Physics: Post your doubts here!

[Tabi Sheikh]

y cant the wavelength b larger than that?

if u will take a larger wave length u can achieve the loud sound but the frequency will be higher therefore u take the long wave length t achieve a low frequency they are lambda is inversely proportional to the frequency......

 

[zain786]

Wrong thread

lol i found dat out after i posted it

 

[zain786]

paper link??

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_32.pdf

QS 4 AND 8 HOW? :'(

 

[ahmed abdulla]

ANY HELPPP ??????????
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_23.pdf
q 6 b(ii)

 

[syed1995]

ANY HELPPP ??????????
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_23.pdf
q 6 b(ii)

From question.

"At point P on the screen, the path difference is zero for light arriving at P from the slits A
and B."

Meaning destructive interference.. So Phase Difference at P is 180 Degrees.

2. Wave A Amplitude 2.. Wave B amplitude 1.4..

Ratio of intensity of light at a bright fringe and intensity of light at a dark fringe ...

At Bright Fringe .. The waves will be in phase and there will be constructive interference (Amplitudes will get added) while at Dark Fringe the waves will be out of phase and there will be destructive interference (Amplitude will be subtracted from each other)

So at Bright A= 2 + 1.4 = 3.4 And at Dark A= 2-1.4 = 0.6

I is directly propotional to A^2 .. so

Ratio of Intensity at bright to dark will be (3.4^2)/(0.6^2) = 32:1 Answer

 

[ahmed abdulla]

From question.

"At point P on the screen, the path difference is zero for light arriving at P from the slits A
and B."

Meaning destructive interference.. So Phase Difference at P is 180 Degrees.

2. Wave A Amplitude 2.. Wave B amplitude 1.4..

Ratio of intensity of light at a bright fringe and intensity of light at a dark fringe ...

At Bright Fringe .. The waves will be in phase and there will be constructive interference (Amplitudes will get added) while at Dark Fringe the waves will be out of phase and there will be destructive interference (Amplitude will be subtracted from each other)

So at Bright A= 2 + 1.4 = 3.4 And at Dark A= 2-1.4 = 0.6

I is directly propotional to A^2 .. so

Ratio of Intensity at bright to dark will be (3.4^2)/(0.6^2) = 32:1 Answer

thanks ... but i got it after trying ...

 

[gary221]

please help me with question number 10(c) how to draw the graph. link to the question is http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf

i have doubt in this question i have drawn the graph below and i think thiere is some mistake in this can some one please check and if theres a mistake rectify it please
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
question number-10(c)
and my attached solved

 

[daredevil]

if u will take a larger wave length u can achieve the loud sound but the frequency will be higher therefore u take the long wave length t achieve a low frequency they are lambda is inversely proportional to the frequency......

why would the frequency b higher?? isn't frequency always smaller with a longer wavelenth?

 

[Tkp]

y cant the wavelength b larger than that?

hw come the wavelength can be larger than that?c an antinode is formed at the open end and at the closed end node is formed.so the distance between adjacent nodes and antinodes is 0.25lambda
so .25 lambda =.45
so lambda =1.8
so 330/1.8 and u will get the ans

 

[daredevil]

hw come the wavelength can be larger than that?c an antinode is formed at the open end and at the closed end node is formed.so the distance between adjacent nodes and antinodes is 0.25lambda
so .25 lambda =.45
so lambda =1.8
so 330/1.8 and u will get the ans

oohhh ryt!! -_- Gosh!! this physics is really starting to get to me now -_- thanks btw

 

[Soldier313]

Reposting my doubts, please can someone help?

Aoa wr wb
Can someone please explain qn 9, the whole of it?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_ms_4.pdf

My previous doubts :

Aoa wr wb
Can someone please explain
5 b ii - ?

11 - i don't understand the last phrase in the ms where it says the crystal is cut.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_42.pdf

Thank you

Thank you

 

[daredevil]

I have 2 variants of the mayy/june 2009 paper 2 but in the past papers section of xtremepapers.com there is only a single variant....
i wanted to ask questions from variant 22 if anyone can answer them for me plz do.... sorry i can't post the link bcz i have it in hard copy: -
M/J 2009 p22
Q6b(ii) Why can't we do it with v=s/t ??
Q2 (whole of it please explain every part.... i got the whole question wrong :O )
Thanks in advance
PlanetMaster can u help me?? I was recommended

 

[Tabi Sheikh]

daredevil ask PlanetMaster

 

[daredevil]

daredevil ask PlanetMaster

ok

 

[syed1995]

I have 2 variants of the mayy/june 2009 paper 2 but in the past papers section of xtremepapers.com there is only a single variant....
i wanted to ask questions from variant 22 if anyone can answer them for me plz do.... sorry i can't post the link bcz i have it in hard copy: -
M/J 2009 p22
Q6b(ii) Why can't we do it with v=s/t ??
Q2 (whole of it please explain every part.... i got the whole question wrong :O )
Thanks in advance
PlanetMaster can u help me?? I was recommended

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Both Variants are present... just scroll down and you will find the variant 22 paper

Q 6 b ii) We only use v = d/t only when velocity is constant and acceleration is 0.

the part i.. shows that the acceleration is not zero so velocity can't be constant.. so can't use that fomula.. we will have to use s = ut + 1/2 at^2

since intial speed is zero .. formula becomes s = 1/2 at^2

Q2.. Well even I am not that good in physics to be able to solve this one All I can do is that the time at which it reaches max height is when v = 0 .. so t = 2.4s

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf

Q2 (bii), how did we find the speed?!

 

[Tabi Sheikh]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf

Q2 (bii), how did we find the speed?!

since the graph is distance time graph take the gradient and it will be the speed of the ball..... but the gradient at time t=0.40 otherwise if u take 0.55/0.4 will be he average speed of the ball......

 

[sma786]

since the graph is distance time graph take the gradient and it will be the speed of the ball..... but the gradient at time t=0.40 otherwise if u take 0.55/0.4 will be he average speed of the ball......

thnks

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
Q3 (a), how did they get 'k' as 180 in the energy formula?!

 

[Tabi Sheikh]

thnks

ur welcome