A2 Physics | Post your doubts here

[princesskt]

initially zero(because zero inside sphere) than it would be with same shape but little more curved i m 100% sure saw it in a book.......

because E=Q/4PIER^2 AND V=Q/4PIEr

 

[Zishi]

the light ray wil travel in straight line and wil reflect the the wall and wil go down at an angle then wil again reflect the downward wall and move upwards and wil do the same to travel to the other side of of the fibre.

the core should be narrow to reduce energY loss as if the core is thicker the light ray has to travel more so enrgy wil be lost

No, The reason it will lose energy does not depend upon travelling more distance... The reason is because of angle... Light ray, when it exceeds a certain degree of Angle from normal, is reflected.. The reflection is called Total Internal Reflection... if TIR doesnot occur, the light will be refracted and energy will be lost..

The thing is that it isn't narrow to keep energy loses very low - it's to avoid multipath dispersion(this makes a pulse spread out in time).

 

[smzimran]

The thing is that it isn't narrow to keep energy loses very low - it's to avoid multipath dispersion(this makes a pulse spread out in time).

AoA,
No offense bro but your signature video, i thought it was an attachment,
wouldn't it be better to just attach the link to the video...

 

[SkyPilotage]

-Units of light intensity candela or lux?
-How does negative feedback increase stability of gain? how does it increase bandwidth and reduce noise?
-June 2009 Paper 4 Number 9.. Cant get my head around it...

 

[smzimran]

-Units of light intensity candela or lux?
-How does negative feedback increase stability of gain? how does it increase bandwidth and reduce noise?
-June 2009 Paper 4 Number 9.. Cant get my head around it...

-candela

-it amplifies the same input twice by adding it back to the input so more stability Noise reduce due to same reason

-Which part of Q9 ?

 

[SkyPilotage]

-candela

-it amplifies the same input twice by adding it back to the input so more stability Noise reduce due to same reason

-Which part of Q9 ?

-candela agreed, but some questions use lux with LDR.
-hmm yeah got that, but somehow, I cant understand why the gain is more stable against heat, light , temperature.
ii) derivation and iii)

 

[leadingguy]

-candela

-it amplifies the same input twice by adding it back to the input so more stability Noise reduce due to same reason

-Which part of Q9 ?

that clears some doubts to me as well!

 

[namename]

because E=Q/4PIER^2 AND V=Q/4PIEr

OK btw when do we use the concept "field strength equals to the negative potential gradient"

 

[smzimran]

-candela agreed, but some questions use lux with LDR.
-hmm yeah got that, but somehow, I cant understand why the gain is more stable against heat, light , temperature.
ii) derivation and iii)

-SI unit is candela, while other units may still be used

-more stable against ???? what?

-
ii)
λ = ln2 / t
λ = ln2 / 5.27
λ = 0.132

A = (Ao) e^(-λt)
A/(Ao) = e^(-λt) [A/(Ao) = 0.92]

0.92 = e^(-λt)
ln(0.92) = (-λt) lne
-λt = ln(0.92)
t = - ln(0.92) / λ
t = - ln(0.92) / 0.132
t = 0.634 years
t= 230 days

 

[SkyPilotage]

-SI unit is candela, while other units may still be used

-more stable against ???? what?

-
ii)
λ = ln2 / t
λ = ln2 / 5.27
λ = 0.132

A = (Ao) e^(-λt)
A/(Ao) = e^(-λt) [A/(Ao) = 0.92]

0.92 = e^(-λt)
ln(0.92) = (-λt) lne
-λt = ln(0.92)
t = - ln(0.92) / λ
t = - ln(0.92) / 0.132
t = 0.634 years
t= 230 days

-yup gain stable against the environment, thats what i dont understand.
- ii) how do you derive it?
iii) uhm how did you get A/Ao = 0.92? why did you subtract the percentages?

