ALL CIE PHYSICS DOUBTS HERE !!!

[Deadly_king]

Refer to the attachments
Look at the paper, for the full question. [Question no. 4]
Thanks in advance.

I assume that you got the answers for part (a)(ii).

You should then use the two answers namely R = L/A and E = WL/Ae.

As you can note, you have been given R = 0.44, = 9.2 x 10^-8, W = 34 and e = 7.7 x 10^-4 and you're asked to find the young modulus E.

But according to the equation you cannot find it since you don't have L and A. However you can use the 1st equation to find it.

1st equation : R = L/A can also be written as R/ = L/A

Now you replace it in the second equation.

E = W/e x L/A. Since L/A = R/, this implies E = W/e x R/ which therefore results in E = WR/e

Hope it helps

[Deadly_king]

W09 P22 Q5b

Thank you

Nov 09 p22

(b) There is no calculation here.

You should have guessed that by the number of marks allocated for the question.

It said state; which means it's pretty obvious from the drawing.

First thing : You should realize that both X and Y are equidistant from a node. Therefore the phase difference should be pie rad or 180^o.

Furthermore it can also be noted that X and Y are in antiphase ( One is above while the other one is below). If Y was 1/8L to the right of P, then Y would have been at a phase difference of 180^o but in phase.

Hope it helps

[TJ-56]

Can sumone pls explain s10 P2 Q7 (b) (ii)
thank you

[$!$RatJumper$!$]

Nov 09 p22

(b) There is no calculation here.

You should have guessed that by the number of marks allocated for the question.

It said state; which means it's pretty obvious from the drawing.

First thing : You should realize that both X and Y are equidistant from a node. Therefore the phase difference should be pie rad or 180^o.

Furthermore it can also be noted that X and Y are in antiphase ( One is above while the other one is below). If Y was 1/8L to the right of P, then Y would have been at a phase difference of 180^o but in phase.

Hope it helps

Hmm im still confused. What i thought was that since X was 1/8th of the wavelength away from the start and Y was 1/8th of the wavelength away from the end, they would each be 45 degrees away from the start and end. Thus 360-(45*2) = 270 degrees.

So what if point X was at the end to the left (by the oscillator) and point Y was at the end to the right (by P), what would the phase diff be then?

[Dania]

I assume that you got the answers for part (a)(ii).

You should then use the two answers namely R = L/A and E = WL/Ae.

As you can note, you have been given R = 0.44, = 9.2 x 10^-8, W = 34 and e = 7.7 x 10^-4 and you're asked to find the young modulus E.

But according to the equation you cannot find it since you don't have L and A. However you can use the 1st equation to find it.

1st equation : R = L/A can also be written as R/ = L/A

Now you replace it in the second equation.

E = W/e x L/A. Since L/A = R/, this implies E = W/e x R/ which therefore results in E = WR/e

Hope it helps

Thanks! Got it.

[Deadly_king]

Hmm im still confused. What i thought was that since X was 1/8th of the wavelength away from the start and Y was 1/8th of the wavelength away from the end, they would each be 45 degrees away from the start and end. Thus 360-(45*2) = 270 degrees.

So what if point X was at the end to the left (by the oscillator) and point Y was at the end to the right (by P), what would the phase diff be then?

Ooh..........i understand what you did. But we don't measure phase difference like this.

If the wave is placed on a graph paper, then the x-axis would be length and not angle. This is why you can't use length as angle.

Phase difference is described by the positions of the wave at the respective points. Whether at X or at Y, the waves undergo similar but opposite motion. This is why they are said to be in anti-phase.

For your question, The phase angle would still be 180^o since even then both points will undergo similar motions but will be in phase as they are not opposite.

[$!$RatJumper$!$]

Ooh..........i understand what you did. But we don't measure phase difference like this.

If the wave is placed on a graph paper, then the x-axis would be length and not angle. This is why you can't use length as angle.

