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having problems staying logged in or are you experiencing issues? please visit our tr section for solutions. the graph shows the electrical resistance of a wire as a function of its length. use the graph to determine the resistance (in milliohms) of an 80.0-m long wire. answer:
Step1: Analyze the graph's trend
The graph is a straight line through the origin, so resistance \( R \) is proportional to length \( L \), i.e., \( R = kL \) (linear relationship).
Step2: Find the slope (proportionality constant)
At \( L = 60 \, \text{m} \), \( R = 1000 \, \text{milliOhm} \). The slope \( k=\frac{R}{L}=\frac{1000}{60}=\frac{50}{3} \, \text{milliOhm/m} \).
Step3: Calculate resistance at \( L = 80 \, \text{m} \)
Using \( R = kL \), substitute \( k=\frac{50}{3} \) and \( L = 80 \):
\( R=\frac{50}{3}\times80=\frac{4000}{3}\approx1333.33 \)? Wait, no—wait, looking at the graph, the line at \( L = 60 \) is 1000, so the ratio of \( 80/60 = 4/3 \). So \( R = 1000\times\frac{4}{3}\approx1333 \)? Wait, no, wait the graph: let's check the spacing. From 0 to 60, resistance is 1000. The line is linear, so the slope is \( 1000/60 = 50/3 \) per meter. Then at 80m: \( R = (50/3)80 = 4000/3 ≈ 1333.33 \)? But wait, maybe easier: the graph's x-axis is length (m), y-axis resistance (milliOhm). The line passes through (60,1000). So the equation is \( y = (\frac{1000}{60})x \). For \( x = 80 \), \( y = \frac{1000}{60}\times80 = \frac{80000}{60} = \frac{4000}{3} ≈ 1333.33 \). But wait, maybe the graph is such that at 60m, it's 1000, so the rate is 1000/60 = 50/3 ≈16.6667 milliOhm per meter. Then 80m: 8016.6667≈1333.33. Alternatively, notice that from 0 to 60, 1000; so 20m increase (from 60 to 80) would add \( \frac{1000}{60}\times20 = \frac{20000}{60}≈333.33 \), so 1000 + 333.33≈1333.33.
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\( \approx 1333 \) (or more precisely \( \frac{4000}{3} \approx 1333.33 \)) milliOhms.