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CHAPTER 1 Tooling Up

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Answers to Understanding the Concepts Questions

1. We can make a vector in any number of dimensions by making sure that it has a component in each of the

dimensions. Thus, if we have a four-dimensional space we can form a set of Cartesian axes, x, y, z, and r w, say, and a vector V will have an x-component Vx , a y-component Vy , a z-component Vz , and a w-component

Vw . The displacement would then be defined as a four-component vector that is the difference between two four-component position vectors.

2. The metric units of distance (cm, m, km, etc) and mass (g, kg, etc.) are based on powers of 10, so it is rather easy to

convert between them (e.g., 1 cm = 10– 2

m = 10– 5

km). The British system units are less convenient to covert

because the conversion factors are not 10 to an integer power. For example, 1 mi = 5280 ft and 1 ft = 12 in, rather

than 1 km = 1000 m and 1 m = 100 cm. Try converting 1 mi2 to in

2 and see for yourself. The metric system of units

is therefore inherently more convenient. Here in the U.S., however, the British system is still widely used because

of tradition, and measuring tools based on it (such as yard sticks and pound scales) are more abundant than those

based on the metric system. This is what makes the metric system less convenient at times. 3. My height is pretty close to 6 feet –– let’s take 6.00 ft as a value to work with. That is equivalent to (6.00 ft)(12

in/ft) = (72.0 in)(2.54 cm/in) = 183 cm. To find the height in meters, we divide the number of centimeters by 100 to

obtain 1.83 m. In every case we have used three significant figures. Incidentally, your height will vary by an

amount of the order of 1 cm over the course of a day –– you are taller in the morning, after a night in bed. So unless

you want to be precise about the time of day, you can reliably state your height to only three significant figures.

4. My mass is 65 kg, which corresponds to a weight of about 145 lb. You can get your own figure by weighing yourself

(usually in lbs) and noting that a weight of 1 lb corresponds to about 0.454 kg of mass on Earth’s surface.

5. An angle, defined as the arc length over the radius, is a dimensionless quantity. So both degree and radian are

dimensionless.

6. An essential feature of any standard is that it stay the same: the human pulse fins this need very poorly. A

pendulum does much better in this regard –– the main reason it might change has to do with changes in its length

caused by temperature changes, and this is relatively easier to control with temperature control and control over

the material and construction of the pendulum. A second feature desirable in a standard is that it be easily

available to all, everywhere. Again, a human pulse serves poorly. Individual pulse rates and steadiness vary

greatly, and it would be inconvenient to choose one person. While not ideal, a pendulum is again fairly

satisfactory. Our ability to construct identical pendulums is good, and variations in the value of g over the surface

of Earth are fairly small. The pendulum clock did in effect form a time standard for many years.

7. Define the speed of light in vacuum as exactly 299,792,458 m/s, and 1 m is then the distance light travels in

vacuum in 1/299,792,458 second.

8. First, we are really asking if three vectors in a plane can form a null vector. That is because we can start with the plane

formed by two of them. If the third vector is not in that plane, then the resultant cannot be zero. So if the vectors do

not all lie in the same plane then we cannot form the null vector. Now

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© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Page 1-1

Fishbane, Gasiorowicz, and Thornton

suppose that they do. Our question can now be rephrased: Can three vectors of equal length in a plane be arranged

with tails touching heads such that they form a closed figure? Imagine three rods of equal length, with the ends of

two of them hinged to the third one, one at each end of the third rod. The two outer rods have one end free to move,

and there is only one position in which their free ends coincide to make a closed triangle. This happens when the

three rods form an equilateral triangle.

9. Yes. They can form a vector triangle shown approximately below:

7 cm 4 cm

5 cm

10. The distance between Los Angeles and New York is 2462 miles, or (2462 mi)(1609 m/mi) = 3.961 × 106 m. Let’s

estimate the width of a human hair as 0.05 mm, or 5 × 10-5

m. then the level of accuracy is

5 × 10– 5

m/(3.961 × 106 m) ≈ 1.3 ×10

–11, or about 1 part per 100 billon. This is 0.01 parts per billion, or 1

× 10– 5

parts per million, or 1 × 10– 8

parts per thousand.

ˆ ˆ r ˆ ˆ r 2 2

) 1/2 2 2

) 1/2

=

11. The vector sum of i and j is the vector V = i + j . The magnitude of V is V = (Vx + Vy = (1 + 1

√2. The result is not unity! For the answer to the second part, read the discussion of Question 8.

12. To communicate with a totally different civilization we must resort to some universal standards. For sizes one can

think of an atom, whose typical size is about 10–10

m. So we might refer to our size as 1010

that of an atom. The

size of the solar system can be expressed the same way.

