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If a star cluster is made up of $10^{6}$ stars whose absolute magnitude is the same as that of the Sun (+5), compute the combined magnitude of the cluster if it is located at a distance of $10 \mathrm{pc}$.
|
At $10 \mathrm{pc}$, the magnitude is (by definition) just the absolute magnitude of the cluster. Since the total luminosity of the cluster is $10^{6}$ times the luminosity of the Sun, we have that
\begin{equation}
\delta m = 2.5 \log \left( \frac{L_{TOT}}{L_{sun}} \right) = 2.5 \log 10^6 = 15.
\end{equation}
Since the Sun has absolute magnitude +5, the magnitude of the cluser is $\boxed{-10}$.
|
Introduction to Astronomy (8.282J Spring 2006)
| 45
|
Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested.
Subproblem 0: Age of our universe when most He nuclei were formed in minutes:
Solution: \boxed{1} minute.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: Age of our universe when hydrogen atoms formed in years:
Solution: \boxed{400000} years.
Final answer: The final answer is 400000. I hope it is correct.
Subproblem 2: Age of our universe today in Gyr:
Solution: \boxed{10} Gyr.
Final answer: The final answer is 10. I hope it is correct.
Subproblem 3: Number of stars in our Galaxy: (Please format your answer as 'xen' representing $x * 10^n$)
|
\boxed{1e11}.
|
Relativity (8.033 Fall 2006)
| 136
|
Subproblem 0: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}$?
Solution: \boxed{1}.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 1: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{O}_{2}$ (in decimal form)?
Solution: \boxed{0.5}.
Final answer: The final answer is 0.5. I hope it is correct.
Subproblem 2: In the balanced equation for the reaction between $\mathrm{CO}$ and $\mathrm{O}_{2}$ to form $\mathrm{CO}_{2}$, what is the coefficient of $\mathrm{CO}_{2}$ (in decimal form)?
Solution: \boxed{1}.
Final answer: The final answer is 1. I hope it is correct.
Subproblem 3: If $32.0 \mathrm{~g}$ of oxygen react with $\mathrm{CO}$ to form carbon dioxide $\left(\mathrm{CO}_{2}\right)$, how much CO was consumed in this reaction (to 1 decimal place)?
|
Molecular Weight (M.W.) of (M.W.) of $\mathrm{O}_{2}: 32.0$
(M.W.) of CO: $28.0$
available oxygen: $32.0 \mathrm{~g}=1$ mole, correspondingly the reaction involves 2 moles of CO [see (a)]:
\[
\mathrm{O}_{2}+2 \mathrm{CO} \rightarrow 2 \mathrm{CO}_{2}
\]
mass of CO reacted $=2$ moles $\times 28 \mathrm{~g} /$ mole $=\boxed{56.0} g$
|
Introduction to Solid State Chemistry (3.091 Fall 2010)
| 177
|
An eclipsing binary consists of two stars of different radii and effective temperatures. Star 1 has radius $R_{1}$ and $T_{1}$, and Star 2 has $R_{2}=0.5 R_{1}$ and $T_{2}=2 T_{1}$. Find the change in bolometric magnitude of the binary, $\Delta m_{\text {bol }}$, when the smaller star is behind the larger star. (Consider only bolometric magnitudes so you don't have to worry about color differences.)
|
\[
\begin{gathered}
\mathcal{F}_{1 \& 2}=4 \pi \sigma\left(T_{1}^{4} R_{1}^{2}+T_{2}^{4} R_{2}^{2}\right) \\
\mathcal{F}_{\text {eclipse }}=4 \pi \sigma T_{1}^{4} R_{1}^{2} \\
\Delta m=-2.5 \log \left(\frac{\mathcal{F}_{1 \& 2}}{\mathcal{F}_{\text {eclipse }}}\right) \\
\Delta m=-2.5 \log \left(1+\frac{T_{2}^{4} R_{2}^{2}}{T_{1}^{4} R_{1}^{2}}\right) \\
\Delta m=-2.5 \log \left(1+\frac{16}{4}\right)=-1.75
\end{gathered}
\]
So, the binary is $\boxed{1.75}$ magnitudes brighter out of eclipse than when star 2 is behind star 1 .
|
Introduction to Astronomy (8.282J Spring 2006)
| 29
|
Preamble: Consider the rotor with moment of inertia \(I\) rotating under the influence of an applied torque \(T\) and the frictional torques from two bearings, each of which can be approximated by a linear frictional element with coefficient \(B\).
