Let's analyze Jeremy's work to determine whether it is correct.
1. Jeremy starts with the ratio [tex]\(\frac{8 m }{3 A }\)[/tex].
2. He then multiplies by [tex]\(\frac{12 m }{1 A }\)[/tex], writing:
[tex]\[
\frac{8 m }{3 A } \times \frac{12 m }{1 A } = \frac{96 m^2}{3 A^2}
\][/tex]
Here, units are multiplied together straightforwardly: meters (m) times meters (m) yields meters squared ([tex]\(m^2\)[/tex]), and amperes (A) times amperes (A) yield amperes squared ([tex]\(A^2\)[/tex]).
3. Jeremy's work simplifies this to:
[tex]\[
\frac{96 m^2}{3 A^2} = \frac{32 m^2}{1 A^2}
\][/tex]
However, at this point, Jeremy makes a conceptual error by simplifying units incorrectly.
To correct Jeremy's work, it's important to recognize the proper conversion and cancellation of the units:
1. The correct conversion factor should be used to cancel out [tex]\(m\)[/tex] from the numerator and denominator effectively. Instead of [tex]\(\frac{12 m }{1 A }\)[/tex], Jeremy should use [tex]\(\frac{1 A }{12 m }\)[/tex].
2. Rewriting the calculation with the correct conversion:
[tex]\[
\frac{8 m }{3 A } \times \frac{1 A }{12 m } = \frac{8 \times 1}{3 \times 12} = \frac{8}{36} = \frac{2}{9}
\][/tex]
Simplifying the numerical fraction gives us [tex]\(\frac{2}{9}\)[/tex], and the units [tex]\(m \cdot A\)[/tex] cancel out correctly.
Thus, the correct answer for simplifying the ratio [tex]\(\frac{8 m }{3 A }\)[/tex] is:
[tex]\[
\boxed{\frac{2}{9}}
\][/tex]
Jeremy's work is incorrect because he chose the wrong conversion factor, [tex]\(\frac{12 m}{1 A}\)[/tex], leading to incorrect unit multiplication and an erroneous final result.
Therefore, the correct answer to the multiple-choice question is:
- Option B: No, Jeremy should have used a conversion factor of [tex]\(\frac{1 A}{12 m}\)[/tex] instead of [tex]\(\frac{12 m}{1 A}\)[/tex]. The correct answer is [tex]\(\frac{2}{9}\)[/tex].
