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Chapter 1
 Description leading to understanding of our
environment.
 Description involves the measurement of the
physical world.
 Understanding our environment demands the
interpretation of accurate measurements (i.e.,
data).
 Therefore, understanding measurement is
essential.
Intro
 Sophisticated methods of measurement have
been developed.
 Measurements – movement, temperature, weather
conditions, time, etc.
 The constant use of measurements are in this
book, including many examples.
 Can everything be measured w/ certainty??
 As smaller and smaller objects were measured it
became apparent that the act of measuring
actually distorted the object.
Intro
 Subset of the Natural Sciences, together with
Biological Sciences
 Physical Sciences = Physics, Chemistry, Geology,
Meteorology, and Astronomy
 This book covers the fundamentals of each of the
five Physical Sciences.
Section 1.1
Section 1.1
 Measurements are the basis of scientific
research/investigation.
 Phenomena are observed, resulting in questions
of how or why these phenomena occur.
 Scientists assume that the universe is orderly and
can be understood.
Section 1.2
 Scientific Method – general methods of
observations, rules for reasoning, and making
predictions
 Can be broken down into:
◦ Observations & Measurements
◦ Hypothesis
◦ Experiments
◦ Theory
◦ Law
Section 1.2
 Quantitative data are gathered
Section 1.2
 Hypothesis – a possible explanation for the
observations
 Example: Matter consists of small particles
(atoms) that simply rearrange themselves
 A tentative answer or educated guess
 New experiments are designed to test the validity
of the hypothesis.
 The Hypothesis is supported if it correctly predicts
the experimental results
Section 1.2
 The testing, under controlled conditions, to
determine if the results support or confirm the
hypothesis
 Experimental results can be duplicated by other
researchers
 No concept or model of nature is valid unless the
predictions are in agreement with experimental
results.
Section 1.2
 Theory – tested explanation for a broad segment
of basic natural phenomena
 Example: Atomic Theory – This theory has
withstood testing for 200+ years.
 Depending on continued experimentation, theories
may be accepted, modified, or rejected.
Section 1.2
 Scientific Law – after a series of experiments a
concise statement (words/math) describes a
fundamental relationship of nature
 Example – Law of Conservation of Mass (no gain
or loss during chemical reaction)
 The law simply states the finding, but does not
explain the behavior.
Section 1.2
Section 1.2
 Sight, Hearing, Smell, Taste, Touch
 Sight and Hearing provide the most information
to our brains about our environment.
 Sensory Limitations – can be reduced by using
measuring devices
 Instruments extend our ability to measure and
learn about our environment.
 Our senses can also be deceived ->
Section 1.3
Lines “a” and “b” are equal in length!
Section 1.3
The lines are all horizontal!
Section 1.3
Section 1.3
Section 1.3
 Expressed in magnitude and units
 Fundamental quantities – length, mass, & time
 The study of Force and Motion requires only
these three quantities.
 Standard Unit – fixed and reproducible value to
take accurate measurements
Section 1.4
 Two major systems of units
 British (English) system – only used widely in the
United States (miles, inches, pounds, seconds,
etc.)
 Metric system – used throughout most of the world
(kilometers, meters, grams, etc.)
 The U.S. “officially” adopted the metric system in
1893, but continues to use the British system.
Section 1.4
Section 1.4
 The measurement of space in any direction
 Space has three dimensions – length, width, and
height.
 Metric Standard Unit = Meter (m), originally
defined as 1/10,000,000 of distance from equator
to north pole
 British Standard Unit = Foot, originally referenced
to the human foot.
Section 1.4
Originally defined as
a physical quantity of
nature.
1/10,000,000 of the
distance from the
equator to the pole.
Section 1.4
The meter is now defined by the distance light
travels in a vacuum/time.
Section 1.4
 The amount of matter an object contains
 An object’s mass is always constant
 Mass is a fundamental unit that will remain
constant throughout the universe.
 Metric Standard Unit = Kilogram (kg) – originally
defined as the amount of water in a 0.1m cube.
Now referenced to a cylinder in Paris
Section 1.4
U.S. Prototype #20 Kilogram, at NIST in Washington, D.C.
Actually – 0.999 999 961 kg of “the” standard in Paris
Section 1.4
 British Standard Unit = Slug (rarely used)
 We use the Pound (lb.)
 The pound is actually not a unit of mass, but rather
of weight, related to gravitational attraction
(depends on where the object is!)
 Object: Earth = 1lb.  Moon = 1/6lb.
