Physics: Post your doubts here!

[The Sarcastic Retard]

QUESTION: An optic fibre has a signal attenuation of 0.20 dB km–1. The input signal to the optic fibre has a power of 26 mW. The receiver at the output of the fibre has a noise power of 6.5 µW. Calculate the maximum uninterrupted length of optic fibre given that the signal-to-noise ratio at the receiver must not be less than 30 dB

ANSWER SCHEME : if P is power at receiver, 30 = 10lg(P / (6.5 × 10–6) C1 P = 6.5 × 10–3 W
loss along cable = 10lg({26 × 10–3} / {6.5 × 10-3})
= 6.0 dB
length = 6.0 / 0.2 = 30 km

I dont understand how to approach this whole problem :/ its from s08 qp 4 of cie a level physics, help will be very very much appreciated.

Which chapter is this?

 

[Salonee B]

Which chapter is this?

It is communicating information

 

[Konstantino Nikolas]

QUESTION: An optic fibre has a signal attenuation of 0.20 dB km–1. The input signal to the optic fibre has a power of 26 mW. The receiver at the output of the fibre has a noise power of 6.5 µW. Calculate the maximum uninterrupted length of optic fibre given that the signal-to-noise ratio at the receiver must not be less than 30 dB

ANSWER SCHEME : if P is power at receiver, 30 = 10lg(P / (6.5 × 10–6) C1 P = 6.5 × 10–3 W
loss along cable = 10lg({26 × 10–3} / {6.5 × 10-3})
= 6.0 dB
length = 6.0 / 0.2 = 30 km

I dont understand how to approach this whole problem :/ its from s08 qp 4 of cie a level physics, help will be very very much appreciated.

Given that signal-to-noise ratio should be no less than 30, we equate it to 30 to get the minimum signal power required to satisfy the condition.

signal-to-noise ratio = 10lg(signal power/noise power)
10lg(P/6.5*10^-3) = 30
--> P = 6.5*10^-3 [This is the minimum signal power required to keep signal-to-noise ratio >/= 30]

We now have to calculate the attenuation with this signal power.

attenuation = 10lg(input power of signal/ output power)
attenuation = 10lg{(26*10^-3)/(6.5*10^-3)} = 6.0205999 dB

We are given the attenuation/unit length = 0.2 dBkm^-1

attenuation/unit length = total attenuation/ length of signal travelled

By substituting the minimum value of signal power, we get the maximum attenuation. This when substituted in attenuation/unit length formula, will give us the maximum length required to satisfy the condition of signal-to-noise ratio >/= 30 dB.

0.2 = 6.0205999 / max length
--> max. length = 30.103 = 30km

Hope you can understand that explanation.

 

[Salonee B]

Given that signal-to-noise ratio should be no less than 30, we equate it to 30 to get the minimum signal power required to satisfy the condition.

signal-to-noise ratio = 10lg(signal power/noise power)
10lg(P/6.5*10^-3) = 30
--> P = 6.5*10^-3 [This is the minimum signal power required to keep signal-to-noise ratio >/= 30]

We now have to calculate the attenuation with this signal power.

attenuation = 10lg(input power of signal/ output power)
attenuation = 10lg{(26*10^-3)/(6.5*10^-3)} = 6.0205999 dB

We are given the attenuation/unit length = 0.2 dBkm^-1

attenuation/unit length = total attenuation/ length of signal travelled

By substituting the minimum value of signal power, we get the maximum attenuation. This when substituted in attenuation/unit length formula, will give us the maximum length required to satisfy the condition of signal-to-noise ratio >/= 30 dB.

0.2 = 6.0205999 / max length
--> max. length = 30.103 = 30km

Hope you can understand that explanation.

Given that signal-to-noise ratio should be no less than 30, we equate it to 30 to get the minimum signal power required to satisfy the condition.

signal-to-noise ratio = 10lg(signal power/noise power)
10lg(P/6.5*10^-3) = 30
--> P = 6.5*10^-3 [This is the minimum signal power required to keep signal-to-noise ratio >/= 30]

We now have to calculate the attenuation with this signal power.

attenuation = 10lg(input power of signal/ output power)
attenuation = 10lg{(26*10^-3)/(6.5*10^-3)} = 6.0205999 dB

We are given the attenuation/unit length = 0.2 dBkm^-1

attenuation/unit length = total attenuation/ length of signal travelled

By substituting the minimum value of signal power, we get the maximum attenuation. This when substituted in attenuation/unit length formula, will give us the maximum length required to satisfy the condition of signal-to-noise ratio >/= 30 dB.

0.2 = 6.0205999 / max length
--> max. length = 30.103 = 30km

Hope you can understand that explanation.

I cannot thank you enough for this explanation!! I finally understand it, god bless you

 

[Konstantino Nikolas]

I cannot thank you enough for this explanation!! I finally understand it, god bless you

You're welcome and thanks.

 

[Pixie Twinkles]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
Question 7 .. how is the correct answer D?

 

[Konstantino Nikolas]

View attachment 59283

If you resolve the tension on both strings in the horizontal direction and take their sum, you get the resultant force acting toward P.

---> [4*(3/5)]+[4*(3/5)] = 4.8 N

Additionally, the vertical components of both tensions is equal and opposite so it is cancelled out.

 

[The Sarcastic Retard]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_42.pdf
Q4b and c detail explanation needed.

 

[ashcull14]

please elaborate the steps...i didnt get the markshceme

 

[ashcull14]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_42.pdf
Q4b and c detail explanation needed.

