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

1. june 13 section a
Question 23
arent b and c also correct as well?
2. Guys make sure to check out the units PA10 from the old spec there are quite a few relevant questions in there you can find them on egsphysics
3. (Original post by oonic0rn)
http://filestore.aqa.org.uk/subjects...1-QP-JAN13.PDF for question 22 i have no idea how to work out the frequency?
By observation, if you compare e=20sin(100pit), to e=BANsin(wt), wt=100pit so w=100pi.

w=2pi/T , so Time period = 1/50 . Freq=1/T , so freq=50Hz.
4. http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF

5. I have two ques:
1) Would you get any marks if you did two distinct working out for one question and one working out was correct although in the answer place you wrote the wrong answer?

2) What if you get a ques wrong like say example with the parcel on the floor van; if you get the equation wrong all the other parts would be affect would ecf still be valid or?
http://filestore.aqa.org.uk/subjects...2-QP-JUN14.PDF

Ohh

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7. Found this on the last past paper I was going to do. Is it a sign?????

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8. (Original post by Kingnig)
Does anyone has a list of past grade boundaries?
on the first page
http://filestore.aqa.org.uk/subjects...2-QP-JUN14.PDF

4) a full circle is 2pi, so using time = distance over speed, 2pi over pi/2 which is 4.
So it takes 4 seconds to go around once. so 6 seconds would be one and a half times so it will be directly opposite P, so a diameter away.

6) you have to make an equation. mv^2/r is equal to mg - T, something along those lines so you get B

8) Use Vmax equation since it has 0 displacement so max speed

13) really tough one, i didnt get this one mate

10. I think it was Newton who said he wasn't sure that change of design to change natural frequency was true but it's here in the mark scheme

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11. One more from me: could anyone help me with this question? Thanks!

12. (Original post by a.a.k)
Ohh

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No worries

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14. What's the best way to revise for 6 markers?

I can look over the bridge applications
Transformers and sending high voltage over long distances

Also, what if the 6 marker is sneakily on momentum again
15. (Original post by aprocrastinator)
what on earth is back emf PLEASE
An emf that opposes the supply voltage (in the opposite direction/ pi out of phase). It opposes the supply voltage because of Lenz' law:

In an electric motor (such as a screwdriver) the supply voltage creates a magnetic field in the coil. If the motor is spinning, it cuts through the field so a change in flux is experienced so an emf is generated in the coil. Because of Lenz' law, this emf must be in the opposite direction to the supply voltage.

A back emf is also generated in a transformer in a similar way: mag field induced in primary, but this time an emf is generated not because the coil is moving but because the voltage alternates. The alternating voltage causes the magnetic field lines to change so the coil experiences a change in flux...
16. What are the differences and similarities between low polar orbits and geosynchronous orbits? Examples for uses of each?
17. (Original post by Dante991)
4) a full circle is 2pi, so using time = distance over speed, 2pi over pi/2 which is 4.
So it takes 4 seconds to go around once. so 6 seconds would be one and a half times so it will be directly opposite P, so a diameter away.

6) you have to make an equation. mv^2/r is equal to mg - T, something along those lines so you get B

8) Use Vmax equation since it has 0 displacement so max speed

13) really tough one, i didnt get this one mate
Thank you so much. The june 14 paper is incredibly difficult, I'm screwed if it's anywhere near as hard tomorrow!
18. (Original post by Dante991)
4) a full circle is 2pi, so using time = distance over speed, 2pi over pi/2 which is 4.
So it takes 4 seconds to go around once. so 6 seconds would be one and a half times so it will be directly opposite P, so a diameter away.

6) you have to make an equation. mv^2/r is equal to mg - T, something along those lines so you get B

8) Use Vmax equation since it has 0 displacement so max speed

13) really tough one, i didnt get this one mate
For 13, you have to equate F and 0.5F for their respective distances.

0.5(QQ/4pi x eo x d ) = QQ/4pi x eo x 20x10-3

All constants cancel, both the charges cancel.
Thank you so much. The june 14 paper is incredibly difficult, I'm screwed if it's anywhere near as hard tomorrow!
Could'nt agree with you more man, way harder than past papers so fingers crossed
20. (Original post by Deddy)
For 13, you have to equate F and 0.5F for their respective distances.

0.5(QQ/4pi x eo x d ) = QQ/4pi x eo x 20x10-3

All constants cancel, both the charges cancel.
Thanks man, I had the right idea but the maths kinda confused me so I couldnt work it out, I hate those sort of questions sometimes

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