13. Let’s compare the estimate with a more accurate calculation of a year, the time it takes for Earth to orbit the sun.

The number of full days in the year is 365.25 –– the extra quarter day is accounted for in our calendars by leap year,

a calendar year with an additional day that occurs once every four calendar years. Thus the number of seconds is

365.25 d = (365.25 d)(24 hr/d)(60 min/hr)(60 s/min) = 3.156 × 107 s. With π = 3.1415..., we see the approximation

is accurate to (3.156 – 3.142)/(3.142) = 0.005 = 0.5%.

r ˆ ˆ

14. v , a i and b j other), with

form the three sides of a right-angled triangle (as iˆ and

ˆj are perpendicular to each r

v the hypotenuse. Thus by Pythagoras’ theorem v2 = a

2 + b

2, or v = (a

2 + b

2)1/2

.

15. A typical scale of reasonable quality probably has an error of plus or minus half a pound. So suppose your weight

is 150 lb, the corresponding percentage uncertainty would be around 0.5 lb/150 lb, or 0.3%.

16. The mass of the worm is proportional to its volume, and if it increases its linear dimensions by a factor of ten, its

mass, and with it its need for oxygen, win increase by a factor of 103, or a thousand. In contrast, its surface area win

increase by a factor of 102, and this is how much the oxygen-absorbing capacity of the worm will increase. Between

the factors of 1000 and 100 there is a deficit of a factor of 10. The giant salamander of China, which can weigh

more than 100 lbs and which spends long periods underwater by absorbing oxygen through its skin, is an extremely

wrinkled animal. In fact, almost all large animals depend on a huge lung or gin area.

17. For an object of linear dimension L, its surface area A is of the order of L2 while its mass m is of the order of L

3, so

the ratio of A/m goes like 1/L, which is larger for smaller objects. The surface area of a mouse is therefore larger

relative to its body size, in comparison with humans. The thermal energy radiated away is proportional to the

surface area of an animal. The relatively high rate of energy loss through the skin of the mouse needs to be

compensated with sufficient food intake. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Page 1-2

Chapter 1: Tooling Up

18. If the number of people on Earth is N and the radius of Earth is R, then the average area available to each person is

A = 4πR2/N. If you were to use a value of R that is 10% larger, i.e., you replaced R by (1+10%)R = 1.1R, then the

corresponding value of a becomes 4πR)2/N = (1.1)

2 A = 1.2A, i.e., it would increase by about 20%.

19. The mass of Earth is of the order of 102 4

kg, so the relative change in its mass due to the introduction of the 1-kg

meteorite is of the order of 1 kg/102 4

kg, or one part in one trillion trillion, which is negligible by any practical measures.

20. This is generally not the case. For the subtraction to yield zero result these two vectors must also have the exact

same orientation.

21. Any point in three-dimensional space is uniquely determined by a vector that runs from the origin of a fixed

coordinate system to that point. Such a vector can always be expressed as the sum of its three component vectors,

each along one of the three coordinate axes. (For example, in the usual xyz coordinate system, rr = xi

ˆ y

ˆj zk

ˆ .)

It therefore takes three vectors to specify a point in three dimensions. There are infinite number of sets of vectors that can do that, as there are infinite number of ways to set up a three-dimensional coordinate system.

r 22. In the vector diagram below v is the velocity of the water (w)

wb r

relative to the bank (b), its magnitude is 6 mph; v sw is the velocity of the swimmer (s) relative to the bank, its magnitude is

r 3 mph; and v is the velocity of the swimmer relative to the

sb r r r

bank. Note that v sb = v sw + v wb. If vwb is increased, the swimmer would end up further downstream, but the time it takes to cross the

river remains the same.

rvsb

rvsw

rvwb


Page 1-3


Solutions to Problems

1. Assuming that the twenty thousand is exact, Ngreen = (15/100)(20,000) = 3.0 × 103 green jelly beans.

2. Product = (105)(10

– 4) = 10

1 = 10; Ratio = 10

– 4/10

5 = (10

– 4)(10

– 5) = 10

– 9 .

3. $356.00 = ($356.00)(100 cents/$) = 35600 cents = 3.5600 × 104 cents.

4. e8 4

= 3.025077322… × 103 6

. (Note that e = 2.718281828… is an irrational number with infinite number of

significant figures.)

5. Cube root = (102 1

)1/3

= 107 ; square = (10

7)2 = 10

1 4 .

6. The height = (1472 ft)(12 in/ft)(2.54 cm/in) = 4.487 × 104 cm = 4.487 × 10

2 m .

7. For a person 6 ft tall, we have h = (6 ft)(0.305 m/ft)/(10–10

m/atom) = 1.8 × 101 0

atoms. 8. The height of the skier is H = [(5 ft)(12 in/ft) + 5 in](2.54 cm/in) = 165 cm. The

length of the skis should be L = H + 5 cm = 165 cm + 5 cm = 170 cm.

9. Price = (1.25 euro/kg)($0.94/euro) = $1.18/kg.

10. Price = (1.20 euro/L)($1/1.04 euro)(3.8 L/gal) = $4.4/gal.

11. g = (9.80 m/s2)(100 cm/m)(1 in/2.54 cm)(1 ft/12 in) = 32.2 ft/s

2 .

