Physics: Post your doubts here!

[mehria]

Okay. So for option A to be correct, it should also have a reading of 6A as per your example. Can you please explain how it will show a reading of 6A if the two resistors are of 2 ohms each and the power supply of 12V? Thanks!

it cnt show the reading of 6A for two resistors...
here we r assumng 6A as I... for option A when S1 is closed then the current won't pass through the second resistor becuz current always chooses low resistance path... so for A our answer will be 12/2=6A i.e. I
n for C it is 3A i.e. I/2

 

[mehria]

Hey guys, can someone please provide me with notes of the AS level topic regarding electric field,electric charge, etc ?

All about Electrostatics..
These r my own notes... hope they help

 

[ZaqZainab]

All about Electrostatics..
These r my own notes... hope they help

I like them .have you made them for other topics if so please post it would be really helpful
Thanks in advance

 

[Thought blocker]

All about Electrostatics..
These r my own notes... hope they help

I like them .have you made them for other topics if so please post it would be really helpful
Thanks in advance

yeap very very helpful

 

[Snackbox86]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
Hey, I am having problems explaining question 30,
The marking scheme gives the answer as B but i still think the answer is D

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
Hey, I am having problems explaining question 30,
The marking scheme gives the answer as B but i still think the answer is D

Its a standing wave, so the 33cm is the distance between two nodes. The wavelength thus becomes .66m. F = 330/. = 500. T = 1/500 = 0.002s = 2ms. B is the right answer

 

[Hijab]

Can someone answer this question especially the b part... I would really be grateful

 

[omarjaved619]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
HELPPP! Paper 4 Question number 11 b (iii)

I managed to figure out the answer, but neither me nor my teacher could understand the workings in the mark scheme AT ALL! I'd be grateful if someone could explain it. Thanks a ton

 

[usama321]

Can someone answer this question especially the b part... I would really be grateful

If i am right, the circuit is like this



The current divides at A. Thus the two resistors from A to B and B to C are in series with each other, and the resistor connected diagonally is parallel with them. So,

1/r = 1/6 + 1/12
r = 4 ohm

Now after this, The next resistor between C and D is in series once again. Thus the resistance till here in the lower two loops become 4 + 6 = 10. Add it with the resistor in the upper loop

1/10 + 1/6
r = 3.75

 

[yousef]

Gamma particles have no mass and that 2 means 2 gamma particles not 2 moles gamma particles like in a a chemical equation. Gamma particles are electromagnetic waves (photons) hence have no rest mass like other electromagnetic waves. You cannot use stoichiometrice ratio like in chemistry and electrons and gamma particles are different. Electrons exhibit wave-particle duality unlike gamma particles.

If so , then why they used mass of photon as mass of meson/2 ?

 

[daredevil]

Use the first law of thermodynamics to explain why the specific latent heat of vaporisation is greater than the specific latent heat of fusion for aparticular substance.

the answer in the ms is:
when evaporating
greater change in separation of atoms/molecules
greater change in volume
identifies each difference correctly with ∆U and w


I don't get it^ :/

 

[NinjaInPyjamas]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf

2nd variant, q7 when switch s1 and s2 both are closed , why is it R?

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf

q7, q11

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

q37

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf

q24, q26

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf

q14, q15, q17, q18, q29, q30, q35, q37

Q: A number of identical springs, each having the same spring constant, are joined in four arrangements. A different load is applied to each arrangement.
Which arrangement has the largest extension?

A: 2 springs in series; 2N
B: 3 springs in series; 1N
C: 2 springs parallel; 6N
D: 3 springs parallel; 8N

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf

q22, q25

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_13.pdf

q10

IMP: http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf

q23

Sorry I know this is a lot of questions! >< please help!

 

[sudeep1097]

If so , then why they used mass of photon as mass of meson/2 ?

I don't have much detail answers about that it's just what i learned in two years. But I think any particle when is at speed of light it will have mass. Hence I said rest mass is zero for gamma particles. In A-levels I think gamma particles not having mass is sufficient.

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

2nd variant, q7 when switch s1 and s2 both are closed , why is it R?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf

q7, q11

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

q37

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf

q24, q26

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

q14, q15, q17, q18, q29, q30, q35, q37

Q: A number of identical springs, each having the same spring constant, are joined in four arrangements. A different load is applied to each arrangement.
Which arrangement has the largest extension?

A: 2 springs in series; 2N
B: 3 springs in series; 1N
C: 2 springs parallel; 6N
D: 3 springs parallel; 8N

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf

q22, q25

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_13.pdf

q10

IMP: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf

q23

Sorry I know this is a lot of questions! >< please help!

