OCR Physics Unit 2 - G482 - (June Exams Preparation) Watch

Joseph-
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#81
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#81
(Original post by precious maro)
who has started past papers for this?
Me, and they are hard!!!
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Boy_wonder_95
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#82
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#82
(Original post by Gotzz)
Well I bet you got near full marks anyway... :rolleyes:
Got 55
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Gotzz
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#83
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#83
(Original post by Boy_wonder_95)
Got 55
See
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Joseph-
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#84
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Anybody have a sheet of all the circuit symbols we need to remember?
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Gotzz
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(Original post by Joseph-)
Anybody have a sheet of all the circuit symbols we need to remember?
They're all in the book?
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Med Student
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#86
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#86
(Original post by Boy_wonder_95)
:no: Doubt it since I struggled! How's G482 going?
didn't you get your results in march for G481 Jan 13?
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Boy_wonder_95
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(Original post by Med Student)
didn't you get your results in march for G481 Jan 13?
Noo I sat G482 in Jan and got that result in March. But I sat Jan 13 G481 as a mock instead.
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Boy_wonder_95
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(Original post by Gotzz)
See
Probably happened cos you said it mate
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brawlerpit
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#89
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(Original post by Boy_wonder_95)
146/150
Hey, can I ask what raw mark you got out of 100 ? :/

Very interested to know what sort of marks gets the higher ums...
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ss2012
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#90
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#90
how's everyone's revision going? less than a month left till the exam

are there any techniques any of you use?
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hosamthemaster
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(Original post by Malawi)
Does anyone know if we need to be able to describe and explain the experiment with the tuning forks to determine the speed of sound?
Yes you have to well it is quite hard because you got to understand the elimination of errors

So you must use a turning fork with a give frequency

Using a open open tube place in a long beaker of water and hit the tuning force to create a sound wave ! This wave will travel through the open end and get reflected by the surface and super poses to create an anti node and a node of the water try to find a resonance sound hence you first harmonic (fundamental) using open closed standing we know that the fundamental is always λ/4 therefore we can calculate wave length by looking at the water level and x it 4 therefore using v=lambda x freq you calculate speed of sound


Repeat for the 3rd harmonic since you can't have the second harmonic therefore you gain a water level reading and use v=lambda x freq

finally the speed of sound is not actually accurate since the wave starts from out of the tube hence "c" correction factor so we solve simultaneously

λ/4 + l1 = C
3/4 λ + l2 =C

FIND C AND THEN PLUS IT TO THE LENGTH AND THEN CALCULATE AGAIN USING V=Fλ TO GET THE SPEED OF LIGHT BLIIIIIIMMMMMMYYYY !!!! HOOOPE THAT HELPED ALL FROM MY BRAIN ( FLASHCARDS ARE THE BEST WAY TO REMEMBER)
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piguy
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In harmonics, how does the waves get reflected if the tubes are open at both ends?
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komalkhan
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Can anyone help me with this Q i dont understand how to get the answer

its Jan 2011 Q7c
Thaaank you
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ss2012
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(Original post by komalkhan)
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Can anyone help me with this Q i dont understand how to get the answer

its Jan 2011 Q7c
Thaaank you
I was stuck on this yesterday too and I couldn't work out why and how to get the colours
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Boy_wonder_95
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(Original post by komalkhan)
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Can anyone help me with this Q i dont understand how to get the answer

its Jan 2011 Q7c
Thaaank you
(Original post by ss2012)
I was stuck on this yesterday too and I couldn't work out why and how to get the colours
I like this question you've just gotta think about it logically.
You know that: d \sin \theta =  n \lambda

From this you can see that the higher the angle, the higher the value of sin theta, hence the higher the wavelength since d and n would remain the same.

Therefore 12 degrees is for the color with the higher wavelength and 7 degrees is for the color with the shorter wavelength. Now the color can only be either Red or Violet as they are at the two extreme ends (ROYGBIV) since it's the first order spectrum. So which out of Red or Violet has the higher wavelength?
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precious maro
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describe the young's double slit experiment?
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ss2012
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(Original post by Boy_wonder_95)
I like this question you've just gotta think about it logically.
You know that: d \sin \theta =  n \lambda

From this you can see that the higher the angle, the higher the value of sin theta, hence the higher the wavelength since d and n would remain the same.

Therefore 12 degrees is for the color with the higher wavelength and 7 degrees is for the color with the shorter wavelength. Now the color can only be either Red or Violet as they are at the two extreme ends (ROYGBIV) since it's the first order spectrum. So which out of Red or Violet has the higher wavelength?

That makes more sense. Is it always red or violet or will we have to know the wavelength of other colours?
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ss2012
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(Original post by precious maro)
describe the young's double slit experiment?
]To produce two rays light is shone through a pair of parallel slits. The light is diffracted by the slits where it falls on a screen beyond the slits, light and dark interference fringes are seen. This is what we call the Young's double slit experiment.

There are usually two slits hence why its known as the double slit experiment, as the light passes through each slit, it spreads into the space beyond (diffraction)

Bright fringes are constructive interference (two waves in phase) and dark fringes are destructive interference where the light cancels each other out (out of phase)

# increasing the slit to screen distance makes the light fringes on the screen wider but DIMMER.
[/B][/B][/B][/B]
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Boy_wonder_95
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(Original post by ss2012)
That makes more sense. Is it always red or violet or will we have to know the wavelength of other colours?
The question says first order so it would be the first two outer colours, if it said second order it would be the next set of order colours and so on... (I think they only ever ask first though). No you don't have to know the wavelengths of over colours, but you know that red has a lower frequency than violet hence a higher wavelength.
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precious maro
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(Original post by ss2012)
]To produce two rays light is shone through a pair of parallel slits. The light is diffracted by the slits where it falls on a screen beyond the slits, light and dark interference fringes are seen. This is what we call the Young's double slit experiment.

There are usually two slits hence why its known as the double slit experiment, as the light passes through each slit, it spreads into the space beyond (diffraction)

Bright fringes are constructive interference (two waves in phase) and dark fringes are destructive interference where the light cancels each other out (out of phase)

# increasing the slit to screen distance makes the light fringes on the screen wider but DIMMER.
[/B][/B][/B][/B]
thank you
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