Solution Manual for Principles and Practice of Physics, 1st Edition

Preview (16 of 898 Pages)

100%

Purchase to unlock

Loading page image...

1-1FOUNDATIONS1Solutions to Developing a Feel Questions, Guided Problems,and Questions and ProblemsDeveloping a Feel1.210m−2.210m3.210m4.1210m5.3106.310kg7.410kg8.1310kg9.510kg10.111011.91012.510Guided Problems1.2 Solar oxygen1.Getting StartedMuch of the plan used in Worked Problem 1.1 can be used here. We wish to use the volume ofthe sun and the percent of the Sun that is made up of oxygen to calculate the mass density and number density ofoxygen atoms. As before we assume the Sun is a perfect sphere.2.Devise PlanAs before, we use the mass density/m Vρ=and the number density/.nN V=We will use343VRπ=for the volume of the Sun. We will use the total mass of the Sun and the fraction of that mass that is madeup of oxygen to calculate the mass of oxygen in the Sun. We will divide by volume to get the mass density. Then,one need only divide by the mass of a single oxygen atom to convert from mass density to number density.3.Execute PlanFirst, the mass density is given byoxygenSun33SunSun30383(0.00970)4433(0.00970)(1.9910kg)13.7 kg/m4(6.9610 m)3mmMVRRρππρπ===×==×We convert from mass density to number density using326326atom13.7 kg/m5.1410atoms/m2.6610kg/atomnmρ−===××4.Evaluate ResultThe calculated mass density of oxygen is approximately two orders of magnitude smaller thanthe mass density of hydrogen calculated in Worked Problem 1.1. This is what we expect, because the oxygenaccounts for only about 1% of the mass of the Sun and hydrogen makes up approximately 70%. Since the oxygenaccounts for two orders of magnitude less mass, it is perfectly sensible that the mass density of oxygen would also betwo orders of magnitude smaller. Similarly, the number density of oxygen is about three orders of magnitude smallerthan the number density of hydrogen. If the only issue were the relative percentages of the Sun’s mass made up byoxygen and hydrogen, we would expect a difference of only two orders of magnitude. However, oxygen is also much

Loading page image...

1-2Chapter 1more massive than hydrogen (around one order of magnitude more massive). Hence, mass densities that differ by twoorders of magnitude mean number densities that differ by three orders of magnitude. Our results are consistent withthose of Worked Problem 1.1.1.4 Box volume1.Getting StartedThis problem is similar to Worked Problem 1.3 in that we are given a mixture of units which we mustconvert to SI units. However, the expressions for volume here is completely different from that of Worked Problem 1.3.2.Devise PlanWe convert from feet to inches using the conversion factor1 ft12 in,=and convert from inches tometers using1 in0.0254 m.=We can convert millimeters and centimeters to meters by simply dividing by theappropriate factor of ten. When all quantities are in units of meters, we proceed to find the volume of the box usingbox,Vwh=Awhere,A,wandhare the length, width, and height of the box, respectively.3.Execute PlanWe first convert the length, width and height to units of meters:31 m1420 mm1.420 m10mm=×=A12 in0.0254 m2.75 ft0.838 m1 ft1 inw=××=21 m87.8 cm0.878 m10cmh=×=Finally the volume is given by3box(1.420 m)(0.838 m)(0.878 m)1.05 m .Vwh===A4.Evaluate ResultA volume of31.05 mis reasonable. The box had a size that was on the order of 1.0 m in eachdimension. One dimension had a length greater than 1.0 m and the other two were smaller than 1.0 m. This is theapproximate size we would expect.1.6 Digits on your own1.Getting StartedThe methods used in Worked Problem 1.5 are essentially the same that will be used here. Theexpressions involve quantities with varying numbers of significant digits. The number of significant digits in theanswer will depend on how many significant digits are in the given in the problem statement, and on the operationcarried out between quantities (subtraction, multiplication, etc).2.Devise PlanAs in Worked Problem 1.5, the number of significant digits in a product or quotient is the same asthe number of significant digits in the input quantity that has thefewestsignificant digits. The number of decimalplaces in a sum or difference is the same as the number of decimal places in the input quantity that has thefewestdecimal places. To express our answers as orders of magnitude, we write each in scientific notation, round thecoefficient either down to 1 (for coefficients3)≤or up to 10 (for coefficients3).>We then write the answer as apower of ten without the coefficient.3.Execute Plan(a) In (205)(0.0041)(489.62), the middle quantity (0.0041) has the fewest significant digits, withonly two. Hence the answer must be given to only two significant digits:24.110 .×Here the coefficient is greaterthan 3, so we round it to 10 and obtain for the order of magnitude23101010 .×=(b) Here, the first factor is given tofour significant digits, which is the fewest of all factors (sinceπis known to many digits). Hence the answer mustbe given to five significant digits:12.47510 .×Here the prefactor is less than 3, so we round it down to 1. This yieldsan order of magnitude of111010 .×=(c) The first term is known to the thousandths place, whereas the second term isknown only to the tenths place. Hence the answer can be given only out to the tenths place:36.980210 .×Theprefactor is greater than 3, so we round it to 10. This yields an order of magnitude of34101010 .×=

Loading page image...

