Author Topic: ALL CIE PHYSICS DOUBTS HERE !!!  (Read 151697 times)

elemis

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #420 on: December 05, 2010, 10:46:04 am »
Why is the answer B and not A ?

The vector triangles for both options are in sequence so whats the deciding factor that makes B the right answer ?

Offline ashish

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #421 on: December 05, 2010, 12:10:59 pm »
Why is the answer B and not A ?

The vector triangles for both options are in sequence so whats the deciding factor that makes B the right answer ?

the answer is B , hmm extend all the line of forces and you will notice that the lines meet at a single place.


elemis

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #422 on: December 05, 2010, 12:49:29 pm »
the answer is B , hmm extend all the line of forces and you will notice that the lines meet at a single place.

Fine, I agree they do meet, but what does that indicate ?

Offline ashish

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #423 on: December 05, 2010, 12:58:03 pm »
Fine, I agree they do meet, but what does that indicate ?

oops i forgot to mention that in my last post ,, when all these lines of forces meet at a single point the means that the body is in equilibrium

elemis

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #424 on: December 05, 2010, 01:14:09 pm »
oops i forgot to mention that in my last post ,, when all these lines of forces meet at a single point the means that the body is in equilibrium

Splendid. Thanks :)

Offline ashish

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #425 on: December 05, 2010, 01:20:28 pm »
Splendid. Thanks :)

you are welcome ..

Offline Monopoly

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #426 on: December 08, 2010, 04:44:36 pm »
this question is from the International AS/A level Physics book, pg - 428

the question:

a sinusoidal carrier wave has a frequency of 750kHz and an unmodulated amplitude of 4.0V. the carrier wave is to be amplitude-modulated by a sinusoidal signal of frequency 3kHz and the amplitude 0.5V. describe the modulated carrier wave.

now the answer at the end of the book says the amplitude varies from 3.5V to 4.0V. Can someone explain why this is so.

elemis

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #427 on: December 12, 2010, 02:25:05 pm »
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w05_qp_2.pdf

Question 7 (b) (ii)

To be honest I dont understand what the voltmeter is measuring i.e. I dont know across what the voltmeter is measuring a voltage.

Can someone please explain both parts in (ii) ?

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #428 on: December 13, 2010, 08:40:16 am »
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w05_qp_2.pdf

Question 7 (b) (ii)

To be honest I dont understand what the voltmeter is measuring i.e. I dont know across what the voltmeter is measuring a voltage.

Can someone please explain both parts in (ii) ?

The voltmeter is measuring the potential difference across the thermistor. When the voltmeter shows zero deflection, it means that the p.d across the thermistor is equal to the p.d across wire MQ. This is why the potential difference is zero as shown by the voltmeter.

7.(b) (ii)

1. Potential difference = 2.70 V -----> Part (a) describes it.

2. Resistance of wire is proportional to both the p.d(V) across wire and the length(L) of the wire.

V = IR -----> R is proportional to V
R = eL/A ----> R is proportional to L.

Hence we can deduce that V is also proportional to L.

This implies that V1/L1 = V2L2 -----> 4.5/100 = 2.7/L2 ----> L2 = 60.0 cm

Hope it helps :)

elemis

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #429 on: December 13, 2010, 08:49:22 am »
I understood everything you explained and thank you for that, but I still dont get how the voltmeter is measuring the p.d. across the thermistor.

Offline Deadly_king

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #430 on: December 13, 2010, 09:56:11 am »
I understood everything you explained and thank you for that, but I still dont get how the voltmeter is measuring the p.d. across the thermistor.

Just a mere precision. ;)

It's not only measuring the p.d across the thermistor but rather the p.d across the circuit BAQM.

When it shows zero deflection ----> p.d across BA = p.d across QM





Offline $H00t!N& $t@r

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #431 on: December 14, 2010, 07:41:52 pm »
can some1 please help me with  q9 and 10 - O/N 2009 P1 variant 11
"If A equals success, then the formula is:
A=X+Y+Z, X is work. Y is play. Z is keep your mouth shut"- Albert Einstein

Freaked12

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #432 on: December 14, 2010, 08:48:41 pm »
I cant remember the formula tbh

as 2m has a greater mass,it will exert more force on the first
sphere
hence the spheres velocity will increse due to collision with a heavier object
and it will move in an opposite direction.

put the values of initial mass and velocity into vf1=vf1= ((m1-m2)/(m1+m2))(v1i)+ (2m2/(mi+m2)vi2

                       u will get the answer
« Last Edit: December 14, 2010, 08:58:50 pm by Requiem~Katie~Paranormal activity »

elemis

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #433 on: December 15, 2010, 03:28:30 am »
can some1 please help me with  q9 and 10 - O/N 2009 P1 variant 11

You dont need to do what requiem said.

A simpler method is to consider the velocities :

U1 - U2 = V2 - V1

Where U = initial velocity before collision
          V = final velocity after collision.

For an elastic collision the left hand side of the above equation should be equal to the right hand side.

Plug in the necessary values and remember velocity is a VECTOR.... you should get the right answer ;)

elemis

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #434 on: December 15, 2010, 03:34:47 am »
Question 10

Consider the picture below.

Using Impulse = Force * time

Impulse = 60*0.5

Impulse = 30Ns

Therefore, the change in momentum of the trolley should be equal to 30Ns.

Hence, using m(v-u) = change in momentum

We get : 30(-v+3) = 30

-v = -2

Hence, v = 2 ms-1