Formulate the state-determined equation of motion for the angular velocity $\omega$ as output and the torque $T$ as input.
|
The equation of motion is
\[
\boxed{I \frac{d \omega}{d t}+2 B \omega=T} \quad \text { or } \quad \frac{d \omega}{d t}=-\frac{2 B}{I} \omega+\frac{1}{I} T
\]
|
Dynamics and Control (2.003 Spring 2005)
| 125
|
Determine the differences in relative electronegativity $(\Delta x$ in $e V)$ for the systems ${H}-{F}$ and ${C}-{F}$ given the following data:
$\begin{array}{cl}\text { Bond Energy } & {kJ} / \text { mole } \\ {H}_{2} & 436 \\ {~F}_{2} & 172 \\ {C}-{C} & 335 \\ {H}-{F} & 565 \\ {C}-{H} & 410\end{array}$
\\
Please format your answer to 2 decimal places.
|
According to Pauling, the square of the difference in electro negativity for two elements $\left(X_{A}-X_{B}\right)^{2}$ is given by the following relationship: $\left(X_{A}-X_{B}\right)^{2}=[$ Bond Energy $(A-B)-\sqrt{\text { Bond Energy AA. Bond Energy } B B}] \times \frac{1}{96.3}$
If bond energies are given in ${kJ}$.\\
$\left(X_{H}-X_{F}\right)^{2}=[565-\sqrt{436 \times 172}] \frac{1}{96.3}=3.02$
\[
\begin{aligned}
& \left({X}_{{H}}-{X}_{{F}}\right)=\sqrt{3.02}=1.7 \\
& \left(X_{C}-X_{H}\right)^{2}=[410-\sqrt{335 \times 436}] \frac{1}{96.3}=0.29 \\
& \left(X_{C}-X_{H}\right)=\sqrt{0.29}= \boxed{0.54}
\end{aligned}
\]
|
Introduction to Solid State Chemistry (3.091 Fall 2010)
| 221
|
Preamble: The Peak District Moorlands in the United Kingdom store 20 million tonnes of carbon, almost half of the carbon stored in the soils of the entire United Kingdom (the Moorlands are only $8 \%$ of the land area). In pristine condition, these peatlands can store an additional 13,000 tonnes of carbon per year.
Given this rate of productivity, how long did it take for the Peatlands to sequester this much carbon?
|
$20,000,000$ tonnes $C / 13,000$ tonnes $C y^{-1}=\boxed{1538}$ years
|
Ecology I (1.018J Fall 2009)
| 59
|
If the Sun's absolute magnitude is $+5$, find the luminosity of a star of magnitude $0$ in ergs/s. A useful constant: the luminosity of the sun is $3.83 \times 10^{33}$ ergs/s.
|
The relation between luminosity and absolute magnitude is: $m - n = 2.5 \log (f_n/f_m)$; note the numerator and denominator: brighter objects have numericallly smaller magnitudes. If a star has magnitude $0$, then since the difference in magnitudes from the sun is $5$, it must have $100$ times the sun's luminosity. Therefore, the answer is $\boxed{3.83e35}$ ergs/s.
|
Introduction to Astronomy (8.282J Spring 2006)
| 10
|
Preamble: The following subproblems refer to the differential equation $\ddot{x}+b \dot{x}+x=0$.\\
What is the characteristic polynomial $p(s)$ of $\ddot{x}+b \dot{x}+x=0$?
|
The characteristic polynomial is $p(s)=\boxed{s^{2}+b s+1}$.
|
Differential Equations (18.03 Spring 2010)
| 103
|
Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \cos (3 t)$, which we will assume is a solution of the differential equation $m \ddot{x}+b \dot{x}+k x=0$.
Subproblem 0: What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$.
Solution: We can write $e^{-t / 2} \cos (3 t)=\operatorname{Re} e^{(-1 / 2 \pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\frac{1}{2} \pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$. Then $b=\boxed{m}$,
Final answer: The final answer is m. I hope it is correct.
Subproblem 1: What is $k$ in terms of $m$? Write $k$ as a constant times a function of $m$.
|
Having found that $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$ in the previous subproblem, $k=\boxed{\frac{37}{4} m}$.
|
Differential Equations (18.03 Spring 2010)
| 104
|
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