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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 !!![]()
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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