12. G = (6.67 × 10–11

m3 · s

– 2 · kg

– 1)(10

2 cm/m)

3(1 kg/10

3 g) = 6.67 × 10

– 8 cm

3 · s

– 2 · g

– 1 .

13. Density = mass/volume = mass/(4πR3/3)

= {(7.35 × 102 2

kg)/[4π(1.74 × 106 m)

3/3]}(10

3 g/kg)(1 m/10

2 cm)

3 = 3.33 g/cm

3 .

14. Density = mass/volume = mass/(4πR3/3)

= {(1.4 × 103 1

kg)/[4π(15 km)3/3]}(1 t/10

3 kg)(1 km/10

3 m)

3(1 m/10

2 cm)

3 = 9.9 × 10

8 t/cm

3 .

15. For the Opel: (100 km/7.0 L)(1 mi/1.61 km)(3.8 L/gal) = 34 mi/gal. For the

Mercedes: (100 km/23 L)(1 mi/1.61 km)(3.8 L/gal) = 10 mi/gal.

16. (a) The distance D = 0.3 mi, with one significant figure.

(b) D = 27.2 mi/100 trips = 0.272 mi, with three significant figures. Since this

is 0.3 to one significant figure, this is consistent with (a).

(c) The uncertainty in the number of trips is ±1/100 or ±1%; thus the uncertainty in the result is ±1%, which is

0.003 mi, so D = (0.272 ± 0.003) mi.

17. Since 22/7 = 3.1428571 and π = 3.1415927, the % error is [(22/7 – π)/π]100% = (0.0012645/π)100% =

0.0402%. Since 355/113 = 3.141592920 and π = 3.141592654…, the % error is (0.000000266/π)100% =

8.49 × 10– 6

% . (Results will depend on precision of your calculator.)


Page 1-4


18. F = mv2/r = (0.00535 kg)(1.1 m/s)

2/(0.3 m) = 2 × 10

– 2 kg · m · s

– 2 .

There is only one significant figure since r has only one significant figure.

19. For each dimension we can find the % uncertainty: w = 0.75 ± 0.02 m = 0.75 m ± (0.02/0.75)(100) = 0.75 m ± 3%; l = 0.5 ± 0.1 m = 0.5 m ± (0.1/0.5)(100) = 0.5 m ± 20%; h = 0.582 ± 0.058 m = 0.582 m ± (0.058/0.582)(100) = 0.582 m ± 10%. Then

the volume is V = wlh = (0.75 m)(0.5 m)(0.582 m) ± (3% + 20% + 10%).

We get V = 0.2 m3 ± 30% = 0.2 ± 0.06 m

3 ≈ 0.2 ± 0.1 m

3. Note that we need to round off the error from 0.06

m3 to 0.1 m

3 in order to match V, since the last digit in the main figure has to be aligned with the first digit in its

error.

20. For the weight, the % uncertainty is [(0.5 lb)/(213 lb)]100% = 0.2%. For the water level change, each reading has an uncertainty of 0.5 cm, so the % uncertainty is [(0.5 cm +

0.5 cm)/(158.5 cm – 152.0 cm)]100% = 15%.

(Keeping two digits is justified since both numerator and denominator have two.) Thus the

total % uncertainty is 15% + 0.2% = 15%.

The density is mass/volume = [(213.0 lb)(1 kg/2.2 lb)]/[(1.5 m2)(6.5 cm)(1 m/100 cm)] = 9.93 × 10

2 kg/m

3. Since

15% of 9.93 is 1.5, the final result is 9.9 × 102 kg/m

3 ± 15%.

21. In terms of the diameter, the area is A = πD2 = πDD. When multiplying, % uncertainties add, so ±%A =

%D + %D = 2(%D) or 10% = 2(%D) and thus %D = 5%.

22. For the infinite series ∑n=0 (x

n/n!) with x = 0.100 000, the first term (n = 0) is (0.100 000)

0/0! = 1.000 00

and the second term (n = 1) is (0.100 000)1/1! = 0.100 000. Since the first term is 1 and succeeding terms decrease,

we need to find the term with a value less than 0.000005, which takes into account rounding. The next terms are (n

= 2): (0.100 000)2/2! = 0.005 000 00, (n = 3): (0.100 000)

3/3! = 0.000 166 667, (n = 4):

(0.100 000)4/4! = 0.000 004 166 67. Thus four terms (n = 0, 1, 2, 3) are needed.

23. Because mv2/2 = p

2/2m, we can write the dimensions as [m][v]

2 = [p

2][m]

– 1.

Putting in the dimensions, we get [M][LT– 1

]2 = [p

2][M

– 1], or [p

2] = [M][L

2T

– 2][M] = [MLT

– 1]2; so

[p] = [MLT– 1

].

24. For E = mc2 we can write the dimensions as [E] = [m][c]

2 = [M][LT

– 1]2 = [ML

2T

– 2].

25. For L = h/mec we can write the dimensions as [L] = [h][me]– 1

[c]– 1

, or [L] = [h][M]– 1

[LT– 1

]– 1

. Thus

[h] = [L][M][LT– 1

] = [ML2T

– 1].

