Physics: Post your doubts here!

unseen95 · Mar 16, 2013

Nazib Intesar said:
Can someone please tell me why the answer is A:-
View attachment 22034

use the formula, Driving force - resistance = mass * acceleration
driving force = component of weight downwards = 2 * 9.8 * sin theta - 5 = 2 * a
theta is given by assuming that the whole figure is a right angled triangle, with theta = sin inverse (3/7)
you get a from here which is 1.7 m per sec square
u = 0
a = 1.7 m per sec square
s = 7 m
using the formula,
v square = u square + 2as
v = 4.9 m per sec.
If it helps like this post

littlecloud11 · Mar 16, 2013

unseen95 said:
please please help me with b(i) and (ii), i was not able to draw both figures in the graph as well as on the diagram. I could not draw using the guidelines given in the marking scheme., so i would be thankful if anyone drew them for me.
The questions are uploaded.

This is how the graph should look.

Output of the comparator = voltage at non-inverting input (V+) - voltage at inverting input (V-)
Whenever the value of (V+) - (V-) is positive the output voltage is +5, when when the value of (V+) - (V-) is -ve the output voltage is -5. There is a change over between +5 and -5 when (V+) = (V-), in this case when both voltages are 2V.

And the diodes should be connected this way-

You can see from the graph of the output that the output is positive for a longer time than it is negative. The question said the red diode remained on for a longer period of time. So, it must conduct when the output is positive, that's why it's connected pointing downwards. The reverse is true for the green diode. It conducts when the output of the comparator is negative, which is for a shorter period of time, hence it points upwards.

unseen95 · Mar 16, 2013

littlecloud11 said:
This is how the graph should look.
View attachment 22052

Output of the comparator = voltage at non-inverting input (V+) - voltage at inverting input (V-)
Whenever the value of (V+) - (V-) is positive the output voltage is +5, when when the value of (V+) - (V-) is -ve the output voltage is -5. There is a change over between +5 and -5 when (V+) = (V-), in this case when both voltages are 2V.

And the diodes should be connected this way-
View attachment 22053

You can see from the graph of the output that the output is positive for a longer time than it is negative. The question said the red diode remained on for a longer period of time. So, it must conduct when the output is positive, that's why it's connected pointing downwards. The reverse is true for the green diode. It conducts when the output of the comparator is negative, which is for a shorter period of time, hence it points upwards.

thank you very much

Heisenberg · Mar 16, 2013

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_41.pdf

Help needed on Q5, part (b)! Why, when calculating the charge on plate X, we use the effective capacitance and not the actual capacitance of X? Then when calculating the charge on Y, we then use the actual capacitance of Y, and a voltage of 3? This is all very confusing

. Help would be greatly appreciated!

Norton · Mar 16, 2013

Help needed on questions 4.)......5.)........6.)........21.)..............24.).......29.)
i need da concept and some explanation PLEASE>>>PLEASE!!!!!
thnk you!

Nab900 · Mar 16, 2013

Assalam O Alaikum
Does anyone have A2 physics Application important notes/points download link???

Ahmedraza73 · Mar 16, 2013

Norton said:
Help needed on questions 4.)......5.)........6.)........21.)..............24.).......29.)
i need da concept and some explanation PLEASE>>>PLEASE!!!!!
thnk you!

Q:6
the area under the graph is triangle so :1/2*b*h
1/2*5*20=50 m
so the answer is C

Q:24 Only transverse wave can be polarized because polarization is said to be occur when oscillation are in one direction in a plane.so the answer will be D

Q:29 E=V/D,
V =9.0
D=4.0*10^-3
PUTTING INTO EQUATION WE GET 9.0/4.0*10^-3=2.3*10^3 NC^-1
so the answer is D

Q:4 UNCERTANITIES ALWAYS ADDED SO IT WILL BE +_0.05mm
so D will be the answer

Norton · Mar 16, 2013

Ahmedraza73 said:
Q:6
the area under the graph is triangle so :1/2*b*h
1/2*5*20=50 m
so the answer is C

Q:24 Only transverse wave can be polarized because polarization is said to be occur when oscillation are in one direction in a plane.so the answer will be D

Q:29 E=V/D,
V =9.0
D=4.0*10^-3
PUTTING INTO EQUATION WE GET 9.0/4.0*10^-3=2.3*10^3 NC^-1
so the answer is D

Q:4 UNCERTANITIES ALWAYS ADDED SO IT WILL BE +_0.05mm
so D will be the answer

thanx alot!!!!!!!!!!!

