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

1. Any websites that go through past papers? Like exam solutions does for maths.
2. (Original post by xMillnsy)
Any websites that go through past papers? Like exam solutions does for maths.
There is a youtube channel which looks like it goes through some papers might wanna take a look if it covers Unit 4 ( I know it covers some Unit 5 options) : https://www.youtube.com/channel/UCy2...6yKm8nrRGjhs-g
3. Can anyone help???
Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
Thanks
4. (Original post by Lili1998)
Can anyone help???
Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
Thanks
was the answer 4 degrees celcius?
5. (Original post by xMillnsy)
Any websites that go through past papers? Like exam solutions does for maths.

Also I will be doing some, they are just waiting to be uploaded

-bills7187

I will also be doing specification overviews for unit 4 and 5 like you can see I have done for AS units on my channel

I have already recorded both overviews for unit 4 and 5 but they just need to be uploaded

Enjoy and good luck in your exam
6. (Original post by boyyo)
was the answer 4 degrees celcius?
That might be the temp change right? ml used or 2.4(75) x 104 seenmc∆θ used or seen
heat lost by cooling water = heat gained in melting ice + heat gained by melted icefinal temp = 3.35 oC-3.38 oC 4.1(4) oC gains 3 marks Thanks
7. (Original post by Lili1998)
That might be the temp change right? ml used or 2.4(75) x 104 seenmc∆θ used or seen
heat lost by cooling water = heat gained in melting ice + heat gained by melted icefinal temp = 3.35 oC-3.38 oC 4.1(4) oC gains 3 marks Thanks
the 4 degrees that I got was the final temp though. So what was your final tempreature?
8. Can anyone help me with this electromagnetic induction question? Topic 8.2, question 4

A small circular coil of diameter 15mm and 25 turns is placed in a fixed position on the axis of a solenoid. The magnetic flux density of the solenoid at this position varies with current according to the equation B=kl, where k=1.2*10^-3TA^-2

(a) calculate the flux linkage through the coil when the current in the solenoid is 1.5A.

I don't understand how I would obtain the magnetic flux density with this data to then calculate the flux linkage?
9. (Original post by Music With Rocks)
Can anyone help me with this electromagnetic induction question? Topic 8.2, question 4

A small circular coil of diameter 15mm and 25 turns is placed in a fixed position on the axis of a solenoid. The magnetic flux density of the solenoid at this position varies with current according to the equation B=kl, where k=1.2*10^-3TA^-2

(a) calculate the flux linkage through the coil when the current in the solenoid is 1.5A.

I don't understand how I would obtain the magnetic flux density with this data to then calculate the flux linkage?
Look at the units of k and see how you can use the given current to isolate simply the magnetic flux density. Then work out the area of one coil, using the given diameter and combine everything with the equation flux linkage = BAN and you should be done

Nice question - where is it from?

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10. Does anyone have a link to, or know where I can get, the June 2015 paper? Really want to have a go at this because of the lack of papers that are available. The model answers would also be fine
11. (Original post by MintyMilk)
Yeah, I was typing from my phone and didn't double check which way around it was. Same logic still applies, just the other way around.

The logic is the important thing
So you always use the shorter time one??

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12. (Original post by kingaaran)
Look at the units of k and see how you can use the given current to isolate simply the magnetic flux density. Then work out the area of one coil, using the given diameter and combine everything with the equation flux linkage = BAN and you should be done

Nice question - where is it from?

Posted from TSR Mobile
Ohhhhhhh... I was reading the I as an L somehow haha, thought it involved the length oops... Thank you for your help

It is from the official AQA Physics A textbook, the one produced by Oxford printing press. It has the guy on the skateboard on the cover.
13. (Original post by Music With Rocks)
Ohhhhhhh... I was reading the I as an L somehow haha, thought it involved the length oops... Thank you for your help

It is from the official AQA Physics A textbook, the one produced by Oxford printing press. It has the guy on the skateboard on the cover.
Oh haha, nice
14. Has everyone done all the papers?
15. (Original post by grace_zzz)
Ohh it now makes a bit more sense. Thanks alot!!
About that question, it came up in one of my lessons and I came up with some sort of explanation afterwards. Here's an extract from the email I wrote to my teacher, and my teacher didn't have any objection so hopefully it makes some sense.

It occurred to me that BANwsinwt gives the emf induced at that particular value of t so you cannot just use that emf value for average because then that means you have disregarded all the emf induced before reaching that final point.*

In order to find average emf induced you have to integrate BANwsinwt with respect to t to find the total emf induced during a period of time, then divide by the time taken, which turns out to be the same as using ∆BAN/∆t.*

This is because the integral of BANwsinwt is -BANcoswt, and if we let t= a to b, the value of the integral is -BAN(coswb - coswa), which is the total emf induced between t= a and b. For average, divide by time taken -> -BAN(coswb - coswa)/t. And when we say ∆BAN/∆t, what we are really saying is (BANcoswb - BANcoswa)/∆t for a turning coil, which is identical to the result obtained through integration. (The negative sign is a little awkward but numerically it works, so never mind!) If we use the values in the past paper question, w=π and t=0.5, so -BAN(cosπ/2 - cos0)/t = BAN/t, so we did not have to bother with cosine and all that business.
16. Need some help again How would I go about doing part (b) and (c). Also will questions like this come up in the exam?

(sorry for the image size lol)
17. Actually managed to solve it eventually, but would still like to know if it is on the spec. I would be very surprised if it was after doing that.
18. (Original post by Music With Rocks)
Can anyone help me with this electromagnetic induction question? Topic 8.2, question 4

A small circular coil of diameter 15mm and 25 turns is placed in a fixed position on the axis of a solenoid. The magnetic flux density of the solenoid at this position varies with current according to the equation B=kl, where k=1.2*10^-3TA^-2

(a) calculate the flux linkage through the coil when the current in the solenoid is 1.5A.

I don't understand how I would obtain the magnetic flux density with this data to then calculate the flux linkage?
I might be being extremely dense but what is K in B=kl? does it stand for fi or something?
19. (Original post by philo-jitsu)
I might be being extremely dense but what is K in B=kl? does it stand for fi or something?
K is just a given constant in this situation. So it doesnt really stand for anything, notice the units for B are T, for K are TA^-1 and for I are A. So the A and the A^-1 cancel leaving you with T on the RHS which matches the LHS.

So k is B/I so multiplying by I gives you B as required, but k is just a constant that is given so this breakdown is not needed when answering the question.

Does that explain it? Or did I just confuse you further? Haha
20. (Original post by Music With Rocks)
K is just a given constant in this situation. So it doesnt really stand for anything, notice the units for B are T, for K are TA^-1 and for I are A. So the A and the A^-1 cancel leaving you with T on the RHS which matches the LHS.

So k is B/I so multiplying by I gives you B as required, but k is just a constant that is given so this breakdown is not needed when answering the question.

Does that explain it? Or did I just confuse you further? Haha
Ohhh so its a way to find the B at a particular point because it varies?.... I was worried I had a huge gap in knowledge lol

also is the answer by any chance 1.06*10^-3T???

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