Two resistors in each of the two wires. So the resistance of each wire seperately is 2R. When connected in parallel

1/2R + 1/2R = 2/2R = 1R

q7) Each div on the main scale represents 0.5mm, and on the secondary scale 0.01mm. There is a positive zero error of 0.14mm. Thus we will subtract it from our reading of 2.59. Answer would be B

q11) F (resultant force) = ma Weight is 20N and as acceleration of freefall is 10, the mass would be 2kg.
6 = 2a a =3

q37) When the lamp is off, it means that there is no current passing through it. This can only happen if they potential difference, that is the voltage difference at the two terminals of the lamp is zero. This would mean that there is the same amount of voltage at both end of the terminals, leading to no current. As a result, the voltage across R1 would be proportional to the length of wire opposite to it, as only the factor of length is alternating here. Thus R1/R2 = L-x/x

24)extension = F *l/A*modulus... Only L and A are changing here. When length increases by a factor of 3, the extension would increase by a factor of 3. Thus it would be 3x now. However, when the diameter is increase to 2D, the area would be (2D/2)^2. It would become Dsquare. Now originally the diameter of the wire was D, thus its area would have been D/2^2 which would become D^2/4. It can be seen that the area of the wire has increased by a factor of four. As extension is inversely proportional to extension, the wire would extend 1/4x. Thus combining the factors of length and area, we get 3/4x

26) Just calculate the frequency of both the waves by 1/T. You will see C to be true

 

[sudeep1097]

Use the first law of thermodynamics to explain why the specific latent heat of vaporisation is greater than the specific latent heat of fusion for aparticular substance.

the answer in the ms is:
when evaporating
greater change in separation of atoms/molecules
greater change in volume
identifies each difference correctly with ∆U and w


I don't get it^ :/

latent heat of vaporisation means heat needed to change into gas and latent heat of fusion to change into liquid, in liquid all the molecules are not free of each other that is they still have molecular interactions i.e. all the atoms are not free only some are free so less energy is required for latent heat of fusion while in gases all the atoms are free so all the bonds between molecules must be broken requiring greater energy.
From point of first law of thermodynamics :
∆U is change in internal energy which is equal to sum of work done(W) on system and heat applied to the system(Q).
For real gas ∆U includes both potential and kinetic energy. But changing state means there should be no increase in temperature hence no increase in average kinetic energy of molecules so ∆U should be zero. To change into vapor molecules have to be further apart, that is volume is more than liquid (since molecules have to be further apart potential energy is also more) . ∆U=Q+W . Most substances expand on heating, and thus external work is done by the system on atmospheric pressure to increase in volume. So work done is -W (refer to note below) and since the ∆U is 0 by first law of thermodynamics heat(Q) equal in magnitude to -W should be added to the system. Since W is more incase of vapor hence Q is also more.
Note: Sign convention for the equation. W means work done on system -W is workdone by system, Q is heat added to system and -Q is heat released into environment.
Everything from point of first law I told is for ideal gases and I think the question is referring to ideal gases as well.

 

[Thought blocker]

latent heat of vaporisation means heat needed to change into gas and latent heat of fusion to change into liquid, in liquid all the molecules are not free of each other that is they still have molecular interactions i.e. all the atoms are not free only some are free so less energy is required for latent heat of fusion while in gases all the atoms are free so all the bonds between molecules must be broken requiring greater energy.
From point of first law of thermodynamics :
∆U is change in internal energy which is equal to sum of work done(W) on system and heat applied to the system(Q).
For real gas ∆U includes both potential and kinetic energy. But changing state means there should be no increase in temperature hence no increase in average kinetic energy of molecules so ∆U should be zero. To change into vapor molecules have to be further apart, that is volume is more than liquid (since molecules have to be further apart potential energy is also more) . ∆U=Q+W . Most substances expand on heating, and thus external work is done by the system on atmospheric pressure to increase in volume. So work done is -W (refer to note below) and since the ∆U is 0 by first law of thermodynamics heat(Q) equal in magnitude to -W should be added to the system. Since W is more incase of vapor hence Q is also more.
Note: Sign convention for the equation. W means work done on system -W is workdone by system, Q is heat added to system and -Q is heat released into environment.

Is this As level ? -_-

 

[sudeep1097]

Is this As level ? -_-

No A2 level thermal physics!

 

[mahabaloch]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q 5 part c and d

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Q2
Q5 part a

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf
Q2 part b2
Q4 part b
Q5 part c
Q6 part b

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Q3 part b
Q7 part b2

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
Q4 part c
Q5 part a 2

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
Q4partb 2
Q6

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_21.pdf
Q3 part c

http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_qp_22.pdf
Q5 partc

I know its Aaaa lot but i really need help
Thank you

 

[sadiaali]

Please help Usama321 & ZaqZainab and others
The electric power consumed by a device may be calculated by using either of the two expressions P=I2R or P=V2/R.
The first expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality.
How can the seemingly different dependence of P or R in these expressions can be explained?

 

[usama321]

Please help Usama321 & ZaqZainab and others
The electric power consumed by a device may be calculated by using either of the two expressions P=I2R or P=V2/R.
The first expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality.
How can the seemingly different dependence of P or R in these expressions can be explained?

This is something that sometimes confuses me too
It all comes to down to the V= IR. We know that if we increase the resistance of a circuit, the pd across that point should increase. But, we also know that the current would decrease overall in the circuit, which would itself result in a lower pd across that point.

In our case, if the voltage across that point is kept the same, than increasing resistance would decrease the current, which would lead to an inverse relationship with power as our main equation is p =IV

However, if the current through the circuit is same, than increasing the resistance would increase the power dissipated across that point, leading to a direct relationship with power.

But, i am still not sure how that would work, i mean keeping the the current in the circuit the same with increasing resistance etc