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

[nafeekhan]

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.

 

[Ahmed Tariq]

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.

 

[nafeekhan]

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 but everyone else said you must substract it

 

[Saad Sarfraz]

NO BUH I ALSO GOT 1000 FREQUENCY and almost all my frnds got the same how can u say it was 500

It was 500 for sure

 

[Saad Sarfraz]

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

 

[raamish]

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​

 

[[D]UNK]

i dunno if they r rite
but i got 13m/s for the speed dunt remember the rest

yea the speed was 13

 

[[D]UNK]

It was 500 for sure

yup it was 500 u had to divide by 2 coz the distance u measerd was between two nodes or anti nodes or wht ever

 

[xxxtoughxxx]

The paper was either TOO long for 1 hour or 1 hour is TOO short for the paper... :/
I cudnt draw the graph

u r variant 22 right?

 

[raamish]

some1 help me plzzz

 

[xxxtoughxxx]

did anyone take 21?

yep apart from electricity it wux ok...

 

[MysteRyGiRl]

ppl wat abt v1???

 

[Rakayz]

sum1 plz explain....... i have a doubt in practicles: w11-qp-34...question 2) f-i & f-ii

thanks

 

[AdeelRox]

something like 1567 or 1576 some thing like that for T

dude it was T=1517

 

[xxxtoughxxx]

graph??????

hahaa dw same as my reaction i think dis variant 2

 

[mustafa arif02]

The paper was either TOO long for 1 hour or 1 hour is TOO short for the paper... :/
I cudnt draw the graph

ya same here

 

[AdeelRox]

I wrote 'Initially, the neutrons have some kinetic energy.'
Meh.

i also wrote the same

 

[00tanveer]

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.

 

[00tanveer]

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.

 

[00tanveer]

Uncertainties must always be quoted to 1 significant figure. So it should be 9x10^-5. And C can be kept to 3 sig. fig.