Solutions Manual For Physics: An Algebra-Based Approach 2nd Edition By O\'Meara PDF

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Test Bank, Solution Manual, eBook For Physics: An Algebra-Based Approach 2nd Edition By O'Meara, Johnson, McFarland, Hirsch, Iqbal ; 9780176817336, 0176817336...

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Chapter 1 Measurement and Types of Quantities Exercises 1-1

Answers will vary. (a) Food can be sweet, salty, bitter, sour, or umami (savory). (b) An odour can be musty, smoky, fruity, etc.

1-2

Answers will vary. Examples of qualitative descriptions are friendly, fun-loving, honest; examples of quantitative descriptions can be height, mass, shoe size.

1-3

(a) quantitative (b) qualitative (c) qualitative (d) quantitative (e) quantitative (f) quantitative

1-4

Answers will vary. 

time: time to recharge a battery; time available between classes



length: height adjustment of a bicycle seat; distance from home to work or school



mass: mass of frozen food defrosting in a microwave oven; mass of a parcel sent by courier



volume: volume of books that a knapsack can hold; volume of water needed to keep hydrated during a long-distance run

1-5

Answers will vary. Electrical voltage: 6.0 V; temperature: 100℃; power: 60 W

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Physics: An Algebra-Based Approach 1-6

Some of the disadvantages are •

Most celestial bodies that are visible at night are not visible during the daytime, and vice versa.

1-7

•

Cloudy conditions interfere with observation of any celestial body.

•

Accuracy is difficult to achieve.

•

Convenience is minimal.

Theoretical aspects of physics involve posing questions, creating ideas to research answers to those questions, experimenting, measuring, analyzing, and collaborating, which leads to theories and more questions. The theoretical research and discovery leads to applications that, in most cases, help to improve our lives. One example is the discovery of current electricity, which has led to countless, very useful, electrical devices.

1-8

The original metre was defined in terms of the distance from the equator to the North Pole, a distance that could only be assumed because it was impossible to measure. The original second was defined in terms of a mean solar day, a quantity that is not constant because Earth’s rotation is very gradually slowing down.

1-9

(a)

Length 2  10 26 m   2  1041 15 length 1 10 m

(b)

Time 5  1017 s   1.7  1042 25 time 3  10 s

(c)

Mass 1  1053 kg   1 1083 31 mass 9 10 kg Mass has by far the greatest range of values.

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Chapter 1—Measurement and Types of Quantities 1-10

In seconds, $1  10 9  1 s  1  10 9 s. $

In years, $1 109  1 s  1 year7  31 years. $

3.2  10 s

8  10 41 kg  4 1011 stars kg 2  1030 star

1-11

# stars =

1-12

# atoms =

1-13

(a) 8.4 × 10 15 (b) 8 × 1036 (c) 8.0 × 108 (d) 1.94 × 105 m/s

1-14

A base unit is a standard unit of measurement from which other units may be derived. In

2 1030 kg  11057 atoms kg 27 1.7  10 atom

the SI, examples are the metre (m), kilogram (kg), and second (s). A derived unit is a measurement unit stated in terms of one or more base units. Examples are a unit for speed (m/s), a unit for surface area (m²), and a unit for solid volume (m³).

1-15

Four examples are watt (W = kg∙m 2∙s–3), pascal (Pa = kg∙s –2), volt (V = kg∙m2∙s–3∙A–1), and becquerel (Bq = s –1).

1-16

Some of the patterns are the prefixes from 10 3 to 10 –3 change by a factor 10 1; the remaining prefixes change by a factor of 10 3; the symbols for the large numbers (from mega upward) are capital letters, and all the other symbols are lower case; the origins of the prefixes are all non-English words; some original meanings relate to the power of 10

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Physics: An Algebra-Based Approach (e.g., Greek femten, or 15, is used for 10 –15), while some others relate to a power of 103 (e.g., Italian setta or 7 is (10 3)7 or 1021).

