•1. Earth is approximately a sphere of radius 6.37 x 10⁶ m. What are (a) its circumference in kilometers, (b) its surface area insquare kilometers, and (c) its volume in cubic kilometers?

A: If we idealize the Earth's geometry as a perfect sphere (it's not, but let's assume it is for the purpose of making an estimate), then we can pretty easily find the circumference, surface area, and volume pretty easily from just the radius. Before we make those calculations, however, note that the problem asks us to give our answer in kilometers, but our radius is only given to us in meters. We can't make calculations with inconsistent units, so let's first convert our radius from meters into kilometers. There are 1000 meters in one kilometer, so let's divide 6.37 x 10^6 m by 1000 to get 6.37 x 10^3 km.

(a) The formula for the circumference of a circle or sphere is C = 2πR_E, where R_E is the radius of the Earth. Let's plug in 6.37 x 10^6 m for the radius and solve.

C = 2πR_E = 2π(6.37 x 10^3 km) = 4.00 x 10^4 km.

(b) The formula for the surface area of a sphere is A = 4πR_E². Let's once again apply our radius to this formula:

A = 4πR_E² = 4π(6.37 x 10^3 km)^2 = 5.10 x 10^8 km^2.

(c) The formula for the volume of a sphere is V = 4π/3 R_E³. You know the drill by now.

V = 4π/3 R_E³ = 4π/3 (6.37 x 10^3 km)^3 = 1.08 x 10^12 km^3.

Notice that the power on the unit is consistent with the dimension. For instance, volume is a three-dimensional unit, so we write its unit as "kilometers cubed". Circumference is a type of length, which is one-dimensional, so the unit doesn't have an exponent (exponent of 1). Realizing this might help you memorize the formulas for the circumference, area, and volume.

FINAL ANSWERS: (a) 4.00 x 10^4 km
(b) 5.10 x 10^8 km^2
(c) 1.08 x 10^12 km^3

•2. A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, defined as 1/12 inch. A common measure for length in the publishing business is a point, defined as 1/72 inch. What is an area of 0.50 gry² in points squared (points²)?

A: We're trying to convert a unit of "gry" into a unit of "points", so let's first find a conversion factor between grys and points using a chain-link conversion:

1 gry = (1/10 line)(1/12 inch)(72 points) = 0.60 point.

Of course, we're dealing with squared units in this case, so let's square both sides of this conversion factor:

1 gry² = (0.60 point)² = 0.36 point²

Now that we have a conversion rate from squared grys to squared points. Now use this conversion to convert 0.50 gry² into point²:

0.50 gry² = (0.50 gry²)(0.36 point² / 1 gry²) = 0.18 point²

FINAL ANSWER: 0.18 point²

•3. The micrometer (1 μm) is often called the micron.

(a) How many microns make up 1.0 km?
(b) What fraction of a centimeter equals 1.0 μm?
(c) How many microns are in 1.0 yd?

A: Know your conversions for this one.

(a) There are 1000 meters in a kilometer, so 1 km = 1000 m. Most people probably know that already. The "micro" prefix is a little trickier, so if you aren't familiar with it already then it's a good thing to memorize. Basically, it's the opposite of the "mega" (million) prefix, so a micrometer / micron is the same as one millionth of a meter. There are one million microns in a meter. Now, let's use this knowledge to get the answer via the basic conversion method:

1.0 km * (1000 m / 1 km) * (1000000 μm / 1 m) = 1000000000 μm

That's 1 with nine 0's after it, or one billion. If you want to make it prettier, you can rewrite it in scientific notation as such:

(1.0 x 10⁹)μm

(b) This time we're converting microns to centimeters. 1 m = 100 cm.

1.0μm * (1 m / 1000000 μm) * (100 cm / 1 m) = 0.0001 cm, or 1/10000ths of a centimeter. This fraction can also be written as 1.0 x 10^-4, or 1/10⁴.

(c) Same as the above two processes. This one is trickier, however, because the yard is an imperial unit (rather than a metric unit) and doesn't mathematically line up nicely with the other units of measurement typically used. You might just want to look up the conversion rate for problems like this, since I personally don't think metric-to-imperial conversion rates are really worth memorizing. 1 yd = 0.9144 m, exactly.

1.0yd * (0.9144 m / 1 yd) * (1000000 μm / 1 m) = 914400 μm, or 9.1 x 10⁵ μm if you prefer to write it more simply and don't mind shaving off a few nonzero digits.

FINAL ANSWERS: (a) 10^9 μm
(b) 10^-4
(c) 9.1 x 10⁵ μm

•4. Spacing in this book was generally done in units of points and picas: 12 points = 1 pica, and 6 picas = 1 inch. If a figure was misplaced in the page proofs by 0.80 cm, what was the misplacement in (a) picas and (b) points?

