http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdfThought blocker
Can you please explain Q5 a ii) and iii)
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdfThought blocker
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf
In question paper 21 can anyone explain Q)5) especially part b)?
For maximum light intensity, the resistance will be lowest i.e. 1.2 kΩ. So total resistance: 1/1200 + 1/600=1/R
Thanks a lot I was a little bit confusedFor maximum light intensity, the resistance will be lowest i.e. 1.2 kΩ. So total resistance: 1/1200 + 1/600=1/R
R is 400 Ω.
It's a pattern, Compression Rarefaction Compression Rarefaction....Can you explain how did you get the wavelength as 2.5 lambda
(i)Total p.d.= 9V.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
Question 6 complete PLEASE!!!
Especially the b part!!! :$
Some One Plz DRAW THEM Plz..!I am on!
(ii)According to Kirchhoff's second law: E1 = І2R2/2 + І1R2/2 + І1R1+ І1r1http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
Can you please explain Q5 a ii) and iii)
The resistance of the first half of the metal wire is R2/2 and so is that of the second part but the current in the first part of the metal wire is I2 because both it gets voltage from E1 and E2 this is not the same as the current in the second half of the metalwire which gets voltage only from E1 so E1= I1r1+I2R2/2+I1R2/2+I1R1http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
Can you please explain Q5 a ii) and iii)
I don't get why are we not including I2R2/2 in iii(ii)According to Kirchhoff's second law: E1 = І2R2/2 + І1R2/2 + І1R1+ І1r1
As I2 and I1 both pass half through the metal wire their p.d. will be divided by two.
(iii) E2 and I3 is in negative direction. p.d. across R2 will be half as the other half is not included in the loop. I1 is common across R1, r1 and R2.
E1 – E2 = –І3r2 + І1 (R1 + r1 + R2 / 2)
The loop does not pass through BJ so I2 isn't included.I don't get why are we not including I2R2/2 in iii
I don't get why are we not including I2R2/2 in iii
As area's formula: pi x r^2, the percentage uncertainty is doubled when squared so 8%.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf
Q1 b part 2 uncertainty never understood it can u explain clearly how to do
ThanksThe loop does not pass through BJ so I2 isn't included.
Thanksbecause the loop HBCDJFGH doesn't include BJ so we will not include I2R2/2
in this diagram the seperation of lines represent how weak the strength is. when the lines are almost parallel as well as the distance is same they are uniform. so by diagram they are uniform at centre of rod.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf
Q5 b ii)
Can anyone please tell which region is C and D with explanation.
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