i don’t know if it’s because i’m a noob, but for literally the first question i did the same calculation as you but got 30% not 20%. 2x(.2/4x100+.3/3x100)
Hi, only the current is multiplied by 2 as equation is P=I^2R, I think I have forgotten to close the bracket in the correct spot in the video. Thanks for spotting!
oh I see what your mean, excellent question!! So we always chose the quantity we measured for the percentage uncertainty calculation, and in this case we have measured the diameter. Actually a great way to make your measurement more accurate is to increase the measurement value as this decreases your percentage uncertainty. Excellent question!
Excellent question. No, because we measured the diameter in the question so we use the diameter value. For instance if you are given a micrometre in the lab you would measure the diameter rather than the radius of the wire.
everytime you raise a quantity to a power, e.g. squaring it, you multiply the percentage uncertainty by that power. ruclips.net/video/Chi-ju6ityc/видео.html
Yep! they also seem to be separate in exam questions from the measurements tables, i.e. if the question gives you the uncertainty up to 2 sig. figs, stick to that.
@@zhelyo_physics ,so if I get my answer to be like 3.20 and it asks for absolute uncertainty or percentage uncertainty, I'm still not sure to what decimal place or significant figure I would write
that would depend on how many sig figs they gave you the uncertainty with. e.g. m=m1+m2 m1=5 kg+-1 kg m2=6 kg +-1 kg the uncertainty in the total mass: 11 kg +- 2 kg , if you were to write 2.0 in this case it will be too many sig. figs.
Hi, for the last question, why don't you do ( (0.01/0.12) / 2 ) x 2 because D is being divided by 2 and we must divide uncertainties by constants too, right?
Excellent questions. We don't do divide uncertainties by constants. We only consider the uncertainties in the quantities we have measured, things like diameters, lengths, voltages etc. If there is a constant like pi, 2, there is no uncertainty in the constant value.
Hi, thank you so much for these videos they are very helpful! I was wondering for the first question, my approach was finding half the range of I^2 which was 1.6 and after that I also squared 4 which was 16 so I had an uncertainty like 16 +- 1.6 . Finally I just calculated the percentage uncertainty for both 16 +- 1.6 and 3.0 +- 0.3 then added them and I still got 20%. Can this be an alternative solution? or was it just pure luck lol.
in the last question, the part of % uncertainty why do we not convert the units of d (diameter) from mm to m as L is in m too and we took r (radius) in m as well
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i don’t know if it’s because i’m a noob, but for literally the first question i did the same calculation as you but got 30% not 20%. 2x(.2/4x100+.3/3x100)
Hi, only the current is multiplied by 2 as equation is P=I^2R, I think I have forgotten to close the bracket in the correct spot in the video. Thanks for spotting!
Hi, thanks for this, I have an assessment coming up and I now fully understand uncertainties.
Fantastic, good luck on the assessment!
thank you so much!! I have a test on this tomorrow!
Anytime! Good luck!
thank you so much for these videos sir. These are just so helpful AND also sir, you are literally the best physics teacher out here.✨
thank you so much for the kind comment!!
Hi - i was just wondering why on the last question you do not divide by 2 on the diameter? Thank you
No worries! I did, the diameter was 0.12 and I have taken the radius - 0.06 mm. Hope this helps!
Thank you, but sorry if i was unclear but i meant the bit with the %uncertainties - why u did not divide from D to R?
oh I see what your mean, excellent question!! So we always chose the quantity we measured for the percentage uncertainty calculation, and in this case we have measured the diameter. Actually a great way to make your measurement more accurate is to increase the measurement value as this decreases your percentage uncertainty. Excellent question!
At 5:43 , I answered the question by saying a vernier capiler should be used - it the same as a micrometer?
yep!
i have A question. for the last one, the area part of the formula, if we use r = 0.06 +- 0.01mm then r^2 % uncertainity would be 0.02 / 0.06 no?
Excellent question. No, because we measured the diameter in the question so we use the diameter value. For instance if you are given a micrometre in the lab you would measure the diameter rather than the radius of the wire.
Why did you work out the percentage uncertainty in d squared in the last question?
Area=(pi*d^2/2) so we need the uncertainty in the diameter
@@zhelyo_physics ok thanks a lot
@@zhelyo_physics you mean pi d^2 / 4
Hi there, I'm just a bit confused on the last question. Why do you multiply the uncertainty I'm the diameter by 2?
everytime you raise a quantity to a power, e.g. squaring it, you multiply the percentage uncertainty by that power.
ruclips.net/video/Chi-ju6ityc/видео.html
Hi why didn't you say 2(0.1/0.06) since its (d/2)² why didn't we consider the over 2 for the percentage uncertainty
So 0.1 is the length, which is not squared, we only multiply by 2 the diameter. Hope this explains it, let me know if not!
when writing out the absolute and percentage uncertainty, should you match the number of decimal places as the answer or the significant figures?
Yep! they also seem to be separate in exam questions from the measurements tables, i.e. if the question gives you the uncertainty up to 2 sig. figs, stick to that.
@@zhelyo_physics ,so if I get my answer to be like 3.20 and it asks for absolute uncertainty or percentage uncertainty, I'm still not sure to what decimal place or significant figure I would write
that would depend on how many sig figs they gave you the uncertainty with. e.g. m=m1+m2 m1=5 kg+-1 kg m2=6 kg +-1 kg the uncertainty in the total mass: 11 kg +- 2 kg , if you were to write 2.0 in this case it will be too many sig. figs.
@@zhelyo_physics ,ok thanks. So basically just try keep uncertainty and the answer to same significant figures?
yep! : )
At 3:13 why didn't you close the brackets? Aren't up only multiply the 2 by the percentage uncertainty in The current?
thanks for spotting! It's a typo, I'm pretty sure the final answer is still correct.
@@zhelyo_physics yeah you get 20% final answer is correct :p
Hi, for the last question, why don't you do ( (0.01/0.12) / 2 ) x 2
because D is being divided by 2 and we must divide uncertainties by constants too, right?
Excellent questions. We don't do divide uncertainties by constants. We only consider the uncertainties in the quantities we have measured, things like diameters, lengths, voltages etc. If there is a constant like pi, 2, there is no uncertainty in the constant value.
Hi, thank you so much for these videos they are very helpful! I was wondering for the first question, my approach was finding half the range of I^2 which was 1.6 and after that I also squared 4 which was 16 so I had an uncertainty like 16 +- 1.6 . Finally I just calculated the percentage uncertainty for both 16 +- 1.6 and 3.0 +- 0.3 then added them and I still got 20%. Can this be an alternative solution? or was it just pure luck lol.
this is typically accepted, but takes a lot longer. Good work!!
thank you for the video
Anytime! Thanks for your comment!
4:48 is it acceptable to say vernier calipers?
yep!
in the last question, the part of % uncertainty why do we not convert the units of d (diameter) from mm to m as L is in m too and we took r (radius) in m as well
excellent question, it is a ratio so the units would cancel as you are dividing by the same factor. Hope this helps!
What is the music playing in the background ?
Hi, it's mentioned in the description. Hope this helps.
@@zhelyo_physics oh oh I didn’t see that. Sorry
Hi, sir. I think you might made a mistake that the last page (0.1/2.1+0.02/1.86+2*0.01/0/12)*100 actually equal to 23%, is not 18%.
Hi, no mistake. The first line is 0.1/21 not 0.1/2.1. Hope this helps!