Physics: Post your doubts here!

[nightrider1993]

Thanks a lot dude !

No problem. Glad i could help

 

[InnocentAngel]

All what u have to do is 0.3/t=0.01
Find t=30 s
so give and estimate of number of oscillations in 30 s....
simple

Hey friend i can't still understand it . Could you elaborate please please please
Thank you.

 

[nightrider1993]

Hey friend i can't still understand it . Could you elaborate please please please
Thank you.

Wa dont u get???

 

[Beca1206]

Can anyone pls upload may/june 2012 paper 2 with the answer sheet pls?

 

[Minato112]

Can anyone pls upload may/june 2012 paper 2 with the answer sheet pls?

Variance 1
Paper : http://olevel.sourceforge.net/papers/9702/9702_s12_qp_21.pdf
Marking Scheme : http://olevel.sourceforge.net/papers/9702/9702_s12_ms_21.pdf

Variance 2
Paper : http://olevel.sourceforge.net/papers/9702/9702_s12_qp_22.pdf
Marking Scheme : http://olevel.sourceforge.net/papers/9702/9702_s12_ms_22.pdf

 

[raamish]

do coherent waves only produce constructive intereference ?

 

[leosco1995]

do coherent waves only produce constructive intereference ?

AFAIK, they can also produce destructive interference.

Can someone help me with Q1 (c) of this paper? I don't get it.

 

[bamteck]

Someone please help me for june 2007 p2 no. 1

 

[Bilal Khan]

Someone please help me for june 2007 p2 no. 1

what help do u need??

 

[bamteck]

what help do u need??

for the no. 1
I can't understand what to do

 

[Albert Einstein]

what help do u need??

Use a compass set mark out angles 40 degrees for 20 then 40 for the appropriate angle from the graph n cntinues until 80

 

[bamteck]

Use a compass set mark out angles 40 degrees for 20 then 40 for the appropriate angle from the graph n cntinues until 80

A bit confusing for me

 

[Albert Einstein]

A bit confusing for me

Confused abt ??.. u knw hw to use a protactor...so refer to the graph at 20 x 10^-3 the angle is 40 degree so using the protactor find the postion for 20 x 10^-3 in the meter ??

 

[bamteck]

Confused abt ??.. u knw hw to use a protactor...so refer to the graph at 20 x 10^-3 the angle is 40 degree so using the protactor find the postion for 20 x 10^-3 in the meter ??

And for the (a)(ii) ?

 

[Albert Einstein]

And for the (a)(ii) ?

See ranging from 20 to 40 difference is 40 degree then frm 40 to 60 is 70 degree...frm 60 to 80 is 90 degree .....what u observ is the difference between the angle isis approximately 2/3 . e.g(70-40=30) and 90-70=20 (2/3 of 30) then 102-90=12 (approximatel 2/3 of 20) therfore whn v= 100 x 10^-3(1.0x10^-5)... the increment is 2/3 of 12 = 8..hence postion for 1.0 x 10^-3 is equal to 102 + 8= 110 degree..refer to markscheme the ans is between 107 and 113 so any doubt in the ans ??

 

[bamteck]

See ranging from 20 to 40 difference is 40 degree then frm 40 to 60 is 70 degree...frm 60 to 80 is 90 degree .....what u observ is the difference between the angle isis approximately 2/3 . e.g(70-40=30) and 90-70=20 (2/3 of 30) then 102-90=12 (approximatel 2/3 of 20) therfore whn v= 100 x 10^-3(1.0x10^-5)... the increment is 2/3 of 12 = 8..hence postion for 1.0 x 10^-3 is equal to 102 + 8= 110 degree..refer to markscheme the ans is between 107 and 113 so any doubt in the ans ??

Thank you very much ! You are a genius !

 

[Albert Einstein]

Thank you very much ! You are a genius ![/qu

Thank you very much ! You are a genius !

Ur welkum dr

 

[bamteck]

June 2007 no.5(b), (c) & 6 as well !
I cant understand

 

[Bilal Khan]

June 2007 no.5(b), (c) & 6 as well !
I cant understand

can u upload the paper??

 

[bamteck]

can u upload the paper??

Please help me for P4 also no. 3