A: (a) We know that 1 inch = 2.54 cm, and 6 picas = 1 inch. We use our chain-link unit conversion to convert 0.80 cm to picas:

0.80 cm = (0.80 cm)(1 inch / 2.54 cm)(6 picas / 1 inch) ≈ 1.9 picas

(b) We now consider that 12 points = 1 pica:

0.80 cm = (0.80 cm)(1 inch / 2.54 cm)(6 picas / 1 inch)(12 points / 1 pica) ≈ 23 points

FINAL ANSWERS: (a) 1.9 picas
(b) 23 points

•5. Horses are to race over a certain English meadow for a distance of 4.0 furlongs. What is the race distance in (a) rods and (b) chains? (1 furlong = 201.168 m, 1 rod = 5.0292 m, and 1 chain = 20.117 m.)

A: Using the conversion factors given to us, we can convert from furlongs to rods and chains using some chain-link conversions. Wow these are getting redundant.

(a) 4.0 furlongs = (4.0 furlongs)(201.168 m / 1 furlong)(1 rod / 5.0292 m) = 160 rods

(b) 4.0 furlongs = (4.0 furlongs)(201.168 m / 1 furlong)(1 chain / 20.117 m) = 40 chains

FINAL ANSWERS: (a) 160 rods
(b) 40 chains

••6. You can easily convert common units and measures electronically, but you still should be able to use a conversion table, such as those in Appendix D. Table 1-6 is part of a conversion table for a system of volume measures once common in Spain; a volume of 1 fanega is equivalent to 55.501 dm³ (cubic decimeters). To complete the table, what numbers (to three significant figures) should be entered in (a) the cahiz column, (b) the fanega column, (c) the cuartilla column, and (d) the almude column, starting with the top blank? Express 7.00 almudes in (e) medios, (f) cahizes, and (g) cubic centimeters (cm³).

A: Oh god, it's one of these tedious things. All right, let's take this one column at a time:

(a) The cahiz column: We can see from the top row that 1 cahiz = 12 fanega, so the inverse of this would be that 1 fanega = 1/12 cahiz. To fill out the first blank in the cahiz columb, we take 1/12 cahiz and write it with three significant figures: 1/12 cahiz = 8.33 x 10^-2 cahiz. We can apply this general process to the rest of the blanks in this column; Since 1 cahiz = 48 fanega, 1 cuartilla = 1/48 cahiz = 2.08 x 10^-2 cahiz. 1 cahiz = 144 almude, so 1 almude = 1/144 cahiz = 6.94 x 10^-3 cahiz. 1 cahiz = 288 medio, so 1 medio = 1/288 cahiz = 3.47 x 10^-3 cahiz.

(b) We switch gears slightly and focus on the fanega column. As we can see from the fanega row, 1 fanega = 4 cuartilla, so 1 cuartilla = 1/4 fanega = 0.250 cahiz. At this point, you probably get the idea, and the process becomes very easy (although it's admittedly monotonous). For the rest of the column, we find 8.33 x 10^-2, and 4.17 x 10^-2.

(c) For the cuartilla column, using the same process as in part (a) and part (b), we find 0.333 and 0.167.

(d) In the almude column, we get 0.500 for the only blank entry.

(e) Now we want to convert 7.00 almudes to medios. Fortunately, the conversion table we just created will help us with that. We know from the table that 1 almude = 2 medio.

7.00 almudes = (7.00 almudes)(2 medios / 1 almude) = 14.0 medios

(f) In this case, we want to use the fact that 1 almude = 6.94 x 10^-3 cahiz.

7.00 almudes = (7.00 almudes)(6.94 x 10^-3 cahiz / 1 almude) = 4.86 x 10^-2 cahiz

(g) The problem gives us the conversion factor between dm and fanegas, so we convert from almudes to fanegas and then dm.

7.00 almudes = (7.00 almudes)(1 fanega / 12 almude)(55.501 dm³ / 1 fanega) = 3.24 x 10⁴ cm³

FINAL ANSWERS: (a) 8.33 x 10^-2, 1/48 cahiz = 2.08 x 10^-2, 6.94 x 10^-3, 3.47 x 10^-3
(b) 0.250, 8.33 x 10^-2, 4.17 x 10^-2
(c) 0.333, 0.167
(d) 0.500
(e) 14.0 medios
(f) 4.86 x 10^-2
(g) 3.24 x 10⁴ cm³

••7. Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumped 2.0 in. of rain in 30 min on a town of area 26 km². What volume of water, in acre-feet, fell on the town?

