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

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Good luckerino to everyone!
Good luck everyone

Wish you all the best :wink:
Don't stress about the exam, I'm in a worse situation than you. Ive just finished the june 14 paper, so yeah, but I'm totally relaxed as i have been with my other exams because least i know I've tried my best as you have.

Don't worry if a question doesn't make sense or wordy, think about it slowly even if you end up working on it for more than expected. Remember, it's not 1.45 mins per mark. If you end up thinking about a question for more than 2 minutes and you still haven't found the answer and that you think you are not close to the answer then you should leave it and carry one with the rest.

Remember you must tick the answer of MC section on the answer sheet so don't try to choose the answers on the question paper and then transfer them to the answer sheet at the end as this would waste your time. Each time you solve a question just quickly tick the answer on the answer sheet, forget the question paper.

If you are starting with multiple choice then once 45 mins is over start the written one as it's has an easier flow. If you are starting with written part and not concerned much about the MC then finish the written part first then spend the remaining time on MC.

I really hope you will all achieve the best you can.
Good luck everyone :smile:


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Original post by dannyoboy007
How do I solve these?





work out potential of all charges at point P and then just add all the potentials together since its a scalar
Original post by Ilovemaths96
how transformers work:
When alternating pd is applied to primary coil, alternating magnetic field is produced in core.
Field passes through secondary
From faraday's law and alternating magnetic flux induces emf in secondary coil

Anything to add/edit?


That's it really :smile:


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Original post by dannyoboy007
How do I solve these?





First one.. i think you take the charges in isolation. Use V=1/4piEo*Q/R
I got 180V for each charge *2 = total of 360V since potentials are a scalar quantity.

Second one.. same thing, but distance between each charge and P is root 2*a using trig.
Hence, the total potential is 4*(1/4piEo*Q/Root 2*a) which simplifies to Q/PiEo root 2 a.. so it should be b.

Not 100% if these are right though
Original post by Mehrdad jafari
That's it really :smile:


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Awesome good luck mate
GOOD LUCK BOYS AND GIRLS! Remember a geosynchronised satellite orbit is one that orbits its planet in the same time period as it takes the planet to rotate (would be 24h time period orbit for one around earth). I only learnt this today as it got me stuck on a past question, it's just there in case anyone needs it! Adios.
Reply 3848
Original post by darkni35
JUNE 13 Unit4.png
Can anyone help with this question? Im sure its quite simple and i might be missing something but when i use F=mw^2r, im getting 19N..

The answer is apparently D, 24N.

(this is from june 13)


The answer is d because t=mw^2r +mg
Reply 3849
Original post by dannyoboy007
How do I solve these?





Usr Pythagoras so. Square root 2^2 + 2^2 which is 2root2 so the answer is a

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Can someone please help me with Q12 JUNE 2011 section A? I really can't get it, thanks!
Original post by MsFahima
It's basically. Total force = Tension - mg.

So

T - mg = mw^2r

Now try. :smile:

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ahh thank you!!
Awesome, thanks for the help!
Original post by tawaz1997
Can someone please help me with Q12 JUNE 2011 section A? I really can't get it, thanks!


Here is the answer!ImageUploadedByStudent Room1433978304.813365.jpg


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Original post by NotLuis
Here is the answer!ImageUploadedByStudent Room1433978304.813365.jpg


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You're the real MVP, thank you so much!!
Original post by tawaz1997
Can someone please help me with Q12 JUNE 2011 section A? I really can't get it, thanks!


The weight of the object on the surface of the earth is R( you don't know what R is) and so when the object is moved to 3r from the surface the total distance from the centre of the earth to the where the object is, is 4R(3R+R). You should realise that the gravitational force on the object, F=GMm/R2, is the weight of the object at the given distance R measured from the centre of the planet.
The same argument goes for the potential but potential in inversely proportional to R and not R2 as with weight.
ImageUploadedByStudent Room1433978553.237327.jpg


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Original post by NotLuis
Here is the answer!ImageUploadedByStudent Room1433978304.813365.jpg


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Just seeing your answer :smile:


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Original post by NotLuis
Here is the answer!ImageUploadedByStudent Room1433978304.813365.jpg


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Actually something like this, i made a mistake thereImageUploadedByStudent Room1433978796.588141.jpg


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Original post by Mehrdad jafari
The weight of the object on the surface of the earth is R( you don't know what R is) and so when the object is moved to 3r from the surface the total distance from the centre of the earth to the where the object is, is 4R(3R+R). You should realise that the gravitational force on the object, F=GMm/R2, is the weight of the object at the given distance R measured from the centre of the planet.
The same argument goes for the potential but potential in inversely proportional to R and not R2 as with weight.
ImageUploadedByStudent Room1433978553.237327.jpg


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thank you!!
Can someone explain why potential is negative? and how many and which satellite equations do we need to know? cheeeeeeerss

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