Big and small numbers in The Physical World

7278 /www/nrich/html/content/id/7278/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"><ul id="stemLinks"> <li><a href="http://nrich.maths.org/7418">Warm-up</a></li> <li><a href="http://nrich.maths.org/6504">Try this next</a></li> <li><a href="http://nrich.maths.org/7428">Think higher</a></li> <li><a href="http://plus.maths.org/content/philosophy-applied-mathematics">Read: mathematics</a></li> <li><a href="http://plus.maths.org/content/bang-crunch-freeze-and-multiverse">Read: science</a></li> <li><a href="http://motivate.maths.org/content/node/37">Explore further</a></li> </ul> <div> </div> Think about how best to approximate these things from the physical world around us. You will need to make some estimations and find information from friends or other sources, as would any scientist! Take care to represent all of your answers using a sensible number of decimal places and be sure to note all of your assumptions clearly. <ol> <li>Light travels at $c=3\times 10^8$ metres per second. How fast in this in miles per hour? How many times faster is this than a sports car?</li> <li>The Milky Way is a spiral galaxy with diamater about 100,000 light years and thickness about 1000 light years. There are estimated to be between 100 billion and 400 billion stars in the galaxy. Estimate the average distance between these stars. </li> <li>Density of lead $11.34$g/cm$^3$. How big would a tonne of lead be?</li> <li>Estimate the mass of ore it takes to produce a roll of aluminum kitchen foil.</li> <li>How many AA batteries contain enough charge between them to run a laptop for an hour?</li> <li>Estimate how many atoms there are in a staple.</li> <li>Einstein's equation tells us that the enegy $E$ stored in matter equals $mc^2$, where $m$ is the mass and $c$ is the speed of light. How much energy is contained in the staple from question 6? How long could this energy run your laptop for?</li> <li>How much energy would it take to raise the air temperature of the room you are in by 1$^\circ$C? How much gas must be burned to produce this much energy? What is the cost of that much gas?</li> </ol> <div class="framework">NOTES AND BACKGROUND An obvious part of the skill with applying mathematics to physics is to know the fundamental formulae and constants relevent to a problem. By not providing these pieces of information directly, you need to engage at a deeper level with the problems. You might not necessarily know all of the required formulae, but working out which parts you can and cannot do is all part of the problem solving process!</div> </mdoxml> <mdoxml version="1.0">Note that some of these questions involve making estimations and assumptions so these answers are not definitive! 1. Required extra data: Typical height a person can raise there centre of mass through when jumping on the Earth $h \approx 0.5 \textrm{m}$ $\therefore$ under a gravitational field of one-sixth of the strength height reached $h \approx 3.0 \textrm{ m}$ 2. Required extra data: None $$ \rho_{Pb} = 11.35\ \textrm{g cm}^{-3} \Rightarrow 1\ \textrm{cm}^3 \textrm{ of lead has mass of } m = 11.35\ \textrm{g}$$ $$ \textrm{Mass of one lead atom } m_{A:Pb} = 3.44 \times 10^{-22}\ \textrm{g}$$ $$ \therefore \textrm{number of atoms in 1 cm}^3\ \ n = \frac{m}{m_{A:Pb}} = \frac{11.35}{3.44 \times 10^{-22}} = 3.30 \times 10^{22}$$ 3. Required extra data: 1 year = 365.25 days 1 year is one full rotation around Sun, therefore period of orbit in seconds is: $$T = 365.25 \times 24 \times 60 \times 60 = 31\ 557\ 600\ \textrm{s}$$ The angular velocity of the Earth's orbit is therefore: $$\omega = \frac{2\pi}{T} = \frac{2\pi}{31\ 557\ 600} = 1.99 \times 10^{-7}\ \textrm{rad s}^{-1}$$ The speed of the Earth relative to the Sun is therefore: $$ v = \omega R = 1.99 \times 10^{-7} \times 149\ 598\ 000\ 000 = 29\ 800\ \textrm{ms}^{-1}$$ 4. Required extra data: Acceleration due to gravity $g = 9.81\ \textrm{ms}^{-2}$ Assuming constant acceleration (i.e. ignoring drag effects), $u = 0 \textrm{ ms}^{-1};\ s = 800\textrm{ m};\ a =g$ $$v^2 = u^2 + 2as \Rightarrow v^2 = 0 + 2(g)(800) \Rightarrow v = 40 \sqrt{g} = 125 \textrm{ ms}^{-1}$$ Note air-resistance is likely to cause the ball to tend towards a terminal velocity below this, typically around $90 \textrm{ ms}^{-1}$ 5. Required extra data: Density of fuel (assume petrol) $\rho_{fuel} \approx 700\ \textrm{kg m}^{-3}$ Typical diameter of fuel tank $D = 4\ \textrm{m}$ Typical length of fuel tank $\ell = 15\ \textrm{m}$ Volume of fuel tank $$V = \pi \left(\frac{D}{2}\right)^2 \ell = \frac{\pi \times 4^2 \times 15}{4} = 60 \pi\ \textrm{m}^3$$ $\therefore$ weight of fuel in tank $$W = V \rho_{fuel} = 60 \pi \times 700 = 132\ 000\ \textrm{kg}$$ This seems to be an overestimate as typical values for capacity of a petrol tanker are between 20000 and 40000 litres. What values might be more sensible for the diameter and length? 