A: We want to get the volume of the water hitting the town in units of acre•ft. Considering the information that the problem gives us, it will be easy to find the volume of the water in ft³, so let's first convert the acre-foot into units of cubed feet. You can look up that 1 acre = 43,560 ft². Therefore,

1 acre•ft = (43,560 ft²)•ft = 43,560 ft³

Now let's find the volume of the water hitting the town. This should be equal to the area of the town (26 squared kilometers) times the "thickness" of the water (2.0 inches of rain). Let's multiply these measurements by one another, also keeping in mind to convert the units properly. The important conversion factors to know is that 1 km = 3281 ft, and 1 ft = 12 inches.

V = (26 km²)(2.0 in.) = (26 km²)(3281 ft / km)²(2.0 in.)(1 ft / 12 in.) = 4.66 x 10⁷

We now have found the volume of the water that hit the town. The problem asks us to give this volume in acre•ft, however, so we'll use the conversion factor that we calculated earlier:

V = (4.66 x 10⁷)(1 acre•ft / 43,560 ft) = 1.1 x 10³ acre•ft

FINAL ANSWER: 1.1 x 10³ acre•ft

••8. Harvard Bridge, which connects MIT with its fraternities across the Charles River, has a length of 364.4 Smoots plus one ear. The unit of one Smoot is based on the length of Oliver Reed Smoot, Jr., class of 1962, who was carried or dragged length by length across the bridge so that other pledge members of the Lambda Chi Alpha fraternity could mark off (with paint) 1-Smoot lengths along the bridge. The marks have been repainted biannually by fraternity pledges since the initial measurement, usually during times of traffic congestion so that the police cannot easily interfere. (Presumably, the police were originally upset because the Smoot is not an SI base unit, but these days they seem to have accepted the unit.) Figure 1-4 shows three parallel paths, measured in Smoots (S), Willies (W), and Zeldas (Z). What is the length of 50.0 Smoots in (a) Willies and (b) Zeldas?

A: Of course MIT of all places is nerdy enough to have one of their most famous inside jokes based around a fake unit of measurement. Good God. Oh wait, I must be an even bigger nerd then, because I'm creating an unofficial solution guide for an obscure physics textbook. Darn.

(a) It's easy to convert from Smoots to Willies, because the diagram clearly shows that 212 S = 258 W. We can use this to convert:

50.0 S = (50.0 S)(258 W / 212 S) = 60.8 W

(b) Converting from Smoots or Willies to Zeldas is trickier, because we don't actually know where the "zero" point on the Zelda path is, so we don't know for sure if 216 Z is equal to 212 S or 258 W. However, we do have the "60" points adjacent to one another, so we can say that 212 S - 32 S = 180 S is equal to 216 Z - 60 Z = 156 Z. In simpler terms, 180 S = 156 Z.

50.0 S = (50.0 S)(156 Z / 180 S) = 43.3 Z

FINAL ANSWERS: (a) 60.8 W
(b) 43.3 Z

••9. Antarctica is roughly semicircular, with a radius of 2000 km (Fig. 1-5). The average thickness of its ice cover is 3000 m. How many cubic centimeters of ice does Antarctica contain? (Ignore the curvature of Earth.)

A: The area of a circle is A = πr². This is a semicircle however, so it should be divided in half. Also, the continent has some thickness to it, so we should multiply the entire area by another variable (let's call it z, in this case) to account for its third dimension.

V = π/2 * r²z

We want our answer in cubic centimeters, so let's convert the radius and the thickness into centimeters:

r = (2000 km)(10³ / 1 km)(10² / 1 m) = 2000 x 10⁵ cm

z = 3000 m = (3000 m)(10² cm / 1 m) = 3000 x 10² cm

Now plug these values into our volume formula above:

V = π/2 * r²z = π/2 * (2000 x 10⁵ cm)²(3000 x 10² cm) = 1.9 x 10²² cm³

FINAL ANSWER: 1.9 x 10²² cm³

•10. Until 1883, every city and town in the United States kept its own local time. Today, travelers reset their watches only when the time change equals 1.0 h. How far, on the average, must you travel in degrees of longitude between the time-zone boundaries at which your watch must be reset by 1.0 h? (Hint: Earth rotates 360° in about 24 h.)

A: Since the Earth goes through 360° of rotation in 24 h, we can use this to find the "rate of change" of "degrees per hour" by dividing the number of degrees by the time period to which it corresponds:

360° / 24 h = 15°

That's 15° per hour. Therefore, a person must travel 15° before they must reset their watch by an hour.

FINAL ANSWER: 15°

•11. For about 10 years after the French Revolution, the French government attempted to base measures of time on multiples often: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?

