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# AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread] watch

1. (Original post by Dann_a)
Yes - whenever you have to talk about the direction of a charge in the magnetic field you can say "Because of Flemings left hand rule it will in go in X direction (or the opposite if its negatively charged"
Right hand rule he meant, that's left :') don't think iv ever seen right come up in an exam tbh
2. (Original post by alevelstresss)
if anyone wants, here are my PHYA4 notes
Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?
3. (Original post by boyyo)
Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?
thats the emf of the cell
4. (Original post by alevelstresss)
Find the speed at which the water moves

speed is equivalent to a length per second, so speed = volume / area

mass = density x volume

momentum = mass x velocity
Still getting it wrong. Is volume cross-sectional area x the rate per second? so 7.2x10^-4 x 2x10^-4 = 1.44x10^-7 ?
5. (Original post by SirRaza97)
Like this. Sorry its in landscape.

Would it look this is if the right one was positive and the left one was negatively charged?
6. (Original post by Ayaz789)
So it'll look like this?
You haven't attached a picture .
7. Can someone show me their working for why the answer is B? Thanks
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Attached Images

8. (Original post by particlestudent)
They are both positive.

Check page 22, I kind of explained it there .
https://kaiserscience.files.wordpres...e-patterns.gif
Would it not be the top right image?
9. (Original post by xMillnsy)
Still getting it wrong. Is volume cross-sectional area x the rate per second? so 7.2x10^-4 x 2x10^-4 = 1.44x10^-7 ?
length = volume / area
length = (2.0x10-4)/(7.2x10-4)
length = (5/18) m

length per second is equivalent to a velocity

v = (5/18) ms-1

mass = volume x density
mass = 1000(2.0x10-4)
mass = 0.2 kg

m = (1/5) kg

therefore using p = mv

p = (1/5) x (5/18)
p = 1/18
p = 0.056 kg ms-1
10. (Original post by boyyo)
Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?
EMF of the power source in the circuit.
11. (Original post by PiTheta97)
Can someone show me their working for why the answer is B? Thanks
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BAN-BANcos(50)
12. (Original post by particlestudent)

BAN-BANcos(50)
Made the mistake of just calculating BANcos50. Cheers

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13. (Original post by PiTheta97)
Can someone show me their working for why the answer is B? Thanks
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Flux linkage is greatest when the coil is perpendicular to the field lines.

When theta=50, Flux Linkage= BAN cos(theta) = 3.8x10^-3
When theta =0, Flux linkage = 5.9x10^-3

5.9x10^-3 - 3.8x10^-3 = 2.1 x10^-3 Wb Turns
14. (Original post by alevelstresss)
length = volume / area
length = (2.0x10-4)/(7.2x10-4)
length = (5/18) m

length per second is equivalent to a velocity

v = (5/18) ms-1

mass = volume x density
mass = 1000(2.0x10-4)
mass = 0.2 kg

m = (1/5) kg

therefore using p = mv

p = (1/5) x (5/18)
p = 1/18
p = 0.056 kg ms-1
ahh I get it now, thanks for your help
15. (Original post by Ayaz789)
Haha its fine, draw it again and show me?
Like this? The repulsion area is closer to the "smaller" charge.

16. (Original post by particlestudent)
You haven't attached a picture .
Well is the picture you attached on page 22?
17. hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?
18. (Original post by abro1089)
hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?
Been trying to get an answer for this....Im still at a loss lol
19. (Original post by SirRaza97)
Like this? The repulsion area is closer to the "smaller" charge.

Ahh okay thanks
20. (Original post by philo-jitsu)
Been trying to get an answer for this....Im still at a loss lol
maybe no one knows xD such a vague topic oh well

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