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Physics: Post your doubts here!

sumaiyarox:)

sumaiyarox:)

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gary221 said:
w12
9 a) Properties of an ideal op-amp:
i) infinite input resistance/impedance.
ii)zero output resistance/impedance.
iii) infinite open-loop gain
iv) infinite slew rate.

b)graph: square wave.
correct cross-over points where V2 = V1.
amplitude 5 V.
correct polarity (positive at t = 0)
View attachment 24443

c) R emits for a longer time ---> R emits when the Vout is positive.
View attachment 24452
Click to expand...
thnnkkkkkk uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu:):):D:):) .....n those do look like arrows:) ... lousy ones at that tho.:) .but arrows alright:p:D
 
Tkp

Tkp

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na dude.check the oct nov 11 one dnt knw which variant they gave 1 ms for direction also
 
sumaiyarox:)

sumaiyarox:)

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gary221 said:
w12
9 a) Properties of an ideal op-amp:
i) infinite input resistance/impedance.
ii)zero output resistance/impedance.
iii) infinite open-loop gain
iv) infinite slew rate.

b)graph: square wave.
correct cross-over points where V2 = V1.
amplitude 5 V.
correct polarity (positive at t = 0)
View attachment 24443

c) R emits for a longer time ---> R emits when the Vout is positive.
View attachment 24452
Click to expand...
y do we get a square graph?
 
Soldier313

Soldier313

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Aoa wr wb
Can someone please explain
qn 9
qn 11 b ii and 11 b iii

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_ms_4.pdf


_________________
5 b ii - ?
11 - i don't understand the last phrase in the ms where it says the crystal is cut.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_42.pdf
Thank you so much.
 
gary221

gary221

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Soldier313 said:
Aoa wr wb
Can someone please explain
qn 9
qn 11 b ii and 11 b iii

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_ms_4.pdf


_________________
5 b ii - ?
11 - i don't understand the last phrase in the ms where it says the crystal is cut.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdfhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_42.pdf
Thank you so much.
Click to expand...


5 b ii) In series, the voltage is divided btw the components (capacitors) ---> V = V1 + V2
V1 across 1 capacitor = 6 V (as mentioned for every 30 microfarads, voltage = 6 V)
V2 across the two capacitors in parallel ---> [ capacitance in parallel = (30 * 30)/ (30 + 30) = 15 microfarad]
So, if voltage across 30 microfarad = 6 V
voltage across 15 V = x
x = 3 V
So, V = V1 + V2
V = 6 +3 = 9V

11. The crystal/transducer detects the reflected ultrasound waves as well.
It can do this bcoz a varying stress is applied across the crystal produces a varying emf.
In order to maximize this effect, the freq of the waves must match the resonant freq of the crystal.
So, the crystal is cut to even the surface so that the complete surface (all d particles on the surface) will then have the same resonating freq.
Hope u gt it!!
 
Soldier313

Soldier313

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gary221 said:
5 b ii) In series, the voltage is divided btw the components (capacitors) ---> V = V1 + V2
V1 across 1 capacitor = 6 V (as mentioned for every 30 microfarads, voltage = 6 V)
V2 across the two capacitors in parallel ---> [ capacitance in parallel = (30 * 30)/ (30 + 30) = 15 microfarad]
So, if voltage across 30 microfarad = 6 V
voltage across 15 V = x
x = 3 V
So, V = V1 + V2
V = 6 +3 = 9V

11. The crystal/transducer detects the reflected ultrasound waves as well.
It can do this bcoz a varying stress is applied across the crystal produces a varying emf.
In order to maximize this effect, the freq of the waves must match the resonant freq of the crystal.
So, the crystal is cut to even the surface so that the complete surface (all d particles on the surface) will then have the same resonating freq.
Hope u gt it!!
Click to expand...

Thank you so much bro! it really helped.....

PS: sorry to bother....but when you do get time could you explain these too please?
qn 9
qn 11 b ii and 11 b iii

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_ms_4.pdf

Thanx a ton!
 
gary221

gary221

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Soldier313 said:
Thank you so much bro! it really helped.....
Click to expand...
Soldier313 said:
PS: sorry to bother....but when you do get time could you explain these too please?
qn 9
qn 11 b ii and 11 b iii

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_ms_4.pdf

Thanx a ton!
Click to expand...


9) For the calibration to have an error of 10%, the source must decay by (10 - 2) = 8%
We know, A = A0 exp^(-λt) ---> We also know, λ = ln 2/ T½ = 0.693/ T½
Substituting the value of λ in ^ equation,
A = A0 exp^(–ln2 t / T½)

(100 - 8) = 100 * exp ^((- o.693 * t)/ 5.27)
0.92 = exp^((-0.693 * t)/5.27)
ln (0.92) = (-0.693 * t )/5.27
t = 0.634 yrs.
t = 230 days.

Hope u gt it!!
:D
 
Soldier313

Soldier313

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gary221 said:
9) For the calibration to have an error of 10%, the source must decay by (10 - 2) = 8%
We know, A = A0 exp^(-λt) ---> We also know, λ = ln 2/ T½ = 0.693/ T½
Substituting the value of λ in ^ equation,
A = A0 exp^(–ln2 t / T½)

(100 - 8) = 100 * exp ^((- o.693 * t)/ 5.27)
0.92 = exp^((-0.693 * t)/5.27)
ln (0.92) = (-0.693 * t )/5.27
t = 0.634 yrs.
t = 230 days.

Hope u gt it!!
:D
Click to expand...
Thanx a lot bruv........my doubt merely is as to why do we subtract 10 and 2 to get an error of 8? i just don't get that bit...:/:(
sorry for the trouble and thanx again.
 
gary221

gary221

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Soldier313 said:
Thanx a lot bruv........my doubt merely is as to why do we subtract 10 and 2 to get an error of 8? i just don't get that bit...:/:(
sorry for the trouble and thanx again.
Click to expand...

bcoz earlier the error in calibration was 2%...ie the source decayed by 2% aftr it was calibrated.
We want an error of 10% ie now the source should decay by 1o% (total) n since it has already decayed by 2%, now it shud only deacy by 8% more.
 
Soldier313

Soldier313

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gary221 said:
bcoz earlier the error in calibration was 2%...ie the source decayed by 2% aftr it was calibrated.
We want an error of 10% ie now the source should decay by 1o% (total) n since it has already decayed by 2%, now it shud only deacy by 8% more.
Click to expand...
Thank youu soo sooo much!!! I understood the whole thing now!
 
H

hela

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Why is it λ / 2 and not λ ... since the distance between the two nodes are λ
 
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