 In fact, the weight of an object will vary slightly
depending on where it is on earth (higher
altitude less weight)
Section 1.4
Section 1.4
 Time - the continuous, forward flowing of events
 Time has only one direction  forward
 Second (s) – the standard unit in both the metric
and British systems
 Originally 1/86,400 of a solar day
 Now based on the vibration of the Cs133
atom
(Atomic Clock)
Section 1.4
Originally
defined as a
fraction of the
average solar
day.
Section 1.4
Defined by the radiation frequency of the Cs133
atom
Section 1.4
 Uses acronym “mks system” from standard units
of length, mass, and time – meter, kilogram,
second
 It is a decimal (base-10) system – this is much
better than the British system
 Administered by -- Bureau International des Poids
et Mesures (BIPM) in Paris
 International System of Units (SI)
 Contains seven base units
Section 1.4
 The fundamental units are a choice of seven well-
defined units which by convention are regarded as
dimensionally independent:
◦ meter, m (length)
◦ kilogram, kg (mass)
◦ second, s (time)
◦ ampere, A (electrical current)
◦ kelvin, K (temperature)
◦ mole, mol (amount of a substance)
◦ candela, cd (luminous intensity)
Section 1.5
 Easy expression and conversion
 Metric examples vs. British examples
◦ 1 kilometer = 1000 meters
◦ 1 mile = 5280 feet
◦ 1 meter = 100 centimeters
◦ 1 yard = 3 feet or 36 inches
◦ 1 liter = 1000 milliliters
◦ 1 quart = 32 ounces or 2 pints
◦ 1 gallon = 128 ounces
Section 1.5
 Mega, M – 106
– 1,000,000 times the base
 Kilo, k – 103
– 1,000 times the base
 Centi, c – 10-2
– 1/100th
of the base
 Milli, m – 10-3
– 1/1000th
of the base
 See Appendix 1 for complete listing
Section 1.5
 Liter – volume of liquid in a 0.1m (10 cm) cube
(10cm x 10cm x 10cm = 1000 cm3
)
 A liter of pure water has a mass of 1 kg or 1000
grams.
 Therefore, 1 cubic cm (cc) of water has a mass of
1 gram.
 By definition 1 liter = 1000 milliliters (ml)
 So, 1 ml = 1 cc = 1 g of pure water.
 1 ml = 1 cc for all liquids, but other liquids do not
have a mass of 1 g
Section 1.5
 A Liter is slightly more
than a quart.
– 1 quart = .946 liter
– 1 liter = 1.06 quart
Section 1.5
 (1 kg = 2.2046
lb on earth)
 The amount of
water in a 0.10m
(10 cm) cube
(0.10m3
)
Section 1.5
 Metric ton -- mass of 1 cubic meter (1 m3
) of water
 1 m = 100 cm
 (100cm)3
= 1,000,000 cm3
 Remember that 1000 cm3
= liter
 Therefore, there are 1000 liters in 1 m3
of water.
 Each liter has a mass of 1 kg.
 1 kg x 1000 = 1 metric ton
Section 1.5
 It is difficult to make all measurements with only
the 7 fundamental units.
 Derived units are therefore used, these are
multiples/combinations of fundamental units.
 We’ve already used derived units Volume 
length3
, m3
, cm3
 Area  length2
, m2
, ft2
, etc.
 Speed  length/time, m/s, miles/hour, etc.
Section 1.6
 Density (r) = mass per unit volume
 r=m/v [or m/length3
(since v = length3
)]
 How “compact” a substance is
 Typical Units used – g/cm3
, kg/m3
 Al = 2.7 g/cm3
,Fe = 7.8 g/cm3
, Au = 19.3 g/cm3
 Average for solid earth = 5.5 g/cm3
Section 1.6
 Hydrometer – a weighted glass bulb
 The higher the hydrometer floats the greater the
density of the liquid
 Pure water = 1g/cm3
 Seawater = 1.025 g/cm3
 Urine = 1.015 to 1.030 g/cm3
 Hydrometers are used to ‘test’ antifreeze in car
radiators – actually measuring the density of the
liquid
Section 1.6
 The denser the liquid the
higher the hydrometer
floats.
