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R = 750 ± 20
find lnR and its absolute error?
How would you find the uncertainty with a logarithm?
find lnR and its absolute error?
How would you find the uncertainty with a logarithm?
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which paper is this ?R = 750 ± 20
find lnR and its absolute error?
How would you find the uncertainty with a logarithm?
Question 2, May/june 2008, paper 5which paper is this ?
where did ln770 come from?Absolute uncertainty should be ln770 - ln750 and take it to 1 sig figure.
we don't use 730 at all ?!Absolute uncertainty should be ln770 - ln750 and take it to 1 sig figure.
this might help.How to calculate absolute uncertainties for divisions?
Example, formula = d/v
V is 10 ± 1
D is 13 ± 0.5
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf
Q7,9,20,22,24,25,27,33.... Please
can some one tell me how to know that this thing is independent variable and dependent mostly in q1 asked
Don't have time will help with 4 .
Just use SI units of the axes.
Power x time
Js^-1 * s
J = joules
There is only one answer for energy so it's A.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf
Q7,9,20,22,24,25,27,33.... Please
Please anyone solve for me.The image is blurred sorry for that. Thanks a lot in advance.
You are life saver! thanksFrom the diagram, we can see that resistors A and B are connected to each other in series. Therefore, their total resistance R(A+B) is equal to R + R = 2R.
Furthermore, together they are connected in parallel to D, so 1/R(A+B+D) = 1/(2R) + 1/(R) = 1/(2R) + 2/(2R) = 3/(2R)
Therefore, R(A+B+D) = 2R/3.
This resistance, R(A+B+D) is connected in a series arrangement to resistor C, so their total resistance R(A+B+C+D) = (2R)/3 + R = (2R + 3R)/3 = 5R/3
This overall is connected in parallel to resistor E, so the resistance R(A+B+C+D+E) is given by
1/R(A+B+C+D+E) = 1/R(A+B+C+D) + 1/R(E) = 3/(5R) + 1/R = 3/(5R) + 5/(5R) = 8/(5R)
Therefore, R(A+B+C+D+E) = 5R/8 ==> C.
Hope this helped!
Good Luck for all your exams!
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