\text{Chapter I}
\text{Chapter I}
\text{Chapter I}
\text{Chapter I}
\text{Dr. Adel Abbout}
\text{Express speed of sound, }330 \mathrm{~m} / \mathrm{s}\text{ in miles}/ \mathrm{h}. \\
(1 \text{ mile}=1609 \mathrm{~m})
\begin{align*}
& \text{A) $147 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{B) $330 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{C) $738 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{D) $0.205 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{E) $980 \ \mathrm{miles}/\mathrm{h}$}
\end{align*}
\text{Problem 1}
\text{Express speed of sound, }330 \mathrm{~m} / \mathrm{s}\text{ in miles}/ \mathrm{h}. \\
(1 \text{ mile}=1609 \mathrm{~m})
v=330 \text{ m/s}=330 \times \frac{1/1609 \text{ miles}}{1/3600 \text{ h}}=330 \times \frac{3600 \text{ miles}}{1609\text { h}}=738 \text{ miles}/\text{h}
\text{Answer C}
\begin{align*}
& \text{A) $147 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{B) $330 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{C) $738 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{D) $0.205 \ \mathrm{miles}/\mathrm{h}$}\\
& \text{E) $980 \ \mathrm{miles}/\mathrm{h}$}
\end{align*}
\text{Problem 1}
$$\left(1\right.\text{ nano } \left.=10^{-9}\right)$$
$$\text{1 shake} =10^{-8}\text{ seconds. } $$
$$\text{Find out how many nano seconds (ns) are there in 1 shake.}$$
\begin{align*}
& \text{A) $0.1 \ \mathrm{ns}$}\\
& \text{B) $0.01 \ \mathrm{ns}$}\\
& \text{C) $100 \ \mathrm{ns}$}\\
& \text{D) $0.001 \ \mathrm{ns}$}\\
& \text{E) $10 \ \mathrm{ns}$}
\end{align*}
\text{Problem 2}
\text{Answer E}
$$\left(1\right.\text{ nano } \left.=10^{-9}\right)$$
$$\text{Find out how many nano seconds (ns) are there in 1 shake.}$$
\begin{align*}
& \text{A) $0.1 \ \mathrm{ns}$}\\
& \text{B) $0.01 \ \mathrm{ns}$}\\
& \text{C) $100 \ \mathrm{ns}$}\\
& \text{D) $0.001 \ \mathrm{ns}$}\\
& \text{E) $10 \ \mathrm{ns}$}
\end{align*}
\text{Problem 2}
$$\text{How many molecules of water are there in a cup containing } 250 \mathrm{~cm}^3 \text{ of water?}$$
$$\text{Molecular mass of $\mathrm{H}_2 \mathrm{O}=18 \mathrm{~g} / \mathrm{mole}$}$$
$$\text{Density of water $=1.0 \mathrm{~g} / \mathrm{cm}^3$}$$
$$\text{Avogadro s number $=6.02 \times 10^{23} \mathrm{molecules} / \mathrm{mole}$}$$
\begin{align*}
& \text{A) $6.0 \times 10^{23}$}\\
& \text{B) $8.4 \times 10^{24}$}\\
& \text{C) $1.9 \times 10^{26}$}\\
& \text{D) $3.7 \times 10^{28}$}\\
& \text{E) $2.5 \times 10^{3}$}
\end{align*}
\text{Problem 3}
\text{Problem 3}
