1st year physics lab help ! (1 Viewer)

[traiwit]

ignore my answers in the picture

seriously, i absolutely have no idea what to do with error stuff
there's no example anywhere, i'm totally lost
can anyone help me for this ?

thank you so much >..<

 

[D94]

P should be give to 2 significant figures, iirc. R has 2 s.f. which means your error can't be more accurate than the value for R. (but you should read on this before taking my word for it)

Also, your percentage error for V² should be 6.0 because you double the error in V for V² (which means your absolute error is wrong too). Percentage error for R should be 5.0

Your lab manual contains info on errors and estimation. There is a formula to determine the error. The errors do not appear to be dependent on each other, therefore you take the square root of the addition of percentage errors in quadrature. Makes no sense in words, you will have to look it up in the manual.

I think you should really read over the lab manual again because these will be tested in almost all the quizzes. I can assure you that these will be the shittest part of Physics.

I will try it when I find my lab manual, that is if it isn't due already.

 

[traiwit]

P should be give to 2 significant figures, iirc. R has 2 s.f. which means your error can't be more accurate than the value for R. (but you should read on this before taking my word for it)

Also, your percentage error for V² should be 6.0 because you double the error in V for V² (which means your absolute error is wrong too). Percentage error for R should be 5.0

Your lab manual contains info on errors and estimation. There is a formula to determine the error. The errors do not appear to be dependent on each other, therefore you take the square root of the addition of percentage errors in quadrature. Makes no sense in words, you will have to look it up in the manual.

I think you should really read over the lab manual again because these will be tested in almost all the quizzes. I can assure you that these will be the shittest part of Physics.

I will try it when I find my lab manual, that is if it isn't due already.

Okay thank you very much, i will read the lab manual again then

edit : all this make sense to me right now, except significant fig stuff could u explain more ?

 

[traiwit]

Another question, if said diameter of sphere is 15.36 +- 0.04 mm
and they ask to find the volume of sphere, what do i have to do with the error value ?

thank you so much

 

[D94]

Okay thank you very much, i will read the lab manual again then

edit : all this make sense to me right now, except significant fig stuff could u explain more ?

Basically, your final value can't be more accurate than the errors given. If one value is accurate to 2 s.f. and your calculate the error to be 0.2 and your calculated value is 5.444, then you somehow have made it more accurate, which isn't correct. You only have accuracy to 2 s.f., so it should be 5.4 ± 0.2.
http://www.animations.physics.unsw.edu.au/jw/graphs.htm#Errors

Also:

When multiplying or dividing measurement figures, the final answer may not have more significant figures than the least number of significant figures in the figures being multiplied or divided. This simply means that an answer cannot be more accurate than the least accurate measurement entering calculation, and that you cannot improve the accuracy of a measurement by doing a calculation (even if you have a 10-digit,scientific calculator).

http://www.physics.purdue.edu/academic_programs/courses/phys220/labfiles/significant.pdf

Another question, if said diameter of sphere is 15.36 +- 0.04 mm
and they ask to find the volume of sphere, what do i have to do with the error value ?

thank you so much

Volume = 4/3 (pi) r³

Even though it's a diameter, the error in calculation is still the same, so the error is essentially tripled for r³ since they dependent on the same source of error, i.e. 0.12 mm.

(although, you should ask around if you can or try and find sites which clarify this)

Edit: fixed equation for volume

 

[traiwit]

Thank you so much D94 for ur explanation

actually it's this question



if u r not busy, could u check my answer please ?

i got 7592.09 +- 64.81 kg/m^3

 

[MiPh]

Thank you so much D94 for ur explanation

actually it's this question

View attachment 27792

if u r not busy, could u check my answer please ?

i got 7592.09 +- 64.81 kg/m^3

I thought my calculations were all wrong, but I also got 7.592+- 3.4 g/cm^3 (and no, I have no idea how to calculate errors and I'm using different units of measurement, or how many sig figs)

 

[D94]

I think it would be 7590 ± 60 kg/m³, although I'm not 100% sure.

It would be 64 but the error in the mass is to 1.s.f, so the overall error shouldn't be more accurate than that.

 

[lolJK]

how do you determine the absolute error of "power"? thanks

 

[traiwit]

i change absolute error to percentage error, then take to the multiplying formula for dependent error
so it's like just * (time) of the percentage error

 

[lolJK]

i change absolute error to percentage error, then take to the multiplying formula for dependent error
so it's like just * (time) of the percentage error

I dont quite understand what you're trying to say. Could you please elaborate, or provide a reference (if any) in the
lab manual.

Thank you

 

[MiPh]

i change absolute error to percentage error, then take to the multiplying formula for dependent error
so it's like just * (time) of the percentage error

You my friend are a legend. 10/10! I understand it all now! Thanks so much, karma be good to you.

 

[traiwit]

You my friend are a legend. 10/10! I understand it all now! Thanks so much, karma be good to you.

cool man ! btw it was really funny that we accidently met hahahaha XD