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# 2016 Official AQA New Spec AS Level Physics Paper 2 - 9th June 2016 watch

1. (Original post by vvlp)
This is a weird topic. When measuring with analogue (e.g. a ruler), the uncertainty is +-0.5mm (assuming the smallest division is 1mm). This taken into account, you're not 100% certain that you're measuring from 0cm, hence:

I've been taught that when using digital equipment the uncertainty is just the smallest division, not halved or anything.
That is true for digital equipment
2. Say we fix the end of a ruler would that get rid of uncertainty where the zero is meaning that the uncertainty is +-0.5mm
3. (Original post by vvlp)
This is a weird topic. When measuring with analogue (e.g. a ruler), the uncertainty is +-0.5mm (assuming the smallest division is 1mm). This taken into account, you're not 100% certain that you're measuring from 0cm, hence:

I've been taught that when using digital equipment the uncertainty is just the smallest division, not halved or anything.
how about a micrometre / caliper assuming that there is no zero error?
4. Anyone please tell me how to do the first question in the specimen paper2? The percentage uncertainty one..

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5. Hi, can someone help me with this question on the specimen paper, mind has just blanked.
6. (Original post by haes)
how about a micrometre / caliper assuming that there is no zero error?
With no zero error there is an uncertainty of +-0.05mm (assuming that the smallest division is 0.1mm).
7. (Original post by Megannk)
Hi, can someone help me with this question on the specimen paper, mind has just blanked.

Here watch this Its how I learnt
8. (Original post by Cinna21)

Here watch this Its how I learnt
Thanks!!
9. (Original post by omer_bawa)
Anyone please tell me how to do the first question in the specimen paper2? The percentage uncertainty one..

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%u.c. = normal u.c. / actual value x100

I struggled to see what the uncertainty actually was as it dosent say but since it has 3.5 I just went with plus or mius 0.1 9the percision basically)
10. (Original post by Kraixo)
Attachment 544417544419Attachment 544417

Hi can someone explain how to answer those 2 questions thanks.
the first one, i would do it like this:

u know the voltage in each branch of the circuit is the same as the source.
then u can invent a voltage for the source - say 1v
then find the current in each branch using ratios
then once u have the current in pq u can find the resistance using v=I*R
11. Ddi anyone compile the topics that didn't come up in paper 1?
12. I got it right when I did the mock exam but can someone explain this question please:

The cell in the following circuit has an emf (electromotive force) of 6.0 V and an internal resistance of 3.0 Ω. The resistance of the variable resistor is set to 12 Ω. How much electrical energy is converted into thermal energy within the cell in 1 minute?

Thanks.
13. (Original post by RKM21)
I got it right when I did the mock exam but can someone explain this question please:

The cell in the following circuit has an emf (electromotive force) of 6.0 V and an internal resistance of 3.0 Ω. The resistance of the variable resistor is set to 12 Ω. How much electrical energy is converted into thermal energy within the cell in 1 minute?

Thanks.
You would calculate charge and use emf=E/Q.
14. (Original post by RKM21)
I got it right when I did the mock exam but can someone explain this question please:

The cell in the following circuit has an emf (electromotive force) of 6.0 V and an internal resistance of 3.0 Ω. The resistance of the variable resistor is set to 12 Ω. How much electrical energy is converted into thermal energy within the cell in 1 minute?

Thanks.
Total resistance of the circuit is 15 ohms (3+12)
So current is 6/15 = 0.4A
The PD across the internal resistance is 0.4*3 = 1.2V
P=IV= 0.4*1.2 = 0.48 J/s
60 seconds in a minute so 0.48*60 = 28.8 J = 29 J
15. (Original post by haes)
Total resistance of the circuit is 15 ohms (3+12)
So current is 6/15 = 0.4A
The PD across the internal resistance is 0.4*3 = 1.2V
P=IV= 0.4*1.2 = 0.48 J/s
60 seconds in a minute so 0.48*60 = 28.8 J = 29 J (C)
thx
16. (Original post by champ_mc99)
Ddi anyone compile the topics that didn't come up in paper 1?
it dosent work like that... anything can come up in paper 2 multiple choice, the experiment is literally random and dosent have to be on the spec... and the section B general questions will be based on experiment so u can work out the possibilites 9obviously not particles)
17. (Original post by maximo17000)
%u.c. = normal u.c. / actual value x100

I struggled to see what the uncertainty actually was as it dosent say but since it has 3.5 I just went with plus or mius 0.1 9the percision basically)
How do u work out the precision? I am so lost in physics.. I am failing tomorrow

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18. (Original post by omer_bawa)
How do u work out the precision? I am so lost in physics.. I am failing tomorrow

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the percision is just the smallest measurement an instrument can do, e.g. a ruler goes up in 1mm increments so the percision is 1mm and the u.c. of a ruler is plus or minus 1mm.

assuming the ruer has mm marks on it of course
19. (Original post by maximo17000)
the percision is just the smallest measurement an instrument can do, e.g. a ruler goes up in 1mm increments so the percision is 1mm and the u.c. of a ruler is plus or minus 1mm.

assuming the ruer has mm marks on it of course
So in that question if u do apply the formula u stated how do u get 2.9% (thats the answer to that question)

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20. (Original post by omer_bawa)
So in that question if u do apply the formula u stated how do u get 2.9% (thats the answer to that question)

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looking at the number of significant figures, it was 3.5x10^3 something, so it must be plus or minus 0.1, if it was 3.55x10^3 it would be plus or minus 0.01

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