Foundations1-34.Evaluate ResultWe could check our answers by getting order of magnitude estimates for each number. (a) Usingorders of magnitude of (205)(0.0041)(489.62) yields2233(10 )(10)(10 )10 ,−=which is consistent with our answer.(b) Similarly, (190.8)(0.407 500)/πbecomes2011(10 )(10 )/(10 )10 ,=which is again consistent with our answer. (c)Finally, to nearest order(6980.035)(0.2)+yields414(10 )(10)10 ,−−=which is also consistent.1.8 Roof area1.Getting StartedWe will approximate the United States as a rectangle 5,000 km wide and 3,000 km high. A verysmall percentage of this area is occupied by buildings. Only in cities can we approximate the surface as beingcovered by structures.2.Devise PlanThe approximate area of the United States is4372(10km)(10km)10km .=Depending on one’sdefinition of a “city” the percentage of this surface area that is occupied by cities could be anywhere between 0.1%and 1%. But since even cities are not completely filled with structures, we will use the smaller of these twopercentages for our estimate (0.1%). These two numbers can be used to find an order of magnitude estimate of thetotal roof area in the United States.3.Execute PlanThe roof area is given by7242roofUS(0.1%)(0.001)(10km )10km .AA===4.Evaluate ResultWe could check our result by estimating the roof surface area in a completely different way. Letus start with the population of the United States (8310people×to one significant digit), and assume that there existsa structure that houses every set of three or four people, and assume further that these structures have a footprint ofapproximately230 m .Of course, many people live alone and many people live in apartment buildings that havemany floors (meaning less of a footprint per person). But these cases might cancel each other out, making ourestimate plausible. This yields a surface area of922.310m .×Let us double this number to account for the places ofbusiness that employ many of these people. This gives a total surface area of order10210m .Using knownconversion factors, one can see that this is equivalent to4210km .This agrees with our previous estimate.Questions and Problems1.1.The word “undetectable” prevents this from being a valid scientific hypothesis. A hypothesis must beexperimentally verifiable.1.2.You assume that the competing product contains non-zero fat, and that the serving sizes of the two are equal.Say the two foods contain the same amount of fat per ounce. The maker of the product being advertised could printhis label showing arecommended serving size50 percent smaller than the recommended serving size of thecompeting product. This makes the claim of 50 percent less fatper servingtrue but of course misleading.1.3.You assume the sequence is linear, meaning each entry is larger than the previous one by a constant amount. Asan alternative, the sequence could be formed by starting with 1, 2 and then setting thethntermncequal to the sumof the previous two terms:12.nnnccc−−=+This would work for321213,ccc=+=+=and would yield 5 as the nextnumber in the sequence.1.4.If you assume that the coins are currently circulated U.S. currency, you would not be able to find a solution. If,however, you consider that the word “cents” is also used to refer to hundredths of other currencies, then all thatwould be required is that increments of 10 and 20 centers exist in some currency. As an example, “cents” may referto hundredths of a Euro. You would say the coins must be worth 10 and 20 Euro cents, respectively (which do exist).

Loading page image...

1-4Chapter 11.5.There are 12 ways:4|3|2|14|3|2|14|3|2|14|3|2|11|2|3|41|2|3|41|2|3|41|2|3|43|1|4|22|1|4|32|4|1|33|4|1|22|4|1|33|4|1|23|1|4|22|1|4|34|3|1|24|3|1|24|3|1|24|3|1|21|2|3|41|2|3|41|2|4|31|2|4|32|1|4|33|4|1|22|1|3|42|4|3|13|4|2|12|1|4|33|4|2|13|1|2|44|3|1|24|3|1|24|3|2|14|3|2|11|2|4|31|2|4|31|2|4|31|2|4|33|1|2|43|4|2|12|1|3|43|4|1|22|4|3|12|1|3|43|4|1|22|1|3|41.6.There is one axis of reflection symmetry. It is marked by the dashed line:1.7.One. An axis of rotational symmetry is an axis about which the object can be rotated (through some angle otherthan a multiple of 360 degrees), that results in an indistinguishable appearance compared to the original orientation ofthe object. For a cone the axis passing through the center of the circular face and through the vertex (point) of thecone is the only axis of rotational symmetry.1.8.One unit down and left from the upper right corner. This way the coins form a square:

Loading page image...

