Qualification > Sciences

A2 physics p5

<< < (7/16) > >>

Eye:
How do you calculate the absolute errors in the second question of p5 and how do you draw the worst acceptable line?

khagga_926:
open this and check dat...start reading from half page 20-22

Eye:
Thanks bud.
(:

taha12:
Ok man this is how you go
in question 2 p5
suppose
I/V         V/V          lg(I/I)       lg(V/V)
2             4+/- 0.1
3            7+/- 0.1

you can simply calculate values of Log I example log 2 =0.301 write it in the column
lg(I/I)

for log V you calculate value  Log V example log 4= 0.602 write it in the column under lg(V/V)

absolute errors in lg (V/V):
error in V/V is +/- 0.1 thus absolute error in lg (V/V) would be *
take the first value in the group V/V which is 4
(log 4)  - (log 3.9) = 0.01 to 1 sig fig
       or
(log 4.1)- (log 4) = 0.01 to 1 sig fig
note that 3.9 or 4.1 value comes from subtracting or adding 0.1 from actual value which was 4 see column above of V/V
both ways are correct

same for the second value 7 +/- 0.1
log 7.1 - log 7.0
or
 log 7 - log 6.9

another example

100 +/- 20

you do for absolute errors

log 100 -log 80

or

log 120 -log 100

just remember to either add or subtract the error from actual value
tricky
if its square of distance

suppose d/cm = 2 cm+/- 0.1
the errors in "d square world be"
(2.1 square ) - 2.0 square
  or

2.0 square - 1.9 square

this would give you error
in this case d square = 4+/-0.4

hope it solves it all for every one

bilalar:
I/V         V/V          lg(I/I)       lg(V/V)
2             4+/- 0.1
3            7+/- 0.1


another way of calculatin the errors is :

lg 4 = 0.602

take (lg 4.1 - lg 3.9)/2
this will give  error +/- 0.01

this is easier and faster method

Navigation

[0] Message Index

[#] Next page

[*] Previous page