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Physics: Post your doubts here!

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unique840

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Ashique said:
^Me too, I, too, need help in that question.

Anyways, can someone help me out in May/June 2007; questions- 8, 13. 24. 28. 33. 34. 36 and 40? I got stuck in these questions..
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_1.pdf
Click to expand...
34)the initial reading is 4 V across the LDR. the resistance decreases when light increases. resistance and voltage are directly proportional. so the possible value of voltage will be less than 4 V that is 3V
 
rz123

rz123

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Ashique said:
^Me too, I, too, need help in that question.

Anyways, can someone help me out in May/June 2007; questions- 8, 13. 24. 28. 33. 34. 36 and 40? I got stuck in these questions..
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s07_qp_1.pdf
Click to expand...
36:I = V/ R , 3 / 6 = .5A. now use this to find the voltage consumed by the internal resistance which is .5 X 2 = 1 V , Terminal voltage will be 3-1= 2 volt, useful output power => p=VI 2 into .5 = 1 W

40: idk
 
XPFMember

XPFMember

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Assalamoalaikum wr wb!!
Just sharing an mcq...i got stuck in it but then understood :D
It's Q:27 of Nov:2011 Paper 13

Ans is D.
​
9702_w11_qp13-q27-jpg.5189

That's bcoz there is interference, so when when one wave is removed, amplitude becomes halved. Since intensity is proportional to square of amplitude, it will become 1/4 as amplitude becomes 1/2
 
abcde

abcde

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Dayyanah

Dayyanah

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Hey i have my P4 mocks tomo... anyone with any last minute notes... its wud b really helpful notes..
thanx!
 
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Anonymous'

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Oliveme said:
o_O umm...i'm sorry but i still don't get it. how do you know there is only one at Y and 3 at X and Z. :confused::unsure:
Click to expand...

Sorry for the late reply, been busy lately.. heheh.. :p hmm.. I'm gonna make it simple.. I have calculated that the no. of maximum order is 3, we know that at position Y, it is zero order, which means, only one image can be seen at position Y, next, at position X, there are 1st order which is right above position Y, 2nd, and 3rd order, therefore, there are three images at position X, same goes to position Z, 1st order, which is right below position Y, then 2nd and 3rd order, thus, there are seven images altogether.. :)

*no. of order = no. of images*
 
Hussnain

Hussnain

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HELP NEEDED.
M/J/04
PAPER 4
Q 2b(ii)
PLEASE REPLY ASAP.
I have my exam tomorrow:(
 
Oliveme

Oliveme

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Anonymous' said:
Sorry for the late reply, been busy lately.. heheh.. :p hmm.. I'm gonna make it simple.. I have calculated that the no. of maximum order is 3, we know that at position Y, it is zero order, which means, only one image can be seen at position Y, next, at position X, there are 1st order which is right above position Y, 2nd, and 3rd order, therefore, there are three images at position X, same goes to position Z, 1st order, which is right below position Y, then 2nd and 3rd order, thus, there are seven images altogether.. :)

*no. of order = no. of images*
Click to expand...
thank you :)
 
U

unique840

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Hussnain said:
HELP NEEDED.
M/J/04
PAPER 4
Q 2b(ii)
PLEASE REPLY ASAP.
I have my exam tomorrow:(
Click to expand...
1/2 m c^2 = 3/2 kT
c^2 = 3k/m T
3k/m is a constant.
c^2 is directly proportional to T
500^2 = constant * 300 (initial temp n rms speed)
constant = 2500/3
now the twice of initial speed would be 1000 so:
1000^2 = 2500/3 * T
T = 1200
 
leadingguy

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Oliveme said:
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w09_qp_11.pdf
please explain these questions ---> 10, 15, 22. Thank you very much.
Click to expand...


10) f=ma
60=30a a= deceleration
a= -2ms-2 (negative sign indicates that itx decelaration)


now apply formula fr acc. >>> a=(Vf- Vi)/t

-2 = (Vf - 3)/0.50

Vf = 2ms-1 ans.

15) according to conservation of momentum same will be the momentum at both sides
so at left side is MV= 2 * 2 =4

so will the momentum at right side MV= 1 * 4 = 4
now we know the velocities of both the objects moving!
take out the kinetic energy of both trolleys movingthese velocities
by the formula 1/2mv^2
the K.E of left will be 8j
of right will be 4j
the total stored in the spring will be 8+4 = 12j
 
leadingguy

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Oliveme said:
one more question paper, please.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w08_qp_1.pdf
questions 27, 34, 36.
Thank you very much :)
Click to expand...


34) q is the charge! it can be found out by Q= IT where I is the current in AMperes and t is the time in sec.

now a car needs 200A till 2 seconds to start means it needs Q=IT 200*2 400C of charge.


now consider car being fully charged means it have 100KC
of charge!
cnvrt kilo coulomb into coulomb
means 100810^3 = 100000C

now divide it by the charge needed per start for the car!


100000/400 = 250 times is the ans!
 
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Anoushay

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hey can anyone please explain how to calculate change in momentum for these values: Mass=0.045 kg, Initial velocity=4.2 m/s Final velocity=-3.6 .The negative sign is creating a confusion. the formula is given : mass*(initial+final) in the mark scheme.it's Nov/2002 P2 Qs3.thank you.
 
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