26. For h2/m

3G we can write the dimensions as

[ h2/m

3G] = [h]

2[m]

– 3[G]

– 1 = [ML

2T

– 1]2[M]

– 3[L

3T

– 2M

– 1]– 1

= [L].

27. The dimensions of the left-hand-side of the formula are [a] = [LT– 2

]. For the right-hand-side [G] = [L3 T

– 2 M

–

1], so the dimensions of the right-hand-side are [MG/r] = [M][G][r

– 1] = [M][L

3 T

– 2 M

– 1][L

– 1] = [L

2 T

–2 ],

which differ from those of [a]. So no, this formula cannot possibly be correct.

28. If we set v = kρα

Bβ

, then [v] = [ρ]α

[B]β

gives

[LT– 1

] = [ML– 3

]α

[ML– 1

T– 2

]β

= [M]α

+

β

[L]–3α

– β

[T]–2β

. By equating the exponents for each dimension, we obtain α + β = 0, – 3α – β = 1, and – 2β = –1.

These equations are satisfied by α = –1/2 and β = 1/2, so v = k(B/ρ)1/2

. Note that we had three equations to solve for two unknowns, so there is redundancy and the results are consistent with all three equations.


Page 1-5


29. (a) For F = Ame–αr

/r4, the exponential function can have no dimension, thus

[αr] = [α][r] = [α][L] = 1 and [α] = [L– 1

] .

(b) Then [F] = [A][m][r]–4

or [MLT– 2

] = [A][M][L]– 4

and [A] = [L5T

– 2].

30. If the quantity is called x, we can set x = hα

mβ

cγ. Then

[x] = [h]α

[m]β

[c]γ

[L] = [ML2T

– 1]α

[M]β

[LT– 1

]γ

[L] = [M]α

+

β

[L]2α

+

γ [T]

–α – γ.

By equating the exponents for each dimension, we obtain α + β = 0, 2α + γ = 1 and – α – γ = 0. These

equations are satisfied by α = 1, β = –1 and γ = –1, so x = h/(mc).

31. (a) [K e2/hc] = [K] [e

2] / [h] [c],

(1) = (1) [e]2/[ML

2T

– 1] [LT

– 1], or [e]

2 = [ML

3T

– 2] and thus [e] = [M

1/2L

3/2T

– 1].

(b) [e2/R] = [ML

3T

– 2]/[L] = [ML

2T

– 2].

32. If we take an average lifetime to be 70 years and the average pulse to be 60 beats/min,

N = (60 beats/min)(70 yr)(365 day/yr)(24 h/day)(60 min/h) ≈ 2 × 109 beats. If we

take the radius of the tire to be 1 ft and the distance to be 3000 mi, we have

N = distance/circumference = (3 × 103 mi)(5 × 10

3 ft/mi)/(2π ft/rotation) ≈ 3 × 10

6 rotations.

33. If we ignore the difference between troy ounces (used for gold) and ordinary ounces, the weight of the gold is W

= ($25 × 106)(1 oz/$400)(1 lb/16 oz)(1 ton/2 × 10

3 lb) = 2 tons.

He probably cannot make it, as station wagons are unlikely able to carry this much load.

34. The tunnel has three circular tubes, two for trains and a smaller one for service. If we approximate the cross-

section of the train tube as a ring with outer radius 4 m and inner radius 3 m, its cross-sectional area is A = π[(4

m)2 – (3 m)

2] ≈ 20 m

2. If we take the service tube to be half of this, the total area is 2.5(20 m

2) = 50 m

2. For a

length of 30 km, the volume of concrete will be

V ≈ (50 m2)(30 km)(10

3 m/km) ≈ 1.5 × 10

6 m

3 .

35. The number of bits is N = (1.44 × 106 byte)(8 bit/byte) = 1.2 × 10

7 bits. We estimate that the circular area

available for storage has an outer radius of 1.5 in and an inner radius of 0.5 in. The area per bit is

A = π(Router2 – Rinner

2)/N = π[(1.5 in)

2 – (0.5 in)

2](2.54 cm/in)

2/(1.2 × 10

7 bits) ≈ 3.5 × 10

– 6 cm

2 ,

which is a square about 19 µm on a side.

36. The weight will be W = (5 gal)(4 L/gal)(1 g/cm3)(10

3 cm

3/L)(1 kg/10

3 g)(2.2 lb/kg) ≈ 40 lb. If a

person could carry 80 lb (properly distributed), that would be about 10 gal.

37. Let’s use estimates of 50 million autos, $500/yr for maintenance, $25,000 annual salary, 4 hours for repair, 5%

of autos in the shop, a population of 300 million, and 1 out of 400 people are mechanics.

(a) N = ($500/yr/auto)(50 × 106 auto)(1 yr/$25,000) ≈ 10

6 mechanics.

(b) N = (50 × 106 auto)(0.05/day)(4 h/auto)(1 day/8 h) ≈ 10

6 mechanics.

