**wangsness electromagnetic fields solutions - Cap - 1**

wangsness electromagnetic fields solutions

(Parte **1** de 2)

1. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2).

(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106µm,

The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109µm.

(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10 −2 m,

We conclude that the fraction of one centimeter equal to 1.0 µm is 1.0 × 10 −4 .

2. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain

(b) With 12 points = 1 pica, we have

(b) and that distance in chains to be

3. Using the given conversion factors, we find (a) the distance d in rods to be

4. The conversion factors 1 gry1/10 line=,1 line=1/12 inchand 1 point = 1/72 inch imply that

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

Thus, 1 gry2 = (0.60 point)2

5. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as

(c) The volume of Earth is ()3

6. From Figure 1.6, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z.

(a) In units of W, we have

(b) In units of Z, we have

V r zpi =

wherez is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

In these units, the thickness becomes

7. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = pir2 /2, where r is the radius. Therefore, the volume is

8. We make use of Table 1-6.

(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega.

Thus, 1 fanega = 1

−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1

−2 cahiz.

Continuing in this way, the remaining entries in the first column are 6.94 × 10 −3 and for the last three entries.

(c) In the third (“cuartilla”) column, we obtain 0.3 and 0.167 for the last two entries.

(d) Finally, in the fourth (“almude”) column, we get 1

2 = 0.500 for the last entry.

(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.0 almudes must be equal to 14.0 medios.

(f) Using the value (1 almude = 6.94 × 10 −3 cahiz) found in part (a), we conclude that

7.0 almudes is equivalent to 4.86 × 10 −2 cahiz.

(g) Since each decimeter is 0.1 meter, then 5.501 cubic decimeters is equal to 0.05501

12 fanega =

Thus,

ft acre ft acre ft.

9. We use the conversion factors found in Appendix D.

10. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so

day s day m s b gc h b gb gµ µ=

seconds | Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, |

1. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600

12. The metric prefixes (micro (µ), pico, nano, …) are given for ready reference on the inside front cover of the textbook (also, Table 1–2).

1 century 1 y 1 day 1 h

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(b) The percent difference is therefore

13. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43.

(b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds. The ratio is therefore 0.864.

(a) Multiplying f by the time-interval t = 7.0 days (which is equivalent to 60480 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations:

which should now be rounded to 3.8 × 108 rotations since the time-interval was specified in the problem to three significant figures.

(b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N=ft, or

which yields the result t = 1557.80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits).

(c) Careful reading of the problem shows that the time-uncertainty per revolution is 17310s−±×. We therefore expect that as a result of one million revolutions, the uncertainty should be 17611 (310)(110)= 310s−−±××±×.

14. We denote the pulsar rotation rate f (for frequency).

3 1 rotation

15. The time on any of these clocks is a straight-line function of that on another, with slopes≠ 1 andy-intercepts ≠ 0. From the data in the figure we deduce

These are used in obtaining the following results.

(a) We find

16. Since a change of longitude equal to 360°corresponds to a 24 hour change, then one expects to change longitude by360/2415°=° before resetting one's watch by 1.0 h.

17. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it would impossible to tell what the correction should be. The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning.

Sun. Mon. Tues. Wed. Thurs. Fri. CLOCK -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat.

Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to +10 s, for clock E it is in the range from -70 s to -2 s. After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E.

18. The last day of the 20 centuries is longer than the first day by ( ) ( )20 century 0.001 s century 0.02 s.=

The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is average increase in length of a day number of days

day y

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or roughly two hours.

19. When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B.

Letd be the distance from point B to your eyes. From Pythagorean theorem, we have 2 2 2 2 2( ) 2d r r h r rh h+ = + = + + or222,drhh=+wherer is the radius of the Earth. Since rh , the second term can be dropped, leading to 22drh≈. Now the angle between the two radii to the two tangent pointsA and B is θ, which is also the angle through which the Sun moves about Earth during the time interval t = 1.1 s. The value of θ can be obtained by using

This yields

Usingtandrθ=, we have 222tan2drrhθ==, or

Using the above value for θ and h = 1.7 m, we have 65.210 m.r=×

(a) We take the volume of the leaf to be its area A multiplied by its thickness z. With densityρ = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be

We convert the volume to SI units:

SinceV = Az with z = 1 × 10 -6 m (metric prefixes can be found in Table 1–2), we obtain

(b) The volume of a cylinder of length A is VA=A where the cross-section area is that of a circle: A = pir2 . Therefore, with r = 2.500 × 10 m3 , we obtain

20. The density of gold is

21. We introduce the notion of density:

and convert to SI units: 1 g = 1 × 10 −3 kg.

(a) For volume conversion, we find 1 cm3 = (1 × 10 in kg/m3 is

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Thus, the mass of a cubic meter of water is 1000 kg.

(b) We divide the mass of the water by the time taken to drain it. The mass is found from M = ρV (the product of the volume of water and its density):

The time is t = (10h)(3600 s/h) = 3.6 × 104 s, so the mass flow rate R is

2. (a) We find the volume in cubic centimeters

1gal 1in and subtract this from 1 × 106 cm3 to obtain 2.69 × 105 cm3

. The conversion gal → in3 is given in Appendix D (immediately below the table of Volume conversions).

(b) The volume found in part (a) is converted (by dividing by (100 cm/m)3 ) to 0.731 m3 , which corresponds to a mass of using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in

after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h).

23. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = Nm or N = ME/m. We convert mass m to kilograms using

N Mm

u kg ub g c h

of this problem is due to the fact that each liter has a mass of one kilogram when water is at its normal density (under standard conditions).

meter of the cloud contains from 50 ×106 to 500 ×106 water drops, then we conclude that the entire cloud contains from 4.7×10 18 to 4.7×10 19 drops. Since the volume of m3 , then the total volume of water in a cloud

(b) Using the fact that 33331 L110cm110m−=×=×, the amount of water estimated in part (a) would fill from 6210× to 7210×bottles.

(c) At 1000 kg for every cubic meter, the mass of water is from two million to twenty million kilograms. The coincidence in numbers between the results of parts (b) and (c)

25. We introduce the notion of density, /mVρ=, and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m.

(a) The density ρ of a sample of iron is

If we ignore the empty spaces between the close-packed spheres, then the density of an individual iron atom will be the same as the density of any iron sample. That is, if M is the mass and V is the volume of an atom, then

(b) We set V = 4piR3 /3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R, we find

The center-to-center distance between atoms is twice the radius, or 2.82 × 10 −10 m.

26. If we estimate the “typical” large domestic cat mass as 10 kg, and the “typical” atom

10 26 atoms. This is close to being a factor of a thousand greater than Avogradro’s number. Thus this is roughly a kilomole of atoms.

27. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km. The difference is 5.95 km.

m V pipiρ ρ − −

We observe that (because a cube has six equal faces) the indicated surface area is 6 m2 .

The number of spheres (the grains of sand) N that have a total surface area of 6 m2 is given by

28. The metric prefixes (micro (µ), pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2). The surface area A of each grain of sand of radius r = 50 µm = 50 × 10 −6 m2 (Appendix E contains a variety of geometry formulas). We introduce the notion of density, /mVρ=, so that the mass can be found from m = ρV, where ρ = 2600 kg/m3 .

Thus, using V = 4pir3 /3, the mass of each grain is

stand for the thickness of the mud after it has (uniformly) distributed in the valley, then its volume there would be (400 m)(400 m)d. Requiring these two volumes to be equal, we can solve for d. Thus, d = 25 m. The volume of a small part of the mud over a patch of area of 4.0 m2 is (4.0)d = 100 m3 . Since each cubic meter corresponds to a mass of

1900 kg (stated in the problem), then the mass of that small part of the mud is

s1000g1 min tdm dt

30. To solve the problem, we note that the first derivative of the function with respect to time gives the rate. Setting the rate to zero gives the time at which an extreme value of the variable mass occurs; here that extreme value is a maximum.

s1000g1 min tdm dt

(d) Similarly, the rate of mass change at 5.00st= is

31. The mass density of the candy is

50.0 mmmV

If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in the container when filled to height h is ,MAhρ= where

2(14.0 cm)(17.0 cm)238 cmA== is the base area of the container that remains unchanged. Thus, the rate of mass change is given by

dM d Ah dh A dt dt dt

32. Table 7 can be completed as follows:

(a) It should be clear that the first column (under “wey”) is the reciprocal of the first row – so that 9

10 = 0.900,

−2 , and so forth. Thus, 1 pottle = 1.56 × 10 −3 wey and 1 gill = 8.32 × 10 −6 wey are the last two entries in the first column.

(b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 caldron (that is, the entries along the “diagonal” in the table must be 1’s). To find out how many chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 1 chaldron = 1 bag. Thus, the next entry in that second column is 1

Similarly, 1 pottle = 1.74 × 10 −3 chaldron and 1 gill = 9.24 × 10 −6 chaldron.

(c) In the third column (under “bag”), we have 1 chaldron = 12.0 bag, 1 bag = 1 bag, 1

(d) In the fourth column (under “pottle”), we find 1 chaldron = 576 pottle, 1 bag = 48 pottle, 1 pottle = 1 pottle, and 1 gill = 5.32 × 10 −3 pottle.

(e) In the last column (under “gill”), we obtain 1 chaldron = 1.08 × 105 gill, 1 bag = 9.02

× 103 gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill.

(f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag. And since each bag is 0.1091 m3 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m3 .

3. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally:

(a) ( ) 2 peck 11tuffets = 11tuffets22pecks

(b) ( ) 0.50 Imperial bushel 11tuffets = 11tuffets5.5Imperial bushels

34. (a) Using the fact that the area A of a rectangle is (width) × (length), we find ( ) ( )( ) total 2

1acre

We multiply this by the perch2→ rood conversion factor (1 rood/40 perch2 ) to obtain the answer:Atotal = 14.5 roods.

(b) We convert our intermediate result in part (a):

total

1perch

Now, we use the feet → meters conversion given in Appendix D to obtain

total

35. (a) Dividing 750 miles by the expected “40 miles per gallon” leads the tourist to believe that the car should need 18.8 gallons (in the U.S.) for the trip.

(b) Dividing the two numbers given (to high precision) in the problem (and rounding off) gives the conversion between U.K. and U.S. gallons. The U.K. gallon is larger than the U.S gallon by a factor of 1.2. Applying this to the result of part (a), we find the answer for part (b) is 2.5 gallons.

36. The customer expects a volume V1 = 20 × 7056 in3 and receives V2 = 20 × 5826 in3 the difference being 3 12=24600 inVVV∆=−, or

where Appendix D has been used.

10 −2 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460 ft2 /gal becomes

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(b) Also, since 1 m3 is equivalent to 1000 L, our result from part (a) becomes

(Parte **1** de 2)