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sorry you're seeing a reply after so long! Anyway i cant post a pic,but the basics of the diagram are what is shown in the question,plus:
-a meter rule placed vertically "behind" the wire,to measure distance
-micrometer to measure diameter of wire
-rheostat in circuit to keep current constant
and ammeter in current
thats it!
hope i helped!
thats decent and nice diagram bro ... thanks for it ....
Quantity doesnt matter if u feel u need to do more u can but i dont think its time to try new things bcoz paper is tomorrow so go through all the ones u have done and u will do great.are we suppose to do such old p.papers? i did from 2007 to 2012
No it's okay! No need to apologize
Thank you so much! I drew the same thing, except for the rheostat. I dont know how to draw a rheostat
Quantity doesnt matter if u feel u need to do more u can but i dont think its time to try new things bcoz paper is tomorrow so go through all the ones u have done and u will do great.
And 2007 to 2012 are enough i suppose
okthis is enough...but if u get time do look at planning excercise of old papers
Hey, i'm really sorry I couldn't post it earlier, I was kinda pressed for time on other matters.
Sorry if the diagram isn't good enough (It looks bad, but in this time, it's the best I could do - sorry!)
View attachment 28405
Good Luck for all your exams!
magnesium how to calculate %age uncertainity? abso/measure value*100?
can u tel me procedure and diagram of oct/nov10 p53 q1?
yes it is correct...uncertainity / value * 100
ok thaaanku god bless
v = s/t where s=distance, t=time
v^2 = (s/t)^2
so v^2 = (0.05/t)^2
then substitute values of time from each row into the eqn
eg for row 1, v^2 = (0.05/0.046)^2
so v^2 = 1.18
then uncertainty is (max value of v^2 - min value of v^2) / 2
so max value of v^2 = (0.051/minimum time)^2
min value of v^2 = (0.049/ max time ) ^2
subtract these two values and then divide by 2 to obtain uncertainty
Hope that helped
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