Foundations1-51.9.“T” and “A” are reflection symmetric across a vertical line passing through the center of the letter. “E” and “B”are reflection symmetric across a horizontal line passing through the center of the letter. “L” and “S” have noreflection symmetry.1.10.Reflection and rotation symmetry. A sphere is reflection symmetric across any plane that passes through itscenter, and rotationally symmetric around any axis passing through its center.1.11.A cube has 9 planes of reflection symmetry and 13 axes of rotational symmetry. 3 planes each bisect four sidesof the cube. The remaining six planes pass through edges that are diagonally opposite each other. 3 of the axes ofrotational symmetry pass orthogonally through the centers of two square faces. 4 of the axes pass through cornersthat are opposite each other along the body diagonal. 6 axes pass through the centers of edges that are opposite eachother along the body diagonal.1.12.There is one axis of reflection symmetry:1.13.The maximum number of axes of reflection symmetry is two. There are two sides about which we have noinformation. Let us assume the side opposite the visible blue side is also blue, and the side opposite the visible redside is also red. In that case, the object would be reflection symmetric about a vertical plane that bisects both bluesides, and a about a vertical plane that bisects both red sides.1.14.We use/xvxt= ΔΔfor constant speed in thexdirection to write:17299,792,458 m3600 s24 h365.25 days78 yrs7.410m1s1 h1 day1 yrxxvtΔ=Δ =××××=×1.15.(a) We convert units using known conversion factors:37141609 m10mm9.310miles1.510mm1 mi1 m×××=×(b) If we divide the distance by the width of Earth, that will tell us how many Earths can fit in that distance.743Earth1.609 km9.310miDistance to Sun1 miNumber of Earths1.210Earths22(6.3810km)R××===××1.16.The blue whale and a human have densities that are the same orders of magnitude, and contain similarconcentrations of different atom types. Hence it is reasonable to say that the ratio of the numbers of atoms in thesetwo species should be roughly equal to the ratio of volumes of these two species:323bw29human1010 .10VV==Since whalesand people are three dimensional, this corresponds to blue whales being roughly one order of magnitude larger ineach of three dimensions. Hence, the blue whale is approximately 10 times longer than a human is tall.1.17.Because the gastrotrich lifetime is given as three days, this should technically be treated as being on the orderof one day. However, this is an approximate value of a lifetime, and if it were slightly higher, it would be treated as

Loading page image...

1-6Chapter 1being on the order of ten days. Either of these is defensible in order of magnitude treatments. One tortoise lifetimecan be related to gastrotrich lifetimes using the following order of magnitude conversions:32510days1 gastrotrich lifetime10yr10 gastrotrich lifetimes1 yr1 day××=Ifwehadused10daysastheorderofmagnitudeforagastrotrichlifetimewewouldhaveobtained410gastrotrich lifetimes. Hence there are 10,000 to 100,000 gastrotrich lifetimes in one tortoise lifetime. Either ofthese is acceptable.1.18.Given the diameter of a droplet of water, we can estimate the volume of a droplet using3drop413Vrπ=≈×338310(10m)10m .−−×=We can estimate the volume of a human body to be on the order of13body10m .V−=Hencean order of magnitude estimate of the number of droplets in the human body is given bybody7drop10 .VV=1.19.Answers may vary by an order of magnitude since some textbooks may be somewhat thicker than 3.0 cm, andothers may be thinner. My textbook has a thickness that is on the order of110m.−The distance to the moon is83.8410 m,×meaning the distance is of order910m. Dividing the distance by the thickness of one textbook yields9101Distance to Moon10Number of books10 .Thickness of book10−===Hence1010copies of my physics textbook could fit betweenEarth and the Moon.1.20.We proceed by finding the total mass of water in the pool, and dividing this by the mass of a single moleculeof water:poolwaterpoolmoleculeA/mVNmMNρ==where3water1000 kg/mρ=is the density of water,3pool(158.51.5) mV=××is the volumeof the swimming pool,0.018kgM=is the molar mass of water, and231A6.0210molN−=×is Avogadro’s number.Using this information we find33323130(1.010kg/m )(191 m ) (6.0210mol)6.410molecules in the pool.(0.018 kg/mol)N−×=×=×1.21.(a) Since numbers are not given, it might be natural to cancel lengths and be left with a factor of32 . In thatcase the volume of the cube would increase by one order of magnitude. But if one were asked to use an order ofmagnitude estimate to first express2Ain terms of1,Aone might find that they have the same order of magnitude andthat the volume therefore does not increase. Either of these answers (one order of magnitude, or zero orders ofmagnitude) is acceptable. (b) Yes, because of the rules of rounding numerical values. For example, if313.5 m ,V=that value would round to an order of magnitude of 10 m3. Then321828 m ,VV==which also rounds to an order ofmagnitude of 10 m3.1.22.The speed of light is of order810m/s.The length of Earth’s trip around the Sun is of order1210m.Hencetheorderofmagnitudeofthetimelightwouldneedtomakethesametriparoundthesunis1248distance10mtime10s.speed10m/s===Hence light would take approximately410 s to complete this trip.1.23.The surface area of the roughly spherical distribution of leaves is24rπ, whereris the radius of the tree’ssphere of leaves. The surface area of an individual leaf is just its lengthAtimes its width.wWe can find the numberof leaves by dividing the total surface area by the area of one leaf:225410(10 m)10(0.1m)(0.1 m)rNwπ===Aleaves.

