ResoluÇÃo halliday (9 ediÇÃo), Exercícios de Física. Universidade Federal do Maranhão (UFMA)
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Lucasramos18 de Março de 2015

ResoluÇÃo halliday (9 ediÇÃo), Exercícios de Física. Universidade Federal do Maranhão (UFMA)

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VOLUME 1.

PART 1.

1 Measurement. 2 Motion Along a Straight Line. 3 Vectors. 4 Motion in Two and Three Dimensions. 5 Force and Motion — I. 6 Force and Motion — II. 7 Kinetic Energy and Work. 8 Potential Energy and Conservation of Energy. 9 Center of Mass and Linear Momentum. 10 Rotation. 11 Rolling, Torque, and Angular Momentum.

PART 2.

12 Equilibrium and Elasticity. 13 Gravitation. 14 Fluids. 15 Oscillations. 16 Waves — I. 17 Waves — II. 18 Temperature, Heat, and the First Law of Thermodynamics. 19 The Kinetic Theory of Gases. 20 Entropy and the Second Law of Thermodynamics.

VOLUME 2.

PART 3.

21 Electric Charge. 22 Electric Fields. 23 Gauss’ Law. 24 Electric Potential. 25 Capacitance. 26 Current and Resistance. 27 Circuits. 28 Magnetic Fields. 29 Magnetic Fields Due to Currents. 30 Induction and Inductance. 31 Electromagnetic Oscillations and Alternating Current. 32 Maxwell’s Equations; Magnetism of Matter.

PART 4.

33 Electromagnetic Waves. 34 Images. 35 Interference. 36 Diffraction. 37 Relativity.

PART 5.

38 Photons and Matter Waves. 39 More About Matter Waves. 40 All About Atoms. 41 Conduction of Electricity in Solids. 42 Nuclear Physics. 43 Energy from the Nucleus. 44 Quarks, Leptons, and the Big Bang.

 

1

Chapter 1 1. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as

( )( )6 3 36.37 10 m 10 km m 6.37 10 km,R −= × = × its circumference is 3 42 2 (6.37 10 km) 4.00 10 km.s Rπ π= = × = × (b) The surface area of Earth is ( )22 3 8 24 4 6.37 10 km 5.10 10 km .A R= π = π × = ×

(c) The volume of Earth is ( )33 3 12 34 4 6.37 10 km 1.08 10 km .3 3V R π π

= = × = ×

2. The conversion factors are: 1 gry 1/10 line= , 1 line 1/12 inch= and 1 point = 1/72 inch. The factors imply that

1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 2 20.50 gry = 0.18 point . 3. 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 × 106 μm,

( ) ( )3 3 6 91km 10 m 10 m 10 m m 10 m.= = =μ μ The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109 μm. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m,

( ) ( )2 2 6 41cm = 10 m = 10 m 10 m m 10 m.− − =μ μ

We conclude that the fraction of one centimeter equal to 1.0 μm is 1.0 × 10−4. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,

CHAPTER 1 2

( ) ( )6 51.0 yd = 0.91m 10 m m 9.1 10 m.= ×μ μ

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

( ) 1 inch 6 picas0.80 cm = 0.80 cm 1.9 picas. 2.54 cm 1 inch

⎛ ⎞⎛ ⎞ ≈⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

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

( ) 1 inch 6 picas 12 points0.80 cm = 0.80 cm 23 points. 2.54 cm 1 inch 1 pica

⎛ ⎞⎛ ⎞⎛ ⎞ ≈⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

5. Given that 1 furlong 201.168 m= , 1 rod 5.0292 m= and 1chain 20.117 m= , we find the relevant conversion factors to be

1 rod1.0 furlong 201.168 m (201.168 m ) 40 rods, 5.0292 m

= = =

and 1 chain1.0 furlong 201.168 m (201.168 m ) 10 chains

20.117 m = = = .

