Does element x present a linear resistance?

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Element x does not exhibit linear resistance as its behavior changes with varying current and voltage. While the initial answer is close, it contains minor inaccuracies in the significant figures of the measurements. The discussion suggests that a graph of resistance calculations at different points would clarify understanding. The term 'linear resistance' is deemed inappropriate, with 'constant resistance' being a more accurate description. Ultimately, the resistance appears constant within the measurement uncertainties.
Purpleshinyrock
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Homework Statement
Consider the element X represented in the figure, for which
know the voltage and current values indicated in the following table:
Relevant Equations
V=RI, R=V/I
ANSWER: element x does not present a linear resistance because it isn't constant as i and v increases.

Screenshot 2021-11-07 at 18-09-20 Microsoft Word - IC_EX_CAP1 - Exer_Cap1 pdf.png
Is my answer correct?
 
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It's close, but one of the readings is a bit off (to the 3 significant figures shown in the measurements). To be sure, it would be nice if they specified an accuracy for those measurements, IMO. It looks like the one flier measurement is only off by about 0.4%...
 
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Perhaps you could show us your calculations of resistance at these various points? I find it easiest to graph these in a spreadsheet, myself. This will help us see if you are understanding the concept of resistance better than a simple yes/no response.
 
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The term ‘linear resistance’ doesn’t seem appropriate. Is this a translation from another language or maybe part of a longer question? Maybe ‘constant resistance’ is what is intended.

The data supplied are given to only 2 or 3 significant figures. Rounding according to the usual convention gives:
R = 1.18/0.25 = 4.72Ω which rounded to 2 sig. figs. gives 4.7Ω
R = 2.35/0.50 = 4.70Ω which rounded to 2 sig. figs. gives 4.7 Ω
R = 5.88/1.25 – 4.704Ω which rounded to 3 sig. figs. gives 4.70 Ω
R = 9.40/2.00 = 4.70Ω which rounded to 3 sig. figs. gives 4.70 Ω

So my answer would be that: within the limits of the measurement-uncertainties, the resistance is constant.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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