Loading page image...

Foundations1-71.24.To find the number of generations since the universe began, we divide the age of the universe by the timerequired for one human generation. We simply have to get both times in the same units. Let us express one humangeneration in seconds:8365.25 days24 h3600 s30 yr9.510s.1yr1 day1h×××=×Hence the number of human generationsNis given by178universe8generation10s109.510stNt===×generations.1.25.Not reasonable. Because light travels much faster than sound, any thunder peal is delayed compared to thelight signal caused by the lightning bolt event. From the principle of causality, the lightning you see after you hearthe peal cannot have caused the peal. The peal you heard must have come from a previous lightning strike.1.26.CalltheperiodoftimerequiredforonesuchoscillationT.Thenthesecondisdefinedsuchthat91.0 s(9.1910 )T=×or1091.0 s1.0910s.(9.1910 )T−==××1.27.That the barrier lowers time after time 30 s before a train passes is consistent with a causal relationshipbetween the two events. The single negative result, however, tells you that the lowering of the barrier cannot be thedirectcause of the passing of the train. More likely, the lowering is triggered when the train passes a sensor quite adistance up the tracks from the barrier and the sensor sends an electrical signal to the lowering mechanism. Amalfunction in either the sensor, the electrical connections, or the lowering mechanism would account for the onenegative result you observed.1.28.The period calculated in Problem 26 was101.0910s.T−=×The speed of light is83.0010m/s.c=×Here thedistance travelled is just the speed times the time:810(3.0010m/s)(1.0910s)0.0326 mdct−=Δ =××=.1.29.2Emc=Here,Eis the type of energy described,mis the mass of the object, andcis the speed of light.1.30.Rectangle, parallelogram, and two equilateral triangles meeting at a vertex forming an hourglass shape.1.31.The problem states that two adjacent sides must make an angle of 30 .°This most likely means the interiorangle between actual sides. This angle is labeledαin the figure below. But one might also describe the exteriorangle in this way. This angle is labeledβbelow. If one accepts this interpretation, some of the 30°angles could beinterior and some could be exterior.Let us first address the case in which the 30°angle refers to the interior angle. Remembering that the segmentscannot cross, there is only one possible arrangement of segments that fits the description:

Loading page image...

1-8Chapter 1Clearly2 sin( )yθΔ=Aand2 (1cos( )).xθΔ=−AThe Pythagorean Theorem tells us that the distance between the twounconnected points must be()1/ 222222(1cos(30 ))sin (30 )1.0 .dxy=Δ+ Δ=−°+°=AASo the distance between unconnected ends is.AIf the 30°angle refers to the exterior angle, we can obtain several possible shapes. The two shapes that give theshortest and longest distances are shown below:We can see in the top figure (that yields the shortest distance)(1cos(30 )cos(60 )0),xΔ=+° +° +Aand since thexandydistances are equal based on symmetry, we find2 (1cos(30 )cos(60 ))3.3 .d=+° +°=AAIn the second figure (that yields the greatest distance) we find2 (1cos(30 ))xΔ=+°Aand2 (sin(30 )).xΔ=°AAgainusing the Pythagorean Theorem yields a distance of1/ 222(2 (1cos(30 )))(2 sin(30 ))3.9 .d⎡⎤=+°+°=⎣⎦AAAIf a combination of interior and exterior angles is used, there are even more possibilities. The shortest of thesedistances is 0 (parallelogram). It can be shown that other possible distances include 2.0 , 2.2 , 2.4 .AAA1.32.Uncle, cousin, grandmother, aunt, grandfather, brother.1.33.Consider the diagram below.

