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What if articles with LaTeX
zjn0505 edited this page Jul 9, 2019
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LaTeX formulas in 2. 73. 119. are not directly wrapped in <p> tag
LaTeX formulas in 124. are presented inside
<span class="ref"> <span class="refnum">[1|2]</span> <span class="refbody"> **LaTeX here** </span> </span>
<p>The probability of getting all 158 questions is: \[\frac{1}{5^{44}}
* \frac{1}{5^{67}} * \frac{1}{5^{47}} \approx \frac{1}{2.7 *
10^{110}}\]</p><p>\[m_{x} = m_{f22} \times \left(\frac{12.5}{19}\right)^{3} \approx 5,600\mathrm{kg}\]</p>
<p>\[\frac{5,600 \mathrm{kg} \times 0.9\mathrm{g} \times 1.4 \text{
meters}}{3.6 \text{ seconds}} = 19.2\mathrm{kW}\]</p><p>\[
(6.022\times10^{23})\times75\mathrm{g}\approx4.52\times10^{22}\mathrm{kg}\]</p><p>\[ \frac{3\text{
trillion}\frac{\mathrm{miles}}{\mathrm{year}}}{30\frac{\mathrm{mph}}{\text{car}}}\approx10\text{
million cars} \]</p><p>\[\text{Gravitational potential energy} = \frac{1}{2} \times 65\mathrm{kg} \times \text{(Earth escape velocity)}^{2}\]</p>
<p>\[\Delta v = v_\text{exhaust} \ln \frac {m_\text{start}} {m_\text{end}}\]</p><p>\[\text{Period}=\frac{1}{\text{Frequency}}=\frac{1}{\frac{300
\text{ billion birds}}{4\pi\times\text{Earth radius}^2}\times
1\frac{\frac{\text{poop}}{\text{bird}}}{\text{hour}}\times16\frac{\text{hours}}{\text{day}}\times
1 \frac{\text{mouth}}{\text{poop}}\times 15 \frac{\text{
cm}^2}{\text{mouth}}}=195\text{ years}\]</p>
<p>\[\frac{300\text{ billion birds} \times 0.5\frac{\text{fluid
oz}}{\text{day}}}{3\text{
trillion}\frac{\mathrm{miles}}{\text{year}}}=0.335 \mathrm{ mm}^2 =
13\text{ miles per gallon}\]</p><p>\[100\textrm{km}\times100\textrm{km}\times6\textrm{cm}=0.6\textrm{km}^3\]</p><p>\[\frac{5\text{ billion people}\times
500\frac{\mathrm{terawatts}}{\text{person}}}{\pi\times\text{Moon
radius}^2}\times20\frac{\mathrm{megajoules}}{\mathrm{kilogram}}\times
3\frac{\mathrm{kilograms}}{\mathrm{liter}}\approx4<br>
\frac{\mathrm{meters}}{\text{second}}\]</p><p>\[ 2\times\text{Depth of Marianas Trench}\times\frac{23.1\textrm{
seconds}}{50\textrm{ meters}}\approx3\textrm{ hours} \]</p>
<p>\[ \frac{\text{Red light wavelength}}{\text{Green light
wavelength}}=\sqrt{\frac{1+\frac{\text{Your speed}}{\text{Speed of
light}}}{1-\frac{\text{Car speed}}{\text{Speed of light}}}} \]</p>
<p>\[ \textrm{Car speed}=\frac{\textrm{c}\times\left ( \textrm{Red
light wavelength}^2-\textrm{Green light wavelength}^2\right
)}{\textrm{Green light wavelength}^2+\textrm{Red light
wavelength}^2}\approx\frac{1}{6}c \]</p><p>\[\text{Radius} = \left (\frac{3}{4\pi}\right )^\frac{1}{3}\left ( \frac{40 \% \times 53\text{ megatons of TNT}}{\text{Mariana Trench pressure}+\text{1 ATM}} \right)^\frac{1}{3}\approx580\text{ meters}\]</p>
<p>\[\text{Radius} = \left (\frac{3}{4\pi}\right) ^\frac{1}{3}\left ( \frac{40 \% \times 53000000\text{ megatons of TNT}}{\text{Mariana Trench pressure}+\text{1 ATM}} \right)^\frac{1}{3}\approx35\text{ miles}\]</p><p>\[\text{Shadow Radius}=\sqrt{-h(h-2r)}\]</p><p>\[66\tfrac{\text{tornadoes}}{\text{year}}\times\frac{1.5\text{
miles}\times50\text{ yards}}{\text{Area of
Florida}}\approx1.4\times10^{-12}\tfrac{\text{tornadoes}}{\text{second}}\]</p>
<p>\[\frac{\tfrac{1}{700,000}\text{deaths}}{70\text{
years}}\approx6.4\times10^{-16}\tfrac{\text{deaths}}{\text{second}}\]</p>
<p>\[\frac{20\text{ bales}}{20\text{ years}}\times\frac{\left
(\tfrac{75\text{ kg}}{\text{Density of cocaine}} \right
)^\tfrac{2}{3}}{\text{Area of
Florida}}\approx2.9\times10^{-21}\tfrac{\text{bales}}{\text{second}}\]</p><p>\[\text{Thrust}=\text{Mass ejection rate}\times\text{Speed of
ejection}\]</p>