Note the cancellation of m (meters), the unwanted unit. Using the given conversion factors, we find (a) the distance d in rods to be

( ) 40 rods4.0 furlongs 4.0 furlongs 160 rods, 1 furlong

d = = =

(b) and that distance in chains to be

( )10 chains4.0 furlongs 4.0 furlongs 40 chains. 1 furlong

d = = =

6. 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 = 112 cahiz, or 8.33 × 10

−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the

already completed part) implies that 1 cuartilla = 148 cahiz, or 2.08 × 10 −2 cahiz.

Continuing in this way, the remaining entries in the first column are 6.94 × 10−3 and 33.47 10−× .

3

(b) In the second (“fanega”) column, we find 0.250, 8.33 × 10−2, and 4.17 × 10−2 for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. (d) Finally, in the fourth (“almude”) column, we get 12 = 0.500 for the last entry. (e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 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 7.00 almudes is equivalent to 4.86 × 10−2 cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 m3 or 55501 cm3. Thus, 7.00 almudes = 7.0012 fanega =

7.00 12 (55501 cm

3) = 3.24 × 104 cm3. 7. We use the conversion factors found in Appendix D. 2 31 acre ft = (43,560 ft ) ft = 43,560 ft⋅ ⋅ Since 2 in. = (1/6) ft, the volume of water that fell during the storm is 2 2 2 7 3(26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft) 4.66 10 ft .V = = = × Thus,

V = × × ⋅

= × ⋅ 4 66 10

4 3560 10 11 10

7

4 3.

. .ft

ft acre ft acre ft.

3

3

8. From Fig. 1-4, 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

( ) 258 W50.0 S 50.0 S 60.8 W 212 S

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

(b) In units of Z, we have

( ) 156 Z50.0 S 50.0 S 43.3 Z 180 S

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

9. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = πr2/2, where r is the radius. Therefore, the volume is

CHAPTER 1 4

2

2 V r zπ=

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

( ) 3 2

510 m 10 cm2000 km 2000 10 cm. 1km 1m

r ⎛ ⎞ ⎛ ⎞

= = ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

In these units, the thickness becomes

( ) 2

210 cm3000 m 3000 m 3000 10 cm 1m

z ⎛ ⎞

= = = ×⎜ ⎟ ⎝ ⎠

which yields ( ) ( )25 2 22 32000 10 cm 3000 10 cm 1.9 10 cm .2V π

= × × = ×

10. Since a change of longitude equal to 360° corresponds to a 24 hour change, then one expects to change longitude by360 / 24 15° = ° before resetting one's watch by 1.0 h. 11. (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. 12. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so

37 10 14 86400

31 6.

. . m m m

day s day m s

b gc h b gb g

μ μ=

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

2 594 33 662, . 7 7 40 5C B B A

t t t t= + = −

These are used in obtaining the following results. (a) We find

( )33 495 s 40B B A A

t t t t′ ′− = − =

when t'AtA = 600 s.

5

(b) We obtain ′ − = ′ − = =t t t tC C B B 2 7

2 7

495 141b g b g s. (c) Clock B reads tB = (33/40)(400) − (662/5) ≈ 198 s when clock A reads tA = 400 s. (d) From tC = 15 = (2/7)tB + (594/7), we get tB ≈ −245 s. 14. The metric prefixes (micro (μ), pico, nano, …) are given for ready reference on the inside front cover of the textbook (also Table 1–2).

(a) ( )6 100 y 365 day 24 h 60 min1 century 10 century 52.6 min.1 century 1 y 1 day 1 hμ − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(b) The percent difference is therefore

52.6 min 50 min 4.9%. 52.6 min

− =

15. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, this is roughly 1.21 × 1012 μs. 16. We denote the pulsar rotation rate f (for frequency).

3

1 rotation 1.55780644887275 10 s

f −= ×

(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations:

( )3 1 rotation 604800 s 388238218.4

1.55780644887275 10 s N

⎛ ⎞ = =⎜ ⎟×⎝ ⎠

which should now be rounded to 3.88 × 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

6 3

1 rotation1 10 1.55780644887275 10 s

t− ⎛ ⎞

× = ⎜ ⎟×⎝ ⎠

CHAPTER 1 6

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

173 10 s−± × . We therefore expect that as a result of one million revolutions, the uncertainty should be 17 6 11 ( 3 10 )(1 10 )= 3 10 s− −± × × ± × . 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.

