Physics: Post your doubts here!

[gary221]

q 9 o/n 2012 pr 41...help plzzz

d whole ques??

 

[sumaiyarox:)]

d whole ques??

yup plzz..n s11 41 too

 

[syed1995]

thnks

taking ----> as + (The direction of velocity)

Negative moment = the moment is in the opposite direction.. in the <---- direction ..

 

[gary221]

yup plzz..n s11 41 too

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)


c) R emits for a longer time ---> R emits when the Vout is positive.

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
Question 1, b(i) in the formula why did the apply π and 4 ?!
Why not only 7.5*(o.38*10^-3)/1/75 ?

Markscheme : http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_ms_21.pdf

 

[Tkp]

taking ----> as + (The direction of velocity)

Negative moment = the moment is in the opposite direction.. in the <---- direction ..

well i knw rhis

 

[sumaiyarox:)]

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

thnnkkkkkk uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu .....n those do look like arrows ... lousy ones at that tho. .but arrows alright

 

[syed1995]

well i knw rhis

sweet. I meant momentum* .. that's why the change can be negative.. but honestly I have never seen them ask for the direction of momentum though.

 

[Tkp]

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

 

[sumaiyarox:)]

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

y do we get a square graph?

 

[Soldier313]

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]

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.

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]

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!!

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]

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!

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!!

 

[Soldier313]

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!!

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]

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.

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]

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.

Thank youu soo sooo much!!! I understood the whole thing now!

 

[gary221]

 

[hela]

sorry to bother you
can you explain this question in http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf

 

[hela]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf
can some one solve and explain this question summer 2012 paper 22 question 6 c i and ii