(c) N = (300 × 106 people)(1 mechanic/400 people) ≈ 10

6 mechanics.

38. Estimate the diameter of a dime as 1 cm. Then

Dsun = (distance to sun/length of arm)Ddime = (93 × 106 mi/100 cm)(1 cm) ≈#9.3 × 10

5 mi.

39. The volume of a droplet is (4π/3)(0.5 × 10– 4

m)3 = 0.5 × 10

–12 m

3. The volume of water needed depends on the size

of the city. If we assume an area of (30 km)2, the volume of water required is

(0.5cm)(30 km)2(10

3 m/km)

2(1 m/100 cm) = 5 × 10

8 m

3.

The number of droplets is N = (5 × 108 m

3)/(0.5 × 10

–12 m

3)= 1 × 10

21 droplets.


Page 1-6


40. The area to be occupied by each tree, 2 meters on a side, is 2 m × 2 m = 4 m2. Thus the total number of trees that

can fit into 4050 m2 is 4050 m

2 / 4 m

2 = 1013. In 100-tree bundles, that would be 1013/100 = 10.13 bundles. So to

fully plant the area you need to order 11 bundles.

41. The surface area of Earth is 4πR2, where R = 6400 km. Let’s estimate the average depth of the oceans and seas to

be h ≈1 mi, or about 2 × 105 cm, then the volume of the water is roughly

V ≈ (2/3)(4π R2 ) h ≈ (2/3) 4π (6400 × 10

5 cm)

2(2 × 10

5 cm) = 7 × 10

23 cm

3 .

One litre is 1000 cm3, so the degree of dilution is 1000 cm

3 / 7 × 10

23 cm

3 = 1 × 10

–19 .

42. Because the size of the disk is small compared to the distance from the source, the fraction of particles falling on

the detector will be the ratio of the area of the detector to the surface area of a sphere centered at the source with a

radius equal to the distance to the detector. Thus

fraction = [π(Rdetector)2]/[4π(ddetector-source)

2] = (2 × 10

– 2 m)

2/[4(2 m)

2] = 2.5 × 10

–

5 . Thus 2.5 × 10

3 particles per second strike the detector.

43. The average speed over a year is (15,000 mi/yr)(1 yr/365 days)(1 day/24 h) = 1.7 mi/h. If we estimate the average

highway speed as 40 mi/h, then the number on the road at any one moment is

N = (200 × 106 automobiles)(1.7 mi/h)/(40 mi/h) ≈ 9 × 10

6 automobiles.

44. If we estimate that in 1000 people there is one driver of an eighteen-wheeler, the number of trucks is N =

population/1000 = 250 × 106/1000 = 2.5 × 10

5 trucks.

If each truck is 25 m long, they would cover a distance of D ≈

(2.5 × 105 trucks)(25 m/truck) ≈ 6 × 10

6 m.

The length of I-80 is

L ≈ (3000 mi)(1.6 × 103 m/mi) ≈ 5 × 10

6 m.

Yes, they can form a continuous line from New York to San Francisco. Note that the numbers are so close that the

answer is questionable. If we had estimated one truck driver for every 2000 people, the answer would change.

45. Because the relationship between the circumference and the radius, C = 2πR, is linear, a change in R of ∆R will

produce a change in C of ∆C = 2π∆R. Thus ∆R = ∆C/2π = 2 m/2π = 0.3 m = 30 cm.

46. If we estimate 100 trees/acre and 500 apples/tree, the number is

N = (10 acre)(100 trees/acre)(500 apples/tree) ≈ 5 × 105 apples.

47. If pure hydrogen, the number is N = (2.0 × 1030

kg)/(1.7 × 10–27

kg) = 1.2 × 1057

hydrogen atoms.

If 70% hydrogen, the number is N = (0.70)(2.0 × 1030

kg)/(1.7 × 10–27

kg) = 8.2 × 1056

hydrogen atoms. 48. If the oil does not spread beyond a maximum area, there is a minimum thickness, which would be the diameter of

a molecule. The number of molecules in the slick can be found from the number in the given mass and from the ratio of the slick

area to the area of a molecule:

N = (0.95 g)(6.02 × 1023

molecule/14 g) = (1.5 × 107 cm

2)/(πd

2/4), from which we get

d ≈ 2 × 10– 8

cm. Note that this is comparable to the thickness of the slick: t = 1 mL/1.5 × 10

7 cm

2 ≈ 7 × 10

– 8 cm. The

difference is probably due to the fact that the slick may not be exactly one molecule thick. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Page 1-7


49. The size of the Manhattan island is about 30 square miles, or (30 mi2)(1.6 × 10

3 m/mi)

2 = 5 × 10

7 m

2. Let’s say that

about a quarter of that area is for buildings, then the combined foot print area of all the buildings in the island is about 1 × 10

7 m

2. If on average the floor-to-floor spacing in a building is about 5 m , then

with an average height of h each building would have h/(5 m) stories, and so the total available space in all the

buildings there would be about (1 × 107 m

2) h/(5 m). On the other hand, assuming that each person takes up about

10 m × 10 m = 100 m2 of space, then the total area needed for 2 million people there would be

(100 m2/person)(2 × 10

6 persons) = 2 × 10

8 m

2 . Thus

(1 × 107 m

2) h/(5 m) ≈ 2 × 10

8 m

2, which gives

h ≈ 100 m. Note that this estimation is very crude, and is probably only good for one significant figure.