A −16 −16 −15 −17 −15 −15 B −3 +5 −10 +5 +6 −7 C −58 −58 −58 −58 −58 −58 D +67 +67 +67 +67 +67 +67 E +70 +55 +2 +20 +10 +10

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

7

( ) ( )

( )

average increase in length of a day number of days

0.01 s 365.25 day 2000 y day y

7305 s

T =

⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ =

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.

Let d be the distance from point B to your eyes. From the Pythagorean theorem, we have 2 2 2 2 2( ) 2d r r h r rh h+ = + = + + or 2 22 ,d rh h= + where r is the radius of the Earth. Since r h , the second term can be dropped, leading to 2 2d rh≈ . Now the angle between the two radii to the two tangent points A and B is θ, which is also the angle through which the Sun moves about Earth during the time interval t = 11.1 s. The value of θ can be obtained by using

360 24 h

tθ =

° .

This yields

(360 )(11.1 s) 0.04625 . (24 h)(60 min/h)(60 s/min)

θ °= = °

Using tand r θ= , we have 2 2 2tan 2d r rhθ= = , or

2 2

tan hr

θ =

Using the above value for θ and h = 1.7 m, we have 65.2 10 m.r = ×

CHAPTER 1 8

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

( ) 33

5 3231 in 2.54 cm193 gal = 193 gal 7.31 10 cm 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

1000 kg m 0.731 m = 731 kg3 2c h c h using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in

5731kg 4.06 10 min = 0.77 y 0.0018 kg min

= ×

after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h). 21. 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 Appendix D (1 u = 1.661 × 10−27 kg). Thus,

N M m

E= = ×

× = ×

598 10 40 1661 10

9 0 10 24

27 49.

. . .kg

u kg ub g c h 22. The density of gold is

3 3

19.32 g 19.32 g/cm . 1 cm

m V

ρ = = =

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

V m= = ρ

1430. .cm3

We convert the volume to SI units:

9

( ) 3

3 6 31 m1.430 cm 1.430 10 m . 100 cm

V − ⎛ ⎞

= = ×⎜ ⎟ ⎝ ⎠

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

A = × ×

= −

1430 10 1 10

1430 6

6

. . .m m

m 3

2

(b) The volume of a cylinder of length is V A= where the cross-section area is that of a circle: A = πr2. Therefore, with r = 2.500 × 10−6 m and V = 1.430 × 10−6 m3, we obtain

4 2 7.284 10 m 72.84 km.

V rπ

= = × =

23. We introduce the notion of density:

ρ = m V

and convert to SI units: 1 g = 1 × 10−3 kg. (a) For volume conversion, we find 1 cm3 = (1 × 10−2m)3 = 1 × 10−6m3. Thus, the density in kg/m3 is

3 3 3 3 3

3 6 3

1 g 10 kg cm1 g cm 1 10 kg m . cm g 10 m

⎛ ⎞ ⎛ ⎞⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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):

( ) ( )3 3 3 65700 m 1 10 kg m 5.70 10 kg.M = × = × The time is t = (10h)(3600 s/h) = 3.6 × 104 s, so the mass flow rate R is

6

4

5.70 10 kg 158 kg s. 3.6 10 s

MR t

× = = =

×

24. 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 m is given by A = 4π(50 × 10−6)2 = 3.14 × 10−8 m2 (Appendix E contains a variety of geometry formulas). We introduce the notion of

CHAPTER 1 10

density, /m Vρ = , so that the mass can be found from m = ρV, where ρ = 2600 kg/m3. Thus, using V = 4πr3/3, the mass of each grain is

( )363 9 3

4 50 10 m4 kg2600 1.36 10 kg. 3 m 3 rm V

ππρ 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

2 8

8 2

6 m 1.91 10 . 3.14 10 m

N −= = ××

Therefore, the total mass M is ( ) ( )8 91.91 10 1.36 10 kg 0.260 kg.M Nm −= = × × = 25. The volume of the section is (2500 m)(800 m)(2.0 m) = 4.0 × 106 m3. Letting “d” 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