1-17

1-18

1

m  1.3 m dam

(a)

1.3 10

(b)

30 nm 

(c)

1.23  10 4 km 

(d)

1.486 103 Tm 

(a)

20 ms 

(b)

1 m 10 μm  8.6 10 4 μm 8.6 cm  2 10 cm m

(c)

3.28 g 

(d)

105 MHz 

(e)

2.4 10 3 MW 103 mW 10 6 W 2.4 10 6mW    2 2 m W MW m

(f)

9.8 m 1 s 1s 9.8 10 12 m    s2 106 μs 106 μs μs 2

(g)

4.7 g 1 kg 106 cm3 4.7  103 kg    cm 3 10 3 g m3 m3

(h)

53 people 1 km2 104 m 2 0.53 people  6 2  km2 10 m ha ha

dam  10 1

1m  3  10  8 m 109 nm 1000 m  1.23  10 7 m km

1012 m 1.486 10 9 m Tm

1s  2 10 2 s 3 10 ms 6

1 Mg 6  3.28 10 Mg 6 10 g 103 kHz  1.05 10 5 kHz 1 MHz

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Chapter 1—Measurement and Types of Quantities 1-19

(a)

280 mm + 37 cm = 28 cm + 37 cm = 65 cm

(b)

9850 mm – 1.68 m = 9.85 m – 1.68 m = 8.17 m

1-20

100 km 1000 m 100 cm 1 in 1 ft 1 mi 62.1 mi       h km m 2.54 cm 12 in 5280 ft h

1-21

(a) L/T (b) L/T (c) L/T2 (d) M/L3

1-22

(a) speed (b) length (c) density (d) acceleration

1-23

(a) MꞏL/T2 (b) MꞏL2 /T3 (c) T 1

1-24

From d = vt, the left side = L and the right side =

L  T = L . Thus, the dimensions are T

equal.

1-25

From d  kt 3 , k  d3 , which has dimensions of L T 3 or L/T3 .

1-26

1 From d  vot  2 at 2 , the left side of the equation has dimension L and the right side has

t

dimension

L L  T  2  T 2  L , so the equation is dimensionally correct. T T

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Physics: An Algebra-Based Approach 1-27

To compare the uncertainties, begin by expressing them in the same unit. Thus, 100 g 

1 kg  0.1 kg . The most expensive scale is likely the first one, 1000 g

42.400 kg  0.005 kg, which provides the greatest number of significant digits and the smallest uncertainty. The least expensive scale is likely the last one, which has the largest uncertainty, 0.1 kg.

1-28

(a) 1 (b) 5 (c) 3 (d) 4

1-29

(a) 3.85 × 104 Gm (b) 9.40 × 10 4 MW (c) 5.51 × 101 dam (d) 8.77 × 10 2 kL (e) 7.66 × 102 g

1-30

percent error 

measured value  accepted value accepted value 3

=

3

3

3

1.08  10 kg/m  1.00 10 kg/m 1.00  103 kg/m3

 100%

 100%

 0.08 100% or 8%

1-31

Perimeter: p  2l  2w  2(1.18 m)+2(0.378 m) =3.116 m or 3.12 m Thus, the perimeter is 3.12 m, rounded off to one estimated digit.

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Chapter 1—Measurement and Types of Quantities Area:

A  lw  1.18 m 0.378 m = 0.446 m 2

Thus, the area is 0.446 m² (rounded off to three significant digits).

1-32

(a)

m n  m p  1.674 927  1027 kg  1.672 621 10 27 kg =2.306  10

(b)

30

kg

m p  m e  1.672 621 10 27 kg  9.109 382 10 31 kg =1.672 621 1027 kg  0.000 910 938 2 1027 kg =1.671 710 10 27 kg

1-33

Greatest distance = 1.495 988 1011 m  3.844 108 m = 1.495 988 1011 m + 0.003 844  1011 m 11

= 1.499 832 10 m Least distance = 1.495 988 1011 m  3.844  108 m = 1.495 988 1011 m  0.003 844  1011 m = 1.492 144  1011 m

1-34

d d 1.495 988 1011 m  499 s. From v  , t   3.00 108 m/s t v Thus, the time to three significant digits is 499 s.

