June 2013 Edexcel exams!

[jtbteddy]

Do you have for physics??? unit 3b? practical? need it !!

I havent made any notes for phys unit 3b you just need to revise units 1 amd 2 and do as many past papers as you can for that however, i have an edexcel document which has a lot of info about unit 3b, ill post that tomorrow morning, mybee it helps you a bitt. Have you begun revising yet?

 

[Unique_39]

Although you figured it out, i just wanted to point out that there are two ways of finding the uncertainty:
1. Highest value - mean
The highest value from the set of values you used to calculate the mean with
2. Range/2
Again, the range should be calculated from the set of values you used to calculate the mean with, dont include any obvious anomalies in your calculations.
Both these calculations should give you the same answer, hope this helped!

So what u mean to say is, the highest value will be subtracted with the lowest value from the set of values used to calculate the mean with? And then this divided by two? Then what does range/2 find? ...I'm sorry , I think I'm confused

 

[abdul aziz123]

I havent made any notes for phys unit 3b you just need to revise units 1 amd 2 and do as many past papers as you can for that however, i have an edexcel document which has a lot of info about unit 3b, ill post that tomorrow morning, mybee it helps you a bitt. Have you begun revising yet?

yeah i did the past papers i know the pattern ! yeah if u can post them tom it would be great! precision erealting to instruments are shit! and graphs

 

[jtbteddy]

So what u mean to say is, the highest value will be subtracted with the lowest value from the set of values used to calculate the mean with? And then this divided by two? Then what does range/2 find? ...I'm sorry , I think I'm confused

Okay let's use question 1 from June 2012 as an example, you have to find the mean and uncertainty of the following values:
0.27 mm, 0.29 mm, 0.26 mm, 0.42 mm, 0.26 mm.
Now, 0.42mm is an obvious anomaly because it is way above the other values, so we're going to discard that and not use it in any of our calculations.
For the mean, you do (0.27 + 0.29 + 0.26 + 0.26) / 4, which is equal to 0.27mm. You should always check that your mean is given to the same number of significant figures as your valaues!
For the uncertainty, you have two options, you can either do range/2 or highest value-mean, they should both give you the same answer:
Option 1: range / 2 --> (0.29-0.26) / 2 = 0.015 which rounds up to 0.02
Option 2: Highest value - mean --> 0.29 - 0.27 = 0.02
Therefore, the final answer is 0.27+-0.02mm, which corresponds to the mark scheme

 

[jtbteddy]

yeah i did the past papers i know the pattern ! yeah if u can post them tom it would be great! precision erealting to instruments are shit! and graphs

Although it's for the AS Practical Assessment, there's a lot of theory involved, so you can use it as a guidance to make notes.

 

[jtbteddy]

no neeed i got them thanx

I updated it, and converted it to pdf, download this one, it's all you need to know for Unit 7:

 

[Unique_39]

Okay let's use question 1 from June 2012 as an example, you have to find the mean and uncertainty of the following values:
0.27 mm, 0.29 mm, 0.26 mm, 0.42 mm, 0.26 mm.
Now, 0.42mm is an obvious anomaly because it is way above the other values, so we're going to discard that and not use it in any of our calculations.
For the mean, you do (0.27 + 0.29 + 0.26 + 0.26) / 4, which is equal to 0.27mm. You should always check that your mean is given to the same number of significant figures as your valaues!
For the uncertainty, you have two options, you can either do range/2 or highest value-mean, they should both give you the same answer:
Option 1: range / 2 --> (0.29-0.26) / 2 = 0.015 which rounds up to 0.02
Option 2: Highest value - mean --> 0.29 - 0.27 = 0.02
Therefore, the final answer is 0.27+-0.02mm, which corresponds to the mark scheme

Thankyou you so much!! I understand it now! Can you help me out with the "relating the gradient of a graph with an equation" questions like the one in Jan 2012 Q8(d)

 

[BeBo^.~]

people i have a question in phy. and i seriously need help
in the specimen phy. unit 3 the first Q. why it is 13.4 ???

 

[Unique_39]

people i have a question in phy. and i seriously need help
in the specimen phy. unit 3 the first Q. why it is 13.4 ???

Which year?

 

[BeBo^.~]

its the sample assessment

 

[jtbteddy]

Thankyou you so much!! I understand it now! Can you help me out with the "relating the gradient of a graph with an equation" questions like the one in Jan 2012 Q8(d)

That's a little bit harder to explain from here, but u'll try! When you get an equation, rearrange it to the form of y=mx+c, in this case, our equation is eV = hf - the work function. Question 8c tells you to plot a graph of V on the y-axis and f on the x-axis. Therefore, in this case, V is y and f is x, so we need to rearrange " eV = hf - the work function" to make V the subject. This could be achieved by dividing "hf - the work function" into e. So you'd get V = (hf-work function) / e. Now, our "c" in this case (the y-intercept) is "-work function/e", and our mx is hf/e. We know from our graph that f is the x, which clearly means that h/e is the gradient. Therefore, to calculate the constant h, we need to find the gradient and multiply it by e.

 

[jtbteddy]

its the sample assessment

Because the wavelength is from one crest to the next, so you have to see the distance between the 2nd "stick" and the 2nd last "stick"

 

[Unique_39]

That's a little bit harder to explain from here, but u'll try! When you get an equation, rearrange it to the form of y=mx+c, in this case, our equation is eV = hf - the work function. Question 8c tells you to plot a graph of V on the y-axis and f on the x-axis. Therefore, in this case, V is y and f is x, so we need to rearrange " eV = hf - the work function" to make V the subject. This could be achieved by dividing "hf - the work function" into e. So you'd get V = (hf-work function) / e. Now, our "c" in this case (the y-intercept) is "-work function/e", and our mx is hf/e. We know from our graph that f is the x, which clearly means that h/e is the gradient. Therefore, to calculate the constant h, we need to find the gradient and multiply it by e.

I think I got a hold of what ur trying to explain...Thanks Alot!!! I really appreciate it! I just hope they don't ask this question in the actual paper

 

[Unique_39]

its the sample assessment

Can you please attach the sample assessment I've been looking for it!

 

[jtbteddy]

Can you please attach the sample assessment I've been looking for it!

 

[BeBo^.~]

Because the wavelength is from one crest to the next, so you have to see the distance between the 2nd "stick" and the 2nd last "stick"

thnx alot know i got it and good luck

 

[heyitsme]

can i have the hodder review questions asap? i have my paper tomorrow!

 

[Manasa]

can someone please teell me what to study for physics unit 3b????????????? notes are most welcome!!!!!!!!

 

[jtbteddy]

can someone please teell me what to study for physics unit 3b????????????? notes are most welcome!!!!!!!!

 

[Manasa]

thnxxx sooo mchh~!!!!