6. Required extra data: Hydrochloric acid is a strong acid (complete dissociation) $\Rightarrow \textrm{pH} = -\log_{10} [\textrm{H}^+]$ Avogadro's constant $N_A = 6.022 \times 10^{23}\ \textrm{mol}^{-1}$ $$\textrm{pH} = 1.0 \Rightarrow [\textrm{H}^+] = 10^{-1.0} = 0.1 \textrm{ mol dm}^{-3}$$ $$\therefore 0.1 \textrm{ mol in } 1 \textrm{ dm}^3 (1\ \ell) \textrm{ which is } 0.1N_A = 6.022 \times 10^{22}\textrm{ hydrogen ions}$$ $$\therefore \textrm{ total sum of positive charges } = 6.022 \times 10^{22} \times 1.6 \times 10^{-19} = 9635 \textrm{ C}$$ 7. Required extra data: Relative molecular mass of water $m_{R} = 18.02\ \textrm{u}$ Avogadro's constant $N_A = 6.022 \times 10^{23}\ \textrm{mol}^{-1}$ 1 mole of water has a mass $m = 18.02\ \textrm{g}$ There are $N_A$ particles in 1 mole of a substance $\therefore$ one molecule of water has mass $$m = \frac{18.02}{6.022 \times 10^{23}} = 2.992 \times 10^{-23}\ \textrm{g}$$ 8. Required extra data: Answer to Q7. (mass of one molecule of water) $m = 2.992 \times 10^{-23}\ \textrm{g}$ Typical volume of an ice cube - assume dimensions $2\ \textrm{cm} \times 2\ \textrm{cm} \times 3\ \textrm{cm} \Rightarrow V = 12\ \textrm{cm}^3$ Density of ice $\rho_{ice} = 0.917\ \textrm{g cm}^{-3}$ Mass of ice cube $$M = \rho_{ice}V = 0.917 \times 12 = 11.004\ \textrm{g}$$ $\therefore$ number of molecules in one ice cube is $$n \approx \frac{11.004}{2.992 \times 10^{-23}} = 3.7 \times 10^{23}$$ 9. Required extra data: Speed of electromagnetic radiation (in free space) $c = 299\ 792\ 458\ \textrm{ms}^{-1}$ $$T = 13.4\ \textrm{billions years} = 13.4 \times 10^9 \times 365.25 \times 24 \times 60 \times 60 = 4.229 \times 10^{17}\ \textrm{s}$$ $$ \therefore d = cT = 299\ 792\ 458 \times 4.229 \times 10^{17} = 1.27 \times 10^{26}\ \textrm{m}$$ 10. Required extra data: Mass of Earth $M_E = 5.9742 \times 10^{24}\ \textrm{kg}$ Speed of electromagnetic radiation (in free space) $c = 299\ 792\ 458\ \textrm{ms}^{-1}$ Einstein's famous formula relating energy to mass is: $$E = mc^2 \therefore E_{Earth} = M_E c^2 = 5.9742 \times 10^{24}\times 299\ 792\ 458^2 = 5.369 \times 10^{41}\ \textrm{J}$$ </mdoxml> <mdoxml version="1.0"><h3>Why do this problem?</h3> The ability to work with big and small numbers is a very important scientific skill. These <a href="http://nrich.maths.org/public/viewer.php?obj_id=6504&part="> questions</a> provide a variety of contexts where handling such numbers is necessary, while offering practice at identifying key information and making decisions about estimates and assumptions. <h3>Possible approach</h3> <div>These questions could be used individually as starters or fillers within lessons. Alternatively, a class could work on different questions in small groups and then present their answers with reasoning to the rest of the class to stimulate discussion. Some questions can be tackled with little specialist knowledge; others may require research and additional information, as well as some assumptions, in order to come up with an answer.</div> <div>It is important to stress that there is no "correct" answer for some of these questions - the value of this task comes from having a go at estimating and calculating with big and small numbers, and discussing assumptions that need to be made and extra data that needs to be sought in order to find a solution.</div> <h3>Key questions</h3> <div>What assumptions have you made?</div> <div>What other information do you need?</div> <div></div>Are there any questions which give an exact answer? Can you say anything about the accuracy of those answers which aren't exact? <h3>Possible extension</h3> This problem could be done in conjunction with <a href="http://nrich.maths.org/public/viewer.php?obj_id=6140&part="> Big and Small Numbers in Biology</a>. Similar questions could be created and shared within the class. <h3>Possible support</h3> Start with those questions which have familiar content, and encourage whole class discussion of the ideas in them, before tackling any of the questions which require extra information. It is well worth suggesting that students read the article <a href="http://nrich.maths.org/public/viewer.php?obj_id=6300&part="> Student Guide to Getting Started with Rich Tasks</a> before tackling a very open problem like this one. </mdoxml> <mdoxml version="1.0"> Note that most of the ideas used here are typically covered at school before the age of 16, although possibly in mathematics, physics or chemistry. In