51.9 10 kg× . 26. (a) The volume of the cloud is (3000 m)π(1000 m)2 = 9.4 × 109 m3. Since each cubic meter of the cloud contains from 50 × 106 to 500 × 106 water drops, then we conclude that the entire cloud contains from 4.7 × 1018 to 4.7 × 1019 drops. Since the volume of each drop is 43 π(10 × 10

− 6 m)3 = 4.2 × 10−15 m3, then the total volume of water in a cloud

is from 32 10× to 42 10× m3. (b) Using the fact that 3 3 3 31 L 1 10 cm 1 10 m−= × = × , the amount of water estimated in part (a) would fill from 62 10× to 72 10× bottles. (c) At 1000 kg for every cubic meter, the mass of water is from 62 10× to 72 10× kg. The coincidence in numbers between the results of parts (b) and (c) 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). 27. We introduce the notion of density, /m Vρ = , and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m. (a) The density ρ of a sample of iron is

11

( ) 3

3 31 kg 100 cm7.87 g cm 7870 kg/m . 1000 g 1 m

ρ ⎛ ⎞ ⎛ ⎞

= =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

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

26 29 3

3 3

9.27 10 kg 1.18 10 m . 7.87 10 kg m

MV

−×= = = × ×ρ

(b) We set V = 4πR3/3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R, we find

( ) 1 329 31 3 103 1.18 10 m3 1.41 10 m. 4 4 VR

− −

⎛ ⎞×⎛ ⎞ ⎜ ⎟= = = ×⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠

The center-to-center distance between atoms is twice the radius, or 2.82 × 10−10 m. 28. If we estimate the “typical” large domestic cat mass as 10 kg, and the “typical” atom (in the cat) as 10 u ≈ 2 × 10−26 kg, then there are roughly (10 kg)/( 2 × 10−26 kg) ≈ 5 × 1026 atoms. This is close to being a factor of a thousand greater than Avogadro’s number. Thus this is roughly a kilomole of atoms. 29. The mass in kilograms is

28 9 100 16 10 10 0 3779. .piculs gin 1picul

tahil 1gin

chee 1tahil

hoon 1 chee

g 1hoon

b g FHG I KJ F HG

I KJ F HG

I KJ F HG

I KJ F HG

I KJ

which yields 1.747 × 106 g or roughly 1.75× 103 kg. 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. (a) Differentiating 0.8( ) 5.00 3.00 20.00m t t t= − + with respect to t gives

0.24.00 3.00.dm t dt

−= −

The water mass is the greatest when / 0,dm dt = or at 1/ 0.2(4.00 / 3.00) 4.21s.t = =

CHAPTER 1 12

(b) At 4.21s,t = the water mass is

0.8( 4.21s) 5.00(4.21) 3.00(4.21) 20.00 23.2 g.m t = = − + = (c) The rate of mass change at 2.00 st = is

0.2

2.00 s

2

g 1 kg 60 s4.00(2.00) 3.00 g/s 0.48 g/s 0.48 s 1000 g 1 min

2.89 10 kg/min. t

dm dt

=

⎡ ⎤= − = = ⋅ ⋅⎣ ⎦

= ×

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

0.2

2.00 s

3

g 1 kg 60 s4.00(5.00) 3.00 g/s 0.101g/s 0.101 s 1000 g 1 min

6.05 10 kg/min. t

dm dt

=

⎡ ⎤= − = − = − ⋅ ⋅⎣ ⎦

= − ×

31. The mass density of the candy is

4 3 4 33 0.0200 g 4.00 10 g/mm 4.00 10 kg/cm .

50.0 mm m V

ρ − −= = = × = ×

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 ,M Ahρ= 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

4 3 2( ) (4.00 10 kg/cm )(238 cm )(0.250 cm/s)

0.0238 kg/s 1.43 kg/min.

dM d Ah dhA dt dt dt

ρ ρ −= = = ×

= =

32. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A = 20 × 12 = 240 m2) in addition to a rectangular box (height h´ = 6.0 m and same base). Therefore,

31 1800 m . 2 2

hV hA h A h A⎛ ⎞′ ′= + = + =⎜ ⎟ ⎝ ⎠

(a) Each dimension is reduced by a factor of 1/12, and we find

Vdoll 3 3m m= FHG I KJ ≈1800

1 12

10 3

c h . .