Loading page image...

Foundations1-9The distance from the Sun to point P is2211211211(1/ 2)(1.5010m)(7.7810m)3.9610mxyΔ+ Δ=×+×=×The time that light would require to cross this distance can be found using1183/(3.9610m) /(3.0010m/s)1.3210stdvΔ ==××=×which is equivalent to about 22.0 minutes.1.34.Time (s)Position (m)0.000.002.000.404.001.596.003.641.35.(a) The position decreases linearly as a function of time, from an initial position of 4.0 m to a final position ofzero at a time of 8.0 s, with a slope of0.5 m/s.−(b)( )xtmtb=+where0.5 m/s, and4.0 m.mb= −=1.36.Forming a regular tetrahedron from these triangles will automatically satisfy all conditions.1.37.1 ft12 inch8.95 m352 inches0.3048 m1 ft××=1.38.We change units using known conversion factors:1 mile35,000 ft6.629 miles5280 ft×=3312 in25.4 mm1 m1 km35,000 ft10.7 km1 ft1 in10mm10m××××=1.39.(a) The density will be the same. (b) The density will be the same.1.40.We change units using known conversion factors:852.997910m1 mile1.86310miles/s1 s1609 m××=×8392.997910m1 s10mm1 in11.8 inches/ns1 s10ns1m25.4 mm××××=

Loading page image...

1-10Chapter 1892.997910m1 km3600 s1.07910km/h1 s1000 m1 hr×××=×1.41.If the two stones are made from the same material, they should have roughly the same density. We calculatethe density of each stone and compare them:23311312.910kg2.910kg/cm10.0 cmmVρ−−×===×23322322.510kg3.310kg/cm7.50 cmmVρ−−×===×No, it is not likely. Stone 2 has considerably higher density.1.42.This length can be expressed as3 ft1 mi1.000 mi440 yd1.250 mi.1 yd5,280 ft+××=We now convert this entirelyto feet:5280 ft1.250 mi6,600 ft.1 mi×=1.43.We can find the units ofAfrom given information:22units[]m/ssm.A=⋅=IfAhas units of meters, and we wishto addAandB, thenBmust also have units of meters.1.44.Mass density is the ratio of mass per unit volume. While one could have any mass of a given substance, or anyvolume of that substance, the density tends to be a constant value for a given material (under certain conditions).1.45.For all cases we find the order of magnitude of the mass of Earth using3E4.3mVRρρπ⎛⎞==⎜⎟⎝⎠(a)3037322airE4(10kg/m )(110(10m) )10kg3mRρπ⎛⎞==××=⎜⎟⎝⎠(b)34373265515E4(10kg/m )(110(10m) )10kg3mRρπ⎛⎞==××=⎜⎟⎝⎠(c)31837340nucleusE4(10kg/m )(110(10m) )10kg3mRρπ⎛⎞==××=⎜⎟⎝⎠1.46.We rearrange the given expression to solve foryand then write all units in terms of SI base units and powersof ten:2 / 32 / 3232 / 31815233612102261.7 Eg fm /ms7.81g/Tm61.7 (10g) (10m) /(10s)7.81 (10g)/(10m)3.9710m /sxyayyμ−−−⎛⎞⋅⎛⎞==⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞⋅=⎜⎟⎝⎠=×1.47.(a)83.0010m/s×(b)16228.9910m /s×(c) No. There is a small difference because the answer to (a) wasrounded before squaring. The answer to (b) was obtained using more digits of the speed of light, and only the resultwas rounded to three significant digits.1.48.The given distance can also be written as 1.25 miles. We now convert to kilometers using known the knownconversion factor:1.609 km1.25 mi2.012 km1 mi×=

Loading page image...