estimation questions don't be afraid to have a go with a guess at some numbers in the problem and then to refine your estimate after checking it makes some sort of sense. Although there is no 'right' answer to an estimation, there are good or bad estimates and sensible or over detailed calculations. Think how you might make your estimation a good one, and think how it makes sense to ignore certain complexities in particular calculations. </mdoxml> <mdoxml version="1.0"> 1. Let's change miles per hour to meters per second: 1 mile = 1.61 km and 1 hour = 3600 seconds. Thus $c=3\times 10^8$m/s = $3\times 10^8$$\times 3600$/$1610$ mph =$6.7\times 10^8$mph. The fastest car in 2010 was the Bugatti Veyron with max speed 250 mph. The speed of light is $6.7\times 10^8$ / $250$ = $2.7\times 10^6$ times bigger than the speed of a sport car. 2. We are given that a diameter of the galaxy is $D = 100,000$ ly and a thickness about $d = 1000 ly$. We can approximate our galaxy as a cylinder then calculate a volume of the galaxy. $V = d\times \pi (D/2)^2$. An average number of starts in the galaxy is $N = 250$ billion stars. Now, we can find out a density of the stars: $N/V$ = $(250\times 10^9) : (1000\times \pi (100,000/2)^2)$ = $0.032 stars/ly^3$. Thus, one star occupies the volume which is equal $1/0.032$ $ly^3$=$31$$ly^3$. In order to find the average distance between stars we take a cube root of 31 to get that the distance is about 3.2 light years. 3. Mass of lead $m = 1 t = 1000 kg = 1\times 10^6 g$. Density of lead $11.34$g/cm$^3$. Thus, the volume of 1 tonne of lead is $V =(1\times 10^6) /(11.34)$ cm$^3$ = 88180 cm$^3$. We can imagine this volume as a cube with side length 44.5 cm or as a ball with diameter $55.2 cm$, for a comparison the diameter of football ball is 14-16 cm. 4. Firstly, lets find a mass of a roll of aluminum kitchen foil. A thickness of foil is about 20 micrometes, lenght is 10 meters and width is about 370 milimetres. Hence, the volume of a roll $V = 20\times 10^{-6} \times 10\times 370\times 10^{-3} m^3= 7.4\times 10^{-5} m^3$. The density of aluminum is $7874 kg/m^3$. The mass of this roll is $7.4\times 10^-5\times 7874 = 0.58 kg$. It takes around 2 kg of bauxite (aluminum ore) to make 1 kg of pure aluminum metal. Thus, we need about 1.2 kg of aluminum ore. 5. One AA zinc-carbon battery has a capacity of 1100 mAh. For example one AA Alkaline battery has much bigger capacity of 2700 mAh. A power being used by a laptop is about 60 watts and AC adapter changes voltage from 240 volts to average 20 volts. The formula which relates power with the voltage and the current is $P = I\times U$ where P is power, I the current and U is the voltage. The electric current is a flow of electric charge. Thus, the charge required is $60/20\times 1 = 3$Ah. We can conclude that we need about three AA zinc-carbon batteries or one good Alkaline battery to run a laptop for one hour. 6. A mass of one staple is about 30 mg. A staple is made of stainless steel which consists mostly of iron. The molar mass of iron is $M = 55.85 g/mol$ and Avogadro's number is $6.02\times 10^23$ 1/mol. Hence, the number of atoms in a staple is $N = 6.02\times 10^{23}\times 0.03/55.85 = 3.2\times 10^{20}$ 7. Energy stored in a staple is $E = mc^2$ where mass of a stample is $m = 30$ mg and the speed of light is $c = 3\times 10^8$ m/s. So, $E = 2.7\times 10^{15}$ J. Power is the rate at which energy is transferred. If we assume that a laptop use 60 W power then it could run for $2.7\times 10^{15} : 60$ s= $4.5\times 10^{13}$ s = 1.4 million years. 8. Suppose a volume of room (class) is $10\times 6\times 3 m^3= 180 m^3$. The specific heat capacity of air when a pressure is constant $c = 1012$ $J/(kg\times K)$. The energy required to raise the temperature by T = 1$^\circ$C could be calculated by $E = mcT$ where m is the mass of air in the room. A density of air is $1.2 kg/m^3$). Thus, $E = 1.2\times 180\times 1012\times 1 = 2.2\times 10^5$ J. The heat of combustion is the energy released when a compound undergoes complete combustion. For the natural gas it is $45 MJ/kg$. The mass of natural gas required to raise the temperature by T = 1$^\circ$C is $2.2\times 10^5 : 45\times 10^6$ = 5 grams. The density of natural gas is about $0.8 kg/m^3$. Thus, we need $V = 6.25\times 10^{-3} m^3$ of natural gas. The current price of a natural gas unit is £ 0.038. 1 unit = 1 cubic foot = 0.283 m^3. Suppose efficiency of heating is 50%, so it would cost about 20p. </mdoxml> 2 3 0 0 0 1 0 Big and small numbers in The Physical World Work with numbers big and small to estimate and calculate various quantities in physical contexts. 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