13

(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore,

Vminiature 3m 6.0 10 m= FHG I KJ ≈ ×

−1800 1 144

3 4 3c h .

33. In this problem we are asked to differentiate between three types of tons: displacement ton, freight ton and register ton, all of which are units of volume. The three different tons are given in terms of barrel bulk, with

31 barrel bulk 0.1415 m 4.0155 U.S. bushels= = using 31 m 28.378 U.S. bushels.= Thus, in terms of U.S. bushels, we have

4.0155 U.S. bushels1 displacement ton (7 barrels bulk) 28.108 U.S. bushels 1 barrel bulk

⎛ ⎞= × =⎜ ⎟ ⎝ ⎠

4.0155 U.S. bushels1 freight ton (8 barrels bulk) 32.124 U.S. bushels 1 barrel bulk

4.0155 U.S. bushels1 register ton (20 barrels bulk) 80.31 U.S. bushels 1 barrel bulk

⎛ ⎞= × =⎜ ⎟ ⎝ ⎠

⎛ ⎞= × =⎜ ⎟ ⎝ ⎠

(a) The difference between 73 “freight” tons and 73 “displacement” tons is

73(freight tons displacement tons) 73(32.124 U.S. bushels 28.108 U.S. bushels)

293.168 U.S. bushels 293 U.S. bushels

VΔ = − = −

= ≈ (b) Similarly, the difference between 73 “register” tons and 73 “displacement” tons is

3

73(register tons displacement tons) 73(80.31 U.S. bushels 28.108 U.S. bushels)

3810.746 U.S. bushels 3.81 10 U.S. bushels

VΔ = − = −

= ≈ × 34. The customer expects a volume V1 = 20 × 7056 in3 and receives V2 = 20 × 5826 in.3, the difference being 31 2 24600 in.V V VΔ = − = , or

( ) 3

3 3

2.54cm 1L24600 in. 403L 1 inch 1000 cm

V ⎛ ⎞ ⎛ ⎞Δ = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

where Appendix D has been used. 35. 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:

CHAPTER 1 14

(a) ( ) 2 peck11 tuffets = 11 tuffets 22 pecks 1 tuffet

⎛ ⎞ =⎜ ⎟

⎝ ⎠ .

(b) ( ) 0.50 Imperial bushel11 tuffets = 11 tuffets 5.5 Imperial bushels 1 tuffet

⎛ ⎞ =⎜ ⎟

⎝ ⎠ .

(c) ( ) 36.3687 L11 tuffets = 5.5 Imperial bushel 200 L 1 Imperial bushel

⎛ ⎞ ≈⎜ ⎟

⎝ ⎠ .

36. 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 910 = 0.900,

3 40 = 7.50 × 10

−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 chaldron (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 112

chaldron = 1 bag. Thus, the next entry in that second column is 112 = 8.33 × 10 −2.

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 pottle = 2.08 × 10−2 bag, and 1 gill = 1.11 × 10−4 bag. (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. 37. The volume of one unit is 1 cm3 = 1 × 10−6 m3, so the volume of a mole of them is 6.02 × 1023 cm3 = 6.02 × 1017 m3. The cube root of this number gives the edge length:

5 38.4 10 m× . This is equivalent to roughly 8 × 102 km. 38. (a) Using the fact that the area A of a rectangle is (width) × (length), we find

15

( ) ( )( )

( ) ( )( ) total

2

2

3.00acre 25.0 perch 4.00 perch

40 perch 4 perch 3.00 acre 100 perch

1acre

580 perch .

