A2 Physics | Post your doubts here

[hassam]

well why we used two different power ratings......to make allowance for heat losses.......so assume that Ploss is same in both cases.....now set up 2 simultaneous eq. and solve

 

[Naruto123]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_qp_4.pdf
question 6b why is it 90 degrees
Can any1 summarize all the graphs for primary current voltage flux and secondary flux current voltage .

 

[prannoy]

Do you mean the root mean squared(rms) voltage? It is simply the amount of direct voltage that gives the same Power to the resistor as the AC. That's how I've understood it..

No i mean mean voltage. I heard its always zero in sinsoidal wave but how??

 

[SkyPilotage]

June 08
P4
Q1 (c)

If centripetal force increases shouldn't the object be more likely to contnue its circular motion ??
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_4.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_ms_4.pdf

Whats holding the object in place is friction.
So centripetal force has to be equal to friction.
If it increases, friction wont be enough to hold the object in place. So as r increases, the centripetal force increases and friction cant hold it in place anymore.

 

[user]

aoa wr wb!

Smoothing..

​

 

[shiningstar]

well why we used two different power ratings......to make allowance for heat losses.......so assume that Ploss is same in both cases.....now set up 2 simultaneous eq. and solve

That is what I did, but it ended up as:
40x190 = 0.85 x 18 x c + Heat lost
7600 = 15.3 x c + Heat Lost (1)
and 6840 = 15.3 x c + Heat Lost (2)
But if I subtract the 2 equations both heat lost and 15.3c will be cancelled :/
What should I do next?
Thank you

 

[user]

Aoa wr wb!

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_43.pdf
What do the last 2 points in q 10 mean?

 

[wahab305]

Never said anything was wrong...It depends on what formula you are using.
Stick to the CIE formula and mentality :- INCREASE IN INTERNAL ENERGY = THERMAL ENERGY SUPPLIED TO SYSTEM +++++ WORK DONE ON THE SYSTEM.
The other formula is mentioned in sources is :- INCREASE IN INTERNAL ENERGY = THERMAL ENERGY SUPPLIED TO SYSTEM -------- WORK DONE BY THE SYSTEM.
P.S:- Capslock was used just to make it clearer, no feelings involved

exactly the other condition is during change of state solid or liquids tend to expand hence they do workdone but internal internal energy increasing good job mate

 

[hassam]

That is what I did, but it ended up as:
40x190 = 0.85 x 18 x c + Heat lost
7600 = 15.3 x c + Heat Lost (1)
and 6840 = 15.3 x c + Heat Lost (2)
But if I subtract the 2 equations both heat lost and 15.3c will be cancelled :/
What should I do next?
Thank you

well heat loss is not same ...its the power loss that is same ....so use Pt instead of HEAT LOOSS IS same

 

[shiningstar]

well heat loss is not same ...its the power loss that is same ....so use Pt instead of HEAT LOOSS IS same

Oh, I got it !
Thank you so much for correction

 

[solidboy123]

May June 2004 paper 4- Question 6 about the internal energy i donot understand anyhing. Plzz anyone can explain me?

 

[shiningstar]

Can someone pls answer the last part of Q4....exam style question in chadha's book...Chapter 24 capacitance??

Total Capacitance in series = 67/2
C decreases to half its value and so will the charge Q as Q=CV and V is constant 12V
So as Q is halved the Current will be halved Q=It
and since P=VI then Power dissipated in the resistor is also halved.

 

[shiningstar]

...so for the last change workdone should be -480 J and increase in internal energy is 0?
Thank You!

No the TOTAL CHANGE in Intenal Energy of WHOLE Cycle PQRP = 0
So: 720-360+inc in internal energy of RP = o
so increase in internal energy of RP = -360 J
so work done = -360 - 480 = -840 J not -480 J
Hope you get it

 

[shiningstar]

May June 2004 paper 4- Question 6 about the internal energy i donot understand anyhing. Plzz anyone can explain me?

1. for an Ideal gas, PE = 0 and since constant temperature thus KE = o so internal energy U = 0. During compression Work is done ON the gas so W is +. Which leads to q is ( - ) as U= w + q
2. No Expansion thus w = 0. Heating means thermal energy is supplied to system so q is +, which leads to U is +
3. during melting heat is supplied to the system so q is +. about the o w I am not sure about it :/

 

[rockerzregmi]

Can anyone post notes for defination . URgent

 

[solidboy123]

1. for an Ideal gas, PE = 0 and since constant temperature thus KE = o so internal energy U = 0. During compression Work is done ON the gas so W is +. Which leads to q is ( - ) as U= w + q
2. No Expansion thus w = 0. Heating means thermal energy is supplied to system so q is +, which leads to U is +
3. during melting heat is supplied to the system so q is +. about the o w I am not sure about it :/

Thanks a lot

 

[Aahliya]

Can anyone solve this question below ??
M/J 2009

9 (a) A sample of a radioactive isotope contains N nuclei at time t. At time (t + Δt), it contains
(N – ΔN) nuclei of the isotope.
For the period Δt, state, in terms of N, ΔN and Δt,
(i) the mean activity of the sample,
activity = ............................................... [1]
(ii) the probability of decay of a nucleus.
probability = ............................................... [1]
(b) A cobalt-60 source having a half-life of 5.27 years is calibrated and found to have an
activity of 3.50 × 105 Bq. The uncertainty in the calibration is ±2%.
Calculate the length of time, in days, after the calibration has been made, for the stated
activity of 3.50 × 105 Bq to have a maximum possible error of 10%.

 

[solidboy123]

Oct nov 2003 question 4 b 1st part. CAN ANYONE HELP ME TO SOLVE THIS PROBLEM

 

[kagome]

M/J 2011 41 Q 8 b(ii)
Please, explaination needed... thanks

 

[hassam]

well in fm......why bandwidth is higher?......is it because frequencies in order of megahertz are used rather than khz as in am......secondly...book mentions bandwidth is 200khz....so max signal frequency cn be 15khz.....now i dint get this......plus do greater sound frequencies imply better quality
and i did not understand the point about spark in car ignition system....book says it produces electromagnetic waves