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Physics: Post your doubts here!

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hey um a little confused wiv effects of differnt changes to da doubles slit exp...if da light intensity increase the ms stated b4 dat brightness of bright fringes increases nd dark ones r darker? but shouldnt da dark ones stay da same bec even if da amplitude increases its still destructive interference nd is cancelled anyways? i mean if u increase da width of slits wiv a constant it remains da same y change here? plz some1 help wiv dis lame part :p.......alsoo i would appreciate it if some1 drew da graph in q(5)a- if da printin wuz right ofcourse nd stress is reduced from 2.5x10^8... tyy nd plzz answerrrr!!
 

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In part a, draw a continue the dotted horizontal line from the original alpha path before it is deviated.Then draw an angle b/w the horizontal and the deviated curve. As for part (b) this alpha particle B is also deviated away from the nucleus but the extent and angle of deviation is less than A because it is further away........
 
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In part a, draw a continue the dotted horizontal line from the original alpha path before it is deviated.Then draw an angle b/w the horizontal and the deviated curve. As for part (b) this alpha particle B is also deviated away from the nucleus but the extent and angle of deviation is less than A because it is further away........



isnt it something like this...
 

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can anyone please post the solutions of these two questions.regarding progressive and stationary waves...do post the answer to the graph as well. Please :)1.png2.png
 
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im having difficulty in finding phase difference and plotting graphs. For Progressive I found phase difference to be= 360*(40/80) and it's graph after 0.5 T is attached below. am i right? and what about the stationary wave?
graph.png
 
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can anyone please post the solutions of these two questions.regarding progressive and stationary waves...do post the answer to the graph as well. Please :)View attachment 11254View attachment 11255
i)Amplitude is 7.6mm which you can see it from the graph (the highest displacement)
ii) 180 (This is 180 because, one graph is completed in 80cm.....So in 80cm 360 degree is completed. So there is a phase difference of 40cm between the point. So it is half of 80com and hence it is half of 360, which is 180.......Or you could use this formula- (40/80)*360 which is the normal formula of phase difference)
iii) V=frequency * wavelength
= 15 * 0.8
=12
(b) This one i have to show you by drawing....if you are interested then i can show you
(c)
i. Zero (Since they are on the same wave and phase)
(ii)
Antinode- A region in which two amplitude are at its highest point in opposite direction
Node- A region where two waves intersect and have no displacement

(iv) Again, i have to draw this...if you are interested then i could give a separate PDF for this by drawing in computer
 
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Can anyone help me with these
m/j 2007- 5b (ii)
m/j 2011 21- 2c(i) 5 (ii)b
m/j 2011 22 5b(ii) 6(ii)1
 
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http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_ms_21.pdf
Q 2(b) and Q 6(a ii) Can someone show exactly how these both are drawn? i have read the mark scheme and examiner report but i might be thinking it is drawn some other way entirely so just to verify my own drawings.

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_21.pdf
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_ms_21.pdf
Q 7(b) in this i also want to confirm if my drawing is correct so if someone else can show the correct one..
 
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i)Amplitude is 7.6mm which you can see it from the graph (the highest displacement)
ii) 180 (This is 180 because, one graph is completed in 80cm.....So in 80cm 360 degree is completed. So there is a phase difference of 40cm between the point. So it is half of 80com and hence it is half of 360, which is 180.......Or you could use this formula- (40/80)*360 which is the normal formula of phase difference)
iii) V=frequency * wavelength
= 15 * 0.8
=12
(b) This one i have to show you by drawing....if you are interested then i can show you
(c)
i. Zero (Since they are on the same wave and phase)
(ii)
Antinode- A region in which two amplitude are at its highest point in opposite direction
Node- A region where two waves intersect and have no displacement

(iv) Again, i have to draw this...if you are interested then i could give a separate PDF for this by drawing in computer
ya i wont da graphs too pls make them !!!
 
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i)Amplitude is 7.6mm which you can see it from the graph (the highest displacement)
ii) 180 (This is 180 because, one graph is completed in 80cm.....So in 80cm 360 degree is completed. So there is a phase difference of 40cm between the point. So it is half of 80com and hence it is half of 360, which is 180.......Or you could use this formula- (40/80)*360 which is the normal formula of phase difference)
iii) V=frequency * wavelength
= 15 * 0.8
=12
(b) This one i have to show you by drawing....if you are interested then i can show you
(c)
i. Zero (Since they are on the same wave and phase)
(ii)
Antinode- A region in which two amplitude are at its highest point in opposite direction
Node- A region where two waves intersect and have no displacement

(iv) Again, i have to draw this...if you are interested then i could give a separate PDF for this by drawing in computer


Yep. i also want the graphs :p ;) :)

thanks alot
 
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please someone answer the question i have posted above on this page its the second time i posted them:( !!
 
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