Section 1.5
 When a combination of units becomes complex
and frequently used –
 It is given a name
◦ newton (N) = kg x m/s2
◦ joule (J) = kg x m2
/s2
◦ watt (W) = kg x m2
/s3
Section 1.6
 Relates one unit to another unit
 Convert British to Metric (1in  cm)
 Convert units within system (1kg  g)
 We use “conversion factors” – many are listed on
inside back cover of book
 1 inch is equivalent to 2.54 centimeters
 Therefore “1 in = 2.54 cm” is our conversion factor
for inches & centimeters
Section 1.6
 Question: How many centimeters are there in 65
inches?
Section 1.6
2.54 cm
= 11 inch
• Since 1 in = 2.54 cm 
1 in
2.54 cm
= 1• Or
2.54 cm
1 in
• 65 in. x = 165 cm (the inches cancel out!!)
 Step 1 - Choose/Use a Conversion Factor,
generally can be looked up.
 Step 2 – Arrange the Conversion Factor into
the appropriate form, so that unwanted units
cancel out.
 Step 3 – Multiply or Divide to calculate answer.
 Use common sense – anticipate answer!
cm
inch
54.2
1
or
inch
cm
1
54.2
for example
Section 1.6
 How fast in mi/h is 50 km/h?
 Conversion Factor is 1km/h=0.621mi/h
50 km/h
Starting Value
hkm
hmi
/1
/621.0
x
Conversion
Factor
= 31.05 mi/h
Result
Section 1.6
Section 1.6
50 km/h
Starting Value
hkm
hmi
/1
/621.0
x
Conversion
Factor
= 31.05 mi/h
Result
 Either Conversion Factor can be used:
 1km/h = 0.621mi/h or 1mi/h = 1.61km/h
 How fast in km/h is 50 mi/h?
hmi
hkm
/621.0
/1
50 mi/h x = 80.5 km/h
Starting Value Conversion
Factor
Same
Result
50 mi/h x
hmi
hkm
/1
/61.1
= 80.5 km/h
Section 1.6
Section 1.6
hmi
hkm
/621.0
/1
50 mi/h x = 80.5 km/h
Starting Value Conversion
Factor
Same
Result
50 mi/h x
hmi
hkm
/1
/61.1
= 80.5 km/h
 22 inches = ?? Meters
 Inches  centimeters  meters
22 in
Starting Value
in
cm
1
54.2
x
Conv. Factor #1
in  cm
cm
m
100
1
x
Conv. Factor #2
cm  m
= 0.56 m
Result
Section 1.6
Section 1.6
22 in
Starting Value
in
cm
1
54.2
x
Conv. Factor #1
in  cm
cm
m
100
1
x
Conv. Factor #2
cm  m
= 0.56 m
Result
 How would one express “First and 10” in meters?
 Conversion Factor is 1yd = 0.914 m
10 yd
Starting Value
Section 1.6
x
Conversion
Factor
yd
m
1
914.0
= ??
Result
10 yd
Starting Value
Section 1.6
x
Conversion
Factor
yd
m
1
914.0
= 9.14 m
Result
Photo Source: Copyright © Bobby H. Bammel. All rights reserved.
Section 1.6
4843 m x
m
ft
1
28.3
= 15,885 feet above SL
(1775 feet higher than the top of Pikes Peak!)
Section 1.6
 Significant figures (“SF”) – a method of expressing
measured numbers properly
 A mathematical operation, such as multiplication,
division, addition, or subtraction cannot give you
more significant figures than you start with.
 For example, 6.8 has two SF and 1.67 has three
SF.
Section 1.7
 When we use hand
calculators we may
end up with results
like: 6.8/1.67 =
4.0718563
 Are all these
numbers
“significant?”
Section 1.7
 General Rule: Report only as many significant
figures in the result as there are in the quantity
with the least.
 6.8 cm/1.67 cm = 4.1(round off 4.0718563)
◦ 6.8 is the limiting term with two SF
 5.687 + 11.11 = 16.80 (round up 16.797)
◦ 11.11 is the limiting term with four SF
Section 1.7
 All non-zero digits are significant
◦ Both 23.4 and 234 have 3 SF
 Zeros are significant if between two non-zero
digits (‘captive’) – 20.05 has 4 SF, 407 has 3 SF
 Zeros are not significant to the left of non-zero
digits – used to locate a decimal point (leading
zeros) – 0.0000035 has 2 SF
 To the right of all non-zero digits (trailing zeros),
must be determined from context – 45.0 has 3 SF
but 4500 probably only has 2 SF
Section 1.7
 Exact Numbers – numbers of people, items, etc.
are assumed to have an unlimited number of SF
 In the process of determining the allowed number
of significant figures, we must generally also
‘round off’ the numbers.
