Finding the gradient question (1 Viewer)

[superbird]

Was the 2nd last question of the exam...
wat did everybody get as the gradient, i got around 3.5...
and for k i think i got around 5.6

 

[xprshn]

Yeah i got gradient as 3.5,
but i got k constant as 1.75x10^-7... <== close to the real value of 2x10^-7 right?

 

[SeDaTeD]

i got gradient 3.6 x10^-6, and k = 1.8 x10^-7
and i put in units too.

 

[acmilan]

Pretty much the same as SeDaTeD

and for the equation of k i got k = (md)/(I(1)l) (gradient*distance/current in 1*length)

 

[Komaticom]

I misinterpretted I2 to be I squared, so I couldn't the question.
For the last part I referred to my data sheet and pulled out the precise k value.

 

[BillyMak]

The gradient wasn't ~3.5*10^-6, it was just ~3.5 since the y- axis was Force (10^-6 N).

 

[acmilan]

I took that into consideration and decided on using just 3.5 as the gradient

 

[BillyMak]

Yes, but you had to use 3.5*10^-6 when calculating k.

 

[Steven12]

Yes, but you had to use 3.5*10^-6 when calculating k.

i Know, when i first got 3.6 as gradient, note how the last question asks to sub in the gradient to get the constant, i got a really large number.

so i put 3.6*10-6 for the gradient.

 

[BillyMak]

I reckon they could accept both answers, it was only worth like 1 or 2 marks.

 

[tinuviel]

Gradient

I got 3.6 and used that to find k....is that wrong?!

 

[BillyMak]

you should have used 3.6*10^-6 to find k.

 

[Shifty]

what did ya's do to get the gradient?

 

[BillyMak]

rise/run

I used all of the points they gave to get about 5 gradients, all ~3.5, then I just took the average.

 

[kunny funt]

yeah i did it from the line of best fit, but i didnt put the x 10^-6 just left it, i only x 10^-6 in the next two parts.

i will lose a mark for that.

 

[BillyMak]

I don't think you'd lose a mark for that. Strictly speaking, since Force was in 10^-6 Newtons, and Current was in single amps, this was marked on the scale. So the gradient of the line was in fact 3.5, i.e. 1 unit across, 3.5 units up, even if the units were tweaked in order to produce a nicer looking graph.

 

[Xayma]

Yes you should have then realised that F/I2=m*10^-6

 

[peeasoup]

what did u fuckheads get for K ?
i got something like 0.18...

 

[Jase]

k , 0.18 x 10^-6 NA^-2

which is about equal to 2 x 10^-7 like in the sheet..

 

[lucyinthehole]

i got what someone on the first page did: gradient=3.5 x 10^-6, (but for units i forgot to put A^2, and so just got N/A) and k=1.75 x 10^whatever the real value is to the power off, can't remember. so it's pretty close to the actual value