ANY DOUBTS HERE!!!

[The SMA]

Optical isomer are possible only in compounds containing a chiral carbon

So it's not possible in the first one.

Here's how the second one would look

nid is there a step/rule on arranging the atoms/groups? whether CH3- group or something else
should be on top the chiral Carbon atom [im taking the second problem H3C-CBr(COOH)-CH2Br as e.g ]

most of my answers for the question of this type does not seem to be similar to the Marking Scheme.
They put something different on top or at any other place. If i don't get the same arrangement as MS
will i be losing marks? :S

[nid404]

nid is there a step/rule on arranging the atoms/groups? whether CH3- group or something else
should be on top the chiral Carbon atom [im taking the second problem H3C-CBr(COOH)-CH2Br as e.g ]

most of my answers for the question of this type does not seem to be similar to the Marking Scheme.
They put something different on top or at any other place. If i don't get the same arrangement as MS
will i be losing marks? :S

I'm really not getting you :/ Rule in arranging? You simply reflect the compound...like you reflect something in a mirror.

the arrangement doesn't matter in this particular case...just that COOH and Br are side chains.

[The SMA]

oh okay. thanks nid! ^^

[halosh92]

physics cie AS
p2 june 2009
variant 1
Q2c, d
for 'c" shouldnt it be -1.6
and for "d"......wats the formula for speed of separation and speed of approach? and do we take the signs into account?

[Meticulous]

Please post the question paper here halosh

[halosh92]

Please post the question paper here halosh

sure
and could u do these as well plz same paper:
3bi .....why is the angle 90 ?
5b
7aii,7aiii

2nd variant:
Q7a
thankyouuuuuuu

[Meticulous]

physics cie AS
p2 june 2009
variant 1
Q2c, d
for 'c" shouldnt it be -1.6
and for "d"......wats the formula for speed of separation and speed of approach? and do we take the signs into account?

c) it will move in the initial direction, which is positive! u see, the MAGNITUDE of force is the same in both, but the direction of force and impulse varies.

d) equation is: u₁ - u₂ = v₂ - v₁ = -(v₁ -v₂  )and yes you do take signs into account

[halosh92]

c) it will move in the initial direction, which is positive! u see, the MAGNITUDE of force is the same in both, but the direction of force and impulse varies.

d) equation is: u₁ - u₂ = v₂ - v₁ = -(v₁ -v₂  )and yes you do take signs into account

thankyou
and could u plz solve the rest?
 

[nid404]

sure
and could u do these as well plz same paper:
3bi .....why is the angle 90 ?
5b
7aii,7aiii

3) b) max value of the function cosx is 1
cos 90=1

max turning effect when theta=90 

5b) there is a formula u need to know

phase difference=( 2pi/ lamda)  X path difference

phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi  X path difference.
path difference is S2M - S1M 
find S2M using Pythagoras = root of 802 + 1002
                                                   =128
path difference= 128-100=28
wavelength= 2pi/pi X 28
               =56cm                 
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz   wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas

7)a)
ii) R between BX implies combined resistance of two parallel resistors in the loop BYCX = R/2
iii) AZ will be the combined resistance of all.
R in BY and CX are paralle...their combined value is R/2  this is in series with 2 more resistors.
So combined resistance = R/2 + 2R = 2.5R

[halosh92]

3) b) max value of the function cosx is 1
cos 90=1

max turning effect when theta=90 

5b) there is a formula u need to know

phase difference=( 2pi/ lamda)  X path difference

phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi  X path difference.
path difference is S2M - S1M 
find S2M using Pythagoras = root of 802 + 1002
                                                   =128
path difference= 128-100=28
wavelength= 2pi/pi X 28
               =56cm                 
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz   wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas

7)a)
ii) R between BX implies combined resistance of two parallel resistors in the loop BYCX = R/2
iii) AZ will be the combined resistance of all.
R in BY and CX are paralle...their combined value is R/2  this is in series with 2 more resistors.
So combined resistance = R/2 + 2R = 2.5R

thankyou so much !!! but for the 5b........wat do we do if its constructive interference? and when do we exactly use that formula u gave, when we have to combine phase difference and path difference??
thxxx
and could u do the 2nd variant of the paper:
Q7a (with details) thxxxx alooot

[nid404]

thankyou so much !!! but for the 5b........wat do we do if its constructive interference? and when do we exactly use that formula u gave, when we have to combine phase difference and path difference??
thxxx
and could u do the 2nd variant of the paper:
Q7a (with details) thxxxx alooot

when it's constructive interference phase difference would be 0/2pi/4pi...etc.
We use the formula when you are given 2 of the values either lambda,path or phase difference and asked to find the other.

yup hold on for a while 

[nid404]

2nd variant:
Q7a
thankyouuuuuuu

7) a

Both are open. No current flows. so max resistance. R=  R=V/I I=0 anything divided by zero is not defined actually but ms says infinity...so we stick to that
S1 open S2 closed. Current will pass only through the wire with S2. So resistance will be 2R (cause there are 2 resistors in series)
When both are close. R= combines resistance of 2R in parallel with each other...so combined resistance is R

[halosh92]

7) a

Both are open. No current flows. so max resistance. R=  R=V/I I=0 anything divided by zero is not defined actually but ms says infinity...so we stick to that
S1 open S2 closed. Current will pass only through the wire with S2. So resistance will be 2R (cause there are 2 resistors in series)
When both are close. R= combines resistance of 2R in parallel with each other...so combined resistance is R

thankyou *bows*

[nid404]

thankyou *bows*

lol..not a problem

[halosh92]

same paper 2nd variant
Q2c
isnt change in momentum= m(v-u)
so isnt it supposed to be = 0.78(-9-4.2)
= -10.3 ??