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physics controlled assessment urgent!

hey i have a controlled assessment in 4 days and i am about to do the evaluative bit and basically we have to learn how to find out percentage error, percentage difference, absolute uncertainty, percentage uncertainty and many other confusing terms like that. i have approached my teacher a couple of times but he just seems to like shouting and answering my questions with questions and has not done any examples of how to go about calculating these values.

if anyone could explain to me these terms and how to calculate them using equations that would be great

i found this link to an old evaluative one and i have been able to do question 1 and 2 to which i got 2% and 3.7%
if someone could confirm if i have done this right and show me how to work out questions 3 and 4 i will eternally be eternally grateful
here is the link:http://www.ocr.org.uk/Images/70734-unit-g483-practical-skills-in-physics-1-evaluative-task-specimen.pdf
Original post by thetraveller98
hey i have a controlled assessment in 4 days and i am about to do the evaluative bit and basically we have to learn how to find out percentage error, percentage difference, absolute uncertainty, percentage uncertainty and many other confusing terms like that. i have approached my teacher a couple of times but he just seems to like shouting and answering my questions with questions and has not done any examples of how to go about calculating these values.

if anyone could explain to me these terms and how to calculate them using equations that would be great

i found this link to an old evaluative one and i have been able to do question 1 and 2 to which i got 2% and 3.7%
if someone could confirm if i have done this right and show me how to work out questions 3 and 4 i will eternally be eternally grateful
here is the link:http://www.ocr.org.uk/Images/70734-unit-g483-practical-skills-in-physics-1-evaluative-task-specimen.pdf


I agree with your answer to question 1 but not to question 2. :no:

The cross sectional area = πr2=πd24\pi r^2 = \dfrac{\pi d^2}{4}.

As you're squaring the diameter, you need to multiply the percentage uncertainty by 2. :smile:
Original post by usycool1
I agree with your answer to question 1 but not to question 2. :no:

The cross sectional area = πr2=πd24\pi r^2 = \dfrac{\pi d^2}{4}.

As you're squaring the diameter, you need to multiply the percentage uncertainty by 2. :smile:


ah yes i see what i did wrong thanks:smile:

could you help out with question 3 and 4 please

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