 

[smzimran]

-yup gain stable against the environment, thats what i dont understand.
- ii) how do you derive it?
iii) uhm how did you get A/Ao = 0.92? why did you subtract the percentages?

Probabilty of decay = change in nuclei / initial number of nuclei
= ΔN / N
I dont think this was difficult


Initially the error is 2% means 0.02
The maximum error should be 10% means 0.10

Increase in error= 0.10 - 0.02 = 0.08
So, if value without error is 1
with error it is (1 - 0.08) = 0.92

Initially activity is A0
Final activity is 0.92 times Ao = 0.92A0
So the ratio = 0.92

 

[namename]

Explain .... please

 

[SkyPilotage]

Probabilty of decay = change in nuclei / initial number of nuclei
= ΔN / N
I dont think this was difficult


Initially the error is 2% means 0.02
The maximum error should be 10% means 0.10

Increase in error= 0.10 - 0.02 = 0.08
So, if value without error is 1
with error it is (1 - 0.08) = 0.92

Initially activity is A0
Final activity is 0.92 times Ao = 0.92A0
So the ratio = 0.92

ah crap, i thought they meant find the decay constant, kept getting the answer as delta N / N x delta T...
hmm abt the percentage, i think they means that there is 0.02 error, but we need to leave a margin 0.08 so that MAX is 0.1.
This means that we did not use the value of activity given in the question right?
I guess I got it now, thanks alot man! much appreciated!

 

[smzimran]

ah crap, i thought they meant find the decay constant, kept getting the answer as delta N / N x delta T...
hmm abt the percentage, i think they means that there is 0.02 error, but we need to leave a margin 0.08 so that MAX is 0.1.
This means that we did not use the value of activity given in the question right?
I guess I got it now, thanks alot man! much appreciated!

Glad I helped!
But you have to solve my june 2010 question !
Dont forget plz

 

[vita199]

can sum1 please explain q10b....i dont get how difference of acoustic impedence has an effect on reflected intensity and also explain part c second part...here is the link for the paper.http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_41.pdf

 

[Dayyanah]

anyone got any notes for charged particles, anything other than the coursebook? i dont get a thing..

 

[princesskt]

OK btw when do we use the concept "field strength equals to the negative potential gradient"[/qu
NOT USED IN CASE OF SPHERE I THINK....DO CORRECT ME IF I M WRONG..........

 

[SkyPilotage]

Glad I helped!
But you have to solve my june 2010 question !
Dont forget plz

Just finished the paper.
For question 2 b)i) internal energyy change is zero since it returns back to its original state. ( ill admit when I saw one mark, It was easier to predict its zero )
for b) ii) from R to P , you will not use the graph. You will find the decrease of internal energy required to make the sum of them all = 0 which is -360 J. Since heat supplied is 480 then work done must be -840 to produce increase of internal energy of -360 hence the total increase of internal energy in one whole cycle is 0 Joules!

 

[smzimran]

Just finished the paper.
For question 2 b)i) internal energyy change is zero since it returns back to its original state. ( ill admit when I saw one mark, It was easier to predict its zero )
for b) ii) from R to P , you will not use the graph. You will find the decrease of internal energy required to make the sum of them all = 0 which is -360 J. Since heat supplied is 480 then work done must be -840 to produce increase of internal energy of -360 hence the total increase of internal energy in one whole cycle is 0 Joules!

Means my guess was correct!
Thanks bro

 

[smzimran]

Just finished the paper.
For question 2 b)i) internal energyy change is zero since it returns back to its original state. ( ill admit when I saw one mark, It was easier to predict its zero )
for b) ii) from R to P , you will not use the graph. You will find the decrease of internal energy required to make the sum of them all = 0 which is -360 J. Since heat supplied is 480 then work done must be -840 to produce increase of internal energy of -360 hence the total increase of internal energy in one whole cycle is 0 Joules!

Bro can you check out june 2009 / p4
Q11 part(b) (iii)