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umm can u explain m/j 2008 p2 question 2..the whole of it?
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thanks! same to uur welcum nd best of luck for ur examz
okumm can u explain m/j 2008 p2 question 2..the whole of it?
2)a)i) the constant k=F/e since the graph is extension over the load u have gotto take the reciprocal of the gradient so to find the gradient just take a co-ordinate i'm takin (20,2.5) gradient=y co-ordinate/x co-ordinate convert 2.5 cm to metres =0.025mumm can u explain m/j 2008 p2 question 2..the whole of it?
okay i'll check it out now and tell and sorry for da delayThanks a lot for the quick answers. One more if you don't mind:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q6 (b) (i) and (ii)
You don't need to tell me the whole experiment, just how it would be done.
ok i'll check it out and tell uOkay so please help me on this sum 09 paper 21 Q5B..
thank u soo much!!!!2)a)i) the constant k=F/e since the graph is extension over the load u have gotto take the reciprocal of the gradient so to find the gradient just take a co-ordinate i'm takin (20,2.5) gradient=y co-ordinate/x co-ordinate convert 2.5 cm to metres =0.025m
gradient =0.025/20 = 1/800 k=reciprocal of gradient so k=800N/m
ii)strain energy = area under the graph
=1/2 x 3.5/100 x 28 =0.49 J
b)i) momentum b4 collision= momentum after collision so in this case momentum b4 collision is zero
0=2400 x V - 800 x v i'm subtracting as both the trolleys will move in opposite direction so 800v=2400V
so v/V=2400/800 = 3/1
ii)0.49 is the total kinetic energy wid which dey movedthe trolley wid mass 2.4kg moves wid a speed of (1/3)v and the trolley wod mass 0.8kg moves wid a speed of v
0.49=(1/2 x 2.4 x (1/3v)^2) + (1/2 x 0.8 x v^2) solve for v and u get 0.96m/s
ok for diffraction of transverse wavesThanks a lot for the quick answers. One more if you don't mind:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q6 (b) (i) and (ii)
You don't need to tell me the whole experiment, just how it would be done.
ok so phase difference is 128-100=28cm this is because s1 is 100cm from M and s2 is 128cm from M u form a triangle and find the hypotenuse which is 128cm so now u find lambda = speed/frequency lambda for frequency of 1000hz=330/1000=0.33m=33cmOkay so please help me on this sum 09 paper 21 Q5B..
mate i did not get that could u pls sketch te old one and new one of both parts in any formats even image in paint will be encouraged pls yar me nt getting it *worrrid* :'( :'(that your new graph should be about quarter the size of your old one
2)a)i) the constant k=F/e since the graph is extension over the load u have gotto take the reciprocal of the gradient so to find the gradient just take a co-ordinate i'm takin (20,2.5) gradient=y co-ordinate/x co-ordinate convert 2.5 cm to metres =0.025m
gradient =0.025/20 = 1/800 k=reciprocal of gradient so k=800N/m
ii)strain energy = area under the graph
=1/2 x 3.5/100 x 28 =0.49 J
b)i) momentum b4 collision= momentum after collision so in this case momentum b4 collision is zero
0=2400 x V - 800 x v i'm subtracting as both the trolleys will move in opposite direction so 800v=2400V
so v/V=2400/800 = 3/1
ii)0.49 is the total kinetic energy wid which dey movedthe trolley wid mass 2.4kg moves wid a speed of (1/3)v and the trolley wod mass 0.8kg moves wid a speed of v
0.49=(1/2 x 2.4 x (1/3v)^2) + (1/2 x 0.8 x v^2) solve for v and u get 0.96m/s
Thanx man but u should knw that i also did it till this part couldn't do the rest of itok so phase difference is 128-100=28cm this is because s1 is 100cm from M and s2 is 128cm from M u form a triangle and find the hypotenuse which is 128cm so now u find lambda = speed/frequency lambda for frequency of 1000hz=330/1000=0.33m=33cm
lambda for frequency 4000hz=330/4000=0.0825m=8.25cm
thts all i have cleared out just wait aliitle and i'll figure out the other part of the question
kay i'll do da rest and tell u tommorowThanx man but u should knw that i also did it till this part couldn't do the rest of it
listen,s09 variant 1 qs 5b....plz help
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
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