A: (a) We want a ratio of decimal weeks to standard weeks. Since a decimal week consists of 10 days and a standard week consists of 7 days,

French week / Standard week = 10/7 = 1.43

(b) One decimal day = 10 * 100 * 100 = 10⁵ s. One standard day = 24 * 60 * 60 = 86400 s. These are not the values we'll use, however - they need to be inverted first. If there are 10⁵ seconds in one decimal day, then one french second is 1/10⁵ of a french day. This same logic applies to the number of seconds in a standard day. Therefore,

French second / Standard second = (10^-5)/(1/86400) = 0.864

FINAL ANSWERS: (a) 1.43
(b) 0.864

•12. The fastest growing plant on record is a Hesperoyucca whipplei that grew 3.7 m in 14 days. What was its growth rate in micrometers per second?

A: As discussed in the previous problem, one day = 86,400 s. We use this to convert from days to seconds, and from meters to micrometers.

(3.7 m / 14 day)(10⁶ μm / 1 m)(1 day / 86,400 s) = 3.1 μm/s

FINAL ANSWER: 3.1 μm/s

•13. Three digital clocks A, B, and C run at different rates and do not have simultaneous readings of zero. Figure 1-6 shows simultaneous readings on pairs of the clocks for four occasions. (At the earliest occasion, for example, B reads 25.0 s and C reads 92.0s.) If two events are 600 s apart on clock A, how far apart are they on (a) clock B and (b) clock C? (c) When clock A reads 400 s, what does clock B read? (d) When clock C reads 15.0 s, what does clock B read? (Assume negative readings for prezero times.)

A: Because all of the clocks run at different rates, and none of the clocks are centered together at any singular point, we'll have to relate the rates of the clocks together using the formula for slopes. We can find the slope of line C in terms of line B by using the points we're given that do match between lines B & C. Since (142 - 92)/(200 - 25) = 2/7, we can say that t_C is proportional to 2/7 * t_B.

Of course, there is still going to be a y-intercept there, which we can find by setting the points equal to each other:

t_C = 2/7 * t_B + b = 92.0 = 2/7 * (25.0) + b --> b = 594/7

So the actual formula would be t_C = 2/7 * t_B + 594/7

We apply this same method to find a formula for clock B in terms of clock A; t_B = 33/40 * t_A - 662/5

(a) Now let's actually begin applying these relationships. If we want to find a difference in time on clock B due to a difference in time on clock A, the relationship can be seen as:

t_B' - t_B = 33/40 * (t_A' - t_A)

Since the difference in time (on clock A) is 600 seconds, then...

t_B' - t_B = 33/40 * (600 s) = 495 s

(b) Now apply this change to clock C:

t_C' - t_C = 2/7 * (t_B' - t_B) = 2/7 * (495) = 141 s

(c) Instead of calculating a difference, now we want to find the actual reading of clock B based on clock A. This time, we'll actually want to consider the y-intercepts we found earlier.

t_B = 33/40 * t_A - 662/5 = (33/40) * (400) - 662/5 ≈ 198 s

(d) Now we want to find the reading on clock B due to a time on clock C.

t_C = 15 = (2/7)t_B + (594/7) --> t_B ≈ -245 s

FINAL ANSWERS: (a) 495 s
(b) 141 s
(c) 198 s
(d) -245 s

•14. A lecture period (50 min) is close to 1 microcentury. (a) How long is a microcentury in minutes? (b) Using

find the percentage difference from the approximation.

A: We want to covert one "microcentury" to minutes. Keep in mind that the prefix "micro" (often represented with a μ) refers to "one millionth" of the unit.

1 μcentury = (10^-6 century)(100 y / 1 century)(365 day / 1 y)(24 h / 1 day)(60 min / 1 h) = 52.6 min

(b) We use the lecture period (50 min) as our approximation, and the value we calculated in part (a) as the "actual" value.

(52.6 min - 50 min)/(52.6 min) = 4.9%

FINAL ANSWERS: (a) 52.6 min
(b) 4.9%

•15. A fortnight is a charming English measure of time equal to 2.0 weeks (the word is a contraction of “fourteen nights”). That is a nice amount of time in pleasant company but perhaps a painful string of microseconds in unpleasant company. How many microseconds are in a fortnight?

A: We are converting one fortnight to microseconds.

1 fortnight = (2 weeks)(7 days / 1 week)(24 h / 1 day)(3600 s / 1 h)(10⁶ μm / 1 s) = 1.21 x 10¹² μs

FINAL ANSWER: 1.21 x 10¹² μs

•16. Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Pulsar PSR 1937 + 21 is an example; it rotates once every 1.557 806 448 872 75 ± 3 ms, where the trailing ±3 indicates the uncertainty in the last decimal place (it does not mean ±3 ms). (a) How many rotations does PSR 1937 + 21 make in 7.00 days? (b) How much time does the pulsar take to rotate exactly one million times and (c) what is the associated uncertainty?

