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

[The SMA]

need help m/j 2007 p2 Q6b

[astarmathsandphysics]

To find the dross section area of metal

[sizbeauty]

hey can sm1 plz tell me tht in nov09/variant 22,q4(partc2(ii))
frm where did v get 0.0362?
2 ?ES = ½ × 1.8 × 10-2 ( 0.0362 – 0.0212)
= 0.077 J


plz sm1 help me!!!!

[sabrina]

hi
i have doubt in O/N2009 var2
q2bi
shouldn't it be like
for 10.7= 6*10^23
      1   = x
n then cross multiply?
i have attached the paper

[sabrina]

n also the separation one. someone plz explain

[sizbeauty]

@ sabrina,
nopz u did it wrong way,
it shuld b lik if for 6*10^23 atoms,u hav 10.7cm^3 vol
so for 1 atom,it shuld b x cm^3 vol..i.e
6*10^23=10.7
1=x
and thn u can get ur ans by cross multipling n for separation,whnver v r givn the vol,v take the cube root of it so it would b (1.78 x 10^-23)1/3 =2.61 x10^-8
hope it helped

cn u plz luk at ma ques???

[ruby92]

m/j 2007
2 d

[The SMA]

m/j 2006 p2 Q6b), why do we need to times 2 the wavelength, using v=f x lambda
and for part a) how to draw the waves?

[sabrina]

@ sabrina,
nopz u did it wrong way,
it shuld b lik if for 6*10^23 atoms,u hav 10.7cm^3 vol
so for 1 atom,it shuld b x cm^3 vol..i.e
6*10^23=10.7
1=x
and thn u can get ur ans by cross multipling n for separation,whnver v r givn the vol,v take the cube root of it so it would b (1.78 x 10^-23)1/3 =2.61 x10^-8
hope it helped

cn u plz luk at ma ques???

thanks a lot now i got it

[halosh92]

m/j 2007
2 d

this is sort of like the projectile
now the electron is exactly in the middle of the plates initally , vertical distance= 1.5cm
so if it has to travel upwards it will travel 1.5/2
right?
which is 0.75 cm basically so....
horizontal velocity is given in the question BUT vertical velocity in projectile cases ( why projectile? because its travelling in a curved path towards the upper plate) the vertical velocity is = 0
so s=ut + 1/2at^2
u=0 as i mentioned
s= 1/2 * (ure "a" value) *( ure "t" value) ^2
u will get an answer which is less than 0.75cm so obviouslly it will hit the plate
got it?
remember to change cms to ms

[halosh92]

m/j 2006 p2 Q6b), why do we need to times 2 the wavelength, using v=f x lambda
and for part a) how to draw the waves?

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?

[halosh92]

could someone briefly explain systematic and random errors?

[Chingoo]

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?

I think you kinda lengthened the whole story...just to add my input (in case it helps), the distance between two consecutive nodes or two consecutive antinodes is lamda/2. See this maxima and minima as a waveform with displacement (y-axis) against distance (x-axis); the minima is the node and maxima the antinode. The antinode can be on either side of the x-axis, because if you remember displacement is taken from a reference point. In case of waves the reference point is the mean position (the node) and the maximum displacement from this mean position is an antinode. If the antinode is a point 2 cm from the mean position, the least displacement is NOT -2 but 0. Hence, like halosh said, simply multiply 32.4 by 2 and you have the lamda.

In times of a sine wave, a node and antinode are 90 degrees out of phase with each other (1/4 of the sine wave graph) and an antinode and another are (270-90=)180 degrees out of phase with each other (1.2 of the sine wave graph).

[halosh92]

I think you kinda lengthened the whole story...just to add my input (in case it helps), the distance between two consecutive nodes or two consecutive antinodes is lamda/2. See this maxima and minima as a waveform with displacement (y-axis) against distance (x-axis); the minima is the node and maxima the antinode. The antinode can be on either side of the x-axis, because if you remember displacement is taken from a reference point. In case of waves the reference point is the mean position (the node) and the maximum displacement from this mean position is an antinode. If the antinode is a point 2 cm from the mean position, the least displacement is NOT -2 but 0. Hence, like halosh said, simply multiply 32.4 by 2 and you have the lamda.

In times of a sine wave, a node and antinode are 90 degrees out of phase with each other (1/4 of the sine wave graph) and an antinode and another are (270-90=)180 degrees out of phase with each other (1.2 of the sine wave graph).

hahah yes my mother always said i talk alot 

[cashem'up]

hey guys one question i had.... if i kept a filament in constant temperature lets say in Ice then would its IV graph be a straight line through origin