Physics: Post your doubts here!

[anonymous123]

make a power time graph. calculate the avg power. equate it to (I^2 * R) or (V^2/R). this current or voltage will be rms.
if the graph is rectified ac with linear change as it is in nov 2003 p4 q4, then (final+initial)/2

all of dis included in AS? dis is so new for me
edit: i just chkd the syllabus and its only in A2.....

 

[Aahliya]

Count each complete gird as 1 grid and each incomplete one as half grid. This is the standard introduced by the marking scheme.

Alright.. thankss dragon

 

[Scafalon40]

Yes its a resistance wire..

Alright, thankyou.
Anyone else wan't to comment too?

 

[anonymous123]

AoA wr wb

i dnt understand the last part
why is it not x-3=6.75

 

[tom ed]

thanku very much...but unfortunately i dont have tht book......i just have a collection of 3 books pacific,cambridge one and bath physics.......

how much does one wave lags behind the other. one wave is equal to 2 pi radians. how much out of phase are two waves on a graph. use chadda's book cambridge endorsed one,
draw a graph , taking displacement on y axis and distance on x axis. draw a sin wave. the wave is 360 degree or 2 pi radians.
take points according on x- axis. take two points suppose A and B , a at start of the wave and b at the end of the wave, both being in same state of motion or at same displacement from mean position. they have a phase difference of 360 degree or 2 pi radians. just like this take any two points and read on the x axis how much they differ in radians or in degrees. this is how we calcular phase angle difference.

 

[waleedsmz]

Can someone please answer me question?

June09) 9) b): http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf

It says: 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%.

I don't get what the Marking Scheme says: http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_ms_4.pdf

 

[omg]

It's a bit confusing T^h/g... Is it T^(h/g) or (T^h)/g ???
If it's the former:
ln R = ln [T^(h/g)] = (h/g) ln T
On a graph of ln R against ln T the gradient is h/g.
If it's the latter:
ln R = ln [(T^h)/g] = h ln T - ln g
On a graph of ln R against ln T, the gradient is h and y-intercept is - ln g.

so in the first case we'll have no y-intercpt??

 

[Hammad Siddiqi]

Alright, thankyou.
Anyone else wan't to comment too?

Do you want to know how to solve it or u were just confirming if its a resistance wire?

 

[hmlahori]

Can anyone please explain this question from paper 2?

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
Q number 5 part B

 

[anonymous123]

AoA wr wb

plz explain this solution

 

[unique840]

all of dis included in AS? dis is so new for me
edit: i just chkd the syllabus and its only in A2.....

ya its for A2 cox rms is in A2. not in As. i didnt knw u were askin 4 As. sorry

 

[anonymous123]

ya its for A2 cox rms is in A2. not in As. i didnt knw u were askin 4 As. sorry

its ok plz try solving the other 2 qs i posted

 

[CaptainDanger]

Two solid substances P and Q have atoms of mass MP and MQ respectively. They have NP and NQ atoms per unit volume such that NP= NQ. It is found by experiment that the density of P is greater than that of Q.

Which of the following deductions from this experiment must be correct?

A MP > MQ

B NP > NQ

C MPNP > MQ NQ

D MP/NP = MQ/NQ

 

[waleedsmz]

its ok plz try solving the other 2 qs i posted

i dnt understand the last part why is it not x-3=6.75

It is called the "Potential Difference across the resistor ". So if 1 end of the resistor has a potential of 3 V and the potential difference is 6.75, then:
3-x= 6.75
x= 3 - 6.75

Concerning the potential divider question:

I'm not entirely sure of this answer but I understand that since the galvanometer is indicating a zero current, therefore the potential difference between the positive plate and B is zero ( same with the negative plate and A ). Why? Because if there was a potential difference there should be a current passing. This does mean that the potential difference across AB is equal to the potential difference across the cell being tested. Thus, you simply need to find the P.D. across AB.

By using ratio of length to resistance, we get the resistance of AB. Then again by using ratio of resistance to P.D. we get the P.D. across AB.

1 ---> 60
0.65 ---> ?

Then

60 + 20 ---> 12
39 ---> ?

Hope this helped. : )

 

[confused123]

hi bro,

AoA wr wb

plz explain this solution

from where did u took this question ? is this answer really correct. i learned the formula in the cambrdige book which says in this case => AB/ AC into Eo i.e .65/1 into 12 = 7.8 Volt.

but here i see a different situation where first we divided ab length with total length and then multiplied with the total resistance of the wire. this gave us resistance across AB wire,

then we used this resistance to find the unknown e.m.f. of the test cell is this whole formula correct in the image?

 

[waleedsmz]

from where did u took this question ? is this answer really correct. i learned the formula in the cambrdige book which says in this case => AB/ AC into Eo i.e .65/1 into 12 = 7.8 Volt.

Well I think your formula only works if there is no other resistance in the circuit ( like the 20 ohms of internal resistance ) ... Since the whole of the line would actually have a P.D. of 12 V so AB/AC into 12... This example is actually somewhere on the site's revision pages. The last one in the D.C. area right here: http://www.xtremepapers.com/revision/a-level/physics/dc_circuits.php

Check my previous reply if you need more details. : )

 

[confused123]

Can anyone please explain this question from paper 2?

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
Q number 5 part B

maximum current will be possible when the voltage across 6 ohm resistor equals to driver cell voltage or when there is least resistance in the circuit. i.e 12, when its 12, I =V/R ,
12/6 = 2 A .
Minimum will be zero . i am not sure bout this but maybe when the jockey is move towards the extreme position then the voltage or resistance might bcm zero then current will be zero as well :/ plz someone else clarify. Thanks

 

[confused123]

Two solid substances P and Q have atoms of mass MP and MQ respectively. They have NP and NQ atoms per unit volume such that NP= NQ. It is found by experiment that the density of P is greater than that of Q.

Which of the following deductions from this experiment must be correct?

A MP > MQ

B NP > NQ

C MPNP > MQ NQ

D MP/NP = MQ/NQ

A? maybe experimental error which caused wrong mass to be calculate which resulted in difference in densities? caz atoms atoms per unit volume will remain same..share the correct option

 

[confused123]

Well I think your formula only works if there is no other resistance in the circuit ( like the 20 ohms of internal resistance ) ... Since the whole of the line would actually have a P.D. of 12 V so AB/AC into 12... This example is actually somewhere on the site's revision pages. The last one in the D.C. area right here: http://www.xtremepapers.com/revision/a-level/physics/dc_circuits.php

Check my previous reply if you need more details. : )

ok you think but you are not sure. i can memorize the situation just like in the questions above. Thank You.

 

[ultraviolet]

Oct/Nov 2011/23 As level
Q1/ d
Q4/ (iv)
Help please :|