A: The frequency of the pulsar's rotation is given by:

f = (1 rotation) / (1.557 806 448 872 75 x 10^-3 s)

The time interval we're given is 7.00 days = 604,800 s. We use this to find the number of rotations at the above frequency:

(a) N = (1 rotation / 1.557 806 448 872 75 x 10^-3 s)(604,800 s) = 3.88 x 10⁸ rotations

(b) In this case, we have the number of rotations and want to find the time. So the formula we used previously can be applied like this:

1 x 10⁶ = (1 rotation / 1.557 806 448 872 75 x 10^-3 s)*t --> t = 1557.80644887275 s

(c) The problem tells us that the uncertainty associated with one revolution is ±3 x 10^-17 s. The uncertainty associated with one million revolutions should then be:

(±3 x 10^-17 s)(1 x 10⁶) = ±3 x 10^-11 s

FINAL ANSWERS: (a) 3.88 x 10⁸ rotations
(b) 1557.80644887275 s
(c) ±3 x 10^-11 s

•17. Five clocks are being tested in a laboratory. Exactly at noon, as determined by the WWV time signal, on successive days of a week the clocks read as in the following table. Rank the five clocks according to their relative value as good timekeepers, best to worst. Justify your choice.

A: If we want to fix any of these clocks, we'll want to apply a certain number of corrections to their time. Correcting any of the clocks is easier if the time discepencies from day-to-day are generally consistent, so we want predictable corrections. Let's make a graph of the corrections that we can apply to each time to make it line up properly with the previous time:

Notice that clocks C and D both seem to have perfectly consistent time discrepencies, meaning they can be fixed with simple and predictable corrections. Since C is closer to zero than D, we say that C is somewhat better than D.

After the two obvious corrections, the other three clocks are pretty distinct in their ranges of correction. A has a very small range of correction, B has a noticeably larger range of correction, and E has a massive range of correction. Our final ranking, then, is C, D, A, B, E.

FINAL ANSWER: C, D, A, B, E

•20. The record for the largest glass bottle was set in 1992 by a team in Millville, New Jersey — they blew a bottle with a volume of 193 U.S. fluid gallons. (a) How much short of 1.0 million cubic centimeters is that? (b) If the bottle were filled with water at the leisurely rate of 1.8 g/min, how long would the filling take? Water has a density of 1000 kg/m³.

A: (a) We convert from 193 U.S. fluid gallons to cubic centimeters. Note that 1 gal = 231 in³, and 1 in = 2.54 cm.

193 gal = (193 gal)(231 in³ / 1 gal)(2.54 cm / 1 in)³ = 7.31 x 10⁵ cm³

Recall that the problem asks us "how much short of 1.0 million cubic centimeters" that the conversion is. So we subtract from 1 x 10⁶ cm³.

1 x 10⁶ cm³ - 7.31 x 10⁵ cm³ = 2.69 x 10⁵ cm³.

(b) First of all, the volume found in part (a) is (7.31 x 10⁵ cm³)(100 cm / 1 m)³ = 0.731 m³. Based on the density of water we're given, this corresponds to a mass of

(1000 kg/m³)(0.731 m³) = 731 kg.

The amount of time it takes for this mass to fill is

(731 kg)(0.0018 kg/min) = 4.06 x 10⁵ min = 0.77 y.

FINAL ANSWERS: (a) 2.69 x 10⁵ cm³
(b) 4.06 x 10⁵ min = 0.77 y

•21. Earth has a mass of 5.98 x 10²⁴ kg. The average mass of the atoms that make up Earth is 40 u. How many atoms are there in Earth?

A: If we define m as the average mass of an atom in the Earth, and N is the number of atoms, then M_E = Nm can be used to represent the mass of the Earth. If we're looking for the number of atoms, then we can find N as N = M_E/m. Note that 1 u = 1.661 x 10^-27 kg. Solve for N:

N = (5.98 x 10²⁴ kg)/(40 u) * (1 u / 1.661 x 10^-27 kg) = 9.0 x 10⁴⁹ atoms

FINAL ANSWER: 9.0 x 10⁴⁹

•22. Gold, which has a density of 19.32 g/cm³, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold, with a mass of 27.63 g, is pressed into a leaf of 1.000 μm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 μm, what is the length of the fiber?

A: (a) If we have a leaf of gold, then its total volume should be equal to the cross-sectional area times the thickness of the leaf. In other words, V = Az, where z represents the leaf's thickness. We want to find A, so let's first calculate the volume of the leaf, and its thickness.

The formula for density is ρ = m/V. Solving for the volume of this gold sheet:

V = m/ρ = (27.63 g)/(19.32 g/cm³) = (1.430 cm³)(1 m / 100 cm)³ = 1.430 x 10^-6 m³.

