A2 Physics | Post your doubts here

[Jonathan127]

Q3 b) iii) I think what you say is right. Work done on the system should be positive. Although I explain this from that perspective, I will try to make some sense by explaining it in an alternative way. If you just ignore the words for a moment,

We know the formula W = -P (dV) , where dV is delta volume.
We can write this as W = -P (V2 - V1)
When V2 > V1, dV is positive, hence Work done is negative.
When V2 < V1, dV is negative, hence Work done is positive.
In this case, V2 > V1, so we may say that the Work done is negative.

I think they did a mistake in typing that as work done on the system.

Oh right! I think I get it. Also, I know this may sound stupid, but for the same question, they said:
- water vapour volume = 2.96 × 10^–2 m3
- volume occupied by liquid water at 100 °C is 1.87 × 10^–5 m3

So, how would the change in volume (dV) = 2.96 × 10^–2 - 1.87 × 10^–5 ?

 

[Jonathan127]

My doubt has been ignored :/ someone please help !

What page is your doubt on?

 

[angelgirl:)]

can any1 plz tell me wen to use cos and wen to use sine in any kind of oscillation...

 

[optimistic]

Hey histephenson007...could you please explain me the answer to MAY/JUNE 2009 Q2 PART a?

 

[ousamah112]

can any1 plz tell me wen to use cos and wen to use sine in any kind of oscillation...

use cos when oscillation starts from its amplitude i.e its heighest position. and sin when oscillation starts from mean position i.e zero

 

[angelgirl:)]

wen to use x.sinwt and wen to suse x.coswt ?

 

[ousamah112]

wen to use x.sinwt and wen to suse x.coswt ?

same answer...

 

[angelgirl:)]

wen to use x.sinwt and wen to suse x.coswt ?

got it...

 

[leadingguy]

can any1 plz tell me wen to use cos and wen to use sine in any kind of oscillation...

for the graph whose starting co-ordinates are (0, 0 ) means start time is zero and displacement at that time is o

finding displacement frm mean postion of such grpah, we use the frmula X = Xosinwt



finding velocity of the same graph we use the frmula V = Xowcoswt


finding acceleration of such graphs we use the formula a = -Xow^2.sinwt


on the contrary, there are graphs whose starting coordiantes are (o , X0) mean that at t= o they are having maximum displacement. for such graphs;



finding displacement frm mean postion of such grpah, we use the frmula X = Xocoswt

finding velocity of the same graph we use the frmula V = -Xowsinwt


finding acceleration of such graphs we use the formula a = -Xow^2.coswt

 

[ousamah112]

got it...

also see cos graphs of oscillation and sin graphs...

 

[KaiserFury]

Hello,

I have a few questions that I'm hoping you guys can answer asap, as my exam is after tomorrow.

I want to know what to write down as my answer when the question is "What is seen on the screen when electrons are diffracted", first as particles and second as waves. As in, concentric circles and stuff. And why, if possible.

I would also like to know, in AM and FM, how to calculate the carrier frequency and the signal frequency, and how to solve part 10(c) in this paper: http://www.xtremepapers.com/CIE/ind... Level/9702 - Physics/&file=9702_s07_qp_4.pdf

http://www.xtremepapers.com/CIE/ind... Level/9702 - Physics/&file=9702_s08_qp_4.pdf
In number 2, part b(iii) of this paper, the mark scheme says the mass melted by heater in 5 minutes is = 64.7 - 0.5 x 16.6. Why is this? Why isn't it just 64.7 - 16.6?

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9702 - Physics/&file=9702_s10_qp_42.pdf
I have no idea how to answer number 10(B)

In a reaction, when one nucleus gives off another nucleus and energy, how can the second one be the one having more binding energy if the first one was able to turn to the 2nd one and give off excess energy?

What's the difference between thermionic emission and the photoelectric effect?

I would greatly appreciate any help with my enquiries.

Thank you.

Any help please?

 

[angelgirl:)]

Any help please?

hey ppl who ever finished their studies plz help her...

 

[abdullah12]

look at the answer here....

dV is delta Volume

P -> Q : P = 4x10^5 , dV = -6x10^-4
............. W = - p(dV) = - (4x10^5)(-6x10^-4) = 240 J (first column)
............. u = q + w = -600 + 240 = -360

Q -> R : +720J

R -> P : omg, how'd they get -840?
.............. by reverse calculation, 840 = (1/2)*P*(dV) which is the area of the triangle in the graph.

