Physics: Post your doubts here!

[zain942010]

I cant open this link hassam

 

[anonymous123]

AoA wr wb
1 what does RMS value of ac voltage mean? plz explain and hw do u find it?
2 whats the difference b/w peak to peak voltage and peak voltage
3 anyone knows an easy mnemonic for the electromagnetic spectrum including the wavelengths?

 

[unique840]

AoA wr wb
1 what does RMS value of ac voltage mean? plz explain and hw do u find it?
2 whats the difference b/w peak to peak voltage and peak voltage
3 anyone knows an easy mnemonic for the electromagnetic spectrum including the wavelengths?

1) rms of an ac is equal to the constant dc that produces same power or heating effect in a identical resistor. for a sinusoidal graph, rms voltage and rms current is (peakvoltage/under root of 2) and (peak current/under root 2) respectively

 

[unique840]

AoA wr wb
1 what does RMS value of ac voltage mean? plz explain and hw do u find it?
2 whats the difference b/w peak to peak voltage and peak voltage
3 anyone knows an easy mnemonic for the electromagnetic spectrum including the wavelengths?

2) peak voltage means the max voltage or the max amplitude in a graph. peak to peak means twice of amplitude. it means from the lowest peak to the highest

 

[anonymous123]

Jazakallah
and the spectrum? any tip on memorizing it?

edit: I got it nvm

 

[unique840]

Jazakallah
and the spectrum? any tip on memorizing it?

sorry i dont have any idea about that

 

[confused123]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
Q 5 please

whatever i got after spending quite a lot time on b part. :|

a ) when the phase angle difference will be pi radians and when path difference is in fraction form i.e n+ 1/2 lambda then the intensity will be zero, no sound will heard on the microphone.
2) if the amplitudes equal to zero then intensity will be zero. not sure bout this one.

b) path difference = S2M - S1M
use pythogrus theorum to find out S2M 100^2 + 80^2 = s2m ...s2m comes 128cm

now 128-100 = 28 cm which is the path difference.

now the change in frequency is from 1khz to 4 khz means there are 4 wavelengths.
now equate 28 = n + 1/2 lambda
n =0 in firdt order so lambda will be 56 cm

n= 1 , 1 + 1/2 lambda = 28cm , so lambda here will be 18.7 cm.

n = 2 , 2 + 1/2 lambda = 28cm , so lambda here will be 11.2 cm

n= 3, 3 + 1/2 lambda = 28cm , so lambda here will be 8 cm

now there r two minimas and i dnt knw which of these 4 are minimas.

 

[confused123]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
Q 5 please

calculate the wavelength and path difference at maximum frequency and minimum frequency. they are 0.85 and 3.4. so after 0.85 1, 1.5, 2, 2.5, 3 comes. so 2 fractional values exist hence 2 minimas. hope it helps.

Thanks to the wonderful unique840!

 

[unique840]

calculate the wavelength and path difference at maximum frequency and minimum frequency. they are 0.85 and 3.4. so after 0.85 1, 1.5, 2, 2.5, 3 comes. so 2 fractional values exist hence 2 minimas. hope it helps.

Thanks to the wonderful unique840!

lol

 

[confused123]

lol

the one and only wonder kid in the country... yes my fellows u r right! miss unique840!! the physics expert ;D

 

[unique840]

the one and only wonder kid in the country... yes my fellows u r right! miss unique840!! the physics expert ;D

bs ab xada makhan mat lagao. extra tareef kar rahe ho wo b jhuti tarif

 

[confused123]

bs ab xada makhan mat lagao. extra tareef kar rahe ho wo b jhuti tarif

makhan toh double rooti pe lagaya jata hay! nai u deserve it, u r helping me from long long ago from stone age till 2012. good work!

 

[unique840]

makhan toh double rooti pe lagaya jata hay! nai u deserve it, u r helping me from long long ago from stone age till 2012. good work!

lol thanku 4 the "extra tareef"

 

[smzimran]

After a week or so, finally i got some spare time to solve AS problems but it seems someone else is doing a wonderful job, nothing here to answer
^ unique840 ! Nice work!

 

[unique840]

After a week or so, finally i got some spare time to solve AS problems but it seems someone else is doing a wonderful job, nothing here to answer
^ unique840 ! Nice work!

lol thanku

 

[Hussnain]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
2(b)(ii).
3(b)(ii)
4(c)(i)
6(b)
7(a),(b) Please answer my questions

 

[omg]

R=T^h/g, where g and h are constants!
how will we open dis equation by applyin log????
please help !!

 

[DragonCub]

Anyone who can solve this :​

M/J 2007​

Question 3(b) The area below the line of the graph of Fig. 3.2 represents the potential difference​

between A and B.​

Use Fig. 3.2 to determine the potential difference between A and B.​

potential difference = …………………………. V​

Count the number of grids in that area and convert to potential value... This should be the most suitable method...

 

[omg]

pls smby ans my qs :/

 

[DragonCub]

R=T^h/g, where g and h are constants!
how will we open dis equation by applyin log????
please help !!

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.