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# AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] watch

1. (Original post by getback339)
Attachment 420169Is this the best way to do these kind of questions? Jan 12, question 15 multiple choice? Thanks
That's how I do them. You can also use the fact that the ratio of the charges/masses is equal to the ratio of the distances squared, then square root (obviously negative root isn't applicable in this case), then you have a linear equation.

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2. (Original post by CD223)
That's how I do them. You can also use the fact that the ratio of the charges/masses is equal to the ratio of the distances squared, then square root (obviously negative root isn't applicable in this case), then you have a linear equation.

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thanks!
3. (Original post by getback339)
thanks!
No problem!

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4. Why do I not include the radius of the earth in this? I thought newtons law was the graviational force of attraction the centre of 2 masses??

Thanks

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5. (Original post by gcsestuff)

Why do I not include the radius of the earth in this? I thought newtons law was the graviational force of attraction the centre of 2 masses??

Thanks

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The orbit distance takes into account the radius.

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6. Hi I don't suppose anyone knows the answer to 2 a i)

The answers aren't in the back of the book. And this one confused me. Is it to do with electric field strength being proportional to potential difference?

Thanks

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Attached Images

7. (Original post by CD223)
The orbit distance takes into account the radius.

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So if I ever see the word orbit, that means from the centre of the earth?

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8. i am had the same dream hahaha
(Original post by gcsestuff)
I thought this too! But surely that would be to easy for a 6 marker ??

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9. (Original post by _Caz_)
Hi I don't suppose anyone knows the answer to 2 a i)

The answers aren't in the back of the book. And this one confused me. Is it to do with electric field strength being proportional to potential difference?

Thanks

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Yup! You assume it's a uniform field between the two plates.

So increasing the potential difference means a greater E.

As the force on the charge (treating the shuttle ball as a point charge here) is proportional to E:

Then that means that a greater force is experienced, causing a larger change in momentum with each impact between the two walls, giving the effect that the shuttle oscillates faster

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10. (Original post by gcsestuff)
So if I ever see the word orbit, that means from the centre of the earth?

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I would interpret that, yeah. But I've emailed my teacher to make sure - I'll let you know what he says!

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11. (Original post by CD223)
Yup! You assume it's a uniform field between the two plates.

So increasing the potential difference means a greater E.

As the force on the charge (treating the shuttle ball as a point charge here) is proportional to E:

Then that means that a greater force is experienced, causing a larger change in momentum with each impact between the two walls, giving the effect that the shuttle oscillates faster

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thank you!
12. (Original post by _Caz_)
thank you!
No worries

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13. (Original post by CD223)
No worries

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Just out of curiosity, what grade are you aiming for in physics (if you don't mind me asking)?
14. (Original post by theoriginalrpr)
i am had the same dream hahaha

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15. (Original post by CD223)
I would interpret that, yeah. But I've emailed my teacher to make sure - I'll let you know what he says!

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Thanks!! You've helped me so much!! If I see a similar question I'll see what it says

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16. (Original post by _Caz_)
Just out of curiosity, what grade are you aiming for in physics (if you don't mind me asking)?
Minimum B

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17. I never knew you could do it using this method.... I always used the rms equations
18. (Original post by _Caz_)
Just out of curiosity, what grade are you aiming for in physics (if you don't mind me asking)?
An A hopefully! Need AAA for uni. You?

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19. (Original post by gcsestuff)
Thanks!! You've helped me so much!! If I see a similar question I'll see what it says

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Sure! No problem.

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20. (Original post by sykik)
I never knew you could do it using this method.... I always used the rms equations
I see! Yeah power as the rate of energy transfer is so useful!

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