Any light bulbs shining now?? Anyone?!
Thank you for asking the question, i thought it was easy

I believe in this question the mark scheme made an error since it is clear that the table in the mark scheme is not identical to the one given in the question.

The correct answer is as follows:

- From P to Q: The work done on the gas is -240, the negative is because work is actually done by the gas, not on it. Therefore the increase in internal energy for this change is = -240 -600 which is equal to -840 J

- From Q to R: This part is straight forward, all you have to do is add 0+720, so it is equal to 720 J

- From R to P: This part requires the use of part (b) (i) in which it is stated that the overall change is equal to 0. Using that, we should arrive at the following equation for the change in internal energy:

-840 + 720 + x = 0 -----> x= +120 J

Using that value, we then work backwards and get the work done on the gas to be equal to (120-480) which is -360 J

Hope that clears it up. If anyone has an alternative solution please explain

 

[JohnChemistry]

I thought the kinetic energy does remain constant during a phase change, cz k.e is proportional to temperature change?

Temperature is only a measure of the average kinetic energy, however you could say that the work done in bringing the molecules fromliquid to gas is much greater than the work done in separating molecules from solid to liquid state.

 

[angelgirl:)]

I believe in this question the mark scheme made an error since it is clear that the table in the mark scheme is not identical to the one given in the question.

The correct answer is as follows:

- From P to Q: The work done on the gas is -240, the negative is because work is actually done by the gas, not on it. Therefore the increase in internal energy for this change is = -240 -600 which is equal to -840 J

- From Q to R: This part is straight forward, all you have to do is add 0+720, so it is equal to 720 J

- From R to P: This part requires the use of part (b) (i) in which it is stated that the overall change is equal to 0. Using that, we should arrive at the following equation for the change in internal energy:

-840 + 720 + x = 0 -----> x= +120 J

Using that value, we then work backwards and get the work done on the gas to be equal to (120-480) which is -360 J

Hope that clears it up. If anyone has an alternative solution please explain

but for P to Q...the volume is decreasing ....so wen ever their is compression it means work is done on the gas...so here work has to be +ve...isnt it...?

 

[abdullah12]

I believe in this question the mark scheme made an error since it is clear that the table in the mark scheme is not identical to the one given in the question.

The correct answer is as follows:

- From P to Q: The work done on the gas is -240, the negative is because work is actually done by the gas, not on it. Therefore the increase in internal energy for this change is = -240 -600 which is equal to -840 J

- From Q to R: This part is straight forward, all you have to do is add 0+720, so it is equal to 720 J

- From R to P: This part requires the use of part (b) (i) in which it is stated that the overall change is equal to 0. Using that, we should arrive at the following equation for the change in internal energy:

-840 + 720 + x = 0 -----> x= +120 J

Using that value, we then work backwards and get the work done on the gas to be equal to (120-480) which is -360 J

Hope that clears it up. If anyone has an alternative solution please explain

but for P to Q...the volume is decreasing ....so wen ever their is compression it means work is done on the gas...so here work has to be +ve...isnt it...?

I'm not sure actually but the formula is Work done = pressure* (Change in volume)

Since the change in volume is = (2-8)*10^-4, I assumed its negative.

Does anyone else have any input.. The question we're referring to is Q2)b)iii) in November 2010 variant 41. The mark scheme appears to be incorrect for this question.

 

[angelgirl:)]

I'm not sure actually but the formula is Work done = pressure* (Change in volume)

Since the change in volume is = (2-8)*10^-4, I assumed its negative.

Does anyone else have any input.. The question we're referring to is Q2)b)iii) in November 2010 variant 41. The mark scheme appears to be incorrect for this question.

there r 2 marking scheme for the same year and paper....so i got really confused that which1 is right ...but since morning iam discussing about this question wit many ppl so now i have little idea of how to handle this type of questions...

 

[ousamah112]

Any help please?

s10 p42 q10 b is like this

 

[ousamah112]

Any help please?

difference bw thermonic emission and photoelectric effect:
thermionic emission is emission of electrons when a metal is heated to a very high temperature.
and photoelectric emission is emission of electrons when a light wave of threshold frequency is incident on a metal surface.

 

[OrlandoBloom<3]

What page is your doubt on?

Oh its on page 42 Thanks !