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Physics: Post your doubts here!
- Thread starter XPFMember
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The graph part is crap! -_- I couldn't do it either.
And Thanks
Thank you but I already checked the booklet. I was looking for a well organised answer. I kinda suck at these kinda questions
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
explain in detail 3 (iii) please?
explain in detail 3 (iii) please?
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K.E=(m*(v)^2)/2
K.E=P^2/2m
Since, p=mv so p^2=(mv)^2
K.E= m^2 * v^2/2m
m cancels out in the numerator and the denominator to give m*v^2/2
Hope i Helped
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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.
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Which type of help ?
I have formulas , may that help you!
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8000 Post complete
Thank you, bro! You giving AS only?
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Yup
I want help in Paper 4. Hope to find someone as helpful as you over here to help me in a-level tooYup
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w07:
q5 a
if you want to know why 0.7 a here is why:
I is directly proportional to A²
I = k A²
we want to make I half times. multiply both sides by ½
½ I = ½(k A²) =k ½(A)²
we want to make that coefficient go inside that square
square of 1/√2 is ½ so
½ I = k( 1/√2 A)²
as 1/√2 ≈ 0.7
½ I = k( 0.7 A)²
and about 60 degree:
the phase difference is sixty degree which is 1/3 of 180. so one big box is equal to sixty degrees. so start the new waves sixty degree left or right from where the previous waves started and the amplitude will be 0.7a and rest of it will be the same i mean the frequency and wavelength.
more detailed description:
i drew the line at 0.7a and -0.7a which will be the point where the crests and trough of the graph are going to be. as the amplitube is 1/sqrt(2) * a
and i dragged every point right by one big box marked by 60 degree, 120 degree and 180 degree(interval of 60) because the complete wave took 6 squares and complete wave is 360 so each box becomes of 60 degree.
drag every point to the right/or left one box.
see this i hope it is clear. i tried my best.
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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
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now i am tired please remind me to solve rest of them in evening.
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Question 6 part b Explain it again please.
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Energy is not conserved.. so inelastic
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Question 6ci and ii)
ci) 1. A= decay constant * N
=> (3.76*10^6) = (0.025)*N... solving for N = 4.67*10^15
2. moles= mass/Mr
moles = no. of paticles /Avogadros constant.
Mass= (4.76*10^15) / (6.02*10^23) = ans * 90/1000 solving mass = 6.98*10^-10 kg
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