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The Physics PHYA2 thread! 5th June 2013

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Original post by usycool1
*or stationary


Hey,

Could you please answer my question above?
Thanks


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Original post by susanjan123


OMG!!! This paper(June 2009) is soooooo easy! I actually got 68 marks in this one when I usually get 64. But, the grade boundaries are also quiet high than usual. 56 for an A.:tongue:
Original post by StalkeR47
OMG!!! This paper(June 2009) is soooooo easy! I actually got 68 marks in this one when I usually get 64. But, the grade boundaries are also quiet high than usual. 56 for an A.:tongue:


Yeah june 2009 is soo easy got 66 is that :smile:

Could you help me on 3e on the attached, could you please draw of what they are looking for? :smile:


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wallaby
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Original post by pilotluke1
Hi June 2012 4aii i got the right answer but by common sense where actually is that angle on the diagram? Also for bi how did they get 65?


Yes, does anyone know, I don't know how they got 65 either? I did 85-30 to get 55 haha.
Original post by Jimmy20002012
Yeah june 2009 is soo easy got 66 is that :smile:

Could you help me on 3e on the attached, could you please draw of what they are looking for? :smile:


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Hi, it is not very complicated. Just draw a plain board and in the board, draw close fringes with middle to be more dark and as you draw fringes away from the middle, their intensity reduces. Does that helps? (Label max and min).
Original post by wallaby
Yes, does anyone know, I don't know how they got 65 either? I did 85-30 to get 55 haha.


Q4aii right? Since the angle is 85 degrees to the left, it is also 85 degrees inside the ray that forms the triangle. (Note: The angle of reflection is always the same as angle of incident). So, to find i, you add 85 + 30 which gives 115 degrees. Since triangle is 180, 180 - 115 gives 65 degrees. Any more help? Ask me :smile:
(edited 10 years ago)
Original post by pilotluke1
Hi June 2012 4aii i got the right answer but by common sense where actually is that angle on the diagram? Also for bi how did they get 65?


Q4aii right? Since the angle is 85 degrees to the left, it is also 85 degrees inside the ray that forms the triangle. (Note: The angle of reflection is always the same as angle of incident). So, to find i, you add 85 + 30 which gives 115 degrees. Since triangle is 180, 180 - 115 gives 65 degrees. Any more help? Ask me
(edited 10 years ago)
Original post by StalkeR47
Hi, it is not very complicated. Just draw a plain board and in the board, draw close fringes with middle to be more dark and as you draw fringes away from the middle, their intensity reduces. Does that helps? (Label max and min).


Thanks that helped :smile:

Finally with this question on 3aiv, I firstly used the equation of P=FV= 96106 x 58= 5574148W, I took that as the useful output power. So using the equation of efficiency= Useful Output Power/Input Power, so the input power= 5574148/0.20(the efficiency)= 2.8 x 10^6 W. however this is not the answer, why?? Thanks


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Here's the question, 3aiv


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Original post by Jimmy20002012
Here's the question, 3aiv


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M8, which past paper is it? I have done that question so I will be able to explain better by looking at the question properly. Thanks.
Original post by StalkeR47
M8, which past paper is it? I have done that question so I will be able to explain better by looking at the question properly. Thanks.


I have just posted it above :smile:


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Original post by StalkeR47
M8, which past paper is it? I have done that question so I will be able to explain better by looking at the question properly. Thanks.


Worked it out yet :smile:


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Original post by Jimmy20002012
Thanks that helped :smile:

Finally with this question on 3aiv, I firstly used the equation of P=FV= 96106 x 58= 5574148W, I took that as the useful output power. So using the equation of efficiency= Useful Output Power/Input Power, so the input power= 5574148/0.20(the efficiency)= 2.8 x 10^6 W. however this is not the answer, why?? Thanks


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Ok, First, equation is-----efficiency = output/input times 100%. Since the question tells you that 20% is the acceleration efficiency. so, (20% = 16.6ms-2). You are asked to calculate the average power input (total power input). So, find the 100% efficiency. 20%=16.6, 1%= 16.6/0.20 = 0.8358ms-2. So, 100% efficiency for the acceleration would be, 100% = 0.83.. times 100 = 83ms-2. The tension is the cable is 9.61x10^4N. so, P=fv can be used to calculate the output power. (v is not given), so use p=f times s/t. This gives, p = (9.61x10^4 x 101)/3.5 = 2.8Mw. Now, since this is only 20% of the total power input, You will have to find 100% because input power means total input power to the machine. So, the final answer would be(lets call 100% power to be x), 2.8/x = 100%. Rearrange, you will find x to be 14Mw. So, 100% input is 14Mw. Does that helps?
(edited 10 years ago)
Original post by StalkeR47
Ok, First, equation is-----efficiency = output/input times 100%. Since the question tells you that 20% is the acceleration efficiency. so, (20% = 16.6ms-2). You are asked to calculate the average power input (total power input). So, find the 100% efficiency. 20%=16.6, 1%= 16.6/0.20 = 0.8358ms-2. So, 100% efficiency for the acceleration would be, 100% = 0.83.. times 100 = 83ms-2. The tension is the cable is 9.61x10^4N. so, P=fv can be used to calculate the output power. (v is not given), so use p=f times s/t. This gives, p = (9.61x10^4 x 101)/3.5 = 2.8Mw. Now, since this is only 20% of the total power input, You will have to find 100% because input power means total input power to the machine. So, the final answer would be(lets call 100% power to be x), 2.8/x = 100%. Rearrange, you will find x to be 14Mw. So, 100% input is 14Mw. Does that helps?


Don't still get how you get 14Mw from that, because you said 2.8/x= 100%, if you rearrange it you get a completely different answer???


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Original post by Jimmy20002012
Don't still get how you get 14Mw from that, because you said 2.8/x= 100%, if you rearrange it you get a completely different answer???


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Ok, 2.8/x=100%. (Convert 100% into decimal which is just 1). x=2.8/1 = 14/5. So, top heavy number is the answer since the denominator is the 5(100%). :smile:
Original post by Jimmy20002012
Don't still get how you get 14Mw from that, because you said 2.8/x= 100%, if you rearrange it you get a completely different answer???


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Check this out! http://www.wolframalpha.com/input/?i=2.8%2Fx%3D100%25+find+x :smile:
Original post by StalkeR47


Got it, thanks :biggrin: :biggrin:


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Original post by Jimmy20002012
Got it, thanks :biggrin: :biggrin:


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No Problem! :smile:
Original post by StalkeR47
No Problem! :smile:


Is there actually anything you don't know for this exam? :smile:


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Original post by eilish1903


Original post by Currentlysmilin
The middle peak refers the to the zero order beam which is the brightest and twice as big as the rest of the peaks, this is because it is the same as the incident beam of light therefore all of the incident light is displayed on the screen. The rest of the peaks are less intense as it had been diffracted more, but they all will have the same width but their peaks would decrease the further away from the zero order beam the less bright they will be.


So the central red graph represents the diffraction grating?

If so, what does the large blue curve and dotted red curve represent? Young's double slit? Really confused as to what each one is.

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