A2 Physics | Post your doubts here

[aliya_zad]

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s05_qp_4.pdf
How do u draw it for Q7)a
And part c why do u take it as 1/10 and not 1/9

 

[SkyPilotage]

Is the potential energy of ideal gases considered to be ZERO or CONSTANT?

 

[leadingguy]

Is the potential energy of ideal gases considered to be ZERO or CONSTANT?

zero

 

[SkyPilotage]

zero

another question,
November 2009 Paper 42 Question 2 cii) Im trying to crack it . Can you help me?
Another question:-
for Internal energy = thermal + work done on system.
So if work done is negative 30 Joules
this means k.e is = -30 joules how can k.e be negative?

 

[SkyPilotage]

C

another question,
November 2009 Paper 42 Question 2 cii) Im trying to crack it . Can you help me?
Another question:-
for Internal energy = thermal + work done on system.
So if work done is negative 30 Joules
this means k.e is = -30 joules how can k.e be negative?

Can we use work done = p x Delta V = nR x Delta T?

 

[leadingguy]

another question,
November 2009 Paper 42 Question 2 cii) Im trying to crack it . Can you help me?
Another question:-
for Internal energy = thermal + work done on system.
So if work done is negative 30 Joules
this means k.e is = -30 joules how can k.e be negative?

yes nov 09 qstn 2ci
work done is P delta V

 

[SkyPilotage]

yes nov 09 qstn 2ci
work done is P delta V

No no C ii)!

 

[leadingguy]

another question,
November 2009 Paper 42 Question 2 cii) Im trying to crack it . Can you help me?
Another question:-
for Internal energy = thermal + work done on system.
So if work done is negative 30 Joules
this means k.e is = -30 joules how can k.e be negative?

work done negative 30 joules means that the wrk is done by the system not On the system ... therefore there is a change in internal energy ... and when work is dfone by the system internal energy is reduced . this reduction is shown by negative sign

 

[SkyPilotage]

work done negative 30 joules means that the wrk is done by the system not On the system ... therefore there is a change in internal energy ... and when work is dfone by the system internal energy is reduced . this reduction is shown by negative sign

I thank you for your reply, but you have not addressed my question.
I believe you will if you tell me whats wrong with c ii)

 

[leadingguy]

I thank you for your reply, but you have not addressed my question.
I believe you will if you tell me whats wrong with c ii)

i think U are aasking fr question 2 cii) here I can conclude that volume of ideal gas is drectly proportional to the temperature change (according to charles law )and temperature change is directly proportional to the change in kinetic energy.

here there is an increase in volume so will the temperature increase


wel this is only what i can conclude but not the exact change as it is also stated in the m.s that there is insufficient data fr the question part to be completed

 

[histephenson007]

June 2009 P4 Q2 a) :- Why is the rate of alpha particles seem constant?
Answer in markscheme:- half life is very long/ but when I calculate half life by ln2 / decay constant its very shoort 1.98 x19^-7...
Am I missing something? I think I am, the constant rate is Delta N / Delta t not the decay constant right?

I made the same mistake. I guess the question has nothing to do with numbers. It just asks us to "suggest" how we can assume the alpha particles to be constant (with reference to the half-life).

That's what I think.
Anybody with a better explanation?!

 

[SkyPilotage]

i think U are aasking fr question 2 cii) here I can conclude that volume of ideal gas is drectly proportional to the temperature change (according to charles law )and temperature change is directly proportional to the change in kinetic energy.

here there is an increase in volume so will the temperature increase

then why is insufficient information quoted in the marking scheme? what is not there?
Question is :- I used the same formula, but what IF i used work done = P x Delta V = n x R x Delta T
The formula we are using is P x Delta V = - work done = decrease in internal energy = decrease in k.e = 3/2 Nx k x T = 3/2 n x R x Delta T
Whats going wrong here?

 

[leadingguy]

then why is insufficient information quoted in the marking scheme? what is not there?
Question is :- I used the same formula, but what IF i used work done = P x Delta V = n x R x Delta T
The formula we are using is P x Delta V = - work done = decrease in internal energy = decrease in k.e = 3/2 Nx k x T = 3/2 n x R x Delta T
Whats going wrong here?

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_42.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_42.pdf

question 4 a ii) please can U explain how can we deduce the ans????

 

[SkyPilotage]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_42.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_ms_42.pdf

question 4 a ii) please can U explain how can we deduce the ans????

