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Physics: Post your doubts here!
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If I remember correctly, we got like 5 or 6 electricity questions.
Were there repeated questions from these electricity questions?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
Can someone please explain this question?
Can someone please explain this question?
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Well, i figured out the answer just now,
For Q24,
Power per unit area is intensity, if you remember! [Intensity = Power/Area] ,
As Intensity (and its energy) is directly proportional to amplitude square,
So, if the wave has an amplitude of A and falls on an area of S,
So, power per unit area [intensity, P] = A^2/S
Thus when A = 2A, and S = 1/3 S
So Intensity = (2A)^2 / (1/3 S),
Thus, after solving it you will find 12A^2/S,
=12P
For Q37:
It will be D not C becoz:
If the variable resistance is zero the
current will be large and the voltmeter reading will be zero. When the variable resistance is 10 Ω the current
will be reduced, but not zero, and the voltmeter reading will be high.
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Which one!
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Someone plz?
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First,
Derive equations of both trains! using s= ut +1/2 at^2
Well first train has constant speed of 10 m/s, thus a = 0 so the equation will be s=10t
For second train it will be s= 0.25t^2 ,
When the trains will pass each other the distance will same so
10t=0.25t^2
=> 0.25t^2 - 10t = 0
solve, you will get t = 0, and t= 40,
Thus t = 40s
since this is a parallel circuit so the emf across 4R will be same as emf across all three remaining resistors (R,2R,3R)
lets say R=1 ohms
since they have given the V across 2R we can find the current in the down branch of the circuit
V=IR
2/2=I
I=1 A
now you can find the V across all the remaining resistors in the down branch (R,3R)
R,
V=1*1=1V
3R,
V=1*3=3V
so total V or emf =1V+2V+3V=6V
Can you please explain how you derived A^2/S PLEASE
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Thank you sooooo much!
Can you explain q21 and 35 of the same paper?
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For 37, the answer is D
When the switch will be connected, the current will pass through both red and green, so the they both will light dimly
I sorry in not giving an excellent explaining but i suck at electricity :/
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You should have read the post many time to understand:
Always remember "Intensity is directly proportional to amplitude square" , Revise the waves chapter,
Intensity proportional to (Amplitude) ^2
So, if the wave has an amplitude of A
The intensity will be proportional to (A)^2
If it has amplitude of 2A
The intensity will be proportional to (2A)^2 = 4A^2
If it has amplitude of 10
The intensity will be proportional to (10)^2 = 100
If it has amplitude of 20
The intensity will be proportional to (20)^2 = 400
I hope you have got it
, If not yet i want you to revise the chapter,Sorry for late reply i have extremely unstable network and having problems in my area!
You should have read the post many time to understand:
Always remember "Intensity is directly proportional to amplitude square" , Revise the waves chapter,
Intensity proportional to (Amplitude) ^2
So, if the wave has an amplitude of A
The intensity will be proportional to (A)^2
If it has amplitude of 2A
The intensity will be proportional to (2A)^2 = 4A^2
If it has amplitude of 10
The intensity will be proportional to (10)^2 = 100
If it has amplitude of 20
The intensity will be proportional to (20)^2 = 400
I hope you have got it
Sorry for late reply i have extremely unstable network and having problems in my area!
yup got this THANKS aLOT
the thing i didn't get is how we changed
I=power/area to I=A^2/S
Sorry for bothering you again
Thought blocker
XPRS Moderator
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First find p.d between S and P then between S and Q
Formula we'll take in account is :
Resistance across p.d / Total resistance * voltage
Case i) - b/w S and P :¬
5000 / 10000 * 2 = 1V1
case ii) - b/w S and Q :¬
3000 / 5000 * 2 = 1.2V2
Now we are asked to find V1 - V2
Hence 1 - 1.2 = -0.2V = C