when you are done, pls teach me how to get the y intercept and why i cant just read it from the graph? thanks a lot! ur helping a lot
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Physics P5 uncertainty help
- Thread starter Richard
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when you are done, pls teach me how to get the y intercept and why i cant just read it from the graph? thanks a lot! ur helping a lot
Y-intercept means where ur line cuts the y-axis when x = 0 if you don't start with a scale of 0 on x-axis, you can never find y0intercept without calculating. So how to you find the y-intercept?when you are done, pls teach me how to get the y intercept and why i cant just read it from the graph? thanks a lot! ur helping a lot
Example:
You plotted your points and made ur line of best fit. Now you take any point on the line of best fit and u replace in the equation y=mx+c where m is the gradient you have calculated in previous part and c is the y-intercept. Now you need to know that you have gradient of best and worst and their difference in magnitude gives the error in gradient. Same thing applies for y-intercept whereby you replace Gbest and Gworst in y=mx+c to get the Ybest and Yworst and the difference in magnitude gives the error in Y-intercept
okay...any1 else can help me pls???
Y-intercept means where ur line cuts the y-axis when x = 0 if you don't start with a scale of 0 on x-axis, you can never find y0intercept without calculating. So how to you find the y-intercept?
Example:
You plotted your points and made ur line of best fit. Now you take any point on the line of best fit and u replace in the equation y=mx+c where m is the gradient you have calculated in previous part and c is the y-intercept. Now you need to know that you have gradient of best and worst and their difference in magnitude gives the error in gradient. Same thing applies for y-intercept whereby you replace Gbest and Gworst in y=mx+c to get the Ybest and Yworst and the difference in magnitude gives the error in Y-intercept
thanks! understood
Oh boy I don't understand how to find V :S I need to print it and focus I guess Brb laterz with a possible solution
are u asking how to calculate the values or the s.f? i got the values bt no idea why they have different sf for different rows :S
how u got values of V ? I don't understand xD
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for June 2012 p52 Q2 d(ii) and e(ii) could u help me please?
Oh boy I don't understand how to find V :S I need to print it and focus I guess Brb laterz with a possible solution
did u see the question diagram? the length of the card is 0.200m, so we just have to use the length and since time is given, sub into formula (distance/time) to get the speed.
ex : (0.02/174*10-3) = 1.149..thats value of V. square it and u'l get 1.32
Does not sound logical but it's the answer loldid u see the question diagram? the length of the card is 0.200m, so we just have to use the length and since time is given, sub into formula (distance/time) to get the speed.
ex : (0.02/174*10-3) = 1.149..thats value of V. square it and u'l get 1.32
Nop u gave 2011
hehehe..yeah sry...act m posting so many dbts in different threads dat i myself got confusedNop u gave 2011
n yeah jarjar is correct...v is to be found lyk dat
I explained that a bit earlier I quote:
Okay I have checked the paper. Well here is the expln:
TO find b it's simple, you take 1 point on your Line of best fit(which you used to calculated gradient) and then you take g best and you replace in euqation to find t
Now for the undertainty, use the equation in the Mark scheme.
When u see t^2 know that error in t = 2Δt/t
if % uncertainty in t the 2Δt/t x 100
E.g Find error in X^9
=> error in X^9 = 9ΔX/X
% error = 9ΔX/X x 100
That's 1 thing. Now in an equation:
b=m/(R^2), find %uncertainty in R
Given the following Data:
b = 3 +/- 1
m=5 +/- 2
Whether its multiplication of division, you ALWAYS add errors.
Let's solve it:
Δb/b x 100 = Δm/m x 100 + 2ΔR/R x 100
You always divide error by true value.
=> Δb/b x 100 = Δm/m x 100 + 2ΔR/R x 100
=> [(1/3) x100] =[(2/5) x 100] + [2(ΔR/R) x 100]
You need to know that % uncertainty in R = ΔR/R x 100 = Let's call it a
=> [(1/3) x100] =[(2/5) x 100] + 2a
Make a S.O.F you get the % uncertainty in R. Now apply this to your question and Voila
I hope it helps u
Demain massacre huehuehue I need to go and work it out I guess brb laterreally? sorry why its not logical eh?
hehehe..yeah sry...act m posting so many dbts in different threads dat i myself got confused
n yeah jarjar is correct...v is to be found lyk dat
hey have u figured out why its different sf for the bottom two rows??
anyonee???
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I explained that a bit earlier I quote:
I hope it helps u
give an example of how you did it please
hey i got it! its simple actually!I need to go and work it out I guess brb later
see..m explaining everything wid respect to the first row!
v=0.2/(174*10^-3) = 1.149
now v need to find the uncertainty in v...
m puting $=delta..so pls undrstnd
$v/v=($d/d) + ($t/t) but ($d/d)=0
so $v=v$t/t = [1.149*(2*10^-3) ]/(174*10^-3) = 0.01
now for V^2 ...
$v^2=[(2)*($V)*(V^2)]/V and substituting values gives 0.03
hope dis helps...!
another doubt hehe. so if lets say i count my gradient and i get something like 123.87 wt uncertainty 5.28. but since uncertainty cn only be one SF, i make it 124 +/- 5. and then the are follow up questions where u have to find uncertainty in one of the unknown in equation, where gradient is involved..so i must use (5.28/123.87) or (5/124) in the calculation?
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wat are u preparing for the planning ??