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Physics: Post your doubts here!

Minato112

Minato112

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geek101 said:
nov 2012..yeah please :p june ofc
Click to expand...
haha srry but forgot that we were in 2012!

Paper 1.
Variant 1: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_11.pdf
Variant 2: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_12.pdf
Variant 3: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_13.pdf

Paper 2.
Variant 1: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_21.pdf
Variant 2: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_22.pdf
Variant 3: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_23.pdf
 
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geek101

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minato112 said:
haha srry but forgot that we were in 2012!

Paper 1.
Variant 1: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_11.pdf
Variant 2: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_12.pdf
Variant 3: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_13.pdf

Paper 2.
Variant 1: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_21.pdf
Variant 2: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_22.pdf
Variant 3: http://olevel.sourceforge.net/papers/9702/9702_s12_qp_23.pdf
Click to expand...

THNX A TONNN!!
 
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raamish

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Qp : http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_2.pdf
M.s: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_ms_1 2 3 4 5 6.pdf

I have an issue in qs 2aiii) The mark scheme is saying that power output will be greater because work would be done against friction but isnt that going to be subtracted from the power output of the car so that the output can be calculated? Also how do we do 3bi)
 
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Kumkum said:
question paper: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_2.pdf
marking scheme: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_ms.pdf
please help me with question 2 (b).:confused:
Click to expand...

Intensity is directly proportional to square of A(A2) so it has double the intensity so it will have 4 times of amplitude. THe phase difference is 180 so the waves are in antiphase so draw the waves opposite each other so that they form like a stationary wave pattern.........................i think :unsure:
 
Kumkum

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raamish said:
Intensity is directly proportional to square of A(A2) so it has double the intensity so it will have 4 times of amplitude. THe phase difference is 180 so the waves are in antiphase so draw the waves opposite each other so that they form like a stationary wave pattern.........................i think :unsure:
Click to expand...
thnx!!
 
Minato112

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raamish said:
hey can u answer my questions plzz :)
Click to expand...
U mean this one?
raamish said:
Qp : http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_2.pdf
M.s: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_ms_1 2 3 4 5 6.pdf

I have an issue in qs 2aiii) The mark scheme is saying that power output will be greater because work would be done against friction but isnt that going to be subtracted from the power output of the car so that the output can be calculated? Also how do we do 3bi)
Click to expand...
 
champ-student

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Minato112

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raamish said:
Qp : http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_2.pdf
M.s: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_ms_1 2 3 4 5 6.pdf

I have an issue in qs 2aiii) The mark scheme is saying that power output will be greater because work would be done against friction but isnt that going to be subtracted from the power output of the car so that the output can be calculated? Also how do we do 3bi)
Click to expand...

The explanation is:
The power output is larger since additional work must be done against friction. The case here, since the speed is constantly increasing, the air resistance force also increases. Thus the case will have to output more power to do so.

The values you calculated for the part 2 (b) (ii) are the power output assuming no additional work is done.

3 b) (i) For the increase in length, U find by how much the pointer moved down.
(6.5/360) * 2 * "pie" * 1.5
= 0.17 cm.

(ii) Increase in strain = Extension / Length
= 0.17 /[ (2.5 * 100) ] --------> Convert to cm
= 6.80 * 10^-4

Hope it helps.
 
Asheeta

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Kirabo Takirambudde said:
please help with part (iii)
Click to expand...

Using power= (V^2/R)
R = (230^2)/1000
therefore R = 52.9 ohm
effective R = [(1/52.9) + (1/52.9) + (1/52.9)]^(-1)
power = (v^2/R)
so power = (230^2)/ effective R
and the ans should be 3KW

Hoped it helped and could you just give me the answers for the above 2 questions as well. Thanks in advance.
 
queen of the legend

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mrnt3250 said:
Please explain this to me.
9702/02/M/J/04
(Paper 2 - M/J 04)
Question 3 - Part b) ii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_2.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_ms_2.pdf

Thanks in advance.
Click to expand...
this is very easy if you were able to do part a , in part a you re asked to calculate the gradient!
you found the gradient =s/t^2 . think of this kinematic equation s=0.5 at^2 . we do not consider s= ut+0.5 a^2 because the car does not have any constant initial velocity before its acceleration .
now to calculate acceleration , a= 2 x s/t^2 that is 2 x the gradient you found . ( i am taking s as the distance here)
 
champ-student

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12 Magnesium chloride (MgCl2) and calcium oxide (CaO) are ionic compounds.
Electricity is passed through the molten compounds, in two separate experiments.
Which are the products at the cathodes?
a magnesium and calcium
b magnesium and oxygen
c oxygen and calcium
d chlorine and oxygen
 
A*(a*)

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COULD SOMEONE ON EARTH PLEASE POST THE END-OF-CHAPTER ANSWERS OF PHYSICS COURSE BOOK?
I AM FINDING THEM FROM YEARS AND WITH NO LUCK! IF ANY ONE HAS TEACHER'S C-D ROM, HE MIGHT HAVE ACCESS TO THAT- AS FAR AS I HAVE HEARD.
 
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