Ahmedraza73 · Mar 16, 2013

Norton said:
thanx alot!!!!!!!!!!!

Your Welcome

Yousif Mukkhtar · Mar 16, 2013

Can anyone explain q 14
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf

Light Yagami · Mar 17, 2013

Why is the answer C:-

Yousif Mukkhtar · Mar 17, 2013

Light Yagami said:
Why is the answer C:-
View attachment 22111

First consider the forces acting on the pivot:
1. Tension in string making an anticlockwise moment
2. Weight of string at 50 cm mark (it is a uniform scale therefore force the centre of gravity is exactly halfway)-making clock wise moment
3. Weight of the 50 g mass to be added making a clockwise moment
Thereforce
Clockwise moment=Anticlock wise moment
50 x09.81 xd + 10 x 100 x 9.81= 60 x 20 x 9.81
Complete that and you will get d as 4 cm
Add 4 to 40 you get 40 cms

Ahmedraza73 · Mar 17, 2013

Yousif Mukkhtar said:
Can anyone explain q 14
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P is acting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B

Light Yagami · Mar 17, 2013

Can someone please explain why options A, B and C are wrong and option D is correct:-

littlecloud11 · Mar 17, 2013

Heisenberg said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_41.pdf

Help needed on Q5, part (b)! Why, when calculating the charge on plate X, we use the effective capacitance and not the actual capacitance of X? Then when calculating the charge on Y, we then use the actual capacitance of Y, and a voltage of 3? This is all very confusing . Help would be greatly appreciated!

You cannot use the actual capacitance of X for the first part because you don't know the don't know the potential difference across X. So it is easier to use the effective capacitance and the total potential difference. Otherwise you can also do it by the longer method if you want to use the actual capacitance.
First calculate the potential difference cross X.
Total resistance in series is 12 micro farad and 24 micro farad from the parallel combination. So, potential across X=
(capacitance of X/ total capacitance in series) * 9
(1/12)/(1/12+1/24) * 9 = 6v
Now use Q =CV for X
Q = 12 micro farad * 6 =7.2* 10^-5 =72 micro C
You get the same answer, just a longer calculation.

For Y you can use the actual capacitance because you already know the potential difference cross Y from your previous calculations.

littlecloud11 · Mar 17, 2013

Light Yagami said:
Can someone please explain why options A, B and C are wrong and option D is correct:-
View attachment 22137

A is wrong because the x-axis is labeled as 'x' which means displacement but the wave has time period labeled on it. B is wrong because amplitude 'a' is the maximum displacement from rest position (0), not the distance between the max and min point. C is wrong because the x-axis is labeled as time period 't' but the wave shows the label of lambda for the horizontal distance. D is correct as the label of 'T' agrees with the x-axis representation and the amplitude is labeled correctly.

Light Yagami · Mar 17, 2013

Why is the answer A:-

littlecloud11 · Mar 17, 2013

Light Yagami said:
Why is the answer A:-
View attachment 22139

I= a^2
so, a^2 = P/ 4pir^2
P, 4 and pi r constant so a^2 = 1/r^2, remove root from both sides and a=1/r

Light Yagami · Mar 17, 2013

Why is the answer D:-

snowbrood · Mar 17, 2013

Light Yagami said:
Why is the answer D:-
View attachment 22147

as we know that frequency of fundamental is f
frequency of second harmonic is 3f
and that of third harmonic is 5f
u have to memorize it
page no 237 chemistry coursebook cie

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