P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE

[Meticulous]

shouldnt we mention something about coherence??
and r u sure about the formulas isnt it for constructive:
path difference: n lamda

and destructive:
path difference= (n + 1/2) lambda and phase differnce = pie

coherent means same frequency

[poisonedrose]

guys how to do o/n2008 q7bii

We know that the 2000 resistor has 3.6volts, so we subtract 6 from 3.6V giving us 2.4V
As the thermistor and 5000R are in parallel, they have the same voltage of 2.4V

Using the ratio method, we can do
3.6   :   2000
2.4   :      x
Finding x- which is 1,333 (the resistance of both of the devices)

From that, we use the combination resistance formula
5000*R2/5000+R2=13333
And with that, we can simply and find out R2- which is the resistance of the thermistor.

[sabrina]

shouldnt we mention something about coherence??
and r u sure about the formulas isnt it for constructive:
path difference: n lamda

and destructive:
path difference= (n + 1/2) lambda and phase differnce = pie

i m sorry abt the constructive interference.forgot to put "n".
abt destructive the formula of my best of knowledge is correct
abt coherent sources = the interference should have constant phase difference to have observable interference. i think that's it.

[sabrina]

We know that the 2000 resistor has 3.6volts, so we subtract 6 from 3.6V giving us 2.4V
As the thermistor and 5000R are in parallel, they have the same voltage of 2.4V

Using the ratio method, we can do
3.6   :   2000
2.4   :      x
Finding x- which is 1,333 (the resistance of both of the devices)

From that, we use the combination resistance formula
5000*R2/5000+R2=13333
And with that, we can simply and find out R2- which is the resistance of the thermistor.

thanks alot

[halosh92]

i m sorry abt the constructive interference.forgot to put "n".
abt destructive the formula of my best of knowledge is correct
abt coherent sources = the interference should have constant phase difference to have observable interference. i think that's it.

alright thx 

[neno]

heyyy guys can u help me out with sum question
#may june 2008:Q6b
#mayjune 2009(21):Q5b

[scorpion9500]

omg my brain just sopped
guys mj 2008 same q
6b
zzzz

[ruby92]

m/j 2008 2 b??

[highly_ambitious]

plz help m/j 2002 question 5.. what is this potential and internal energy ..how to increase or decrease them through piston mechanism. plz explain this question

[The SMA]

ok the wavelength they gave is basically the length between one antinode and another successive antinode ok?
(two positions loud voice as they mentioned)
so...
if u draw the wave  u will first have an antonide then node then antinode the normal wave
so the distance from one antinode to the node is 32.4/2 ok?
always remember distance from node to node = half a wavelength
and wavelength from node to antinode= quarter a wavelength
so wat we found is quarter a wavelength now multiply by 4
got it?

thanks halosh and chingoo, i think i finally got it.

[scorpion9500]

hey every1 about 2008 6b
i got it
i will just give u the hint and u figure it
if s2 is closed then electricity will pass through it and wont pass through B as there is less resistance in that route
so imagine that there is no B when s2 is closed

[zxcvbnm]

hey so i came across a few questions regarding momentum (i dont remember the year) but they mainly dealt with the final momentum not being the same as the initial momentum, and asked for the reason for this...i think it was in november 09, paper 22, the question realted to projectiles...wait i'll try finding it...please help me with this asap!

[zxcvbnm]

no,it's not from 09....so basically the question asked about a ball hitting a plate and bouncing, but the final momentum not being the same as the initial momentum....

[zxcvbnm]

oh momentum is conserved but i still dont get it
here
Q4, c

[Chingoo]

Momentum of the ball before and after collision is not the same, because the momentum of an isolated system remains conversed, not of a particular object. Hence the plate gains the momentum in the direction lost by the ball and so, the momentum of the system (ball and plate) is conserved.