Foundations1-111.49.Your answer has four significant digits. When dividing a quantity by an integer, the number of significantdigits should not change.1.50.Yes, there is a difference in the precision. You will calculate your gas mileage by dividing the number of milesyou travel by the gallons of gasoline used. Since the gas pump gives you thousandths (and most vehicles take 10 galor more) you know the fuel used to five significant digits. Neither distance given has this many, making the precisionof the distance value the limiting factor. Thus you can calculate mileage to three significant digits when you use40.0 mi for distance and to four significant digits when you use 400.0 mi.1.51.The odometer should still say 35,987.1 km. 47.00 m is only 0.04700 km. Because the odometer reading is onlygiven to the nearest tenth of a kilometer, the sum (final odometer reading) must also be given only to the tenth of akilometer.1.52.We convert the given amount of ingested caffeine using known conversion factors:2322334 mg2 servings365.25 days1g1 mol6.0210molecules7.710molecules/yrservingday1 yr10 mg194.19 g1mol××××××=×1.53.The molarity of the solution must be known to a precision of 1 part in 15. Since we can measure volume toarbitrary precision and we have the molar mass to very high precision, the limiting factor on the precision of ourmolarity is the precision of the mass we use. Hence the precision of the mass we use must be at least as good as therequired precision of the molarity. We can only measure tenths of a gram, so we must have at least 1.5 grams in orderto know the precision of our mass to 1 part in 15 or better. Mathematically:(0.150.01) mol/L0.1 g0.010.1 g(0.01 mol/L)(58.44 g/mol)0.17 L0.15()1.5 gmmMVMVmMVVVmMVΔ±=±≤ Δ=≥≥==As can be seen from the above calculation, the answer could also be given in terms of the minimum volume ofsolution prepared. That minimum volume is 0.17 L.1.54.The mass density is given by/.m Vρ=The volume is given in milliliters, which are equivalent to cubiccentimeters. Hence363333325.403 g10kg10cm1.08510kg/m23.42 cm1 g1 mρ−=××=×1.55.The mass evaporated each second is given bydishliquidfinal3(145.67 g)(0.335 g)(145.82 g)(25.01 s)7.410g/smmmmtt−+−Δ=ΔΔ+−==×Note that we have only two significant digits. Because two of the masses were only known to the hundredths place,we only had two digits of precision after taking the sum (and difference) of masses.1.56.There are too many significant digits. The volume is only given to two significant digits. The answer couldhave at most two significant digits.

Loading page image...

1-12Chapter 11.57.We find the distance travelled by each hand in a day and we take the difference. The minute hand makes 24full revolutions in one day, such that the distance travelled by its tip is24 rev224 rev2(0.0113 m)1.70 m1 day1 rev1 day1 revrππ×=×=The hour hand makes only two revolutions, such that the distance travelled by its tip is2 rev22 rev2(0.0080 m)0.10 m1 day1 rev1 day1 revrππ×=×=Clearly, the minute hand travels farther in one day by a distance of 1.60 m.1.58.The path lengths are exactly the same along the straight sections of track. We need only consider the twocurved ends. Along those sections, the runner in lane 1 moves through a circle of radius136.80 m.R=A runner inlane 8 moves through a circle of radius836.80 m7(1.22 m)45.34 m.R=+=The total path length difference willthen be given by the difference in the circumferences of these two circles:812()2(8.54 m)53.7 m.xRRππΔ=−==1.59.Place two coins on the balance, and hold the third in your hand. If they are balanced, the one coin in your handis the counterfeit. If the two coins are not balanced, then one must be the counterfeit. Swap the lighter of the coins onthe balance for the one in your hand. If the two coins now have equal mass, then the counterfeit coin is the one youjust removed, and it is lighter than real coins. If they are still unbalanced, then the counterfeit is the one that left inplace on the scale, and it is heavier than real coins.1.60.(a) We proceed by assuming nearly perfect packing of the grains of rice (no unoccupied volume in the cup).This would almost certainly not be exactly true, but since the rice is not rigid after cooking, deformations could allowthe rice to be tightly packed almost to this limit. With that assumption, we find the number of grains by dividing thetotal volume by the volume of a single grain:3cupcup422grain250 cm110grains(0.6 cm)(0.1 cm)VVNVRππ====×A(b) We simply divide the given energy by the number of grains we found in part (a):4785 Cal0.06 Cal/grain.1.310grains=×(c) The total food calories needed would be 8,000. Dividing this by the 785 food calories in one cup, we find that10.2 cups of such rice would be required.1.61.FromPrinciplesFigure 1.9 we see that there are approximately8010atoms in the universe. We also knowthat the number of atoms in a mole is of order2410atoms/Mole.Hence the number of Moles is given by80562410atoms10Moles10atoms/Mole=1.62.We divide the volume of the bread by the number of raisins in the bread and find:33bread2raisin(1)(10 in)10 in /raisin10raisinsVN==This volume could be treated as a spherical or cubic region of bread surrounding each raisin. The side length of sucha cubic region would be of order 1 inch. This also means that 1 inch would be the order of magnitude of the distancebetween raisins.1.63.We could divide the volume of the tree by the volume of the board to obtain the number of boards. But if ahalf-integer number of boards fits along the height of the tree, this does not help us and we have to round down. So,let us first find how many boards fit along the height.3tree10mm1 in1 ft32 m105 ft or 17.5 boards1 m25.4 mm12 inh=×××=

Loading page image...

