problem
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Suppose air molecules have a collision cross section of $10^{-16} \mathrm{~cm}^{2}$. If the (number) density of air molecules is $10^{19} \mathrm{~cm}^{-3}$, what is the collision mean free path in cm? Answer to one significant figure.
|
\[
\ell=\frac{1}{n \sigma}=\frac{1}{10^{19} 10^{-16}}=\boxed{1e-3} \mathrm{~cm}
\]
|
Introduction to Astronomy (8.282J Spring 2006)
| 47
|
Preamble: You are given an equation of motion of the form:
\[
\dot{y}+5 y=10 u
\]
What is the time constant for this system?
|
We find the homogenous solution, solving:
\[
\dot{y}+5 y=0
\]
by trying a solution of the form $y=A \cdot e^{s, t}$.
Calculation:
\[
\dot{y}=A \cdot s \cdot e^{s \cdot t} \mid \Rightarrow A \cdot s \cdot e^{s t}+5 A \cdot e^{s t}=0
\]
yields that $s=-5$, meaning the solution is $y=A \cdot e^{-5 \cdot t}=A \cdot e^{-t / \tau}$, meaning $\tau = \boxed{0.2}$.
|
Dynamics and Control (2.003 Spring 2005)
| 128
|
Calculate the volume in mL of $0.25 \mathrm{M} \mathrm{NaI}$ that would be needed to precipitate all the $\mathrm{g}^{2+}$ ion from $45 \mathrm{~mL}$ of a $0.10 \mathrm{M} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ solution according to the following reaction:
\[
2 \mathrm{NaI}(\mathrm{aq})+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{HgI}_{2}(\mathrm{~s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})
\]
|
\[
\begin{aligned}
&2 \mathrm{NaI}(\mathrm{aq})+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{HgI}_{2}(\mathrm{~s})+\mathrm{NaNO}_{3}(\mathrm{aq}) \\
&\frac{0.10 \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}}{1 \mathrm{~L}} \times 0.045 \mathrm{~L}=4.5 \times 10^{-3} \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} \\
&4.5 \times 10^{-3} \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} \times \frac{2 \mathrm{~mol} \mathrm{NaI}}{1 \mathrm{~mol} \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}}=9.00 \times 10^{-3} \mathrm{~mol} \mathrm{NaI} \\
&\frac{9.00 \times 10^{-3} \mathrm{~mol} \mathrm{NaI}}{0.25 \frac{\mathrm{mol} \mathrm{NaI}}{\mathrm{L}}}=3.6 \times 10^{-2} \mathrm{~L} \times \frac{1000 \mathrm{ml}}{1 \mathrm{~L}}=\boxed{36} \mathrm{~mL} \mathrm{NaI}
\end{aligned}
\]
|
Introduction to Solid State Chemistry (3.091 Fall 2010)
| 224
|
A cubic metal $(r=0.77 \AA$ ) exhibits plastic deformation by slip along $<111>$ directions. Determine its planar packing density (atoms $/ \mathrm{m}^{2}$) for its densest family of planes. Please format your answer as $n \times 10^x$ where $n$ is to 2 decimal places.
|
Slip along $<111>$ directions suggests a BCC system, corresponding to $\{110\},<111>$ slip. Therefore:
\[
\begin{aligned}
&a \sqrt{3}=4 r \\
&a=\frac{4 r}{\sqrt{3}}=1.78 \times 10^{-10} \mathrm{~m}
\end{aligned}
\]
Densest planes are $\{110\}$, so we find:
\[
\frac{2 \text { atoms }}{a^{2} \sqrt{2}}=\boxed{4.46e19} \text { atoms } / \mathrm{m}^{2}
\]
|
Introduction to Solid State Chemistry (3.091 Fall 2010)
| 215
|
Rewrite the function $\cos (\pi t)-\sqrt{3} \sin (\pi t)$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$.
|
The right triangle has hypotenuse of length $\sqrt{1^{2}+(-\sqrt{3})^{2}}=2$. The circular frequency of both summands is $\pi$, so $\omega=\pi$. The argument of the hypotenuse is $-\pi / 3$, so $f(t)=\boxed{2 \cos (\pi t+\pi / 3)}$.
|
Differential Equations (18.03 Spring 2010)
| 83
|
Given the ordinary differential equation $\ddot{x}-a^{2} x=0$, where $a$ is a nonzero real-valued constant, find a solution $x(t)$ to this equation such that $x(0) = 1$ and $\dot{x}(0)=0$.
|
First, notice that both $x(t)=e^{a t}$ and $x(t)=e^{-a t}$ are solutions to $\ddot{x}-a^{2} x=0$. Then for any constants $c_{1}$ and $c_{2}$, $x(t)=c_{1} e^{a t}+c_{2} e^{-a t}$ are also solutions to $\ddot{x}-a^{2} x=0$. Moreover, $x(0)=c_{1}+c_{2}$, and $\dot{x}(0)=a\left(c_{1}-c_{2}\right)$. Assuming $a \neq 0$, to satisfy the given conditions, we need $c_{1}+c_{2}=1$ and $a\left(c_{1}-c_{2}\right)=0$, which implies $c_{1}=c_{2}=1 / 2$. So $x(t)=\boxed{\frac{1}{2}(\exp{a*t} + \exp{-a*t})}$.
|
Differential Equations (18.03 Spring 2010)
| 95
|
Preamble: Here we consider a system described by the differential equation
\[
\ddot{y}+10 \dot{y}+10000 y=0 .
