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Physics: Post your doubts here!

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a021011 said:
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s09_ms_2.pdf

Paper 21, Please Explain Question 5b
path difference is 28 cm.
take the first n last frequency and the calculate the corresponding lambda and path difference. then insert the values of 'n' and 'n/2' which determines the max n min. so there are 2 min
 

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AkhlaqAhmed said:
People!!!!!!!!!!!!! I have a doubt in O/N 2010 paper 43, question 3a(i). how do we find the displacement when there's no value given for "t" in the given equation!!
HELP....i'm dead in this question!!!!!!! :O:

i think from the equation, amplitude is 4.0cm, do the displacement is 8.0cm
 
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Wud anybody help me, please!! Well, i got a problem with a quesion.. It's QN. 25.. May/june 2004, paper one..
I'd be very glad if smbdy'd help me... :)
 
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^well its A as the q point is stationary , its at the maximum displacement position. resting position u can say. and P is downwards as the wave profile behind is downwards. the wave pattern behind the point p is negative !
 
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DaEconomist said:
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w09_qp_12.pdf

Question 21

any help will be greatly appreciated.

thanks

To find the percentage of length that contracted, you need to get the ratio of the extended length to the original length, and that's strain.strain= dL/L

Young Modulus=stress/strain
=F/A by dL/L

Replacing
2.0 × 10^11=(20/pie(2.5×10^-4))/ (dL/L)
dL/L = 5.1×10^-4
Percentage=5.1×10^-4 ×100
=5.1×10^-2
Ans B
 
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Distance between maxima = Half wavelength
so lamda/2 = 15mm
lamda=30mm=.030m
microwave is an electromagnetic wave, meaning that its speed is 3.0*10^8.
v=f(lamda)
f=v/ (lamda)
=3×10^8 / 0.03
= 1.0×10^10 Hz
Ans C 1.0 × 10^10 Hz
 
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Hey, Can anyone post expected question for physcics practical papers 33/34 and Physics paper 2 ?
 
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^^22. Its D because you have removed one spring and the weight and now when you put a 2W weight so the extension would have to be more then twice as u took off some weight before. hence the value 3x will be correct.
 
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can anyone show me which finger pointing which side for left hand rule in this question? Thanks

paper42/june 2010
 

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I am having problem with following question./?
Please solve it with explanation...
Physics 9702(paper2) may/june 2007 question 3c ii and 4d i
 

XPFMember

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Assalamoalaikum!!

ching293 said:
can anyone show me which finger pointing which side for left hand rule in this question? Thanks

paper42/june 2010


Remember when using Fleming's Left Hand Rule, Second finger is the CONVENTIONAL current direction! so the electron movement is opp. to the conventional current..
in this case since the direction of electron movement is left, the direction of conventional current will be to the right...
now try it, you'll get the movement towards QR :)
 
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XPFMember said:
Assalamoalaikum!!

ching293 said:
can anyone show me which finger pointing which side for left hand rule in this question? Thanks

paper42/june 2010


Remember when using Fleming's Left Hand Rule, Second finger is the CONVENTIONAL current direction! so the electron movement is opp. to the conventional current..
in this case since the direction of electron movement is left, the direction of conventional current will be to the right...
now try it, you'll get the movement towards QR :)

Thank you so much, very helpful :oops:
 
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