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HOW WAS PHYSICS PAPER 2 AS!!!!!!!!!!!

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you are correct...dont worry, no need to take inverse, you are just dividing by 1.2 * 10^-6 on other side.....so take into account the simple uncertainti +-0.01.....no need for minus sign

I know, but most of my friends tell me that since we are taking inverse of V/t, the power will come infront and it will become negative :S..
I hope you and I are right. My uncertainty value was 8.9 X 10^-5

also , my value of C came 1.035X10^-3
in the next question after uncertainty, the significant figure one,
I wrote it as : 1.04 X10^-3 + or - .089 X 10-3

is it right? some of them are telling I have to round up .089 to .09 :S.. what are your remakrs about that?


BTW, if my uncertainty value is wrong, will they allow ecf for the appropriate significant figure question ?

ANYONE LOOKING AT THIS, WOULD BE NICE IF YOU ANSWERED .. :) thanks.
 
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I know, but most of my friends tell me that since we are taking inverse of V/t, the power will come infront and it will become negative :S..
I hope you and I are right. My uncertainty value was 8.9 X 10^-5

also , my value of C came 1.035X10^-3
in the next question after uncertainty, the significant figure one,
I wrote it as : 1.04 X10^-3 + or - .089 X 10-3

is it right? some of them are telling I have to round up .089 to .09 :S.. what are your remakrs about that?


BTW, if my uncertainty value is wrong, will they allow ecf for the appropriate significant figure question ?

ANYONE LOOKING AT THIS, WOULD BE NICE IF YOU ANSWERED .. :) thanks.

Since the errors in the raw data were to 1 significant figures each, the error of 8.9x10^-5 had to be written as 9x10^-5, i.e. 1 significant figure. You would probably lose only the last 1 mark.
 
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wh
Since the errors in the raw data were to 1 significant figures each, the error of 8.9x10^-5 had to be written as 9x10^-5, i.e. 1 significant figure. You would probably lose only the last 1 mark.
at about the uncertainty value?
Do you think it is right?
Since we were taking inverse of V/t, were we suppose to subtract its uncertainty? I added it :p but everyone else said you must substract it :(
 
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Guys if you get the First part wrong in Q1 will ecf be used. Caz i think a made a mistake in the calculation i got
C= 1.30 x 10^-3
Uncertanitny= 0.11 x 10^-3
T= 1.3x10^-3 +- 0.1x10^-3
 
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hey guys can any1 help me my physics practical ppr is tomorrow and i have some qs like:

hmm guys need help in significant figures:

my readings are:
V/V:8.0 V
I/A:0.32
P/W:2.56
R/ohm:25.0
R4/ohm: 3.9*10^5

Guys please check my readings and tell me how many significant figures i should give for each reading
 
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sum1 plz explain....... i have a doubt in practicles: w11-qp-34...question 2) f-i & f-ii

thanks
 
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It was in quilibrium because there was no resultant force acting on the log as suggested by its motion with a constant velocity. This answer should relate to the definition you had written for Newton's 1st Law earlier in the question. :)
 
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No it was not because some work is also done against friction. (This should be sufficient for 2 marks) So output power would be rather the sum of gain in potential energy per unit time and work done against friction per unit time. :)
 
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Uncertainties must always be quoted to 1 significant figure. So it should be 9x10^-5. And C can be kept to 3 sig. fig. :)
 
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