\]
Subproblem 0: What is the value of the natural frequency \(\omega_{n}\) in radians per second?
Solution: $\omega_{n}=\sqrt{\frac{k}{m}}$
So
$\omega_{n} =\boxed{100} \mathrm{rad} / \mathrm{s}$
Final answer: The final answer is 100. I hope it is correct.
Subproblem 1: What is the value of the damping ratio \(\zeta\)?
Solution: $\zeta=\frac{b}{2 \sqrt{k m}}$
So
$\zeta =\boxed{0.05}$
Final answer: The final answer is 0.05. I hope it is correct.
Subproblem 2: What is the value of the damped natural frequency \(\omega_{d}\) in radians per second? Give your answer to three significant figures.
|
$\omega_{d}=\omega_{n} \sqrt{1-\zeta^{2}}$
So
$\omega_{d}=\boxed{99.9} \mathrm{rad} / \mathrm{s}$
|
Dynamics and Control (2.003 Spring 2005)
| 133
|
Preamble: A formation energy of $2.0 \mathrm{eV}$ is required to create a vacancy in a particular metal. At $800^{\circ} \mathrm{C}$ there is one vacancy for every 10,000 atoms.
At what temperature (in Celsius) will there be one vacancy for every 1,000 atoms? Format your answer as an integer.
|
We need to know the temperature dependence of the vacancy density:
\[
\frac{1}{10^{4}}=A e^{-\frac{\Delta H_{v}}{k T_{1}}} \quad \text { and } \frac{1}{10^{3}}=A e^{-\frac{\Delta H_{v}}{k T_{x}}}
\]
From the ratio: $\frac{\frac{1}{10^{4}}}{\frac{1}{10^{3}}}=\frac{10^{3}}{10^{4}}=\frac{\mathrm{Ae}^{-\Delta \mathrm{H}_{v} / \mathrm{k} T_{1}}}{\mathrm{Ae}^{-\Delta \mathrm{H}_{v} / \mathrm{kT} \mathrm{x}}}$ we get $-\ln 10=-\frac{\Delta \mathrm{H}_{\mathrm{v}}}{\mathrm{k}}\left(\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{\mathrm{x}}}\right)$
\[
\begin{aligned}
&\therefore \quad\left(\frac{1}{T_{1}}-\frac{1}{T_{x}}\right)=\frac{k \ln 10}{\Delta H_{v}} \\
&\frac{1}{T_{x}}=\frac{1}{T_{1}}-\frac{k \ln 10}{\Delta H_{v}}=\frac{1}{1073}-\frac{1.38 \times 10^{-23} \times \ln 10}{2 \times 1.6 \times 10^{-19}}=8.33 \times 10^{-4} \\
&T_{x}=1200 \mathrm{~K}= \boxed{928}^{\circ} \mathrm{C}
\end{aligned}
\]
|
Introduction to Solid State Chemistry (3.091 Fall 2010)
| 209
|
Preamble: In Cambridge, shoppers can buy apples from two sources: a local orchard, and a store that ships apples from out of state. The orchard can produce up to 50 apples per day at a constant marginal cost of 25 cents per apple. The store can supply any remaining apples demanded, at a constant marginal cost of 75 cents per unit. When apples cost 75 cents per apple, the residents of Cambridge buy 150 apples in a day.
Assume that the city of Cambridge sets the price of apples within its borders. What price should it set, in cents?
|
The city should set the price of apples to be $\boxed{75}$ cents since that is the marginal cost when residents eat at least 50 apples a day, which they do when the price is 75 cents or less.
|
Principles of Microeconomics (14.01 Fall 2011)
| 257
|
At $100^{\circ} \mathrm{C}$ copper $(\mathrm{Cu})$ has a lattice constant of $3.655 \AA$. What is its density in $g/cm^3$ at this temperature? Please round your answer to 2 decimal places.
|
$\mathrm{Cu}$ is FCC, so $\mathrm{n}=4$
\[
\begin{aligned}
&\mathrm{a}=3.655 \AA=3.655 \times 10^{-10} \mathrm{~m} \\
&\text { atomic weight }=63.55 \mathrm{~g} / \mathrm{mole} \\
&\frac{\text { atomic weight }}{\rho} \times 10^{-6}=\frac{N_{\mathrm{A}}}{\mathrm{n}} \times \mathrm{a}^{3} \\
&\rho=\frac{(63.55 \mathrm{~g} / \mathrm{mole})(4 \text { atoms } / \text { unit cell })}{\left(6.023 \times 10^{23} \text { atoms } / \mathrm{mole}\right)\left(3.655 \times 10^{-10} \mathrm{~m}^{3}\right)}= \boxed{8.64} \mathrm{~g} / \mathrm{cm}^{3}
\end{aligned}
\]
|
Introduction to Solid State Chemistry (3.091 Fall 2010)
| 190
|
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