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Physics: An Algebra-Based Approach 1-35

In all cases, the actual numerical values will vary. (a)

# cells =



mass mass/cell 60 kg 12 110 kg/cell

 about 1 1014 cells

Thus, the number would be about 1014 cells. (b)

First, we must make some assumptions, all of which are approximations. North American population (including Canada, the USA, and Mexico) is about 500 million; the average number of patties consumed per year per person is about 30; and the average mass of a patty is about 0.2 kg.

estimated mass 

=

mass patties  # of people patty person 0.2 kg 30 patties   5 108 people patty person

= 3 10 9 kg Thus, the total mass is about 10 9 kg to 1010 kg. (c)

Assume that the diameter of the Ferris wheel is about 20 m to 30 m, which means that the circumference, C  d , is about 100 m. Assume that the straight line distance from Calgary to Winnipeg is about 1000 km. # of rotations =

distance distance/rotation



1 103 km 1 103 m/rotation

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Chapter 1—Measurement and Types of Quantities



1 10 3 km 1 101 km/rotation

 1 104 rotations Thus, the number of rotations is about 10 000 or 10 4 rotations. (d)

Assume Canada’s population (including children) is about 3.6 × 10 2 people with an average arm span slightly more than 1 m, so the total “population arm span” is about 4 × 107 m or 4 × 104 km. 5

# of populations =

1-36

coastline length 2 10 km   5 populations length/population 4 104 km/population

Answers will vary. Some examples of scalar quantities are area, volume, speed, density, energy, power, and frequency.

1-37

Answers will vary. Two examples of displacement are a ball tossed 10 m west and a walk of 100 m south from the bus stop to the residence. Two examples of velocity are a motorbike travelling at 50 km/h north and a jogger running along a path at 5 m/s southeast.

CHAPTER REVIEW Multiple-Choice Questions 1-38

(c)

1-39

(e)

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Physics: An Algebra-Based Approach 1-40

(a)

1-41

(c)

1-42

(c)

1-43

(c)

Review Questions and Problems 1-44

Measurement is important (a) in society in order to have efficient communication in numerous aspects of our lives, including manufacturing, building infrastructure, selling, and buying (b) in physics in order to design and perform experiments and develop theories and applications resulting from experiments and discoveries

1-45

The base units are the metre (m) for length, the second (s) for time, and the kilogram (kg) for mass.

1-46

It isn’t necessary to have a base unit for area because area can be expressed in terms of the base unit for length.

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Chapter 1—Measurement and Types of Quantities 1-47

Assume each storey is 3 m high. Then the height of the building is height =30 storeys 

3m storey

 90 m = 90 m 

1 dam 10 m

 9 dam

Thus, the height is about 90 m, or 9 dam.

1-48

1-49

(a)

8.85 × 103 m, 8.85 × 104 dm, 8.85 × 10 5 cm

(b)

1.90 × 105 kg, 1.90 × 108 g, 1.90 × 1010 cg

(c)

6.9 × 1015 s, 6.9 × 10 21 s, 6.9 × 10 3 Es

Answers depend on each student’s mass. Using a mass of 60 kg, the ratios would be (a)

16 Mg 10 3 kg 3  10 2   60 kg 1 Mg 1

(b)

0.26 kg 4  103  60 kg 1

(c)

5.2  102 kg 9  60 kg 1

(d)

6.7 106 dg 1 kg 1g 1101  3   60 kg 10 g 10 dg 1

(e)

28 g 1 kg 5  104   60 kg 103 g 1

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Physics: An Algebra-Based Approach 1-50

mass = 3.0 kg  8.6 10 4  2.6  10 5 kg = 2.6  102 Mg

1-51

1-52

(a)

2.00  10 2 km  3.00  10 3 m = 2.00  10 2 km  3.0 km  1.97  10 2 km

(b)

4.4  10 3 m m  2.2 10 2 2 5 2 2.0  10 s s

(c)

4.4 103 m 4.4 103 m 7 m  5 2 10 2  1.1 10 (2.0 10 s) 4.0 10 s s2

(a)

r is the radius; b is the base; h is the height.

(b)

The dimension of πr ² is L2 and the dimension of bh/2 is L 2. So, we conclude that area has the dimension L 2.

1-53

(a)

The symbols are dimensions: M for mass, L for length, and T for time.

