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The graph part is crap! -_- I couldn't do it either.Quoted :
11b(ii)
sampling frequency = 44.1 Khz
Since each sample consists of 16- bit word, bit rate(bits per second) = 44.1 x 1000 x 16
Now,
time for recording = (5 x 60) + 40 s = 340 s
number of bits generated in recording time = 340 x bit rate
The writing question , I really memorize them ,
can you draw the graph part in Q9 of same yr ?
Thank you but I already checked the booklet. I was looking for a well organised answer. I kinda suck at these kinda questionshttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_41.pdf
question 13 b
For this see the Application Booklet..
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_41.pdf
question 10, 11bii, 12b
For Q10: Application Booklet.
Q11 bii : 1. Shoter aerial is required.
2. Less attenuation occurs. 3. Less distortion.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf
explain in detail 3 (iii) please?
K.E=(m*(v)^2)/2http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf
explain in detail 3 (iii) please?
Thank you but can you please properly reply coz i really need helpw07/2
Q5(c)As stationary wave doesn't travel, the speed calculated is speed of the incident and reflected waves. these two waves superimpose/interfere to create the stationary wave
(d) there is no d part of this question
Q2 (a) electric field strength is the strength of electric field. it can be measured by putting a +ve charge in the field and determine the force it experiences. it is a vector quantity. its formal definition will be "force per unit positive charge in an electric field"
Q(5)(a)
in a transverse wave the particles of medium(like water molecules in water waves) travel perpendicular to the direction in which the energy propagates.
in polarisation we restrict the direction of oscillation of particles in one only one direction which is perpendicular to the direction of propagation of energy.
remind me rest of them tomorrow i am having a severe headache now.
Which type of help ?I seriously need help inP4 physics! any smarties out there? Please help me.
Which type of help ?
I have formulas , may that help you!
YupThank you, bro! You giving AS only?
I want help in Paper 4. Hope to find someone as helpful as you over here to help me in a-level too
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q 5 part c and d
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2
Q5 part a
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf
Q2 part b2
Q4 part b
Q5 part c
Q6 part b
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q3 part b
Q7 part b2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
Q4 part c
Q5 part a 2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
Q4partb 2
Q6
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_21.pdf
Q3 part c
http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_qp_22.pdf
Q5 partc
I know its Aaaa lot but i really need help
Thank you
Some body please help
s08:http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf
Q2 part b2
Q4 part b
Q5 part c
Q6 part b
now i am tired please remind me to solve rest of them in evening.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q 5 part c and d
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2
Q5 part a
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf
Q2 part b2
Q4 part b
Q5 part c
Q6 part b
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q3 part b
Q7 part b2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
Q4 part c
Q5 part a 2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
Q4partb 2
Q6
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_21.pdf
Q3 part c
http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_qp_22.pdf
Q5 partc
I know its Aaaa lot but i really need help
Thank you
Some body please help
Question 6 part b Explain it again please.s08:
Q2 (b)
from graph, there is force of 28 N for extension of 3.5 cm
E=1/2 eF = 1/2 * 3.5X10^-2 m * 28 N
=0.49 J
Q4 (b)
P/2.2X10^9 = change in V /V
as 2.2X10^9 is a very big number, the change in v/V is very large number, the change in V will be very small.
Q5(c)
first find λ:
difference between two consecutive nodes(or antinodes)= 1/2 λ so λ = 2*17.8 cm = 35.6 cm .
convert λ into m: λ = 0.356 m
v=fλ so v= 125 Hz * 0.356m = 44.5 m/s
v=√(T/m) so v²=T/m so m=T/v²
m=4 N/ (44.5 m/s )² = 2.0 X 10^-3 kg/m
Q6 (b)
there is a long way this question can be done. but i have a shortcut which always works. i don't have any proof for that. if you want to use it then here is how it works:
power add up in parallel and add like this in series 1/Pc = 1/P1 + 1/P2
(1) circuit is not complete so there will be no current at all. so power = 0
(2) as there is shortcircuit no current will pass through resistor B. only current will pass through B. so P=1.5 KW
(3) again resistor B is short-circuit so no current through it. A and C are in parallel. by the formula i gave you, the power will add up in parallel. so P=3 KW
(4) now there is no current through C but A and B are in series. so power will be calculated through the formula i gave you:
1/P = 1/1.5 + 1/1.5 ==> 1/P = 2/1.5 ==> p=1.5/2 = 0.75 KW
(5) now power of A and B are in parallel with C. so power of A and B will add up with C. 0.75 KW + 1.5 = 2.25 KW
Energy is not conserved.. so inelasticThat's what I meant. Why is the collision inelastic when an atom collides with a moving piston?
Question 6ci and ii)sweetjinnah sitooon
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Question 2b)
Question 6ci and ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Question 6b) and 7a, 7c
Plz help someone?
sweetjinnah sitooon
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Question 2b)
Question 6ci and ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Question 6b) and 7a, 7c
Plz help someone?
sweetjinnah sitooon
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_4.pdf
Question 2b)
Question 6ci and ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Question 6b) and 7a, 7c
Plz help someone?
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