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physics p5 discussion

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ok mate look
My Readings are
36 +/-4.8
100 +/-8
196 +/-11.2
324 +/-14.4
etc
Now on the graph the scale is 10 units per a small square

how can I draw the 4.8 bar (which is really small) on such a tiny scale (4.8 cm when 10 cm is small square)
 
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okay well that's freaky, 4.8 Is definitely not going to be visible with a naked eye =P
Here's the thing, our teacher told us to draw the error bars of 0.4 only(for all values of l^2), the one's that have been provided in the table already.. so what you're saying is a totally new concept to me. Lets see how others respond to it because since P5 ain't that old all the techniques are yet to be learned
 
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beacon_of_light said:
fluffycloud said:
Heyyy!
Paper 5 Summer 08 Q2 the last partt.
I don't get how to do?

you have the formula R=Ro e^-ρ η x ...& R= 10% or 0.1 ... solve it and you'll get the ans
What was the answer to yours?
I got, -0.042+/-0.004
 
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srukhan said:
okay well that's freaky, 4.8 Is definitely not going to be visible with a naked eye =P
Here's the thing, our teacher told us to draw the error bars of 0.4 only(for all values of l^2), the one's that have been provided in the table already.. so what you're saying is a totally new concept to me. Lets see how others respond to it because since P5 ain't that old all the techniques are yet to be learned
damn
my teacher gave COMPLETELY different idea :S
he said find uncertainty for EACH reading and use it for Error Bar
 
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How many papers have you tried that way? If you're getting answers close to the mark scheme then that could too be a valid way of doing the grapghs..
My queries: Nov 2009/Q2
-My besf fit line(joining the 4 dots that fall perfectly in line) does not pass through the points mentioned in the mark scheme?
-also, June 10 variant 1/Q2 uncertainity in Xc?
-variant 2/Q2 part e
 
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umm @yasser37
if u dont mind me rplyin
i hv done dat ppr n yeah da bars r real small but u can draw dem
n i think ur tchr is rite cux i got da seme error n ma ans r correct
da 4.8 bar u r talkin bout is very very small but u CAN draw it
its xaminers headache 2 locate it...
agree?!
n whoevr was asking bout CRO u dont need 2 no da details but as a voltmeter u can use it by reading the y-axis as v read it in waves graf ... amplitude !
its normally used in ac currents bcux dey hv da wave like curves
hope dat ans u!!!!
@srukhan which yr grafs u ve drawn can u kindly tell me..>?!
 
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srukhan it worked out well for me
varien 52
@workinghard
I see
can u upload urz so that I have a look at it if possible
Thanks for the support
 
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srukhan said:
beacon_of_light said:
fluffycloud said:
Heyyy!
Paper 5 Summer 08 Q2 the last partt.
I don't get how to do?

you have the formula R=Ro e^-ρ η x ...& R= 10% or 0.1 ... solve it and you'll get the ans
What was the answer to yours?
I got, -0.042+/-0.004
I got +0.042 ... how come you get a negative value or that's a typing mistake?
 
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i ll try 2 yasser37
u ll hv 2 wait doh!!!
sorry i didnt in da end... but ill try ma lvl best!!!
 
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@workinghard: November 2009/52
I effing love your signature, by the way !

@yasser37: Cool then please take some time out and upload it for me, would really appreciate?
that way I would also get a glimpse of your way of doing the error bars

@beacon_of_light: I did it this way,
ln0.1=(-11300).(-0.00485).(x)
so the answer was a -ve value, how about you?
 
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lnR=lnRo - pnx ... i solved it using this formula and got a positive value ... ln0.1 = - ( 11300 x 0.00485) x ... so x is a positive value ...
 
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@syazapion- independent variable is the one that we are not making a change too,which stays the same,the dependent variable is the one which we change to change the independent variant,which is like dependent on us.Juse see the graph we draw to represent the independent and dependent variable.so always the indipendent variable is on the x-axis where as the dependent on the y axis as we always change the value in the axis.hope that cleared ur doubt.
and for uncertainity just subtract the the max value from the original value,like think that value of l is given 6+-0.4 so for finding untertinity for this just use this formula +-(lmax-l) which comes out to be +-(6.4-6)=+-.4 , and as u asked for log ,it all goes the same, +-(log6.4-log6),hope it helped u out.:)
 
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syazapion said:
Uhh.....I know this isnt a part of answering your question but since this is a p5 discussion tread i thought i could ....well....ask a question myself. :sorry: (lol this icon is cute...*ahem* anyway...)

The point im getting through here is that, when it comes to error bars, right..Uh...how many units does the error bar has to have? is there some sort of calculation to tell how many units it suppose to have?

And some how i got confuse on with independent and dependent variable now. Can someone give me a short brief on that?

I only had a little much practice with physics paper 5. I know the idea and everything. But Its been a long time since i last answered paper 5.
@syazapion- independent variable is the one that we are not making a change to,which stays the same,the dependent variable is the one which we change to change the independent variant,which is like dependent on us.Juse see the graph we draw to represent the independent and dependent variable.so always the indipendent variable is on the x-axis where as the dependent on the y axis as we always change the value in the Y axis.hope that cleared ur doubt.
and for uncertainity just subtract the the max value from the original value,like think that value of l is given 6+-0.4 so for finding untertinity for this just use this formula +-(lmax-l) which comes out to be +-(6.4-6)=+-.4 , and as u asked for log ,it all goes the same, +-(log6.4-log6),hope it helped u out.
 
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Guys respond to my query on page 5, I cant do any further and it's freaking me out yaar
 
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listennn paper 53 oct nov 10 quest 1 ....how will we make electricla contact contact with the aluminium strip...MS sayx using of electrodes...how is dat??
 
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