\text{Answer B }
$$\text{How many molecules of water are there in a cup containing } 250 \mathrm{~cm}^3 \text{ of water?}$$
$$\text{Molecular mass of $\mathrm{H}_2 \mathrm{O}=18 \mathrm{~g} / \mathrm{mole}$}$$
$$\text{Density of water $=1.0 \mathrm{~g} / \mathrm{cm}^3$}$$
$$\text{Avogadro s number $=6.02 \times 10^{23} \mathrm{molecules} / \mathrm{mole}$}$$
\text{$n$: number of moles}, \text{and $M$: Molar mass}
n=\frac{m}{M}=\frac{\rho V}{M}=\frac{1 \text{ g/$cm^3$} \times 250\text{ $cm^3$}}{18 \text{ g/mole}}=13.89 \text{ moles}
N=n A=13.89\times6.02\times10^{23}=8.36\times 10^{24} \text{ molecule}
\text{Problem 4}
\begin{align*}
& \text{Using the fact that the speed of light in space is about $3.00 \times 10^{8}$ m/s,}\\
& \text{determine how many miles light will travel in one hour.}\\
& \text{(1 mile = 1.61 km)}\\[1em]
& \text{A) $6.71 \times 10^{8}$ miles}\\
& \text{B) $2.50 \times 10^{6}$ miles}\\
& \text{C) $5.40 \times 10^{9}$ miles}\\
& \text{D) $8.32 \times 10^{3}$ miles}\\
& \text{E) $4.83 \times 10^{2}$ miles}
\end{align*}
\text{Problem 4}
\begin{align*}
& \text{Using the fact that the speed of light in space is about $3.00 \times 10^{8}$ m/s,}\\
& \text{determine how many miles light will travel in one hour.}\\
& \text{(1 mile = 1.61 km)}\\[1em]
\end{align*}
d=v t= 3\times 10^8 m/s \times 3600 s=1.08\times 10^{12} m
d=(1.08\times 10^{12} /1610) \times \text{ miles}=6.7 \times 10^8 \text{ miles}
\text{Answer A}
\text{Problem 5}
\begin{align*}
& \text{Suppose}\\[0.5em]
& \quad A = \frac{B^{n}}{C^{m}} \\[0.5em]
& \text{where $A$ has dimensions $LT$, $B$ has dimensions $L^{2}T^{-1}$, and $C$}\\
& \text{has dimensions $LT^{2}$. Then the exponents $n$ and $m$ have the values:}\\[1em]
& \text{A) $n = \tfrac{1}{5} \ ; \ m = \tfrac{3}{5}$}\\
& \text{B) $n = 2 \ ; \ m = 3$}\\
& \text{C) $n = \tfrac{4}{5} \ ; \ m = -\tfrac{1}{5}$}\\
& \text{D) $n = \tfrac{1}{5} \ ; \ m = -\tfrac{3}{5}$}\\
& \text{E) $n = \tfrac{1}{2} \ ; \ m = \tfrac{1}{2}$}
\end{align*}
\text{Problem 5}
\begin{align*}
& \text{Suppose}\\[0.5em]
& \quad A = \frac{B^{n}}{C^{m}} \\[0.5em]
& \text{where $A$ has dimensions $LT$, $B$ has dimensions $L^{2}T^{-1}$, and $C$}\\
& \text{has dimensions $LT^{2}$. Then the exponents $n$ and $m$ have the values:}\\[1em]
\end{align*}
\begin{align*}
& [A] = \frac{[B]^n}{[C]^m} = \frac{(L^{2}T^{-1})^n}{(LT^{2})^m}
= L^{2n-m}\,T^{-n-2m}. \\[1em]
& \text{Since we have } [A] = L^{1}T^{1}: \\[0.5em]
& \quad \begin{cases}
2n - m = 1, \\
-n - 2m = 1.