A = +

⎛ ⎞ = +⎜ ⎟

⎝ ⎠ =

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):

( ) 2

2 5 2 total

16.5ft580 perch 1.58 10 ft . 1perch

A ⎛ ⎞

= = ×⎜ ⎟ ⎝ ⎠

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

( ) 2

5 2 4 2 total

1m1.58 10 ft 1.47 10 m . 3.281ft

A ⎛ ⎞

= × = ×⎜ ⎟ ⎝ ⎠

39. This problem compares the U.K gallon with U.S. gallon, two non-SI units for volume. The interpretation of the type of gallons, whether U.K. or U.S., affects the amount of gasoline one calculates for traveling a given distance. If the fuel consumption rate is R (in miles/gallon), then the amount of gasoline (in gallons) needed for a trip of distance d (in miles) would be

(miles)(gallon) (miles/gallon) dV

R =

Since the car was manufactured in the U.K., the fuel consumption rate is calibrated based on U.K. gallon, and the correct interpretation should be “40 miles per U.K. gallon.” In U.K., one would think of gallon as U.K. gallon; however, in the U.S., the word “gallon” would naturally be interpreted as U.S. gallon. Note also that since 1 U.K. gallon 4.5460900 L= and 1 U.S. gallon 3.7854118 L= , the relationship between the two is

1 U.S. gallon1 U.K. gallon (4.5460900 L) 1.20095 U.S. gallons 3.7854118 L

⎛ ⎞= =⎜ ⎟ ⎝ ⎠

(a) The amount of gasoline actually required is

750 miles 18.75 U. K. gallons 18.8 U. K. gallons 40 miles/U. K. gallon

V ′ = = ≈

CHAPTER 1 16

This means that the driver mistakenly believes that the car should need 18.8 U.S. gallons. (b) Using the conversion factor found above, the actual amount required is equivalent to

( ) 1.20095 U.S. gallons18.75 U. K. gallons 22.5 U.S. gallons 1 U.K. gallon

V ⎛ ⎞′ = × ≈⎜ ⎟ ⎝ ⎠

.

40. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0 u, but converted to kilograms), then the computation reduces to simply taking the reciprocal of the number given in Eq. 1-9 and rounding off appropriately. Thus, the answer is 6.0 × 1026. 41. Using the (exact) conversion 1 in = 2.54 cm = 0.0254 m, we find that

0.0254 m1 ft 12 in. (12 in.) 0.3048 m 1in.

⎛ ⎞ = = × =⎜ ⎟

⎝ ⎠

and 3 3 31 ft (0.3048 m) 0.0283 m= = for volume (these results also can be found in Appendix D). Thus, the volume of a cord of wood is 3(8 ft) (4 ft) (4 ft) 128 ftV = × × = . Using the conversion factor found above, we obtain

3 3 3 3

3

0.0283 m1 cord 128 ft (128 ft ) 3.625 m 1 ft

V ⎛ ⎞

= = = × =⎜ ⎟ ⎝ ⎠

which implies that 3 11 m cord 0.276 cord 0.3 cord 3.625

⎛ ⎞= = ≈⎜ ⎟ ⎝ ⎠

.

42. (a) In atomic mass units, the mass of one molecule is (16 + 1 + 1)u = 18 u. Using Eq. 1-9, we find

( ) 27

261.6605402 10 kg18u = 18u 3.0 10 kg. 1u

− −⎛ ⎞× = ×⎜ ⎟

⎝ ⎠

(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of water molecules:

21 46

26

1.4 10 5 10 . 3.0 10

N − ×

≈ ≈ × ×

43. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3 × 106 mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thus, (2.3 × 106 mg/week)/(604800 s/week) = 3.8 mg/s. 44. The volume of the water that fell is

17

( ) ( ) ( ) ( ) ( ) ( )

2 2 2

6 2

6 3

1000 m 0.0254 m26 km 2.0 in. 26 km 2.0 in. 1 km 1 in.

26 10 m 0.0508 m 1.3 10 m .

