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geek101 THANK U VERY MUCH!!!!!!!!!!!!!!!
Thank you so muchnow see, for the error bars lets say the error is +- 0.5 and the point is 22 for the x axis
then the error bar must be from >> 21.5 and 22.5
how do you make the error bar now, for this you must see what is the value of one square on the x-axis. Lets assume one box on the x axis is 0.05. Which means and error of 0.5 on each side will cover 10 boxes on the x axis. To draw the error bar you will draw a horizontal line (because error is for a value on the x-axis) which will be 10 boxes behind 22 and 10 boxes ahead of it!
If the point 22 was for the y axis, then the error bar would be a vertical line of 10 boxes above and below 22.
hope this can help
but sometimes they tell to join all the error bar peaks. how can it be done?now see, for the error bars lets say the error is +- 0.5 and the point is 22 for the x axis
then the error bar must be from >> 21.5 and 22.5
how do you make the error bar now, for this you must see what is the value of one square on the x-axis. Lets assume one box on the x axis is 0.05. Which means and error of 0.5 on each side will cover 10 boxes on the x axis. To draw the error bar you will draw a horizontal line (because error is for a value on the x-axis) which will be 10 boxes behind 22 and 10 boxes ahead of it!
If the point 22 was for the y axis, then the error bar would be a vertical line of 10 boxes above and below 22.
hope this can help
can anybody please fill out the errors section in this paper's question 2 so that i can match ...... MS doesnt give the error values :\ :|
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_52.pdf
thank you
Find the gradient first and then round it. For part d, h is the same as gradient. For g its not the same. log g = y intercept. so to find g u have to inverse log or 10^ ur y intercept value. Then to find error u use the same way of log. Find inverse log highest value of y intercept - inverse log y intercepthey guys in qs 2 ofn09/51 in calculating gradient im calculating gradient by using these values= 2.795-2.988/2.519-2.47. So should i round these values to 3 sig figures or should i round the answer of gradient that comes to 3 s.f. If i round these values before calculating the gradient the value of gradient is changed a little. WHat is the procedure for this can anybody tell me. And in d part the error for h and g should be carried from gradient and y intercept right. for e.g if error of intercept 0.05 then for g it is also 0.05?
Dont worry ! You can score a minimum of 5-10 marks by just writing basic lines. Just write as much as u know about the given experiment. The variables, graph drawing, safety precautions and anything related to your experimentsMy problem is I don't know how to describe experiments or how to draw an appropriate diagram !!
Soldier313
Can you help me with the question i posted above?
in j10/52 the answer to 2d is coming 987+- 24.1 This absolute uncertainty is very strange to me. Can anybody check their answers to this part and confirm that the uncertainty i have calculated is correct or not . to calculate uncertainty i used formula= g from best fit line grad- g from worst fit line gradient to calculate the absolute uncertainty or error. Can anybody confirm plz
and can anybody also tell what is their percentage uncertainty coming. Mine is 4.44% I am having trouble taking it out
i got an error of 19.5, i think what you are doing is correct
i did it this way though :
[ (error in gradient)/ (gradient of best fit) ] x g
and my percentage error value is 3.99 %
I'd prefer if someone else confirms with this though
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