50. y y

x x

51. Because she is always a constant distance from the center of the lake, her position can be described by the clockwise

angle φ from the north-south line drawn from the center to the starting point (south end of the lake). Her direction of

travel will be tangent to the circle with a constant speed of 3 m/s at an angle φ clockwise to the south direction.

r r

ˆ ˆ ˆ ˆ ˆ ˆ .

52. A + B = (6i – 5 j ) + (8i + 3 j ) = 14i − 2j

r r

ˆ ˆ ˆ ˆ ˆ ˆ .

A – B = (6i – 5 j ) – (8i + 3 j ) = −2i − 8j

53. If we take iˆ as east and

ˆj as north, the four displacements in paces are 4

ˆj , 6

cos 45° iˆ + 6 sin 45°

ˆj , 2i

ˆ and – 5i

ˆ .

The resultant displacement is their sum: r

R = 4 ˆj + 6(0.707)i

ˆ + 6(0.707)

ˆj + 2i

ˆ – 5i

ˆ

= 1.2ˆi + 8.2

ˆj paces (8.3 paces, 82

o north of east).

54. The magnitude of the first vector in meters is (50 ft)(0.3048 m/ft) = 15 m, and that of the second one is (24 ft)(0.3048

m/ft) = 7.3 m. If we take iˆ as east and

ˆj as north, the two displacements are

ˆ ˆ ˆ ˆ and 15 m (cos 135° i + sin 135° j ) = (−11i + 11j ) m

7.3 m (cos 45° iˆ + sin 45°

ˆj ) = (5.2i

ˆ + 5.2

ˆj ) m .


Page 1-8


55. We take the origin where the catch was made, with the x-axis to the right and

the y-axis upfield.

Catch: 0iˆ + 0

ˆj ;

first turn: 0iˆ + 0

ˆj + 15

ˆj = 15

ˆj ;

second turn: 0iˆ + 15

ˆj – 15i

ˆ = –15i

ˆ + 15

ˆj

; third turn: –15iˆ + 15

ˆj + 10

ˆj = –15i

ˆ + 25

ˆj ; fourth turn: –15i

ˆ + 25

ˆj + 20i

ˆ = 5i

ˆ + 25

ˆj ; fifth turn: 5i

ˆ + 25

ˆj – 5i

ˆ = 25

ˆj ;

touchdown: 25 ˆj + 65

ˆj = 90

ˆj .

y (yd)

10

5

5 10 x (yd)

56. The x-component is the horizontal component: Vx = – V sin α.

57. The x-component is the vertical component: Vx = + V cos α.

58. The angle between the vector and the x-axis is now α + 45°, so the x-component is

Vx = + V cos(α + 45°). [If α > 45°, (α + 45°) > 90°, so the cosine will be negative and the same expression will be valid.]

59. AB = (7, 0) = 7 at an angle of 0°. C

BC = (– 7, 7) = 7√2 at an angle of 135°.

CA = (0, – 7) = 7 at an angle of 270°.

A B

r r r r

60. (a) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ .

A + B + C + D = (– 2i – 3 j ) + (i + 2 j + 3k ) + (3 j + 3k ) + (– 2i – k ) = −3i + 2j + 5k r r

(b) A – D ˆ ˆ ˆ ˆ ˆ ˆ . = (– 2i – 3 j ) – (– 2i – k ) = − 3j + k

r r r

(c) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ .

A + D – B = (– 2i – 3 j ) + (– 2i – k ) – (i + 2 j + 3k ) = −5i − 5j − 4k

r r ˆ ˆ

ˆ ˆ ˆ

ˆ

ˆ r r

2

2

2

(d) A – C

| A – C | = 2 + 6 + 3 = 7.

= (– 2i – 3 j ) – (3 j + 3k ) = – 2i – 6 j – 3k ;


Page 1-9


61. (a) r ˆ ˆ r ˆ ˆ r ˆ ˆ r ˆ ˆ A = – 4i + 2 j , B = – i + 4 j , C = 2i + 2 j , D = 5i – 3 j .

rr r

(b) 2 A + C ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ . – D = 2(– 4i + 2 j ) + (2i + 2 j ) – (5i – 3 j ) = −11i + 9j

r r

B ˆ ˆ ˆ ˆ = ˆ . + C /2 = (–i + 4 j ) + (2i + 2 j )/2 5j

r

r

ˆ ˆ ˆ ˆ ˆ −

ˆ .