(b)

speed (v): [ v] = L/T acceleration (a ): [a ] = L/T2 area (A ): [ A] = L2

1-54

(c)

[ P] = MꞏL2/T3, [ p] = M/LT 2

(d)

[ E] = Mꞏ[ v]2

(e)

[ P] = Mꞏ[ v]ꞏ[a]

It is possible for the measurements to be multiplied (e.g., area × length = volume, or L² × L = L³), but it is not possible to add them (e.g., you can’t add area and length).

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Chapter 1—Measurement and Types of Quantities 1-55

Wood contracts as it dries, so it must be fully dried (cured) and totally contracted before being marked to ensure accurate scale divisions.

measured value  accepted value

1-56 percent error 

=

=

accepted value 1.82 10 8 m/s 1.86 108 m/s 1.86  108 m/s

 100%

 100%

0.04 m/s  100% 1.86 m/s

 2%

1-57

length remaining = 500 m  3(120 cm) = 500 m  3(1.20 m) = 500 m  3.60 m = 496 m

1-58

Let M represent the men’s discus, W represent the women’s discus, l represent length or distance, and d represent diameter. lM  lW 

=

=

lM l  25.0 rev  W  25.0 rev rev rev

 (d M ) rev

 25.0 rev 

 (0.221 m) rev

 (d W ) rev

 25.0 rev 

 25.0 rev

 (0.182 m) rev

 25.0 rev

= 3.06 m

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Physics: An Algebra-Based Approach 1-59

A  lw  7.32 m 2.44 m =17.9 m 2

1-60

V  πr 2h 2

 π  4.2 cm  22.8 cm   1.3 10 3 cm 3

1-61

Answers will depend on the assumptions made. (a)

Assume a normal pace is 0.6 m or 6 10  4 km.

#paces =



distance distance/pace 1.6 km 6 104 km/pace

 2.7 10 3 paces Thus, the number of paces is likely between 2 10 3 and 3  10 3. (b)

Assume that the size of a kernel of corn is about 3 mm by 2 mm by 2 mm. The kernel’s volume is V  lwh  (0.3 cm)(0.2 cm)(0.2 cm) 3

= 0.012 cm

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Chapter 1—Measurement and Types of Quantities

# kernels =



volume volume/kernel (10 cm)3 0.012 cm 3 / kernel 4

 8.3 10 kernels

Thus, the estimated number is between 1 104 and 1 105 kernels. (c)

Above-ground pools are often circular. Assume the diameter of the pool is 4 m or 40 dm and the height is 0.6 m or 6 dm.

V  area  height = r 2  h   (20 dm) 2 6 dm = 8 10 3 dm 3 or 8 10 3 L Thus, the volume would be about 1  104 L. (d)

Assume a heartbeat of 72 beats per minute, or 1.2 beats/s. For this example, assume an age of 19 years.

# beats = heart rate  age (in seconds) = 1.2

beats days h s  19 years  365  24  3600 s year day h

 7.2 108 beats Thus, the number of beats would be about 1  109 . (e)

Let’s assume that the average person sleeps about 8 h each day. Let’s also make the very simplistic assumption that the human population of about 7 billion is distributed approximately equally around Earth’s time zones.

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Physics: An Algebra-Based Approach

# people = fraction sleeping  total population =

1  7 109 people 3

= 2.3 109 people or about 2 10 9 people

1-62 (a) vector (b) scalar (c) scalar (d) scalar (e) vector (f) scalar

Applying Your Knowledge 1-63

The meanings are not exactly the same, but they are close. The SI is based on multiples of 10, whereas computers and data storage are based on powers of 2. For example, in the SI, kilo means 103 , or 1000, and mega means 106, but a kilobyte is 2 10 bytes, or 1024 bytes, and a megabyte is 2 20 bytes or 1.05 10 6 bytes.

1-64

Assume the finger is a cylinder of diameter 1 cm and length 5 cm. # cells =





Vfinger V cell πrfinger 2 hfinger 4 πrcell3 3 (0.5 cm) 2 (5 cm) 4 (5  10 4 cm) 3 / cell 3

 7.5 109 cells

Thus, there are about 109 to 10 10 cells.

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