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Ocr Physics B Paper 2 June 8th 2018
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poseidon820
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#1
How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.
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ETbuymilkandeggs
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#2
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#2
Never been so defeated by a paper. Both the fact that I found the questions difficult, and that everyone found them relatively easy. I got about 50%.
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in a fish bowl
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#3
Not too bad apart from that last 6 marker and the banana question. A lot of people found it difficult though
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poseidon820
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#4
(Original post by ETbuymilkandeggs)
Never been so defeated by a paper. Both the fact that I found the questions difficult, and that everyone found them relatively easy. I got about 50%.
Never been so defeated by a paper. Both the fact that I found the questions difficult, and that everyone found them relatively easy. I got about 50%.
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ETbuymilkandeggs
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#5
(Original post by poseidon820)
That's pretty rough, but I thought last year's paper 2 was easier, however that had lower boundaries, so I think they may be in your favour.
That's pretty rough, but I thought last year's paper 2 was easier, however that had lower boundaries, so I think they may be in your favour.
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Tom19800
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#6
There were a few quite hard questions, but overall I think it was easier than paper 1. Guesses for grade boundaries?
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Lavalamp1999
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kathschof1
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#8
(Original post by Lavalamp1999)
What was the last 6 marker again?
What was the last 6 marker again?
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Lavalamp1999
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#9
(Original post by kathschof1)
About the non-reflective coating on the solar cells. It gave you their refractive indexes and the wavelength of light going in. I had no idea what to say lol.
About the non-reflective coating on the solar cells. It gave you their refractive indexes and the wavelength of light going in. I had no idea what to say lol.
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kathschof1
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#10
(Original post by Lavalamp1999)
Oh yeah. The wavelength decreases by the same proportion as the refractive index increases. So by the time the phasor passes through the coating and bounces back it has rotated by pi radians, destructively interfering with the top surface > less reflecting.
Oh yeah. The wavelength decreases by the same proportion as the refractive index increases. So by the time the phasor passes through the coating and bounces back it has rotated by pi radians, destructively interfering with the top surface > less reflecting.
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Vanilla Cupcake
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#11
what we thinking in terms of grade boundaries? found it a lot harder than last year's paper 2 so im really hoping for it to be lower lol. there were a lot of calculations i was surprised but that one about the bananas over 20 years was tragic. hoping they're generous with method marks
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Dahen2323
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#12
(Original post by Lavalamp1999)
Oh yeah. The wavelength decreases by the same proportion as the refractive index increases. So by the time the phasor passes through the coating and bounces back it has rotated by pi radians, destructively interfering with the top surface > less reflecting.
Oh yeah. The wavelength decreases by the same proportion as the refractive index increases. So by the time the phasor passes through the coating and bounces back it has rotated by pi radians, destructively interfering with the top surface > less reflecting.
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Dahen2323
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#13
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#13
(Original post by poseidon820)
How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.
How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.
(Original post by poseidon820)
How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.
How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.
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Tom19800
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#14
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#14
(Original post by Dahen2323)
What did everyone
What did everyone get for the efficiency on the the last equation
What did everyone
What did everyone get for the efficiency on the the last equation
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Lavalamp1999
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#15
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#15
(Original post by Dahen2323)
What did everyone
What did everyone get for the efficiency on the the last equation
What did everyone
What did everyone get for the efficiency on the the last equation
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Vanilla Cupcake
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Dahen2323
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#17
(Original post by Tom19800)
I just ran out of time half way through the question, so annoyed. I was going to do the intensity is proportional to 1/r^2, so 1/5.2^2 = the intensity at 5.2AU. Then I'd find the efficiency using the difference in power from the value i'd calculated? I might be remembering the question incorrectly though.
I just ran out of time half way through the question, so annoyed. I was going to do the intensity is proportional to 1/r^2, so 1/5.2^2 = the intensity at 5.2AU. Then I'd find the efficiency using the difference in power from the value i'd calculated? I might be remembering the question incorrectly though.
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Dahen2323
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#18
(Original post by poseidon820)
How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.
How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.
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Lavalamp1999
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#19
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#19
(Original post by Dahen2323)
What about the question about does the solar pulse have enough energy to move from 1500 m to 2300m or whatever it was. Could you use mgh for gravitational potential energy change as you didn’t have the mass of earth and the radius ?
What about the question about does the solar pulse have enough energy to move from 1500 m to 2300m or whatever it was. Could you use mgh for gravitational potential energy change as you didn’t have the mass of earth and the radius ?
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Dahen2323
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#20
(Original post by Lavalamp1999)
Yeah you can use mgh. It rose from 1500m to 8500m and had to charge the batteries.
Yeah you can use mgh. It rose from 1500m to 8500m and had to charge the batteries.
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