Young's Modulus of Copper - PRACTICAL - A Level Physics
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- เผยแพร่เมื่อ 1 ต.ค. 2024
- In this video I go through an AQA Physics A Level Required Practical that uses a Young's Modulus experiment to calculate the Young's Modulus of copper wire. (This is AQA Required Practical 4.) This is also the OCR A Level Physics PAG 2.1 Practical.
This is a straight-forward way to measure the Young's Modulus of a material. Copper wire is often used as it is stretched by weights hung over the edge of the bench, with the small extension carefully measured with a ruler. Wear eye protection in-case the wire snaps.
See instructions for all the practicals at www.alevelphysi...
Thanks for watching,
Lewis
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The 9 dislikes were the people whose wire snapped and hit their eyes.
@Matteo Zachariah I don't give 3 shits
@@djfish3099 damn these bots have come a long way
turn on subtitles at 1:04
Edit: please like or comment when you see this so that I can revisit this clip😂
lol
😂😂😂😂
I LAUGHED SO HARD MY ANUS FELL OUT
😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂🤣😂😂😂😂😂😂
3:29 sir did you really need to say all that 😂😂😂 my eyes are feeling weird
Surely as you are stretching the wire, the cross-sectional area will get smaller?
Wouldn't this throw off results, as another variable is changing?
Matt Whitelock You're right, the assumption we make is that this decrease is very small and can be ignored from our calculations for working out the stress.
A Level Physics Online Ah okay, thank you that makes sense
Since YM= stress/strain could you not plot a graph of stress against strain, where the gradient is now the YM?
It's a lot simpler to do it that way. I guess it was just to show the relationships
Because the stress (adjusted using the masses) will be the independent variable, and independent variables are usually plotted on the x-axis.
@@ghgjftrh8144 force extension graphs are the exception tho
Cheers mate helped a lot all these videos with these TAG assessments coming up
Isnt the length L from the fixed end at the G clamp right until the end of the wire? Just need to clarify please
quality video as always :)
Thanks - more coming along soon.
Yes get it lad
WIIIIIRRRRRRRE ~ Jessie
mucho graciesas me amigo
Why do you use pi*D^2/4 and how does that differ from using PI*r^2
It is the same - but I used the diameter in the equation as that’s what you actually measure.
why do you measure the young modulus for copper in this way and not in the same way as steel?
steel doesnt stretch (as easily)
Hi 👋
I just had a question, how would you calculate the percentage uncertainty in the force if you have multiple different forces that you are applying to the object?
Secondly, once you work out the percentage uncertainty in the force, area, extension and original length, how can you calculate the percentage uncertainty in the Young Modulus altogether?
Find the abs uncertainty for the masses (tbh this should be given) then use it to calculate the percent uncertainty. Even when you add more masses the percent uncertainty should remain the same since the masses are uniform
And to work out ym uncertainty in percent, add the force, area, extension and length percent uncertainties together
Probably not useful for you anymore lol
@@boingk2585 cheers mate, it’s useful for me :)
@@anonymooseuser2150 np 👍
why does l have to be as long as possible you never said why
I think it’s so you can monitor as much extension as possible
It’s to reduce any uncertainties
It reduces percentage uncertainty
I dont care anymore Im done with physics lol
Calculate the value of Young's modulus of
Brass wire by hanging the weight 0g, 500g,
1000g. Perform the whole experiment in
Uranus.
how would you convert mm^2 to m when calculating the wire cross sectional area?
well mili = 10^-3, so 1 milimetre is 10^-3 metres. When that is squared the exponent doubles, so it would be 10^-6. So we take the value in milimetres and multiply by 10^-6
Convert the number in the beginning to make them all into meters then work out the area