Physics: Post your doubts here!

[bhagwan dhakal]

can i get help for q no 5 iv of 9702/23/m/j 16

 

[Clark20]

Can anyone explain this question?

 

[Garry Soloan]

Does anyone have notes for Communications Systems?

 

[Mushraf altaf]

can somebody pls help me put with this??

 

[Mushraf altaf]

 

[Ebrahim12]

View attachment 64541 can
somebody pls help me put with this??

(i)
ratio of input to output = (64)(0.21) dB
ratio of output to noise = 25 dB

ratio of input to noise = 25 + 13.44 = 26.88 dB
Normally you would multiply ratios to combine them, but since the decibel is in a logarithmic scale you add the ratios.

(ii)
26.88 = 10 log(P/9.2μ)
P = 4.5x10^-3

 

[Ebrahim12]

View attachment 64542

The frequencies in the graph are F-f, F, F+f
where F is the freq of carrier wave and f is the freq of the signal wave

(i)
F = 909kHz
lambda = 330 m

(ii) and (iii)

Bandwidth is the range of frequencies used for transmission
In general, bandwidth = (F+f) - (F-f) = 2f

Bandwidth = 918 - 900 = 18 kHz
f = 9000 Hz

 

[Jane_Kim]

Can anyone help me with this question? Pls...

 

[Jane_Kim]

oh thanks. It helped me. What about this question???

 

[Donenr]

Hello, I would like about Q22 in the 9702/11/m/j/18 I want to ask for Why is the question B and not A?

Here is the examiner's report on the question

 

[PlanetMaster]

Hello, I would like about Q22 in the 9702/11/m/j/18 I want to ask for Why is the question B and not A?
View attachment 64548
Here is the examiner's report on the question
View attachment 64550

Hi Donenr,

Welcome to XtremePapers Community.

Let us look at the air molecules in their undisturbed positions and compare them as a sound wave passes by.



If we plot displacement on the y-axis and distance on the x-axis, we get a sine wave.

If we plot a displacement-time graph for a single particle we see:



This is true whether we have a longitudinal or transverse wave.

Hope this answers your question.

 

[Omkar Kerkar]

From O/N 2017 paper 11
The circuit diagram shows four resistors of different resistances P,Q,R and S connected to a battery. (SEE IMAGE) The voltmeter reading is zero. Which equation is correct?
A) P - Q = R - S
B) P - S = Q - R
C) PQ = RS
D) PS = QR

THANKS FOR THE HELP IN ADVANCE

 

[PlanetMaster]

View attachment 64586

From O/N 2017 paper 11
The circuit diagram shows four resistors of different resistances P,Q,R and S connected to a battery. (SEE IMAGE) The voltmeter reading is zero. Which equation is correct?
A) P - Q = R - S
B) P - S = Q - R
C) PQ = RS
D) PS = QR

THANKS FOR THE HELP IN ADVANCE

Since V=0 therefore V1=V2.

Since PQ resistor system is essentially a potential divider, we have for V1:



and similarly for V2:



Since V=0 therefore V1=V2. Substituting V1 and V2 gives us:



And equating these gives us:



Hope this helps.

 

[Omkar Kerkar]

View attachment 64587

Since V=0 therefore V1=V2.

Since PQ resistor system is essentially a potential divider, we have for V1:

View attachment 64588

and similarly for V2:

View attachment 64591

Since V=0 therefore V1=V2. Substituting V1 and V2 gives us:

View attachment 64592

And equating these gives us:

View attachment 64593

Hope this helps.

RIGHT! MAkes sense, tysm

 

[MShaheerUddin]

https://www.google.com/url?sa=t&sou...FjAAegQIARAB&usg=AOvVaw39LCX2gD7-Xn14i_ibN8k0

Please help me understand Q1 part ii
Why did we take potential -4.4?

 

[hellodjfos;s'ff]

https://www.google.com/url?sa=t&sou...FjAAegQIARAB&usg=AOvVaw39LCX2gD7-Xn14i_ibN8k0

Please help me understand Q1 part ii
Why did we take potential -4.4?

You have either posted the wrong link or posted in the wrong thread.

Edit: Regardless of that, I assume you are asking about s16 qp41 q1b(ii) - physics.
If that's the case, in part b(ii) we are trying to find the minimum speed required for the rock to reach star B, so the rock must reach the point where the gravitational attraction is zero. This is the point where the forces of attraction on the rock by stars A and B are equal and after this point, the rock will get pulled by star B. If you observe the graph, you will notice that the point where the potential begins to change direction is near x=0.52 or 0.54. At this point the potential is -4.4, so if you find the change in GPE between potential=-10 and potential=-4.4, you will get the change in GPE needed to reach star B. And since change in GPE=change in KE, you can now find the velocity.

 

[MShaheerUddin]

You have either posted the wrong link or posted in the wrong thread.

Edit: Regardless of that, I assume you are asking about s16 qp41 q1b(ii) - physics.
If that's the case, in part b(ii) we are trying to find the minimum speed required for the rock to reach star B, so the rock must reach the point where the gravitational attraction is zero. This is the point where the forces of attraction on the rock by stars A and B are equal and after this point, the rock will get pulled by star B. If you observe the graph, you will notice that the point where the potential begins to change direction is near x=0.52 or 0.54. At this point the potential is -4.4, so if you find the change in GPE between potential=-10 and potential=-4.4, you will get the change in GPE needed to reach star B. And since change in GPE=change in KE, you can now find the velocity.

Thanks bro but still i am not clear that why for minimum velocity it has to reach to point where gravitational attraction is zero

 

[PlanetMaster]

Thanks bro but still i am not clear that why for minimum velocity it has to reach to point where gravitational attraction is zero

Please provide a link to the question. The link you provided earlier seems to be for a different subject.

 

[hellodjfos;s'ff]

Thanks bro but still i am not clear that why for minimum velocity it has to reach to point where gravitational attraction is zero

Star A Rock Zero GPE point Star B
I'll assume that above are the positions of the objects. So Between Star A and zero GPE point, the star A is trying to pull the rock towards itself so the rock must have enough KE to reach zero GPE point for it to leave star A's gravitational attraction. After zero GPE point, the star B will automatically try to pull the rock towards itself so the rock can even be stationary after zero GPE point and it will reach star B. So we need to find the change in GPE from Star A to zero GPE point because that is the change in KE that will take place for the rock to come within reach of star B's gravitational pull.

 

[MShaheerUddin]

Star A Rock Zero GPE point Star B
I'll assume that above are the positions of the objects. So Between Star A and zero GPE point, the star A is trying to pull the rock towards itself so the rock must have enough KE to reach zero GPE point for it to leave star A's gravitational attraction. After zero GPE point, the star B will automatically try to pull the rock towards itself so the rock can even be stationary after zero GPE point and it will reach star B. So we need to find the change in GPE from Star A to zero GPE point because that is the change in KE that will take place for the rock to come within reach of star B's gravitational pull.

Thanks a lot