Dr – Br = (5i – 3 j ) – (–i + 4 j ) = 6i 7 j

| D – B |= 62 + 7

2 = 9.2.

r r r

r

− d c /2

d

r r

r

c

2 a + c − d

r

r r

r

ra

r

b +

rc/2

d − b

− b

ra

b

62. For the first sequence:

r r ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ A + B = (3i + 4 j ) + (2i – 2 j + 4k ) = 5i + 2 j + 4k ;

r r r

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ .

( A + B ) + C = (5i + 2 j + 4k ) + (–i + 5 j – 3k ) = 4i + 7j + k

For the second sequence:

r r ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ B + C = (2i – 2 j + 4k ) + (–i + 5 j – 3k ) = i + 3 j + k ;

r r r

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ .

( B + C ) + A = (i + 3 j + k ) + (3i + 4 j ) = 4i + 7j + k

63. The components are shown in the figure. Here β = 120° and α = 60° – θ , so

from the sine theorem

A x / sin (60° – θ ) = A / sin 120° , or Ax = A sin (60° – θ )/sin 120° ;

A y / sin θ = A / sin 120° , or Ay = A sin θ /sin 120° .

y ra

ray α

θ β x r

ax

64. If we estimate that the area of the head on which hair grows is 1/3 of the surface area of a sphere of diameter 20

cm, then the fraction = area of lock/scalp area: fraction = π(R )

2/[π(D )

2/3] = (2 cm)

2/[(20 cm)

2/3] = 0.03 .

lock head

The number of hairs is N = area of lock/area of one hair: N = [π(D )

2/4]/[π(D )

2/4] = [(4 cm)/(10

– 4 m)(10

2 cm/m)]

2 ≈ 2 × 105

hairs . lock hair

65. (a) (1 kg/103 g)(1 metric ton/10

3 kg) = 10

– 6 metric tons/g.

(b) (1 m/102 cm)

3 = 10

– 6 m

3/cm

3 .

66. We estimate the mass to be 2.5 kg.

Density = mass/volume = (2.5 kg)(103 g/kg)/[(25 cm)(22 cm)(4 cm)] ≈ 1.1 g/cm

3.

This is greater than the density of water (1 g/cm3), so we would not expect the book to float.

67. The time is distance traveled/speed:

t = d/c = (1.5 × 108 mi)(1.6 × 10

3 m/mi)/(2.998 × 10

8 m/s) = 8.0 × 10

2 s , or 13 minutes.

68. m = 18 g (1 kg/103 g)/(6.02 × 10

2 3 molecules) = 3.0 × 10

–26 kg/molecule.



69. If we assume 70 kg mass for the human body, we get

N = 70 kg/(3.0 × 10–26

kg/molecule) = 2 × 102 7

molecules. 70. Because the number density is assumed the same,

N star = (N

silver/V

silver)

V

star =

N

silver (

V

star/V

silver) =

N

silver (R

star/R

silver)3;

N star = 108 nucleons [(12 × 103 m)/(5 × 10

–15 m)]

3 = 1.5 × 10

5 7 nucleons.

71. Because 18 g of water occupies 18 cm3, the volume of one molecule is

18 cm3/(6.02 × 10

2 3 molecules) = 3.0 × 10

–23 cm

3.

This must be L3, so L = (3.0 × 10

–23 cm

3)1/3

≈ 3 × 10– 8

cm.

72. The gasoline usage ∝ mass ∝ volume ∝ (dimension)3. A 20% decrease in each dimension means the volume

changes by (0.80)3 = 0.512. Thus there would be a 49% savings.

73. m = (2 × 103 0

kg/star)(101 1

star/galaxy) = 2 × 104 1

kg/galaxy.

N = m/mH = 2 × 104 1

kg/(1.67 × 10–27

kg/H atom) = 1.2 × 106 8

H atoms. 74. Because the number of pages can be counted (remember each page is printed on both sides), we need the thickness

of the book to better than 10%. If we measure the book’s thickness to ± 1 mm, it will be 4.0 ± 0.1 cm or 4.0 cm ± 3 %. Allowing for some miscellaneous pages, the number will be 650 ± 20 pages = 650 pages

± 3%. Therefore the thickness of a page

= (4.0 cm ± 3%)/(650 pages ± 3%) = 6.2 × 10– 3

cm ± 6% = (6.2 ± 0.4) × 10– 3

cm.

75. For the proposed relation [τ] = [l]rg]s, we have [T] =

[L]r[LT

– 2]s [L]

r s[T]

–s.

By equating the exponents for each dimension, we obtain r + s = 0, and – 2s = 1. These equations are

satisfied by r = 1/2, and s = – 1/2, so τ = (dimensionless constant)(l/g)1/2

.

76. If the side length of the room is L and the area is A, then A = L2 and

A ≈ (dA/dL) L = (dL2/dL) L = 2 L L, or A/A = 2 L L/L

2 = 2 L/L. Thus L / L ≈

A/A)/2 = (5%)/2 = 2.5%.

77. We will estimate 1 × 106 persons with each person using 10 gal/day and each truck carrying 1 × 10

4 gal. Then N =

(106 persons)(10 gal/day/person)/(10

4 gal/truck) ≈ 10

3 trucks/day.