The thickness of the leaf is given as z = 1 x 10^-6 m. Now we solve for A:

A = (1.430 x 10^-6 m³)/(1 x 10^-6 m) = 1.430 m²

(b) Now the geometry is a cylinder rather than a gold leaf. If ℓ is the length of the cylinder, then the volume of the cylinder is V = Aℓ. A is the cross-sectional area of the cylinder, and since the cross-sectional area of the cylinder is a circle, A = πr². Now we solve for ℓ:

V = πr²ℓ --> ℓ = V/(πr²) = (1.430 x 10^-6 m³)/[π(2.500 x 10^-6 m)²] = 7.284 x 10⁴ m = 72.84 km

FINAL ANSWERS: (a) 1.430 m²
(b) 7.284 x 10⁴ m = 72.84 km

•23. (a) Assuming that water has a density of exactly 1 g/cm³, find the mass of one cubic meter of water in kilograms. (b) Suppose that it takes 10.0 h to drain a container of 5700 m³ of water. What is the "mass flow rate," in kilograms per second, of water from the container?

A: (a) Let's first correct the units on the water's density, since "g/cm³" is gross. (More serious version of this explanation: We want to change the g to kg, and cm³ to m³, since those units are more common and will actually be consistent with the latter half of the problem).

Probably the hardest part of this problem for people is realizing that the conversion from cm³ to m³ is not the same as from cm to m. These are 3-dimensional units, after all. Typically, the conversion rate is (100 cm / 1 m), but because this expansion is being made from three directions, we'll cube this conversion entirely: (100 cm / 1 m) * (100 cm / 1 m) * (100 cm / 1 m) = (1000000 cm³ / m³), or ((1.0 x 10⁶ cm³) / m³).

1g/cm³ * (1 kg / 1000 g) * (1.0 x 10⁶ cm³/m³) = 1000 kg/m³

Great. Now that we have the water's density in a manageable format, let's actually solve the problem.

The density formula is Density = Mass/Volume. Solve for Mass and it becomes Mass = Density * Volume. Plug in our values and...

(1000kg/m³)(1 m³) = 1000 kg

(b) Once again, this problem becomes very straightforward once we have the units converted accordingly. Let's convert the hours into seconds, and m^3 into kilograms. 1 m^3 of water = 1000 kg, assuming the water perfectly weighs 1kg/L in this scenario.

10.0h * (3600s/1h) = 36000s

5700 m³ * (1000 kg / 1 m³) = 5700000 kg

Now to divide to solve for kg/s:

(5700000 kg)/(36000 s) = 158 kg/s

And we're all set.

FINAL ANSWERS: (a) 1000 kg
(b) 158 kg/s

48. The common Eastern mole, a mammal, typically has a mass of 75 g, which corresponds to about 7.5 moles of atoms. (A mole of atoms is 6.02 x 10²³ atoms.) In atomic mass units (u), what is the average mass of the atoms in the common Eastern mole?

A: Let's first find the mass of the mole in atomic mass units, and the number of atoms in 7.5 moles.

(75 g)(1 kg/ 1 g)(1 u / 1.660 538 86 x 10^-27 kg) = 4.52 x 10²⁵ u

(7.5 moles)(6.02 x 10²³ atoms / 1 mole) = 4.515 x 10²⁴ atoms

Now we divide the mass by the number of atoms, to find the average mass per atom.

(4.52 x 10²⁵ u / 4.515 x 10²⁴ atoms) ≈ 10 u

FINAL ANSWER: 10 u

56. The corn–hog ratio is a financial term used in the pig market and presumably is related to the cost of feeding a pig until it is large enough for market. It is defined as the ratio of the market price of a pig with a mass of 3.108 slugs to the market price of a U.S. bushel of corn. (The word “slug” is derived from an old German word that means “to hit”; we have the same meaning for “slug” as a verb in modern English.) A U.S. bushel is equal to 35.238 L. If the corn–hog ratio is listed as 5.7 on the market exchange, what is it in the metric units of

(Hint: See the Mass table in Appendix D.)

A: The hog-corn ratio as it's defined by the problem is

hog/corn = (price of 3.108 slugs)/(price of U.S. bushel)

Let's convert into metric units, and turn it into the value for a price ratio of 1. Keep in mind that 1 slug = 14.59 kg.

1 hog/corn = (45.346 kg)/(35.238 L) = 1.287 kg/L

This is the ratio of a 3.108 slug pig to a bushel of corn, if their prices are both equal (in other words, with a ratio of 1 between prices). To find this value at a ratio of 5.7, we multiply this value by 5.7.

5.7 (1.287) = 7.3 (price of 1 kilogram of pig)/(price of 1 liter of corn)

FINAL ANSWER: 7.3 (price of 1 kilogram of pig)/(price of 1 liter of corn)

58. A standard interior staircase has steps each with a rise (height) of 19 cm and a run (horizontal depth) of 23 cm. Research suggests that the stairs would be safer for descent if the run were, instead, 28 cm. For a particular staircase of total height 4.57 m, how much farther into the room would the staircase extend if this change in run were made?