Phase difference is described by the positions of the wave at the respective points. Whether at X or at Y, the waves undergo similar but opposite motion. This is why they are said to be in anti-phase.

For your question, The phase angle would still be 180^o since even then both points will undergo similar motions but will be in phase as they are not opposite.

Alright i get it now thank you +rep

Can you please do S09 P21 Q5b

Thanks

[Deadly_king]

Can sumone pls explain s10 P2 Q7 (b) (ii)
thank you

7.(b)(ii)

1. 1.1 MeV is supplied in the form of kinetic energy of the alpha-particle.

2. First you should convert 1.1 MeV into joules. => K.E = (1.1 x 10⁶) x (1.6 x 10^-19) = 1.76 x 10^-13 J

Now you equate K.E = 0.5mv² = 1.76 x 10^-13

Mass of an alpha-particle is 4u = 4(1.67 x 10^-27)kg

Hence speed of alpha-particle is found to be the square root of 2(1.76 x 10^-13)/4(1.67 x 10^-27)

v = 7.3 x 10⁶ ms^-1

Hope it helps

[Dania]

Help! Please.
The whole question.

[$!$RatJumper$!$]

Help! Please.
The whole question.

Haha looks like the same question is getting to all of us I also need help on that

[TJ-56]

Haha looks like the same question is getting to all of us I also need help on that

5(a)
1. The waves should meet in antiphase at M. ( phase difference = pie)
2. Sources must emit waves having same amplitude.

(b)
So the wavelength ranges from 330/4000  to 330/1000 (v/f) which is 8.25 to 33 cm
The distance from S2 to M is 128 (root(100sqr +80sqr))
So the phase difference is 128-100 which is 28
using the equation
phi/2pi = AB/lambda
as phi must be pi rad, 1/2 = AB/lambda
AB= 28
so making lambda the subject: lambda= (28*2)/n , where n could be 1,3,5,7 (to be out of phase)
so trying out 1,3,5,7 give us 56, 18.7, 11.2, and 8 respectively, as 18.7 and 11.2 are the only 2 values inside the range i calculated at the beginning, then the number is 2
hope that was helpful

[TJ-56]

7.(b)(ii)

1. 1.1 MeV is supplied in the form of kinetic energy of the alpha-particle.

2. First you should convert 1.1 MeV into joules. => K.E = (1.1 x 10⁶) x (1.6 x 10^-19) = 1.76 x 10^-13 J

Now you equate K.E = 0.5mv² = 1.76 x 10^-13

Mass of an alpha-particle is 4u = 4(1.67 x 10^-27)kg

Hence speed of alpha-particle is found to be the square root of 2(1.76 x 10^-13)/4(1.67 x 10^-27)

v = 7.3 x 10⁶ ms^-1

Hope it helps

Thanx for the answer +rep
But can u explain further part 1 of the question
thanx in advance

[thecandydoll]

O/N 2002 Q3 C

[ashish]

O/N 2002 Q3 C

according to me the time in contact is 0.15 ( from the graph)
 F= change in momentum/time
= 0.35/0.15
=2.33N

[$!$RatJumper$!$]

5 (b)

So the wavelength ranges from 330/4000  to 330/1000 (v/f) which is 8.25 to 33 cm
The distance from S2 to M is 128 (root(100sqr +80sqr))
So the phase difference is 128-100 which is 28
using the equation
phi/2pi = AB/lambda
as phi must be pi rad, 1/2 = AB/lambda
AB= 28
so making lambda the subject: lambda= (28*2)/n , where n could be 1,3,5,7 (to be out of phase)
so trying out 1,3,5,7 give us 56, 18.7, 11.2, and 8 respectively, as 18.7 and 11.2 are the only 2 values inside the range i calculated at the beginning, then the number is 2
hope that was helpful

What is this equation?? I've never heard about it. What is it used to calculate and when should we use it?

By the way, whats phi?

phi/2pi = AB/lambda

Thankx