If a bath requires 30 gal, this would require an additional average of 10 gal/day, so the number of trucks would double

to 2 × 103 trucks/day.

78. We will estimate 30 m between houses, the mail carrier takes 1.5 min for each house and delivers mail for 4 hours

(the rest of the time is for sorting, etc.). The number of houses = (4 h)(60 min/h)/(1.5 min) = 160 houses.

The distance walked = (160 houses)(30 m/house) ≈ 5 × 103 m (≈ 3 mi).

The amount of mail = (160 houses)(4 pieces/house) ≈ 640 pieces. If we assume an average of 2 ounces/piece,

the total mass = (640 pieces)(2 oz/piece)(1 lb/16 oz)(1 kg/2.2 lb) ≈ 36 kg.

79. From Kepler’s third law: (TV/TE)2 = (DV/DE)

3, or DV = (TV/TE)

2/3 DE , so DV =

(0.61 yr/1 yr)2/3

(1.5 × 108 km) = 1.1 × 10

8 km.

For Saturn we get TS = (DS/DE)3/2

TE = (14 × 108 km/1.5 × 10

8 km)

3/2 (1 yr) = 29 yr.

80. η = Fy/vA, so

[η] = [F][ y]/([v][ A]) = [MLT–2

][L]/([LT–1

][L2]) = [ML

–1T

–1].


Page 1-11


81. The number of molecules in cylinder = total mass of air/mass of a molecule, or N = [(30 ft)(12 in/ft)(2.54 cm/in)(1 cm

2)(1 g/cm

3)]/[(1.6)(18g)/6.0 × 10

2 3 molecules] = 1.9

× 102 5

air molecules above 1 cm2.

The total number is the number above 1 cm2 times the surface area of the earth:

N total = N(4πR2) = (1.9 × 10

2 5 molecules/cm

2){4π[(6.4 × 10

6 m)(10

2 cm/m)]

2}

≈ 104 4

molecules. 82. If we restrict the vectors to two dimensions, this can be shown graphically:

cr

r

ra + b +

rc r

r r b

c b r

r

a

a

83. (a) The two vectors can be obtained from the diagram:

r ˆ ˆ u = u cos θ i + u sin θ j

r ˆ + v cos θ

ˆ v = – v sin θ i j , or

r ˆ – v cos θ

ˆ v = + v sin θ i j

(b) vxuy + vyux = (– v sin θ)(u cos θ) + (v cos θ)(u sin θ)

= uv(–#sin θ cos θ + sin θ cos θ) = 0, or

vxuy + vyux = (v sin θ)(u cos θ) + (– v cos θ)(u sin θ)

= uv(sin θ cos θ – sin θ cos θ) = 0.

r a

cr

r r r b + c + a

r

b

y r

u r v

θ

θ

θ

r x

v

84. (a) iˆ r = (1) cos θ i

ˆ + (1) sin θ

ˆj = cos θ i

ˆ + sin θ

ˆj .

(b) iry = sin θ.

(c) iˆ t = (1)(– sin θ)i

ˆ + (1) cos θ

ˆj = − sinθ

î + cos θ

ˆj .

85. (a) The component of rr in the xy-plane is r sin θ at an angle of φ with

the x-axis.

Thus the x-component of rr is r sin θ cos φ.

(b) Since rr is at an angler of θ from the z-axis, the z-component is r cos θ.

(c) The y-component of r is ry = r sin θ sin φ.

y îr ît

r P

θ

O

x


Page 1-12


86. If we assume 20 breaths/min and that each breath involved different molecules, then we can find the number of

molecules Michelangelo breathed during his lifetime from the product of the number of molecules in each breath

and the number of breaths in 91 years:

N = (2.5 L)(6.0 × 102 3

/22.4 L)(20/min)(91 yr)(365 day/yr)(24 h/day)(60 min/h)

≈ 6.4 × 1031 molecules.

We assume that these molecules are dispersed uniformly in the volume of the atmosphere: V =

4πR2H ≈ (4)(3)(6.4 × 10

6 m)

2(8 × 10

3 m) ≈ 5 × 10

1 8 m

3.

If the lungs contain 2.5 L of air, the number of these molecules in the lungs is

N(lungs) = [(6.4 × 103 1

molecules)/(5 × 101 8

m3)](2.5 L)(10

– 3 m

3/L) ≈ 3 × 10

1 0 molecules.

87. With the three physical attributes we can form the following relation:

t0 = (constant) λα

l βτγ, which leads to [t0] = [λ]

α[l]β[τ]

γ , or [T]

= [ML– 1

]α

[L]β

[MLT– 2

]γ = [M]

α γ [L]

–α β

γ [T]

–γ.

By equating the exponents for each dimension, we obtain α + γ = 0, – α + β + γ = 0, and – 2γ = 1. These

equations are satisfied by α = 1/2, β = 1, and γ = – 1/2, so

t0 = (a constant) l (λ/τ)1/2

.


Page 1-13

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