Q4b= V=kQ/r .. Q is same at P and Q (6.4*10^-19 C) AND r is same therefore, V at P and Q is same ..so ∆V = 0...as work done = Q*∆V ,it is zer0
Q4c= for finding the work done we need to know the potential difference between the midpoint and point P
potential at any point is the sum of potentials for each charge
potential at mid point would be between A and B taken from the midpoint
using V= kQ/r and k = 4πε0
(potential of A from midpoint) k*6.4*10^-19/ 6*10^-6 + (potential of B from midpoint) k*6.4*10^-19 / 6*10^-6
potential at point P would be potential between A and B considered from point P
potential at A from P= V = k *6.4*10^-19 /3*10^-6 + (potential of B from P) = k*6.4*10^-19 /9*10^-6
POTENTIAL DIFFERENCE = would b the difference between both values = k*6.4 *10^-19 / 9*10^-6
Workdone = q∆V
q= charge on an electron = 1.6*10^-19C
so, 1.6*10^-19 * k*6.4*10^-19/ 9*10^-6
=1.0 *10^-22 J

 

[ashcull14]

PLEASE HELP!!!!

 

[The Sarcastic Retard]

Q4b= V=kQ/r .. Q is same at P and Q (6.4*10^-19 C) AND r is same therefore, V at P and Q is same ..so ∆V = 0...as work done = Q*∆V ,it is zer0
Q4c= for finding the work done we need to know the potential difference between the midpoint and point P
potential at any point is the sum of potentials for each charge
potential at mid point would be between A and B taken from the midpoint
using V= kQ/r and k = 4πε0
(potential of A from midpoint) k*6.4*10^-19/ 6*10^-6 + (potential of B from midpoint) k*6.4*10^-19 / 6*10^-6
potential at point P would be potential between A and B considered from point P
potential at A from P= V = k *6.4*10^-19 /3*10^-6 + (potential of B from P) = k*6.4*10^-19 /9*10^-6
POTENTIAL DIFFERENCE = would b the difference between both values = k*6.4 *10^-19 / 9*10^-6
Workdone = q∆V
q= charge on an electron = 1.6*10^-19C
so, 1.6*10^-19 * k*6.4*10^-19/ 9*10^-6
=1.0 *10^-22 J

Thanks.
k = 1/4πε0 not 4πε0. ^_^

 

[The Sarcastic Retard]

I am facing lots of difficulties in electric field chapter.
I can never solve a full question.
Help required.

 

[Catherine_1]

Hey

Could anyone please help me out with the following questions it would be of immense help:-



I have no idea why both are 3A :3

The b) part of the following is my doubt





In the ms for b part where did the 25J come from?


In this my doubt is the 5 c ii part



So for averaging why are we multiplying the flux linkage by 2?

Please help, if possible.

 

[Konstantino Nikolas]

I am facing lots of difficulties in electric field chapter.
I can never solve a full question.
Help required.

^^ His videos are pretty good ... helps you understand the concept.

Or these from Khan Academy

 

[The Sarcastic Retard]

^^ His videos are pretty good ... helps you understand the concept.

Or these from Khan Academy

I will get back after watching this video and again solving few questions.
Thanks.

 

[The Sarcastic Retard]

Hey

Could anyone please help me out with the following questions it would be of immense help:-
View attachment 59324

View attachment 59325
I have no idea why both are 3A :3

The b) part of the following is my doubt
View attachment 59326
View attachment 59327
View attachment 59329

View attachment 59328
In the ms for b part where did the 25J come from?


In this my doubt is the 5 c ii part
View attachment 59330View attachment 59331
View attachment 59332

So for averaging why are we multiplying the flux linkage by 2?

Please help, if possible.

Q646 http://physics-ref.blogspot.in/2015/05/physics-9702-doubts-help-page-128.html

 

[The Sarcastic Retard]

Hey

Could anyone please help me out with the following questions it would be of immense help:-
View attachment 59324

View attachment 59325
I have no idea why both are 3A :3

The b) part of the following is my doubt
View attachment 59326
View attachment 59327
View attachment 59329

View attachment 59328
In the ms for b part where did the 25J come from?


In this my doubt is the 5 c ii part
View attachment 59330View attachment 59331
View attachment 59332

So for averaging why are we multiplying the flux linkage by 2?

Please help, if possible.

last questions, its coz we are reversing.

 

[The Sarcastic Retard]

Hey

Could anyone please help me out with the following questions it would be of immense help:-
View attachment 59324

View attachment 59325
I have no idea why both are 3A :3

The b) part of the following is my doubt
View attachment 59326
View attachment 59327
View attachment 59329

View attachment 59328
In the ms for b part where did the 25J come from?


In this my doubt is the 5 c ii part
View attachment 59330View attachment 59331
View attachment 59332

So for averaging why are we multiplying the flux linkage by 2?

Please help, if possible.

http://physics-ref.blogspot.in/2015/04/physics-9702-doubts-help-page-114.html - Q580

 

[Charlotte20102013]

Hi can someone help me with this please:

An iron block of 30 kg mass and 90°C temperature is dropped into an insulated tank that
contains 50 kg of water at an unknown temperature. At the same time a paddle wheel is driven
by a 400 W motor is activated to stir the water. It is observed that thermal equilibrium is
reached after 20 minutes with a final temperature of 28°C.

Determine: a) The initial temperature of the water. b) The entropy generated by this process. Explain the significance of the entropy change obtained.

Your help will be greatly appreciated! Thanks!