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Resistivity ($\rho$) is the reciprocal of conductivity ($\sigma$).
$\sigma = ne\mu_e$
Given: $n = 10^{19} \text{ m}^{-3}$, $\mu_e = 1.6 \text{ m}^2/\text{Vs}$, and $e = 1.6 \times 10^{-19} \text{ C}$.
$\sigma = 10^{19} \times (1.6 \times 10^{-19}) \times 1.6 = 1.6 \times 1.6 = 2.56 \ \Omega^{-1}\text{m}^{-1}$.
$\rho = \frac{1}{\sigma} = \frac{1}{2.56} \approx 0.39 \ \Omega\text{m}$.
The closest value provided is $0.4 \ \Omega\text{m}$.
Find all $x$ satisfying $(\cot^{-1}x)^2 - 7(\cot^{-1}x)+10>0$.
Let $u=\cot^{-1}x$. Solve $u^2-7u+10>0$, i.e., $(u-2)(u-5)>0$.
This gives $u<2$ or $u>5$.
Since $\cot^{-1}x$ is a decreasing function with range $(0,\pi)$:
- $u<2 \Rightarrow \cot^{-1}x<2 \Rightarrow x>\cot2$ (decreasing function reverses inequality)
- $u>5 \Rightarrow \cot^{-1}x>5 \Rightarrow x<\cot5$
Solution: $x\in(-\infty,\cot5)\cup(\cot2,\infty)$
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