V ⎛ ⎞ ⎛ ⎞

= = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

= ×

= ×

We write the mass-per-unit-volume (density) of the water as:

3 31 10 kg m .m V

= = ×ρ

The mass of the water that fell is therefore given by m = ρV:

( ) ( )3 3 6 3 91 10 kg m 1.3 10 m 1.3 10 kg.m = × × = × 45. The number of seconds in a year is 3.156 × 107. This is listed in Appendix D and results from the product

(365.25 day/y) (24 h/day) (60 min/h) (60 s/min). (a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than there are seconds per year. (b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by

10 10

10 6

10 4= − u - day,

which may also be expressed as 10 86400 1

8 64− F HG

I KJ =u - day

u - sec u - day

u - sec.c h . 46. The volume removed in one year is

V = (75 10 m ) (26 m) 2 10 m4 2 7 3× ≈ ×

which we convert to cubic kilometers: V = × F HG

I KJ =2 10

1 0 0207 3

m km 1000 m

km3 3c h . . 47. We convert meters to astronomical units, and seconds to minutes, using

CHAPTER 1 18

8

1000 m 1 km 1 AU 1.50 10 km 60 s 1 min .

=

= × =

Thus, 3.0 × 108 m/s becomes

8

8

3.0 10 m 1 km AU 60 s 0.12 AU min. s 1000 m 1.50 10 km min

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞× ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎟⎜⎟ ⎟ ⎟ =⎜ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟⎟⎜ ⎝ ⎠×⎝ ⎠ ⎝ ⎠⎝ ⎠

48. Since one atomic mass unit is 241 u 1.66 10 g−= × (see Appendix D), the mass of one mole of atoms is about 24 23(1.66 10 g)(6.02 10 ) 1g.m −= × × = On the other hand, the mass of one mole of atoms in the common Eastern mole is

75 g 10 g 7.5

m′ = =

Therefore, in atomic mass units, the average mass of one atom in the common Eastern mole is

23 23

10 g 1.66 10 g 10 u. 6.02 10A

m N

−′ = = × = ×

49. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain

1 1

1 97 1

3 88 2ken

m m

m

2

2

2

2= = . . .

(b) Similarly, we find 1 1

197 1

7 653 3ken

m m

m

3 3

3= = . . .

(c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus,

( ) ( )22 33.00 5.50 156 ken .r hπ π= = (d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) = 1.19 × 103 m3. 50. 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. 51. (a) For the minimum (43 cm) case, 9 cubits converts as follows:

19

( ) 0.43m9cubits 9cubits 3.9m. 1cubit

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

And for the maximum (53 cm) case we obtain

( ) 0.53m9cubits 9cubits 4.8m. 1cubit

⎛ ⎞ = =⎜ ⎟

⎝ ⎠

(b) Similarly, with 0.43 m → 430 mm and 0.53 m → 530 mm, we find 3.9 × 103 mm and 4.8 × 103 mm, respectively. (c) We can convert length and diameter first and then compute the volume, or first compute the volume and then convert. We proceed using the latter approach (where d is diameter and is length).

( ) 3

2 3 3 3 cylinder, min

0.43m28 cubit 28 cubit 2.2 m . 4 1 cubit

V dπ ⎛ ⎞

= = = =⎜ ⎟ ⎝ ⎠

Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, max = 4.2 m3. 52. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn along with appropriate conversion factors:

( ) ( ) ( ) ( ) ( ) ( )

2

28 2 36

100 hide 110 acre 4047 m 1 wp 1acre1 hide

1 10 m 1 barn

25 wp 1 10 .

11 barn −×

≈ ×

53. The objective of this problem is to convert the Earth-Sun distance to parsecs and light-years. To relate parsec (pc) to AU, we note that when θ is measured in radians, it is equal to the arc length s divided by the radius R. For a very large radius circle and small value of θ, the arc may be approximated as the straight line-segment of length 1 AU. Thus,

( ) 61 arcmin 1 2 radian1 arcsec 1 arcsec 4.85 10 rad 60 arcsec 60 arcmin 360

θ − ⎛ ⎞⎛ ⎞° π⎛ ⎞= = = ×⎜ ⎟⎜ ⎟⎜ ⎟°⎝ ⎠⎝ ⎠⎝ ⎠

Therefore, one parsec is

5o 6 1 AU1 pc 2.06 10 AU

4.85 10 sR θ −

= = = = × ×

Next, we relate AU to light-year (ly). Since a year is about 3.16 × 107 s, we have

( ) ( )7 121ly 186,000mi s 3.16 10 s 5.9 10 mi= × = × .

