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Absolute error in gradient= gradient of best fit line- gradient of worst acceptable
absolute error in y intercept= (y intercept calculated using best fit line)- (Y intercept found using worst acceptable)
Tip- To find the y intercept
1. You will probably have the equation ordered in y=mX+C form
2. Substitute a coordinates for y and x in y=mx+c from best fit line for x and y and also the gradient of the best fin line and find C of best fit
3. Similarly find the y intercept of worst acceptable using a coordinate in WA and gradient of WA
4. the difference of the 2 intecpts give their absolute uncertinity
Good Luck and Best regards (SRI LANKA ROX)
Absolute error in gradient= gradient of best fit line- gradient of worst acceptable
absolute error in y intercept= (y intercept calculated using best fit line)- (Y intercept found using worst acceptable)
Tip- To find the y intercept
1. You will probably have the equation ordered in y=mX+C form
2. Substitute a coordinates for y and x in y=mx+c from best fit line for x and y and also the gradient of the best fin line and find C of best fit
3. Similarly find the y intercept of worst acceptable using a coordinate in WA and gradient of WA
4. the difference of the 2 intecpts give their absolute uncertinity
Good Luck and Best regards (SRI LANKA ROX)
Absolute error in gradient= gradient of best fit line- gradient of worst acceptable
absolute error in y intercept= (y intercept calculated using best fit line)- (Y intercept found using worst acceptable)
Tip- To find the y intercept
1. You will probably have the equation ordered in y=mX+C form
2. Substitute a coordinates for y and x in y=mx+c from best fit line for x and y and also the gradient of the best fin line and find C of best fit
3. Similarly find the y intercept of worst acceptable using a coordinate in WA and gradient of WA
4. the difference of the 2 intecpts give their absolute uncertinity
Good Luck and Best regards (SRI LANKA ROX)
Suppose your uncertainty is .007 for .123 value of t^2 and according to the scale of y-axis one small square is equal to (.12-.10)/10= .002.So u divide .007 by .002=3.5.Now u draw a vertical line from the point of .123 covering 3.5 block in upward and downward direction.And where that vertical line finishes u draw a small (almost of the size of smallest square of the graph)horizontal line to show its limits both upward and downwards.hey thaaank u soo much, and can u tel me or post picture of how to draw error bars like in quest 2 oct/nov 2009 p52?
Suppose your uncertainty is .007 for .123 value of t^2 and according to the scale of y-axis one small square is equal to (.12-.10)/10= .002.So u divide .007 by .002=3.5.Now u draw a vertical line from the point of .123 covering 3.5 block in upward and downward direction.And where that vertical line finishes u draw a small (almost of the size of smallest square of the graph)horizontal line to show its limits both upward and downwards.
Its not the graph of the A level question but give u rough idea of how a error bar looks.
Yes 3 box vertically up and downwards also.oh thaaaaank u soo much, great ambiguity cleared, llike in(0.12+- 0.006) if 1 ssmall box = 0.002 then we will draw error bars covering 3 boxes right?
god bless you!
oh thaaaaank u soo much, great ambiguity cleared, llike in(0.12+- 0.006) if 1 ssmall box = 0.002 then we will draw error bars covering 3 boxes right?
god bless you!
The pic is to blur to see where the worst fit is joining the top error bar.is my graph and error bars correct?
Read my comment# 86 https://www.xtremepapers.com/community/threads/physics-paper-5-tips.12941/page-5is my graph and error bars correct?
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/...nd AS Level/Physics (9702)/9702_w11_qp_52.pdf
thank you
Hey SIS,
I cant read your values. Please can you upload a better picture.
It seems that you error bars is okay.
Here are some tips
1- always draw your graphs using a very fine pencil ie Sharpen your pencil when plotting a graph)
2- Don't just limit your graph to fist and last point, extend it
3- i was amazed to hear that the line of best fit don't necessarily have to pass through all the point BUT should have an even spread. That is equal number of points on either side of the graph.
PS - your WA line is no clear Please see my diagram
HOPE all this will be help ful to you,
Posted by a proud lankan lion
Hey SIS,
I cant read your values. Please can you upload a better picture.
It seems that you error bars is okay.
Here are some tips
1- always draw your graphs using a very fine pencil ie Sharpen your pencil when plotting a graph)
2- Don't just limit your graph to fist and last point, extend it
3- i was amazed to hear that the line of best fit don't necessarily have to pass through all the point BUT should have an even spread. That is equal number of points on either side of the graph.
PS - your WA line is no clear Please see my diagram
HOPE all this will be help ful to you,
Posted by a proud lankan lion
fb.junks can u tel me how to calculate uncertainities in y intercept and in constants? also how to determine y intercept as in paper http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_51.pdf ? my both WA and best fit intercept is 0
use the formula y=mx+c . c is y-intercept. first use the gradient and x and y values f best fit line gradient. Then do it with the worst fit line gradient. The y intercept of the best fit line gradient is the answer and the error will be the difference between the 2 y intercepts.
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