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# Unit 4 Physics Edexcel A2 and Edexcel IAL Watch

1. (Original post by pinksmartiee)
Also.. Q15)a)iii) and
Q16b)ii) I don't know where the markscheme got p = E/c from..

http://qualifications.pearson.com/co...e_20110127.pdf
For 15(aiii), it's linked with the previous parts above. When it was in position Y, the capacitor was discharging. In position X, what's happening is that the capacitor begins to charge as it's now connected to the battery.
For 16(bii), you have to calculate the momentum first so as to apply the equation lambda=plank's constant divided by momentum. You have the energy and speed of light. Simply substitute them.
2. Do you by any chance know why we have to times E by cos49 ?

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3. (Original post by pinksmartiee)
Can someone help me with Q13? I don't even understand what this is asking.. Thanks!

https://60abffc9b401b1c0936e01291c15...%20Physics.pdf
I think this is something to do with the inverse square law, we know that the force between two charged objects is inversely proportional to the square of distance between them. So, with the relationship that's given to us, i.e 20mm distance->0.5 N Force...We have to make some link. It's given that the point charges exert a force F and distance is given d. If you notice that F is actually the double of 0.5 F, so this means the distance must decrease by 4 times to produce double the force.
4. (Original post by ayvaak)
Do you by any chance know why we have to times E by cos49 ?

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Yeah,I guess so because you need the resultant i.e overall electric field strength so that's the only way you will get it because at B,both charges are producing an effect.
5. (Original post by sabahshahed294)
Yeah,I guess so because you need the resultant i.e overall electric field strength so that's the only way you will get it because at B,both charges are producing an effect.
so the result E would be the horizontal line between them?
6. (Original post by ayvaak)
so the result E would be the horizontal line between them?
Yes, I believe so.
7. Does anyone know why the answer is B? I keep gettig answers that arent the options
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8. (Original post by ayvaak)
Does anyone know why the answer is B? I keep gettig answers that arent the options
Force= Charge multiplied by Electric Field Strength so Force= Mass into Acceleration meaning ma= qE so you know the mass of proton, charge and electric field strength. You can just rearrange the formula.
9. Is physics unit 4 anyone else's last exam?
10. (Original post by sabahshahed294)
Force= Charge multiplied by Electric Field Strength so Force= Mass into Acceleration meaning ma= qE so you know the mass of proton, charge and electric field strength. You can just rearrange the formula.

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Thank you again.

Why cant you do these calculations in radians as apposed to degrees mode on the calc?
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11. (Original post by sabahshahed294)
For 15(aiii), it's linked with the previous parts above. When it was in position Y, the capacitor was discharging. In position X, what's happening is that the capacitor begins to charge as it's now connected to the battery.
For 16(bii), you have to calculate the momentum first so as to apply the equation lambda=plank's constant divided by momentum. You have the energy and speed of light. Simply substitute them.
(Original post by sabahshahed294)
I think this is something to do with the inverse square law, we know that the force between two charged objects is inversely proportional to the square of distance between them. So, with the relationship that's given to us, i.e 20mm distance->0.5 N Force...We have to make some link. It's given that the point charges exert a force F and distance is given d. If you notice that F is actually the double of 0.5 F, so this means the distance must decrease by 4 times to produce double the force.
12. (Original post by ayvaak)
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Thank you again.

Any idea on why this is cos 65 instead of sin 65?
That is sin 65, I think you typed it into your calculator wrong.
13. (Original post by ayvaak)
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Thank you again.

Why cant you do these calculations in radians as apposed to degrees mode on the calc?
I'm not really sure on that reason tbh. I guess for accuracy purposes maybe. (I'm not 100% sure. Not really very good in Physics Unit 4 lol. )
14. (Original post by ayvaak)
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Thank you again.

Why cant you do these calculations in radians as apposed to degrees mode on the calc?
because the angle is measured in degrees
15. (Original post by klosovic)
because the angle is measured in degrees
Ahaha. Sorry that was really stupid of me

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Anyone have an explanation for why a force is created in the opposite direction to the magnetic field in this question?
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16. Can anyone explain to me question 18 (b)(ii) http://qualifications.pearson.com/co...e_20130613.pdf . Isn't max emf induced when the coil is perpendicular (0 degrees on the graph?).
Here is the mark scheme : http://qualifications.pearson.com/co...c_20130815.pdf
17. (Original post by Moe21)
Can anyone explain to me question 18 (b)(ii) http://qualifications.pearson.com/co...e_20130613.pdf . Isn't max emf induced when the coil is perpendicular (0 degrees on the graph?).
Here is the mark scheme : http://qualifications.pearson.com/co...c_20130815.pdf
I believe the reason why is because on any graph the maximum point will be a stationary point, and therefore the gradient at that point will be 0. So if you plot a graph of flux linkage, the point where the flux linkage is greatest i.e. when the plain of the coil is perpendicular to the magnetic field, this point will have a gradient of 0. As induced emf is proportional to the rate of change of flux, which is the gradient of the graph, then the emf induced when the flux linkage is at it's maximum is 0.
18. Hey can anyone help with question 5 and question 15. c)

http://qualifications.pearson.com/co...e_20100128.pdf
19. Nm on 15.c)
20. (Original post by Dr.strange12)
Hey can anyone help with question 5 and question 15. c)

http://qualifications.pearson.com/co...e_20100128.pdf
Q5 use F= BILsin30

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