Depends on what option you did. I found last year's astro paper difficult
what was last year's astro grade boundary
25 for an A on the astro section. I've noticed that the astrophysics boundary is typically the highest. A* was 62-60 depending on option, whereas usually it's about 57-59 for an A* aside from astrophysics which always seems to be about 60
I didn't find it hard at all but that's how low they've been in one period. June 2010 was 56 for an A* and that one was much easier than this one. Again June 2011 was 57 for an A* and it was very easy.
5 sub questions from June 2012 was repeated (closely) on this paper, and 4 from June 2013. The parts that weren't repeated were very weird unusual calculations
June 2012 boundaries were 53 for an A* June 2013 boundaries were 54 for an A*
That's why I think it will be 53. June 2014 was 52 for an A* and it wasn't even that hard
what did people put for where fusion and fission nuclei are placed on the graph?
i just said that fusion nuclei are found when A<50 and fission when A>50, because fusion nuclei are smaller and more stable and fission nuclei are large and unstable?
3)c) I got 0.0375m, as I got the wavelength as 0.15m and it asked for the distance between a maximum and a minimum recording, i.e. the distance between an antinode and a node, which is a quarter of a wavelength (0.15/4) = 0.0375m
Anyone else do it this way?
The wavelength wasn't 0.15m. They gave you 2.2GHz as the frequency (2.2*10^9), so 3*10^8 / 2.2*10^9 = wavelength - this gives 0.136m, which you then divide by 4 to get 0.341m between a node and antinode - your method was right, I think you must have miscalculated wavelength though.
The wavelength wasn't 0.15m. They gave you 2.2GHz as the frequency (2.2*10^9), so 3*10^8 / 2.2*10^9 = wavelength - this gives 0.136m, which you then divide by 4 to get 0.341m between a node and antinode - your method was right, I think you must have miscalculated wavelength though.
the distance between a maxima and minima is not the same as the distance between a node and antinode.
The wavelength wasn't 0.15m. They gave you 2.2GHz as the frequency (2.2*10^9), so 3*10^8 / 2.2*10^9 = wavelength - this gives 0.136m, which you then divide by 4 to get 0.341m between a node and antinode - your method was right, I think you must have miscalculated wavelength though.
I think it's half a wavelength because you were moving the reflector not the detector. So the position of the nodes changes due to more nodes being made. This caused the difference to be half a wavelength from the original node to a newly formed antinode. Not entirely sure though, it was a weird question.
I think it's half a wavelength because you were moving the reflector not the detector. So the position of the nodes changes due to more nodes being made. This caused the difference to be half a wavelength from the original node to a newly formed antinode. Not entirely sure though, it was a weird question.