<p>\[10\tfrac{\text{bullets}}{\text{second}}\times8\tfrac{\text{grams}}{\text{bullet}}\times715\tfrac{\text{meters}}{\text{second}}=57.2\text{
N}\approx13\text{ pounds of force}\]</p><p>\[1\text{ cent}\times\left (
\frac{30\frac{\text{miles}}{\text{gallon}}} {\frac{2.73\text{
grams}}{50\text{
pounds}}\times0.5%\times\frac{$3.50}{\text{gallon}}} \right
)\approx140,000\text{ miles}\]</p>
<p>\[\frac{70\text{ kg}\times\text{Earth Gravity}\times50\text{
centimeters}}{20\%}=1.72\text{ kJ}=0.41\text{ food calories}\]</p><p>\[1500\text{ft}^2 \times 4\tfrac{\text{
meters}}{\text{year}}\times1\tfrac{\text{kg}}{\text{liter}}
\times9.81\tfrac{\text{m}}{\text{s}^2}\times5\text{
meters}\times15\tfrac{\text{cents}}{\text{kWh}}=\frac{$1.14}{\text{year}}\]</p><p>\[\frac{210,000,000 \text{ cars} \times 4100
\frac{\text{lbs}}{\text{car}} \times 75\text{mph}}{2.3 \times
10^{21}\text{kg}} \approx 18 \frac{\text{cm}}{\text{year}} \gg
0.69 \frac{\text{cm}}{\text{year}}\]</p><p>\[1\text{ cup}\times\text{Water heat capacity}\times\tfrac{100^\circ\rm{C}-20^\circ\rm{C}}{2\text{ minutes}}=700\text{ watts}\]</p><p>A fatal radiation dose is about 4 sieverts. Using the inverse-square law, we can calculate the radiation dose:
\[ 0.5\text{ nanosieverts} \times\left ( \frac{1\text{ parsec}}{x}\right )^2 = 5\text{ sieverts} \]
\[ x=0.00001118\text{ parsecs}=2.3\text{ AU} \]
2.3 AU is a little more than the distance between the Sun and Mars.</p><p>\[ \Delta v = v_\text{exhaust} \ln \frac{m_\text{initial}}{m_\text{final}} \]</p>
<p>\[ \frac{\text{Mass of ship plus golf balls}}{\text{Mass of ship alone}} = e ^ \left ( \frac{\text{Ship's change in speed}}{\text{Speed of golf ball}} \right ) \]</p><p>First, let's look at the basic energy requirements. Vaporizing a liter of water takes about 2.6 megajoules,<span class="ref"><span class="refnum">[1]</span><span class="refbody">It takes more energy if the water is colder, but not <em>much</em> more. Heating the water up to the edge of boiling only takes a little of the 2.6 megajoules. Most of it goes into pushing it over the threshold from 100°C water to 100°C vapor.</span></span> and a big rainstorm might drop half an inch of rain per hour. This is one of those places where the equation isn't complicated—you just <a href="http://www.wolframalpha.com/input/?i=2.6+megajoules%2Fliter+*+1+cm%2Fhour">multiply</a> the 2.6 megajoules per liter by the rainfall rate and you get laser umbrella power requirement (watts per square meter protected). It's weird when units work out so straightforwardly:
\[2.6\tfrac{\text{megajoules}}{\text{liter}}\times0.5\tfrac{\text{inches}}{\text{hour}}=9200\tfrac{\text{watts}}{\text{square meter}}\]
9 kilowatts per square meter is an order of magnitude more power than is delivered to the surface by sunlight, so your surroundings are going to heat up pretty fast. In effect, you're creating a cloud of steam around yourself, into which you're pumping more and more laser energy.</p><p>This<span class="ref"><span class="refnum">[2]</span><span class="refbody">Not this one. The other one.</span></span>​<span class="ref"><span class="refnum">[3]</span><span class="refbody">The simplest approach, which gives us an approximate answer, is to treat the swimmer as a simple projectile. The formula for the height of a projectile is:<br /><br />\( \frac{\text{speed}^2}{2\times\text{gravity}} \)<br /><br />... which tells us that a champion swimmer moving at 2 meters per second (4.5 mph) would only have enough kinetic energy to lift their body <a href="http://www.wolframalpha.com/input/?i=%282+m%2Fs%29%5E2+%2F+%282+*+earth+gravity%29">about 20 centimeters</a> against gravity. <br /><br />That's not totally accurate, although it's enough to tell us that dolphin jumps on Earth probably aren't in the cards for us. But to get a more accurate answer (and an equation we can apply to the Moon), we need to account for a few other things.