Foundations1-13So we can only fit 17 entire boards along the height of the tree. Now if we look at a cross-section of the tree, we cancalculate how many boards can fit across the tree. We will treat the tree as though it has a circular cross section.22233tree2board(0.40 m)10mm1 in190 boards(2 in)1 m25.4 mmAAπ⎛⎞⎛⎞=××=⎜⎟⎜⎟⎝⎠⎝⎠Clearly, this overestimates the number of boards that can actually be cut, because of the curvature of the tree trunk.We might estimate how many boards will be affected by the curvature by determining how many boards would liealong the circumference of the trunk. We do this by dividing the circumference of the tree by the width of one board.Some might lie with an edge parallel to the trunk, whereas most will lie with their edges at some angle to the trunk.Still, as this is only an estimate, let us approximate the width of a single board as 2 in. Then the number of boardsthat would lie along the edge is3treetreeboardboard22(0.40 m)10mm1 in49 boards2 in1 m25.4 mmCrNwwππ===××=So we estimate that we can fit a cross section of 140 complete boards in the cross section of the tree.Finally, we have a total number of boards given by the number of boards that fit length-wise along the height of thetree times the number of complete boards in a cross section:3totalhighacross17140210NNN==×=×In the last step we have rounded to one significant figure. In addition to being given only one significant figure in thewidth of the board, we have also ignored secondary corrections such as wood that is lost to the sawing process.Hence our best estimate is3210boards.×1.64.Estimates will vary, as there are many methods of estimating. One method of estimation would be to count thenumber of times it occurs on a few lines to obtain an average number of occurrences per line of text. One could thenmultiply by the number of lines in a book. One could also estimate the number of letters in the book and assume that“d” occurs just as often as all other letters (not strictly true, but correct to an order of magnitude). The former methodis crude, but requires no prior knowledge of the frequency of letters’ occurrence in the English language. For thisreason, we proceed with the former method. After counting the number of times “d” occurs on a few lines, I estimatethat it occurs approximately twice on each line (of order 1). Each page of this book contains a number of lines that isof order210 . The book has a number of pages that is of order310 . Multiplying yields:2351 "d"10lines10pages10"d"/book1 line1 page1 book××=One could obtain a much more reliable estimate if one happens to know that the letter “d” accounts for approximately 4%of the letters in a typical English text. But this method agrees with our crude method to the nearest order of magnitude.1.65.The average full head of human hair has on the order of510hairs. Shoulder-length hair is right on theboundary between being of order 1 m and 0.1 m, but let us assume that the average human has hair that is shorterthan 0.30 m. This yields a total length of all hairs of410m.1.66.Not feasible during a four year stay at university, but perhaps possible by starting at age 5. $200K ofuniversity cost and 5 cents per can means the student must collect 4 million cans. Doing this during a 4-yearuniversity career would require almost 3000 cans every day. Because these must be pulled out of waste bins, even atone can every 10 seconds that uses more than 8 hours each and every day. This is not feasible. But if the studentstarted collecting at age 5 and spent 12 full years at the job, at about 3 hours per day, by the time she reacheduniversity she would have her tuition money. Not a fun childhood, but perhaps feasible.1.67.CallSthe storage capacity andAthe area required. We wish to obtain an order of magnitude by which thefraction/SAhas increased. Hence we estimate()()121228ff7202ii(10bytes) / (10 plates)(10m /plates)/10/(10bytes) / (10 plates)(10m /plates)SASA−==Hence storage per unit area has increased by a factor of approximately810 .

Loading page image...

