From fb02e1039531e21730924588fef5e42092f62c6b Mon Sep 17 00:00:00 2001
From: "W. Trevor King" <wking@drexel.edu>
Date: Fri, 7 Jan 2011 11:51:31 -0500
Subject: [PATCH] Add week 1 problems for Winter 2011 edition of Physics 101.

---
 .../Serway_and_Jewett_8/problem01.10.tex      | 19 +++++++++++++
 .../Serway_and_Jewett_8/problem01.11.tex      | 28 +++++++++++++++++++
 .../Serway_and_Jewett_8/problem01.13.tex      | 11 ++++++++
 .../Serway_and_Jewett_8/problem01.15.tex      | 15 ++++++++++
 4 files changed, 73 insertions(+)
 create mode 100644 latex/problems/Serway_and_Jewett_8/problem01.10.tex
 create mode 100644 latex/problems/Serway_and_Jewett_8/problem01.11.tex
 create mode 100644 latex/problems/Serway_and_Jewett_8/problem01.13.tex
 create mode 100644 latex/problems/Serway_and_Jewett_8/problem01.15.tex

diff --git a/latex/problems/Serway_and_Jewett_8/problem01.10.tex b/latex/problems/Serway_and_Jewett_8/problem01.10.tex
new file mode 100644
index 0000000..3cf0c87
--- /dev/null
+++ b/latex/problems/Serway_and_Jewett_8/problem01.10.tex
@@ -0,0 +1,19 @@
+\begin{problem*}{1.10}
+Newton's law of universal gravitation is represented by
+\begin{equation}
+F = \frac{GMm}{r^2}
+\end{equation}
+where $F$ is the magnitude of the gravitational force exerted by one
+small object on another, $M$ and $m$ are the masses of the objects,
+and $r$ is a distance.  Force has the SI units
+$\bareU{kg$\cdot$m/s$^2$}$.  What are the SI units of the
+proportionality constant $G$?
+\end{problem*} % Probem 1.10
+
+\begin{solution}
+\begin{align}
+  F &= \frac{GMm}{r^2} \\
+  \bareU{kg$\cdot$m/s$^2$} &= G\frac{\bareU{kg$^2$}}{\bareU{m$^2$}} \\
+  G &= \ans{\bareU{m$^3$/kg$\cdot$s$^2$}}
+\end{align}
+\end{solution}
diff --git a/latex/problems/Serway_and_Jewett_8/problem01.11.tex b/latex/problems/Serway_and_Jewett_8/problem01.11.tex
new file mode 100644
index 0000000..9dff725
--- /dev/null
+++ b/latex/problems/Serway_and_Jewett_8/problem01.11.tex
@@ -0,0 +1,28 @@
+\begin{problem*}{1.11}
+Kinetic energy $K$ (Chapter 7) has dimensions
+$\bareU{kg$\cdot$m$^2$/s$^2$}$.  It can be written in terms of the
+momentum $p$ (Chapter 9) and mass $m$ as
+\begin{equation}
+K = \frac{p^2}{2m}
+\end{equation}
+\Part{a} Determine the proper units for momentum using dimensional
+analysis.  \Part{b} The unit of force is the newton $\bareU{N}$, where
+$1\U{N} = 1\U{kg$\cdot$m/s$^2$}$.  What are the units of momentum $p$
+in terms of a newton and another fundamental SI unit?
+\end{problem*} % Probem 1.11
+
+\begin{solution}
+\Part{a}
+\begin{align}
+  K &= \frac{p^2}{2m} \\
+  \bareU{kg$\cdot$m$^2$/s$^2$} &= \frac{p^2}{\bareU{kg}} \\
+  p^2 &= \frac{\bareU{kg$^2\cdot$m$^2$}}{\bareU{s$^2$}} \\
+  p &= \ans{\bareU{kg$\cdot$m/s}}
+\end{align}
+
+\Part{b}
+\begin{equation}
+  p = \bareU{kg$\cdot$m/s} = \bareU{kg$\cdot$m/s$^2$}\cdot\bareU{s}
+    = \ans{\bareU{N$\cdot$s}}
+\end{equation}
+\end{solution}
diff --git a/latex/problems/Serway_and_Jewett_8/problem01.13.tex b/latex/problems/Serway_and_Jewett_8/problem01.13.tex
new file mode 100644
index 0000000..945b23b
--- /dev/null
+++ b/latex/problems/Serway_and_Jewett_8/problem01.13.tex
@@ -0,0 +1,11 @@
+\begin{problem*}{1.13}
+A rectangular building lot has a width of $75.0\U{ft}$ and a length of
+$125\U{ft}$.  Determine the area of this lot in square meters.
+\end{problem*} % Probem 1.13
+
+\begin{solution}
+\begin{equation}
+A = 75.0\U{ft} \cdot 125\U{ft} \cdot \p({\frac{1\U{m}}{3.28\U{ft}}})^2
+  = \ans{871\U{m$^2$}}
+\end{equation}
+\end{solution}
diff --git a/latex/problems/Serway_and_Jewett_8/problem01.15.tex b/latex/problems/Serway_and_Jewett_8/problem01.15.tex
new file mode 100644
index 0000000..07882f1
--- /dev/null
+++ b/latex/problems/Serway_and_Jewett_8/problem01.15.tex
@@ -0,0 +1,15 @@
+\begin{problem*}{1.15}
+A solid piece of led has a mass of $23.94\U{g}$ and a volume of
+$2.10\U{cm$^3$}$.  From these data, calculate the density of lead in
+SI units (kilograms per cubic meter).
+\end{problem*} % Probem 1.15
+
+\begin{solution}
+\begin{equation}
+  \rho = \frac{m}{V}
+    = \frac{23.94\U{g}}{2.10\U{cm$^3$}}
+      \cdot\frac{1\U{kg}}{10^3\U{g}}
+      \cdot\p({\frac{10^2\U{cm}}{1\U{m}}})^3
+    = \ans{11,400\U{kg/m$^3$}}
+\end{equation}
+\end{solution}
-- 
2.26.2