Examples of H.L. Mencken in the following topics:
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- In the 1920s, the satirist H.L.
- Mencken led the attack on the genteel
tradition in American literature, ridiculing the provincialism of American
intellectual life.
- Many emerging Southern writers, however, were
already highly critical of contemporary life in the South were emboldened by
Mencken's essay.
- H.L.
- Mencken was an influential American writer and social critic who unwittingly helped to launch the Southern Renaissance literary movement.
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- . ~ H.L.
- Mencken, in response to every item of hate mail he received.
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- $\displaystyle \Psi({\bf r}, t) = \sum_j \sum_{l=0}^\infty A_{jl}(t) \lambda^l \psi_j({\bf r}) e^{-iE_j t/\hbar}$
- $\displaystyle i \bar{h} \sum_j \sum_{l=0}^\infty \left [ -\frac{i E_j}{\bar{h}} A_{jl}(t) + {A_{jl}(t)}{t} \right ] \lambda^l \psi_j({\bf r}) e^{-iE_j t/{\bar{h}}} \nonumber \\ = \displaystyle \sum_j \sum_{l=0}^\infty \left [ E_j + \lambda H'({\bf r},t) \right ] \lambda^l \psi_j({\bf r}) e^{-iE_j t/{\bar{h}}}.$
- $\displaystyle i \bar{h} \sum_{l=0}^\infty \left [-\frac{i E_f}{\bar{h}} A_{fl}(t) + \frac{A_{fl}(t)}{t} \right ] \lambda^l {-2in}e^{-iE_f t/ \bar{h}} \nonumber \\ \displaystyle =\sum_{i} \sum_{l=0}^\infty A_{il}(t) \lambda^l \Biggr[ \langle \psi_f | \psi_j \rangle E_j + \lambda \langle\psi_f | H'({\bf r},t) |\psi_j \rangle \Biggr] e^{-iE_j t/ \bar{h}} \\ \displaystyle = \sum_{l=0}^\infty \lambda^l \Biggr [ A_{fl}(t) E_f e^{-iE_f t/ \bar{h}} + \lambda \sum_{j} A_{jl}(t) \langle\psi_f | H'({\bf r},t) |\psi_j \rangle e^{-iE_j t/ \bar{h}} \Biggr ]$
- $\displaystyle i\bar{h} \left [ -\frac{i E_f}{\bar{h}} A_{f0}(t) + {A_{f0}(t)}{t} \right ] e^{-iE_f t/\\bar{h}} = A_{f0}(t) E_f e^{-iE_f t/\bar{h}}.$
- $\displaystyle i\bar{h} \left [ -\frac{i E_f}{\bar{h}} A_{f1}(t) + {A_{f1}(t)}{t} \right ] e^{-iE_f t/\bar{h}} = A_{f1}(t) E_f e^{-iE_f t/\bar{h}} + \nonumber \\ \displaystyle {-1in} \sum_{j} A_{j0}(0) \langle\psi_f | H'({\bf r},t) |\psi_j \rangle e^{-iE_j t/\bar{h}} $
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- Limits of functions can often be determined using simple laws, such as L'Hôpital's rule and squeeze theorem.
- In this atom, we will study two examples: L'Hôpital's rule or the squeeze theorem.
- Let $f$, $g$, and $h$ be functions defined on $I$, except possibly at $a$ itself.
- Suppose that for every $x$ in $I$ not equal to $a$, we have $g(x) \leq f(x) \leq h(x)$, and also suppose that $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.
- Calculate a limit using simple laws, such as L'Hôpital's Rule or the squeeze theorem
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- $6e^- + 14 H^+(aq) + Cr_2O_7^{2-} \rightarrow 2 Cr^{3+} + 7 H_2O(l)$
- Balanced reduction half-reaction: $6e^- + 14 H^+(aq) + Cr_2O_7^{2-} \rightarrow 2 Cr^{3+} + 7 H_2O(l)$
- $6e^- + 14 H^+(aq) + Cr_2O_7^{2-} \rightarrow 2 Cr^{3+} + 7 H_2O(l)$
- $6e^-+3H_2O(l) + 3NO_2^-(aq) + 14 H^+(aq) + Cr_2O_7^{2-} \rightarrow 2 Cr^{3+} + 7 H_2O(l) + 3NO_3^-(aq)+6H^+(aq)+6e^-$
- $3NO_2^-(aq) + 8 H^+(aq) + Cr_2O_7^{2-} \rightarrow 2 Cr^{3+} + 4 H_2O(l) + 3NO_3^-(aq)$
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- A strong acid will react with a strong base to form a neutral (pH = 7) solution.
- $HCl (aq) + NaOH (aq) \rightarrow H_2O (l) + NaCl (aq)$
- What is the unknown concentration of a 25.00 mL HCl sample that requires 40.00 mL of 0.450 M NaOH to reach the equivalence point in a titration?
- $HCl (aq) + NaOH (aq) \rightarrow H_2O (l) + NaCl (aq)$
- Step 3: Calculate the molar concentration of HCL in the 25.00 mL sample.
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- $\displaystyle {\bf B} = - \frac{1}{c} {\bf v} \times {\bf E} = \frac{{\bf l}}{mecr} \frac{dU}{dr}$
- $\displaystyle H_{so} = \frac{1}{2m^2 c^2} {\bf s}\cdot{\bf l} \frac{1}{r} \frac{d U}{dr}$
- Let's focus on states with the same values of $S$ and $L$ but different values of $J$.
- $\displaystyle {\bf J}^2 = ({\bf L} + {\bf S}) \cdot ({\bf L} + {\bf S}) = {\bf L}^2 + {\bf S}^2 + 2{\bf L}\cdot{\bf S} $
- $\displaystyle H_{so} = \frac{1}{2} \xi \left ({\bf J}^2 - {\bf L}^2 - {\bf S}^2 \right ) = \frac{1}{2} C \left [ J(J+1) - L(L+1) - S(S+1) \right ]$
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- P(G|H) = [P(G AND H)]/[P(H)] = 0.3/0.5 = 0.6 = P(G)
- For practice, show that P(H|G) = P(H) to show that G and H are independent events.
- Let L be the event that the student has long hair.
- Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45 ; but P(F)P(L) = (0.60)(0.50)= 0.30 The events of being female and having long hair are not independent because P(F and L) does not equal P(F)P(L).
- check whether P(L|F) equals P(L): We are given that P(L|F) = 0.75 but P(L) = 0.50; they are not equal.
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- $\displaystyle l = \Omega r^2 = \left ( GM r \right )^{1/2}.$
- $\displaystyle \dot L^+ = \dot M \left ( GM r \right )^{1/2}.$
- $\displaystyle \dot L^- = \beta \dot M \left ( GM r_I \right )^{1/2}$
- $\displaystyle \tau = f_\phi \left ( 2 \pi r \right ) \left ( 2 h \right ) ( r ) = \dot L^+ - \dot L^- = \dot M \left [ \left ( GM r \right )^{1/2} - \beta \left ( GM r_I \right )^{1/2} \right ]$
- $\displaystyle 2 h Q \approx 2 h \frac{\left(f_\phi\right)^2}{\eta} = \frac{3 \dot M}{4\pi r^2}\frac{GM}{r} \left [1 - \left (\frac{r_I}{r} \right )^{1/2} \right ]
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- Since $e^{x}$ does not depend on $h$, it is constant as $h$ goes to $0$.