1-14Chapter 11.68.We convert to units of m/s using known conversion factors:1991.0824310nm1 m1 yr1 day1 h343.009 m/s1 yr10nm365.242 days24.0 h3600 s×××××=1.69.We first convert from inches to millimeters:25.4 mm2.75 in69.9 mm1 in×=We know this thickness is from 200 sheets, meaning 20069.9 mmt=or69.9 mm0.349 mm/sheet.200 sheetst==Hencethe thickness of each sheet is 0.349 mm.1.70.(a) We use known conversion factors:33334103331 m(10 )mm5.010L5.010mm10L1 m×××=×(b) The density of water is36333931.010kg10mg1 m1.0 mg/mm .m1kg10mm×××=From here it follows trivially that themass of the water is105.010mg.×(c) Answers will vary based on assumptions about the size of a standard glass. A glass that holds a full liter would bevery large, but a significant fraction of a liter would be plausible. Let us assume that 8 averaged sized glasses isapproximately 5 L. Then we find41 day1 yr5.010L30 yrs5.0 L365 days×××=Hence the water would last about 30 years. Answers up to 45 years would also be plausible given differentassumptions about the glass volume.1.71.No, the model will not fit. Atoms are typically510times larger than nuclei. In order for the nucleus to be that large,the atom itself would need to be about 50 km across.1.72.We calculate both sides of the equation and check for agreement.(a) Initially keeping all the significant digits provided:226211322414321432(9.80665 m/s )(6.37814010m)(6.673810m /(kg s ))(5.973610kg)3.9894110m /s3.9866610m /sEEgRGM−=×=×⋅××=×The results agree to three significant digits.(b) Proceeding in exactly the same way as in part (a) one can round each quantity to various numbers of significantdigits and calculate the two sides of the equation. We do this and look for a point at which the two answers no longeragree to three significant digits. One obtains:143214321432143214321432143214324 sig. dig. : 3.9893810m /s3.9870510m /s3 sig. dig. : 3.9931010m /s3.9819910m /s2 sig. dig. : 4.0140810m /s4.0200010m /s1 sig. dig. : 3.610m /s4.210m /s×=××=××=××=×Clearly, the two sides of the equation agree to three significant digits unless the physical values are all rounded toonly three significant digits or fewer.

Loading page image...

Foundations1-151.73.When the tire swing moves through an angleθfrom vertical, it will have risen a vertical distance(1cos( )),Lθ−andit will have moved over a horizontal distancesin( ).LθCall the angle that the ground makes with the horizontal.φThengroundgroundtan( )/.yxφ= ΔΔHence we can write the amount by which the ground has risen, when the tire swingmoves through a certain angle:groundgroundtan( )sin( ) tan( ).yxLφθφΔ= Δ=The total vertical distance from the groundto the tire swing is theniswingground(1.0 m)(1cos( ))sin( ) tan( )yyyLLθθφ+ Δ− Δ=+−−Not surprisingly, the maximum of this height occurs when the swing is at its maximum angle. This height is onlyabout 1.1 m, which is a perfectly safe height. Of course, swinging out over the water (where the ground does not riseup to meet the swing) will result in a greater height above the water. But this is also only about 1.7 m, which is alsonot a dangerous height for falling into water. You might also have a concern that the tire swing may drag a childacross the ground. We can find the minimum height of the tire swing above the ground by differentiating with respecttoθand requiring that the derivative be equal to zero. This yields the conditiontan( )tan( ),θφ=or.θφ=This tellsus that the minimum height of the tire swing above the ground is about 0.89 m. This is high enough that it is notlikely that a child would unexpectedly strike the ground. There is no need for concern. The swing is perfectly safe.1.74.We begin by calculating how much oxygen is actually needed per breath. Since oxygen is only 20.95% of theair we breathe, the volume of oxygen taken in during each breath is only 0.943 L. Since each breath only absorbs25% of the oxygen present, the oxygen that actually gets used in each breath is only 0.236 L. This corresponds toapproximately42.3610kg−×of oxygen per breath. Finally we calculate how much oxygen is needed for an entireyear using simple conversion factors:42.3610kg15 breaths60 min24 h365 days1900 kgbreath1 min1 h1 day1 yr−×××××=The mass of oxygen required is around 1900 kg. This is reasonable. However, at atmospheric pressure and roomtemperature this air would occupy 1900 cubic meters, or a cubic room of side length 12 m. In an environment inwhich space is a precious commodity, this is probably not feasible. Alternatives would be to use highly compressed,cooled air to decrease storage requirements, or rely on plant life in the ship to replenish oxygen.1.75.(a) The density is given by301734310kg10kg/m(1)(10)(10m)mV==(b) To the nearest order of magnitude, the density of Earth is4310kg/m ,and the density of water is3310kg/m .Hence the neutron star is 13 orders of magnitude greater than the density of Earth and 14 orders of magnitude greaterthan the density of water. (c) The mass of a liter of water is 1.0 kg. Since the neutron star is 14 orders of magnitudedenser than water, the mass contained in a 2-L container would be of the order1410kg.

Preview Mode

This document has 898 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Exploring Light Intensity Through Polarizers

IPL Lab Report

Distance Time Velocity Gizmo Handout-2

Energy Skate Park Activity

Exploring Density Measuring Mass and Volume of Obj

Student Exploration Distance Time Graphs

Barron's Physics Practice Plus: 400+ Online Questions and Quick Study Review (2022)

Test Bank for Conceptual Integrated Science, 3rd Edition

College Physics: A Strategic Approach, 4th Edition Test Bank

Practical Applications Of Density Calculations In Everyday Scenarios

Company

Explore

Study Tools