Physics: Post your doubts here!

[Rizwan Javed]

View attachment 59281

how is the phase difference between P and Q= 0??

In stationary waves, all the points in between the two adjacent nodes are in phase. Since they are in phase the phase difference between them will be zero.

 

[Rizwan Javed]

View attachment 59280

the (ii) part pls

First reason would be that there is no work done against air resistance.
Second could be that since the speed is constant, there is no gain or loss of Kinetic Energy.
Therefore the rate of increase of P.E is equal to power calculated.

 

[amal sharkawi]

 

[amal sharkawi]

View attachment 59282

 

[amal sharkawi]

 

[Salonee B]

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.

 

[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.