(a) Since 51 pc 2.06 10 AU= × , inverting the relationship gives

CHAPTER 1 20

( ) 65 1 pc1 AU 1 AU 4.9 10 pc.

2.06 10 AU R

⎛ ⎞ = = = ×⎜ ⎟×⎝ ⎠

(b) Given that 61AU 92.9 10 mi= × and 121ly 5.9 10 mi= × , the two expressions together lead to

6 6 5 5 12

1 ly1 AU 92.9 10 mi (92.9 10 mi) 1.57 10 ly 1.6 10 ly 5.9 10 mi

− −⎛ ⎞= × = × = × ≈ ×⎜ ⎟×⎝ ⎠ .

Our results can be further combined to give 1 pc 3.2 ly= . 54. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3, and 1 in.3 = 1.639 × 10−2 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460 ft2/gal becomes

22 2 2460 ft 1 m 1 gal460 ft /gal 11.3 m L.

gal 3.28 ft 3.79 L ⎛ ⎞⎛ ⎞ ⎛ ⎞

= =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

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

2 2 4 1

3

11.3 m 1000 L11.3 m /L 1.13 10 m . L 1 m

−⎛ ⎞⎛ ⎞= = ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

(c) The inverse of the original quantity is (460 ft2/gal)−1 = 2.17 × 10−3 gal/ft2. (d) The answer in (c) represents the volume of the paint (in gallons) needed to cover a square foot of area. From this, we could also figure the paint thickness [it turns out to be about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part (b)].

21

Chapter 2 1.The speed (assumed constant) is v = (90 km/h)(1000 m/km) ⁄ (3600 s/h) = 25 m/s. Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s) ≈ 13 m. 2. (a) Using the fact that time = distance/velocity while the velocity is constant, we find

avg 73.2 m 73.2 m 3.05 m1.22 m/s

73.2 m 73.2 m 1.74 m/s.v += = +

(b) Using the fact that distance = vt while the velocity v is constant, we find

vavg m / s)(60 s) m / s)(60 s)

s m / s.= + =( . ( . .122 3 05

120 2 14

(c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer.

3. Since the trip consists of two parts, let the displacements during first and second parts of the motion be Δx1 and Δx2, and the corresponding time intervals be Δt1 and Δt2, respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is Δx = Δx1 + Δx2, and the total time for the trip is Δt = Δt1 + Δt2. Using the definition of average velocity given in Eq. 2-2, we have

1 2 avg

1 2

.x xxv t t t

Δ + ΔΔ = =

Δ Δ + Δ

To find the average speed, we note that during a time Δt if the velocity remains a positive constant, then the speed is equal to the magnitude of velocity, and the distance is equal to the magnitude of displacement, with | |d x v t= Δ = Δ .

CHAPTER 2

22

(a) During the first part of the motion, the displacement is Δx1 = 40 km and the time interval is

t1 40 133= =( . km)

(30 km / h) h.

Similarly, during the second part the displacement is Δx2 = 40 km and the time interval is

t2 40 0 67= =( . km)

(60 km / h) h.

The total displacement is Δx = Δx1 + Δx2 = 40 km + 40 km = 80 km, and the total time elapsed is Δt = Δt1 + Δt2 = 2.00 h. Consequently, the average velocity is

avg (80 km) 40 km/h. (2.0 h)

xv t

Δ = = =

Δ

(b) In this case, the average speed is the same as the magnitude of the average velocity: avg 40 km/h.s = (c) The graph of the entire trip is shown below; it consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to (Δt1, Δx1) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (Δt1, Δx1) to (Δt, Δx) = (2.00 h, 80 km).

4. Average speed, as opposed to average velocity, relates to the total distance, as opposed to the net displacement. The distance D up the hill is, of course, the same as the distance down the hill, and since the speed is constant (during each stage of the motion) we have speed = D/t. Thus, the average speed is

D D t t

D D v

D v

up down

up down

up down

+

+ =

+

2

which, after canceling D and plugging in vup = 40 km/h and vdown = 60 km/h, yields 48 km/h for the average speed. 5. Using x = 3t – 4t2 + t3 with SI units understood is efficient (and is the approach we

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