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What is the answer ?
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What is the answer ?
What is the answer ?
AWhat is the answer ?
w12_12Aoa everyone.
Im REALLY stuck on this one specific type of question in all my pastpapers and they are the diffraction grating ones. I havent even gotten a single one right yet and I would EXTREMELY appreciate if someone could just explain 2 of these so I'll understand the rest myself. JazakAllah
Q. 30 from O.N/2012
Q. 27 from O.N/2010
10)Q1 answer is C
as c= Wavelenght/ time
t=wavelenght/c= 2x10^15
waves in one sec is frequency so f= 1/2x10^15= 5x10^14
Q5
Well I got it wrong at first... But it's just
look at the actual true current .8 and look at faulty ammeter its also .8, so both the readings are closest
Q10
M not completely sure of the formula but in A's We use the formula
initial + final velocity of sphere 1= initial +final velocity of sphere 2
by arranging them we will have
u1+v1= -u2 + v2
rearrange them
u1+u2=v2- v1
About the formula Thought blocker M I correct???
Ahahaha..... Thanks... I know about relative speeds... Add Maths RV.... Bro u have amazing concepts.... (Y)10)
look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..
Can someone help me with JUN 11 P11 Qs 27?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.Can someone help me with JUN 11 P11 Qs 27?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
aah now i get itfind d which is 1/n -->1/500 * 10^ (-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
thanks my problem is now solved and the planet master has received enough lectures from me."there are two vectors, P and P' where P is initial momentum and P' is final momentum.
P=mv
P'=-mv
change in momentum is P'-P which is same as P + (-P)
just add them like this.
-mv + (-mv) = -2mv"
Haris Bin Zahid
This answer is for you from Suchal.. He is unable to post here..
hahathanks my problem is now solved and the planet master has received enough lectures from me.
but its not tomorrow. o_oBest ov luck for tomorrow every1 <3 may Allah be wid u all
Best ov luck for tomorrow every1 <3 may Allah be wid u all
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