Was thinking about it to, still didnt get to it, but what I think is that since the electric field lines of SIMILAR charges are in opposite directions, then the sum of electric field strengths has to be zero at some point which is show on the graph. But if they are opposite charges, the electric field strength is the sum of both at that point which have the same firection. So if the charges are opposite , then the electric field strength can never be zero between both of them!
Tell me if you agree?

 

[leadingguy]

Was thinking about it to, still didnt get to it, but what I think is that since the electric field lines of SIMILAR charges are in opposite directions, then the sum of electric field strengths has to be zero at some point which is show on the graph. But if they are opposite charges, the electric field strength is the sum of both at that point which have the same firection. So if the charges are opposite , then the electric field strength can never be zero between both of them!
Tell me if you agree?

yes I wil go with u, that sum of electric fields are zero at some point, shown on the graph ... so same charges are there ..

same thing is mentioned in ms too. and yes if seen diagramaticaly same charges have a neutral point

 

[namename]

Help:

The specific impedance of fat, muscle and bone are 1.4x10^6 , 1.6x10^6, and 6.5x10^6 respectively. The linear absorption coefficients in fat and in muscle are 0.24 cm-1 and 0.23cm-1 respectively.
Discuss quantitatively, in terms of I, the reflection and the transmission of the beam of ultrasound as it passes through the layer of fat of thickness 4mm, into the muscle of thickness 43.5mm and finally into the bone.

In terms of I
a) Incident on fat-muscle boundary
b) transmitted through fat/muscle boundary
c) received at muscle-bone boundary
d) transmitted into bone

 

[histephenson007]

Help:

The specific impedance of fat, muscle and bone are 1.4x10^6 , 1.6x10^6, and 6.5x10^6 respectively. The linear absorption coefficients in fat and in muscle are 0.24 cm-1 and 0.23cm-1 respectively.
Discuss quantitatively, in terms of I, the reflection and the transmission of the beam of ultrasound as it passes through the layer of fat of thickness 4mm, into the muscle of thickness 43.5mm and finally into the bone.

In terms of I
a) Incident on fat-muscle boundary
b) transmitted through fat/muscle boundary
c) received at muscle-bone boundary
d) transmitted into bone

A) = I e^-ux = I e^-(0.24)(0.4) = 0.9I

B) I (transmitted) = I (incident) - I (reflected)
I (transmitted) = 0.9I - ((Z2 - Z1) ^2 )/(Z2+Z1) ^2)) I
I (transmitted) = 0.9I - ((1.6-1.4) ^2)/(1.6+1.4) ^2)) I
So, I (transmitted) = 0.9I - (1/225) I = 0.896 I
Which is again, almost equal to 0.9I

C) = (0.896)I e^-ux = I e^-(0.23)(4.35) = 0.37 (0.896)I = 0.33I

D) I (transmitted) = I (incident) - I (reflected)
I (transmitted) = 0.33I - ((Z2 - Z1) ^2 )/(Z2+Z1) ^2)) I
I (transmitted) = 0.33I - ((6.5-1.6) ^2)/(6.5+1.6) ^2)) I
So, I (transmitted) = 0.33I - 0.37 I = -0.04I
Which is almost equal to 0 I.

Maybe the significant digits aren't exactly equal, but the process should be correct

 

[hassam]

Why are you complicating the problem.
Work done is P x Delta V.
If it asks for change in K.e then you can use 3/2 Pv = 3/2 (NkT)

change in k.e is equal to change in internal energy for ideal gas....ain't it?

 

[SkyPilotage]

yes I wil go with u, that sum of electric fields are zero at some point, shown on the graph ... so same charges are there ..

same thing is mentioned in ms too. and yes if seen diagramaticaly same charges have a neutral point

yay!
But pleaase do help me crack Nov 09 P42 Question 2 c ii)
I cant seem to figure out why this question was omitted.
Plus Im confused of which way to solve it. By charle's law, volume increase hence temperature increases.
But by first law of thermodynamics , work is done by gas hence internal energy decreases, hence kinetic energy decreases so temperature decreases.
This is a bit confusing?

 

[SkyPilotage]

change in k.e is equal to change in internal energy for ideal gas....ain't it?

that is correct, but I kind of have a similar problem if you notice Nov 09 P 42 Q 2 cii)
But when it asks for work done, just use P x delta V.
as for the other part, im not sure.
P.S:- your diode graphs you showed were incorrect for full wave rectification.