Minor formatting changes

Chapter1.tex

@@ -39,7 +39,7 @@
     \begin{enumerate}
         \item
         The total current required by New York City that will flow through the cable is $I = \frac{P}{V} = \frac{10^{10}\;\W}{115\;\V} = 869.6\;\M\A$. Therefore, the total power lost per foot of cable can be calculated by:
-        \[P = I^2R = (869.6 \times 10^6\;\A)^2 \times (5\times10^{-8}\;\frac{\Ohm}{\ensuremath{\text{ft}}}) = \mans{3.78\times10^{8} \frac{\W}{\ensuremath{\text{ft}}}} \] 
+        \[P = I^2R = (869.6 \times 10^6\;\A)^2 \times \left(5\times10^{-8}\;\frac{\Ohm}{\ensuremath{\text{ft}}}\right) = \mans{3.78\times10^{8} \frac{\W}{\ensuremath{\text{ft}}}} \] 
         \item
         The length of cable over which all $10^{10}\;\W$ will be lost is:
         \[L = \frac{10^{10}\;\W}{3.78 \times 10^{8}\;\frac{\W}{\ensuremath{\text{ft}}}} = \mans{26.45\;ft}\]
@@ -48,7 +48,7 @@
         \[A = \pi DL = \pi \times 30.48\;{\ensuremath{\text{cm}}}\;\times 806.196\;{\ensuremath{\text{cm}}} = 7.72 \times 10^4\;{\ensuremath{\text{cm}}}^2\]
         Therefore,
         \[T = \sqrt[4]{\frac{P}{A\sigma}} = \sqrt[4]{\frac{10^{10}\;\W}{7.72 \times 10^4\;{\ensuremath{\text{cm}}}^2 \times 6 \times 10^{-12}\;\frac{\W}{\text{K}^4\text{cm}^2}}} = \mans{12,121\;K} \]
-        This is indeed a preposterous temperature, more than twice that at the surface of the Sun! The "solution" to this problem is to look at the melting point of copper, which is $\sim$1358 K at standard pressure. The copper cable will melt long before such a temperature is reached.
+        This is indeed a preposterous temperature, more than twice that at the surface of the Sun! The ``solution'' to this problem is to look at the melting point of copper, which is $\sim$1358 K at standard pressure. The copper cable will melt long before such a temperature is reached.
     \end{enumerate}
 
     \ex{1.10}