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OCR Physics A G485 - Frontiers of Physics - 18th June 2015

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Reply 300
Original post by Elcor
Cheers buddy. Hope adjustment works out!


Thank you!
Original post by Tiwa
Yep! I firmed Teessdide University for Chemical Engineering Meng for 4 years. Hoping to go for adjustment on results day though. What about you?


Good luck with adjustment, off to UCL here (hopefully)!
Reply 302
Original post by Tom102
Good luck with adjustment, off to UCL here (hopefully)!


Thank you! Hope you make it to UCL!
Original post by Tom102
Good luck with adjustment, off to UCL here (hopefully)!


Ahh nice, my firm is Cambridge but my insurance is UCL, what course are you going for in UCL?
Reply 304
I hope there isn't a really general MRI question, there's so much to talk about!

Does anyone know if we shoud say 'high energy state' or 'anti-parallel alignment' when referring to the protons adsorbing radio photons, or will either suffice?
Original post by Elcor
I hope there isn't a really general MRI question, there's so much to talk about!

Does anyone know if we shoud say 'high energy state' or 'anti-parallel alignment' when referring to the protons adsorbing radio photons, or will either suffice?


From what I've seen, high energy state is the one to go for.


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Reply 306
Original post by Hilton184
From what I've seen, high energy state is the one to go for.


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Appreciated. I wonder if we're even required to know about the protons' alignment.

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Original post by Elcor
Appreciated. I wonder if we're even required to know about the protons' alignment.

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Seeing as I've no clue what that is and never been taught it, I should hope not
Reply 308
Original post by L'Evil Fish
Seeing as I've no clue what that is and never been taught it, I should hope not


You might have to know that the protons align with the magnetic field and precess about the field lines at the Larmor frequency
The G484 forum isn't so active so I going to post a G484 question here. So does anyone know if Kepler's third law, the T^2 = blah blah what ever, should only be used for geostationary? Or can it be used for anything? I don't know when to use it :C
Original post by sagar448
The G484 forum isn't so active so I going to post a G484 question here. So does anyone know if Kepler's third law, the T^2 = blah blah what ever, should only be used for geostationary? Or can it be used for anything? I don't know when to use it :C


Considering Kepler's Laws were known before that of Arthur Clarke's geostationary orbit idea, I assume it can be used for anything, whenever appropriate of course.
Original post by randlemcmurphy
Considering Kepler's Laws were known before that of Arthur Clarke's geostationary orbit idea, I assume it can be used for anything, whenever appropriate of course.


But then why sometimes in the mark scheme when you can use the kepler's law they show that you needed to combine equation for centripetal motion and use those to work out the orbit rather than the geostationary equation? I mean it does not say anywhere in the question that it is not geostationary or it is...
Reply 312
Original post by sagar448
The G484 forum isn't so active so I going to post a G484 question here. So does anyone know if Kepler's third law, the T^2 = blah blah what ever, should only be used for geostationary? Or can it be used for anything? I don't know when to use it :C


Use it for literally any orbit

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Original post by sagar448
The G484 forum isn't so active so I going to post a G484 question here. So does anyone know if Kepler's third law, the T^2 = blah blah what ever, should only be used for geostationary? Or can it be used for anything? I don't know when to use it :C


Just as randlemcmurphy said, it can be used to calculate the period or the distance from the planet to the moon of any moon orbiting a planet


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Original post by sagar448
But then why sometimes in the mark scheme when you can use the kepler's law they show that you needed to combine equation for centripetal motion and use those to work out the orbit rather than the geostationary equation? I mean it does not say anywhere in the question that it is not geostationary or it is...


Yeah the equation doesn't apply for objects in circular motion when the weight of the object is not considered


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Original post by Mehrdad jafari
Yeah the equation doesn't apply for objects in circular motion when the weight of the object is not considered


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But orbits are circular motion....... :/
Original post by sagar448
But orbits are circular motion....... :/


That's correct but the centripetal force keeping planets in their orbits is their weight which is the gravitational force, that is GMm/r2. But in simple circular motions as in rotating an object in a horizontal plane with a string the centripetal force is no longer the weight of the object but simply mv2/r


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Original post by sagar448
But orbits are circular motion....... :/


If you still don't get it let me know


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Original post by Mehrdad jafari
If you still don't get it let me know


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Oh yes! I got it thank you. So basically you can use the Kepler's equation when orbits with weights are involved. So most of the time. :tongue: Thanks.
Original post by sagar448
Oh yes! I got it thank you. So basically you can use the Kepler's equation when orbits with weights are involved. So most of the time. :tongue: Thanks.


Im glad you got it :smile:
Yeah that's it. Because in each case the proportionality of T is different
ImageUploadedByStudent Room1432393928.847383.jpg


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