A: If we're extending the horizontal run of each step slightly, then the total change in horizontal length is equal to the change in length of a single step times the number of steps (approximately). The increased change can be found as N_stepsΔx, where Δx is 0.05 m (the change in length of a single step). The number of steps is equal to the total height of the stairs divided by the "rise" (height) of each step.

N_stepsΔx = [(4.57 m)/(0.19 m)](0.05 m) = 1.2 m

And this is a pretty good approximation. We can make it slightly better, however, by realizing that the "run" on the heighest step doesn't really count as a step (since it's part of the next level) and likely won't contribute to the overall change in horizontal depth of the stairs. We can correct this by simply subtracting 1 from the number of steps.

(N_steps - 1)Δx = [(4.57 m)/(0.19 m) - 1](0.05 m) = 1.15 m ≈ 1.2 m

Rounded to two significant figures, it ends up being the same value anyway.

FINAL ANSWER: 1.2 m

(MC) 1. (5.0 x 10⁴) x (3.0 x 10^-6) =

Option 1: 1.5 x 10⁵
Option 2: 1.5 x 10^-3
Option 3: 1.5 x 10³
Option 4: 1.5 x 10¹
Option 5: 1.5 x 10^-1

A. This isn't an equation (as in, there are no variables here, so there's nothing to solve for), so a calculator will do the trick just fine on this problem as long as you can identify which answer in scientific notation equals the output that your calculator will give you. However, a calculator shouldn't be necessary for solving this expression.

Thanks to the Commutative Property of Multiplication (which states that any set of numbers can be multiplied in any order you want, just as long as multiplication is the only relevant operation being used), technically method of multiplication you choose will get a "correct" answer, but judging by the options we're given it looks as though we'll want to preserve the syntax of scientific notation. The reason why we would want to do this usually has to do with keeping a consistent number of significant figures. Whatever the case may be, a specific process will need to be used here.

The most viable way to compute a scientific notation problem in which both terms are being multiplied by one another is to simply multiply the parts in one term by their corresponding bits in the opposite term. Let's elaborate on this:

In the problem above, the two bits being multiplied by the "ten-power" term are 5.0 and 3.0. Because we want to preserve the ten-power terms, let's just multiply these standard integers on their own. 5.0 * 3.0 = 15. Cool. Now let's multiply the ten-power terms. (10^4) * (10^-6) can be calculated easily enough using the concept that two of the same base numbers raised to different powers multiplied by one another is equal to that same base number raised to the power of the sum of the powers... Well, that might seem like a lot to swallow. It's easier-done-than-said, fortunately. The 10's are the same, so ignore them for a second and focus on the powers: 4 and -6. So, add them together and we get (4) + (-6), which equals (-2).

Our current answer is now (15 x 10^-2). This is improper, however. According to the actual rules of scientific notation, the number that ISN'T the 10^n (which, fun fact, is known as the "significand" in the context of scientific notation) must be a number from 1.0 to 9.9. There can not be more than one digit left of the decimal point. To fix this, just shift the decimals accordingly. Move the decimal point on the "15" to the left one space to become "1.5". In order to keep the same overall value of the number, we must alter the order of magnitude (the power that 10 is raised to) as well. Because we lowered 15 by a power of 10, the 10^-2 must be raised by a power of 10 equally to counteract this. It then becomes 10^-1.

Therefore, our final answer is:

FINAL ANSWER: 1.5 x 10^-1

(MC) 2. The SI standard of length is based on:

Option 1: a precision meter stick in Paris
Option 2: the distance from the north pole to the equator along a meridian passing through Paris
Option 3: the speed of light
Option 4: wavelength of light emitted by Hg¹⁹⁸
Option 5: wavelength of light emitted by Kr⁸⁶

A: A pretty straightforward question with a straightforward answer. However, because this is more of a "fact" question than a logic/concept question, you can't really figure this out unless you already know the answer - so make sure you know it.

Now, let's take a look at the SI definition of what the standard of length (AKA .. A meter, or metre if you're not American) actually is. According to the BIPM, the French conference that decides what the standard of length should be, the formal definition is as follows:

"The metre is the length of the path travelled by light in vacuum during a time interval of 1/(299 792 458) of a second."

Only one of our options fits into that, so our answer is:

FINAL ANSWER: the speed of light

(MC) 3. A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:

OPTION 1: 100 km/hr
OPTION 2: 50 km/hr
OPTION 3: cannot be calculated without knowing the acceleration
OPTION 4: 0 km/hr
OPTION 5: 200 km/hr

A: Ahh, a classic question that inevitably comes up every single time a professor wants to quiz their students on the basic concept of velocity. If you understand velocity, then this problem can be solved conceptually easily enough, but I'll break it down anyway:

The intuitive formula for velocity is V = (Xf - Xi)/(t) -- Where "V" equals the average velocity, "Xf" equals the final position relative to our starting point (also known as our "displacement"), "Xi" equals the initial position, and "t" equals the time elapsed during the event described. Plug our numbers into this formula and we get (0 - 0)/(2hr), which obviously equals 0 km/hr.