<br /><br />When a swimmer first breaks the surface, they don't have to lift their full weight; they're partially supported by buoyancy. As more of their body leaves the water, the force of buoyancy decreases, since their body is displacing less water. Since the force of gravity isn't changing, their net weight increases. <br /><br />You can calculate how much potential energy is required to lift a body vertically through the surface to a certain height, but it's a complicated integral (you integrate the displacement of the submerged portion of their body over the vertical distance they travel) and depends on their body shape. For a human body moving fast enough to jump most of the way out of the water, this effect probably adds about half a torso-length to their final height—and less if they're not able to make it all the way out.<br /><br />The other effect we have to account for is the fact that a swimmer can continue kicking as they start to leave the water. When a swimmer is submerged and moving at top speed, the drag from the water is equal to the thrust they generate by kicking and ... whatever the gerund form of the verb is for the things your arms do while swimming. My first thought was "stroking," but it's definitely not that.<br /><br />Anyway, once the jumping swimmer breaks the surface, the drag almost vanishes, but they can keep kicking for a few moments. To figure out how much energy this adds, you can multiply the thrust from kicking by the distance over which they're kicking after breaking the surface, since energy equals force times distance. The distance is most of a body length, or 1 to 1.5 meters. As for the force from kicking, random Google results for a search for lifeguard qualifications suggest that good swimmers might be able to carry 10 lbs over their heads for a short distance, which means they're generating a little more than 10 pounds-force (50+ N) of kicking thrust.<br /><br />We can combine all these together into a big ol' equation:<br /><br />\[ \text{Jump height}=\left(\frac{\tfrac{1}{2}\times\text{body mass}\times\left(\text{top speed}\right)^2+\text{kick force}\times\text{torso length}}{\text{Earth gravity}\times\text{body mass}}\right)+\left(\text{buoyancy correction} \right) \]</span></span> footnote contains some detail on the math behind a dolphin jump. Calculating the height a swimmer can jump out of the water requires taking several different things into account, but the bottom line is that a normal swimmer on the Moon could probably launch themselves a full meter out of the water, and Michael Phelps may well be able to manage 2 or 3.</p><p>\[ \text{People required}=\frac{\pi\times\left(\tfrac{1}{4}\text{ inch} \right )^2\times200\text{ MPa}}{102.5\text{ kg}/\text{person}}\approx25\text{ people} \]</p><p>\[55\text{ mph}\times1\text{ inch}\times9\text{ feet}\times\text{water density}\times335\tfrac{\text{J}}{\text{gram}}=574\text{ megawatts}\]</p><p>\[ \frac{100,000\text{ }\tfrac{\text{cubic feet}}{\text{second}}}{\pi\left ( \tfrac{7\text{ mm}}{2} \right )^2}=73,600,000\text{ }\tfrac{\text{meters}}{\text{second}}= 0.25c \]</p><p>\[ \text{Minimum slope} = \text{Coefficient of rolling resistance} \]</p>
<p>\[ \text{Forward pull from gravity} = \text{Rolling resistance} + \text{Drag force} \]</p>
<p>\[ m g \sin(\theta) = g \cos(\theta) C_r m + \tfrac{1}{2} C_d \rho A V^2 \]</p>
<p>\[ V = \sqrt{\frac{m g \sin(\theta) - g \cos(\theta) C_r m}{ \tfrac{1}{2} C_d \rho A}} \]</p>