\end{cases} \\[1em]
& \text{The solution is:} \\[0.5em]
& m = 2n - 1 \;\;\Rightarrow\;\; -n - 2(2n-1) = 1
\;\;\Rightarrow\;\; -5n + 2 = 1
\;\;\Rightarrow\;\; n = \tfrac{1}{5}, \\[0.5em]
& m = 2\left(\tfrac{1}{5}\right) - 1 = -\tfrac{3}{5}. \\[1em]
& \boxed{n = \tfrac{1}{5}, \;\; m = -\tfrac{3}{5}}
\quad
\end{align*}
\text{Answer D}
\text{The velocity of a particle is given by $\mathbf{v}=\mathbf{A t}^2+(\mathbf{B} / \mathbf{A}) \mathbf{t}$, where $\mathbf{v}$ is in $\mathrm{m} / \mathrm{s}$ and $\mathbf{t}$}
\text{ is in seconds. The dimension of B is:}
\begin{align*}
& \text{A) $\mathrm{L}^2 \mathrm{~T}^{-5}$}\\
& \text{B) $\mathrm{L}^3 \mathrm{~T}^{-3}$}\\
& \text{C) $\mathrm{L}^2 \mathrm{~T}^3$ }\\
& \text{D) $\mathrm{L}^4 \mathrm{~T}^4$ }\\
& \text{E) $L^5 T^{-6}$}
\end{align*}
\displaystyle [v]=[A][t]^2\text{ and }[v]=\frac{[B]}{[A]}[t]
\displaystyle LT^{-1}=[A]T^2\text{ and }[v]=\frac{[B]}{[A]}T
\text{In a sum or an equality, all the terms have the same dimension}
\displaystyle LT^{-1}=[A]T^2
\displaystyle LT^{-1}=\frac{[B]}{[A]}T
\text{ and }
\Rightarrow
\displaystyle LT^{-3}=[A]
\Rightarrow
\displaystyle [A]LT^{-2}=[B]
\displaystyle L^2T^{-5}=[B]
\Rightarrow
\text{in a sum X+Y, each term has the same dimension}
\text{The velocity of a particle is given by $\mathbf{v}=\mathbf{A t}^2+(\mathbf{B} / \mathbf{A}) \mathbf{t}$, where $\mathbf{v}$ is in $\mathrm{m} / \mathrm{s}$ and $\mathbf{t}$}
\text{ is in seconds. The dimension of B is:}
\text{Problem 7}
\begin{align*}
& \text{The transverse displacement of an oscillating string is given by }
y = y_m \cos(\omega t + \phi) \\[0.5em]
& \text{where $y$ is in cm and $t$ in ms.} \\[1em]
& \text{What is the dimension of $\omega$?}
\end{align*}
\text{The term inside a trigonometric function has no dimension.}
[\omega][t]=1\Rightarrow [\omega]=T^{-1}
\text{Therefore, we can write:}
\text{Problem 7}
\begin{align*}
& \text{The transverse displacement of an oscillating string is given by }
y = y_m \cos(\omega t + \phi) \\[0.5em]
& \text{where $y$ is in cm and $t$ in ms.} \\[1em]
& \text{What is the dimension of $\omega$?}
\end{align*}
\text{Problem 8}
\begin{align*}
& \text{From the fact that the average density of Earth is }
5.50 \ \mathrm{g/cm^3} \text{ and its mean radius is } \\
& 6.37 \times 10^{6} \ \mathrm{m},\text{the mass of the Earth is:} \\[1em]
& \text{A) $7.01 \times 10^{17} \ \mathrm{kg}$} \\
& \text{B) $3.98 \times 10^{21} \ \mathrm{kg}$} \\
& \text{C) $5.95 \times 10^{24} \ \mathrm{kg}$} \\
& \text{D) $2.80 \times 10^{18} \ \mathrm{kg}$} \\
& \text{E) $5.50 \times 10^{23} \ \mathrm{kg}$}
\end{align*}
\text{Problem 8}
\begin{align*}
& \text{From the fact that the average density of Earth is }
5.50 \ \mathrm{g/cm^3} \text{ and its mean radius is } \\
& 6.37 \times 10^{6} \ \mathrm{m},\text{the mass of the Earth is:} \\[1em]
\end{align*}
\rho=5.50 \frac{\mathrm{~g}}{\mathrm{~cm}^3}=5.50 \times 10^3 \frac{\mathrm{~kg}}{\mathrm{~m}^3}=5500 \frac{\mathrm{~kg}}{\mathrm{~m}^3} .
\begin{aligned}
&\text {Volume of a sphere: $V=\frac{4}{3} \pi r^3$ }\\
\end{aligned}
V=\frac{4}{3} \pi r^3 \approx \frac{4}{3}\pi \times 6.37^3\times10^{18}=1.0827 \times 10^{21} \mathrm{~m}^3 .
