OCR Physics A G484 - The Newtonian World - 11th June 2015 Watch

fnatic NateDestiel
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#81
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#81
(Original post by rachelc142)
I miss fp1! You doing fp2 as well this year?
yep!
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rachelc142
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#82
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(Original post by fnatic NateDestiel)
yep!
i think its a bit of a ***** so far , hate maclaurin n further complex numbers, quite like polar co-ords so far, not startin dif. equations yet in maths or further
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twisted
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#83
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How long would it take to self tutor this?
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Ki Yung Na
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#84
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Do you guys find the textbook sufficient enough? I've found it to be very mixed (same with AS) sometimes it does enough but other times it's just awkward. Any books/websites/resources in particular you guys think is effective?
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fnatic NateDestiel
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(Original post by rachelc142)
i think its a bit of a ***** so far , hate maclaurin n further complex numbers, quite like polar co-ords so far, not startin dif. equations yet in maths or further
Haven't started it yet only done the bits that linked in FP1.
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L'Evil Fish
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#86
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(Original post by twisted)
How long would it take to self tutor this?
Probably a good solid day.
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Makashima
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#87
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Hi guyssssssss! I am taking this toooooo :P

By the way guys, so vmax = 2πf gives you the t/4?
Anyone can explain me to this, my note is not very clear...
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L'Evil Fish
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(Original post by Makashima)
Hi guyssssssss! I am taking this toooooo :P

By the way guys, so vmax = 2πf gives you the t/4?
Anyone can explain me to this, my note is not very clear...
V max occurs at 0 displacement from equilibrium

So if you start a cycle from max displacement

T/4 will be initial 0 displacement which is v max

Although I've never thought of it like that
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Makashima
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(Original post by L'Evil Fish)
V max occurs at 0 displacement from equilibrium

So if you start a cycle from max displacement

T/4 will be initial 0 displacement which is v max

Although I've never thought of it like that

Ahhh I kinda get what you mean...It's because Ive seen questions that require you to know that vmax only give you t/4 therefore to get whole t you would have to multiple it by 4...Thanks anyway :3
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Octane
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(Original post by fnatic NateDestiel)
Do you want it? Topic by topic exam questions for all the sciences. Mine are ocr a though.

Would an A* in Physics be conditional for you?
May I have ExamQuest for OCR A too? Thanks.
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NeroM
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#91
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Can i have ExamQuest for OCR A too? Thank you.
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Makashima
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#92
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Anyone have past paper exam for 2014 summer?
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L'Evil Fish
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#93
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(Original post by Makashima)
Anyone have past paper exam for 2014 summer?
No

And for good reason. If your school use it as a mock, you want it to be accurate
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Makashima
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(Original post by L'Evil Fish)
No

And for good reason. If your school use it as a mock, you want it to be accurate
LOLOLOLOL I like your thinking BUT
I am definitely sure that they won't use it for the mock as if other students wouldn't have a mind like mine?

Also because last year for AS mock jan, all they did was random mixing of all past paper exams. So they will do this again

And I have went through all papers listed on the OCR web
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MysteriousKnight
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Hi guys!

I have a question from the January 2010 G484 paper. For question 1B. i) You are asked to calculate the area under the graph, what I done was calculate the area as if its a triangle but on the mark scheme that scores one mark. The mark scheme says to count the boxes and I got 510 boxes-ish which is within their range of 500-600 boxes, then I got confused on how they converted this area to the impulse of 2.2 Ns. Can someone help me plz? :s

Thanks!
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Makashima
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(Original post by MysteriousKnight)
Hi guys!

x!
I hope I am not too late! :P

500 individuals boxes! NO, too effort!
2 marks usually mean worth 2 minutes or so...thus the method must be quick/easy

Instead of counting individual box
Do this:
Therefore each boxes = 500 x 0.2x10^-3 = 0.1N.s

so here there is 16 complete boxes
what about the uncompleted boxes? well using your own judgement, how much complete boxes can be made from those pieces?

For me I say maybe 4-6 boxes but Im going to go with 4...
4+16 = 20 boxes

20 ( 500 x 0.2 x10^-3) = 2Ns



MS: 2.2 Ns {allow 2.0 to 2.4}
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MysteriousKnight
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(Original post by Makashima)
I hope I am not too late! :P

500 individuals boxes! NO, too effort!
2 marks usually mean worth 2 minutes or so...thus the method must be quick/easy

Instead of counting individual box
Do this:
Therefore each boxes = 500 x 0.2x10^-3 = 0.1N.s

so here there is 16 complete boxes
what about the uncompleted boxes? well using your own judgement, how much complete boxes can be made from those pieces?

For me I say maybe 4-6 boxes but Im going to go with 4...
4+16 = 20 boxes

20 ( 500 x 0.2 x10^-3) = 2Ns



MS: 2.2 Ns {allow 2.0 to 2.4}
Its not too late I guess ... xD

I knew it couldnt be that hard thanks!
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MysteriousKnight
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Hi guys!

Another q but this time from the june 13 paper. Basically the last question is asking you to find the pressure after an additional 1.5m^3 air is added to the cylinder on top of 0.05m^3 of air that's already in the cylinder.

My question is this when I worked out the pressure I used the total volume i.e. 1.55m^3 but the mark scheme says the volume is 0.05m^3 that you have to use. Could someone please explain this to me ... ?

Thanks!
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Elcor
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(Original post by MysteriousKnight)
Hi guys!

Another q but this time from the june 13 paper. Basically the last question is asking you to find the pressure after an additional 1.5m^3 air is added to the cylinder on top of 0.05m^3 of air that's already in the cylinder.

My question is this when I worked out the pressure I used the total volume i.e. 1.55m^3 but the mark scheme says the volume is 0.05m^3 that you have to use. Could someone please explain this to me ... ?

Thanks!
Imagine forcing a bunch of air into a really strong metal cylinder. Is the cylinder going to expand massively and have the air pressure stay the same?

The cylinder's volume will remain constant, but the number of moles of gas inside it will increase, making pressure increase. Work out this number of moles and use an equation to work out the pressure.
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Tiwa
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#100
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Anyone have the Newtonian world June 2014 paper?
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