Physics: Post your doubts here!

[..sacrifice4Revenge..]

http://physics-ref.blogspot.com/2014/08/9702-june-2011-paper-42-43-worked.html

thanks but I actually need the concept and not the answer.
why is the Potential taken across the 1kOhm resistor and not the strain gauge? how does that work every time, how do we decide across which resistor to take potential.

 

[mehria]

check at
http://physics-ref.blogspot.com/2014/07/9702-june-2013-paper-42-worked.html

bcoz at t=0, the displacement is not zero. if you plot a graph, the oscillations would correspond to a cos graph, not a sine.

for sine, when t=0, displacement is also zero


...well, this should also have been in your notes

http://maxpapers.com/wp-content/uploads/2012/11/9702_s12_qp_42.pdf
Q1 (c)(ii) i didnt get the ans in ms... wat do they mean by " ΔEP = Fx because F constant as x ! radius of orbit"

 

[farhan980]

The width depends on where Vin is 0, so everytime the graph of Vin is 0 you switch the polarity try sketching it and upload it here so I check how you sketched it

is this how i should do it ?

 

[mehria]

I really need help with this concept:

http://prntscr.com/75vq5i

may june 11 42 q9 b.

how do we identify across which resistor is the input going to be?
for example, in this case. is the 4.5V going to go to the Non-iverting input from across the 1kV resistor or the strain gauge?

A generalised explanation is needed on this thing,not this question specific. since I struggle on every question of this kind.


Thanks a lot and God Bless!

it is going across 1 kΩ resistor.. in such questions u'll always apply the concept of potential divider... the resistor n strain gauge acts as potential divider to the non-inverting input... that's y u cn see that the voltage across 1.0 kΩ is V1 ..ths is clearly indicated in the diagram...


just lyk the potential divider lyk above... instead of Vout lets suppose that it's V2
so the value of V2 is calculated as=> V2 = [(R2) / (R1 + R2) ] x Vin

the concept is very easy... just ignre the rest of the connections n focus on the wire that has strain gauge n resistor across it...

 

[mehria]

Could someone just give me a quick explanation with tips on how I should find the area under a graph when I'm asked to do so? Do I count the squares? Then what? Thank you

cn u post an example so that explaining it gets easier...

n in general u dnt need to count the squares... count the area of 1 cm square n then count the number of the squares having area of 1cm^2 .. then multiply the number of squares with the value of area of 1 square... u'll get ur ans...

 

[..sacrifice4Revenge..]

it is going across 1 kΩ resistor.. in such questions u'll always apply the concept of potential divider... the resistor n strain gauge acts as potential divider to the non-inverting input... that's y u cn see that the voltage across 1.0 kΩ is V1 ..ths is clearly indicated in the diagram...
View attachment 53800



just lyk the potential divider lyk above... instead of Vout lets suppose that it's V2
so the value of V2 is calculated as=> V2 = [(R2) / (R1 + R2) ] x Vin

the concept is very easy... just ignre the rest of the connections n focus on the wire that has strain gauge n resistor across it...

thank you soo much!

btw.. there's another thing, you got sometime?

 

[mehria]

thank you soo much!

btw.. there's another thing, you got sometime?

my pleasure (^_^)
yup

 

[..sacrifice4Revenge..]

1 http://prntscr.com/75zebj
2- http://prntscr.com/75zdy9

can you do this question,
and also explain this a bit. I kind of have missed this topic altogether.
like if for side AB force, we use length of BC, why's that.
and if the coil was rotated and now is perpendicular, how will we calculate force on AB now.

mehria

 

[TheJDOG]

is this how i should do it ?

Yes

 

[farhan980]

Could someone just give me a quick explanation with tips on how I should find the area under a graph when I'm asked to do so? Do I count the squares? Then what? Thank you

Yes

thank you very much . Oh and for the area under the graph, yes u have to count the number of squares and multiply by the area of one square .

 

[..sacrifice4Revenge..]

1eV=1.6x10^-19 J
Why in some calculations is 1.6x10^-13J used?:S

 

[farhan980]

1eV=1.6x10^-19 J
Why in some calculations is 1.6x10^-13J used?:S

1eV=1.6x10^-19
1MeV=(1.6x10^-19)x(10^6)=1.6x10^-13

 

[Mahnoorfatima]

https://docs.google.com/viewerng/vi...wp-content/uploads/2012/11/9702_s12_qp_33.pdf
Question 1 part e. Can someone PLEASE explain how to find units ?????

 

[mehria]

1 http://prntscr.com/75zebj
2- http://prntscr.com/75zdy9

can you do this question,
and also explain this a bit. I kind of have missed this topic altogether.
like if for side AB force, we use length of BC, why's that.
and if the coil was rotated and now is perpendicular, how will we calculate force on AB now.

mehria

cn u post the ms as well?

 

[Komail Sabba']

http://maxpapers.com/wp-content/uploads/2012/11/9702_w14_qp_41.pdf

Question 5bii please.

 

[..sacrifice4Revenge..]

1 http://prntscr.com/75zebj
2- http://prntscr.com/75zdy9

can you do this question,
and also explain this a bit. I kind of have missed this topic altogether.
like if for side AB force, we use length of BC, why's that.
and if the coil was rotated and now is perpendicular, how will we calculate force on AB now.

mehria

Physicist can you please explain :|

mehria ms has used BC length for Force on AB
BC force is zero

 

[..sacrifice4Revenge..]

Pg 2
http://www.directupload.net/file/d/3990/w2ndknja_jpg.htm

Last part
Sorry for the bad angle

 

[..sacrifice4Revenge..]

http://imageshack.us/photo/my-images/540/3MXWb3.jpg

Is this correct? Except for the arrows..

 

[Debonny1]

Help? It's a non-inverting circuit btw with a gain of 10

 

[Shadow]

Can anybody provide me notes of telecommunication? I really suck at this topic And i'll be losing about 10 marks straight if i don't study it all. Anybody?
Also i want some notes on A2 core topics of line spectrum and temperature. It would be great if anybody would help me with this.