M=\rho V=5500 \times 1.0827\times 10^{21} \mathrm{~Kg}
\text{Answer C}
\text{Problem 9}
\begin{align*}
& \text{An aluminum cylinder of density $2.70 \ \mathrm{g/cm^3}$, a radius of $2.30 \ \mathrm{cm}$, and a height of $1.40 \ \mathrm{m}$} \\
&\text{has the mass of:}\\[1em]
& \text{A) $25.0 \ \mathrm{kg}$}\\
& \text{B) $45.1 \ \mathrm{kg}$}\\
& \text{C) $13.8 \ \mathrm{kg}$}\\
& \text{D) $8.50 \ \mathrm{kg}$}\\
& \text{E) $6.28 \ \mathrm{kg}$}
\end{align*}
\text{Problem 9}
\begin{align*}
& \text{An aluminum cylinder of density $2.70 \ \mathrm{g/cm^3}$, a radius of $2.30 \ \mathrm{cm}$, and a height of $1.40 \ \mathrm{m}$} \\
&\text{has the mass of:}\\[1em]
\end{align*}
\begin{align*}
& \rho = 2.70\,\mathrm{g/cm^3} = 2700\,\mathrm{kg/m^3},\quad
r = 2.30\,\mathrm{cm} = 0.0230\,\mathrm{m},\quad
h = 1.40\,\mathrm{m}.\\[0.5em]
& V = \pi r^{2} h = \pi(0.0230)^{2}(1.40) \approx 2.327 \times 10^{-3}\ \mathrm{m^{3}}.\\[0.5em]
& m = \rho V \approx 2700 \times 2.327 \times 10^{-3}\ \mathrm{kg} \approx 6.28\ \mathrm{kg}.\\[0.6em]
\end{align*}
\text{Answer E}
\text{Problem 10}
\begin{align*}
& \text{A cylindrical can, $6.00$ inches high and $3.00$ inches in diameter is filled with water.}\\
& \text{Density of water is $1.00 \ \mathrm{g/cm^3}$. What is the mass of water in the can in gram?}\\
& \text{(1 inch = $2.54$ cm).}\\[1em]
& \text{A) $277 \ \mathrm{g}$}\\
& \text{B) $695 \ \mathrm{g}$}\\
& \text{C) $182 \ \mathrm{g}$}\\
& \text{D) $107 \ \mathrm{g}$}\\
& \text{E) $2780 \ \mathrm{g}$}
\end{align*}
\text{Problem 10}
\text{Answer B}
\begin{align*}
& \text{A cylindrical can, $6.00$ inches high and $3.00$ inches in diameter is filled with water.}\\
& \text{Density of water is $1.00 \ \mathrm{g/cm^3}$. What is the mass of water in the can in gram?}\\
& \text{(1 inch = $2.54$ cm).}\\[1em]
& \text{A) $277 \ \mathrm{g}$}\\
& \text{B) $695 \ \mathrm{g}$}\\
& \text{C) $182 \ \mathrm{g}$}\\
& \text{D) $107 \ \mathrm{g}$}\\
& \text{E) $2780 \ \mathrm{g}$}
\end{align*}
\text{Problem 11}
\begin{align*}
& \text{A drop of oil (mass $= 0.90$ milligram and density $= 918 \ \mathrm{kg/m^3}$)}\\
& \text{spreads out on a surface and forms a circular thin film of radius $= 41.8 \ \mathrm{cm}$}\\
& \text{and thickness $h$. Find $h$ in nanometer (nm).}\\
& \text{(1 nano $= 10^{-9}$).}\\[1em]
& \text{A) $0.60 \ \mathrm{nm}$}\\
& \text{B) $0.00060 \ \mathrm{nm}$}\\
& \text{C) $0.15 \ \mathrm{nm}$}\\
& \text{D) $1.8 \ \mathrm{nm}$}\\
& \text{E) $0.030 \ \mathrm{nm}$}
\end{align*}
\text{Problem 11}
\begin{align*}
& \text{A drop of oil (mass $= 0.90$ milligram and density $= 918 \ \mathrm{kg/m^3}$)}\\
& \text{spreads out on a surface and forms a circular thin film of radius $= 41.8 \ \mathrm{cm}$}\\
& \text{and thickness $h$. Find $h$ in nanometer (nm).}\\
& \text{(1 nano $= 10^{-9}$).}\\[1em]
& \text{A) $0.60 \ \mathrm{nm}$}\\
& \text{B) $0.00060 \ \mathrm{nm}$}\\
& \text{C) $0.15 \ \mathrm{nm}$}\\
& \text{D) $1.8 \ \mathrm{nm}$}\\
& \text{E) $0.030 \ \mathrm{nm}$}
\end{align*}
\text{Answer D}
Chapter 01-Phys101
By smstry