If this answer comes as a shock to you, I suggest doing a bit of research on vectors and velocity before attempting the problems that follow this one. It's a pretty fundamental concept that's important to get down - As I hinted in the last paragraph, the velocity is dependent NOT on the total distance actually travelled during a timeframe, but is dependent on the total DISPLACEMENT during a timeframe. Displacement is based purely on the distance from the starting point to the end point, regardless of what it did in between. Since the starting point and the end point are the same in this case, there is no displacement at all. The displacement is zero, and as such the velocity is zero as well.

FINAL ANSWER: 0km/hr

(MC) 4. Which of the following five coordinate versus time graphs represents the motion of an object moving with a constant nonzero speed?


OPTION 1: I
OPTION 2: II
OPTION 3: III
OPTION 4: IV
OPTION 5: V

A: Another pretty benign concept question. The only important thing here is the phrase "constant nonzero speed". Because of 'nonzero', we can immediately cross off Graph III because it clearly shows the coordinates of an object that isn't moving at all (make sure you're aware of the fact that all of these are plots of position vs. time!), as its x-coordinate never changes. Avoid graphs that have a slope of zero.

The word 'constant' tells us that there's no acceleration, meaning that the line we're looking for shouldn't be curved at all, because this would imply that the rate of change of the object's movement is altered in some way throughout the course of the time interval.

Graph II is the only option we're given that has a completely straight line without a slope of zero, so that's our answer.

FINAL ANSWER: II

(MC) 5. This graph shows the position of a particle as a function of time. What is its instantaneous velocity at t = 7 s?


OPTION 1: 3m/s
OPTION 2: 12m/s
OPTION 3: -12m/s
OPTION 4: Need additional information.
OPTION 5: -3m/s

A: This is a position-time graph, so the "instantaneous velocity" at any given point will be equal to the instantaneous rate of change of this graph at that point (also known as a derivative, if you know anything about calculus). The point at t = 7 is on the same interval as the constant slope from t = 5 to t = 9, so let's use the slope formula on that interval.

Slope formula = (y2 - y1)/(x2 - x1). Plug in our values: (0 - 12)/(9 - 5) = -12/4 = -3

And the slope over that point is -3. Because we know that velocity is the slope of the position graph, we now know that the instantaneous velocity at this point is -3 m/s.

FINAL ANSWER: -3 m/s

(MC) 6. Can an object have positive acceleration and decreasing speed?

OPTION 1: Yes, this is possible if the initial velocity is negative.
OPTION 2: Yes, this is possible but only if the object is moving in two dimensions.
OPTION 3: Yes, this is possible if the initial velocity is zero.
OPTION 4: No, this is not possible.
OPTION 5: Yes, speed will always decrease if acceleration is positive.

A: This question mainly just determines whether or not you know what "speed" is. It's important to know that speed is a scalar (doesn't take direction into account) as opposed to acceleration, which is a vector (does take direction into account). When acceleration is "increasing", it's common to intuitively believe that this is simply just "the speed is increasing", but what increasing acceleration really implies is that the speed -or rather, the velocity- is increasing in the "positive" direction.

In other words, if the velocity is positive initially, and suddenly the speed starts increasing, then acceleration is increasing. However, if velocity is negative initially, then an increase in acceleration causes the velocity to become "less negative" and approach zero until it becomes positive like in our previous example. In this case, although the velocity is "increasing" in the sense that we're at a negative number and approaching the positives, the speed is decreasing. If you're still not understanding this, it might help to think of speed as the absolute value of velocity -- Although the velocity increases as it approaches zero from the negative side, the speed is only decreasing because the absolute value approaches zero either way.

FINAL ANSWER: Yes, this is possible if the initial velocity is negative.

(MC) 7. Consider the following five graphs (note the axes carefully). Which of these represent(s) motion at constant speed?


OPTION 1: I, II, and III only
OPTION 2: I and IV only
OPTION 3: I and II only
OPTION 4: IV and V only
OPTION 5: IV only

A: As stated earlier (and as you should know already), motion at "constant speed" implies zero acceleration. Graphs III and V can be ruled out right away because they're both acceleration graphs with nonzero points. Out of the three graphs we have left, it should be clear that the correct answer can't be Graph II either, since it shows the velocity increasing at a constant acceleration.

On the other hand, Graph I shows the position increasing on a linear slope (meaning constant velocity), and Graph IV shows a velocity graph with a slope of zero (meaning, the velocity isn't changing). There you go.

FINAL ANSWER: I and IV only