
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Q13:CLOCK WISE moment = anticlockwise moment so FxD=FxD and the rule have a mass of 100g in the centre at 50cm mark so 10 cm from the 40 mark so it is (20g x 60)+(100g x 10)/50 = 44 which is C
Q15: i spend days trying to do and i couldn't do it
Q25:also i didnt know it
Q27: find d which is (1/500) time ten to the power of minus 3 = 2 x 10^-6...then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D
Q37: he want the ratio of the V1/V2 so we need the distance of R1 from x divided by the distance of R2 from x...the distance from R1 from X is x and the distance of R2 from X is the total wire length - the distace of R1 from X so it is L-x so the answer is D x/L-x
 
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I have a question from June 2007 paper 4...........
this question is number 7 part (c).........why will increasing the load of the resistor reduces damping of the oscillation?
 

Jaf

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Q25)
Let's say the length of the trumpet is L and the speed of sound in the horn is c.
At the first frequency, only one note and one antinode are produced. This is 1/4 the wavelength of the sound wave. So L = 1/4 lambda and hence lambda is 4L. c = f x lambda = 75 x 4L = 300L

In the next frequency, there HAS to be a node at the mouthpiece and an antinode at the bell. This means there's another node inside the horn. So now L = 3/4 lambda and lambda = 4/3 L. We also know the speed of the wave remains constant.
c = f x lambda = f x 4L/3
300L/(4L/3) = 225. There's only one such answer with this as the first frequency and that's D.

I'll do 15 later. I'm kind of in a hurry.
 
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