Section 1.7
 If the first digit to be dropped is less than 5, leave
the preceding digit unchanged.
◦ Round off to 3 SF: 26.142  26.1
 If the first digit to be dropped is 5 or greater,
increase the preceding digit by one.
◦ Round off to 3 SF: 10.063  10.1
Section 1.7
 Round off 0.0997 to two SF
 0.0997  0.10
 What about this? 5.0 x 356 = 1780
 Round off 1780 to 2 SF
 1780  1800
Section 1.7
 Many numbers are very large or very small – it is
more convenient to express them in ‘powers-of-10’
notation
 1,000,000 = 10x10x10x10x10x10 = 106
Section 1.7
000,000,1
1
6
10
1
= = 0.000001 = 10-6
Section 1.7
Standard HMCO copyright line
 The distance to the sun can be expressed many
ways:
◦ 93,000,000 miles
◦ 93 x 106
miles
◦ 9.3 x 107
miles
◦ 0.93 x 108
miles
 All four are correct, but 9.3 x 107
miles is the
preferred format.
Section 1.7
 The exponent, or power-of-10, is increased by
one for every place the decimal point is shifted to
the left.
◦ 360,000 = 3.6 x 105
 The exponent, or power-of-10, is decreased by
one for every place the decimal point is shifted to
the right.
◦ 0.0694 = 6.94 x 10-2
Section 1.7
Rules for Scientific Notation
 5.6256 x 0.0012 = 0.0067507
  round to 2 SF
 0.0067507 rounds to 0.0068
  change to scientific notation
 0.0068 = 6.8 x 10-3
Section 1.7
Example
Rounding/Scientific Notation
 0.0024/8.05 = 0.0002981
  round to 2 SF
 0.0002981 rounds to 0.00030
  change to scientific notation
 0.00030 = 3.0 x 10-4
 **Note that the “trailing zero” is significant**
Section 1.7
Example
Rounding/Scientific Notation
 Read the problem, and identify the chapter
principle that applies to it. Write down the given
quantities w/ units. Make a sketch.
 Determine what is wanted – write it down.
 Check the units, and make conversions if
necessary.
 Survey equations – use appropriate one.
 Do the math, using appropriate units, round off,
and adjust number of significant figures.
Section 1.7
Problem Solving
 The earth goes around the sun in a nearly circular
orbit with a radius of 93 million miles. How many
miles does Earth travel in making one revolution
about the sun?
Section 1.7
Problem Solving
 The earth goes around the sun in a nearly
circular orbit with a radius of 93 million miles.
How many miles does Earth travel in making one
revolution about the sun?
 Determine what parts of the question are
important and how to attack the problem.
Section 1.7
Problem Example
 The earth goes around the sun in a nearly
circular orbit with a radius of 93 million miles.
How many miles does Earth travel in making one
revolution about the sun?
 In order to solve this problem notice that you
need an equation for a circular orbit
(circumference)
 The radius of 93,000,000 miles is given
 Our answer also needs to be in miles
(convenient!)
 Equation: c = 2pr (p = 3.14159…)
Section 1.7
Problem Example
 Circumference = c = 2pr (p = 3.14159…)
 c = 2 x 3.14159 x 93,000,000 miles
 or
 c = 2 x 3.14159 x 9.3 x 107
miles
 c = 58.433574 x 107
miles
 round off and adjust to two SF
 c = 5.8 x 108
miles
 5.8 x 108
miles = distance that the earth travels in
one revolution around the sun
Section 1.7
Problem Solving
 Density: r = m/V
Section 1.7

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Digital-Transformation-Roadmap-for-Companies.pptx

Sci 1010 chapter 1

  • 2.  Description leading to understanding of our environment.  Description involves the measurement of the physical world.  Understanding our environment demands the interpretation of accurate measurements (i.e., data).  Therefore, understanding measurement is essential. Intro
  • 3.  Sophisticated methods of measurement have been developed.  Measurements – movement, temperature, weather conditions, time, etc.  The constant use of measurements are in this book, including many examples.  Can everything be measured w/ certainty??  As smaller and smaller objects were measured it became apparent that the act of measuring actually distorted the object. Intro
  • 4.  Subset of the Natural Sciences, together with Biological Sciences  Physical Sciences = Physics, Chemistry, Geology, Meteorology, and Astronomy  This book covers the fundamentals of each of the five Physical Sciences. Section 1.1
  • 6.  Measurements are the basis of scientific research/investigation.  Phenomena are observed, resulting in questions of how or why these phenomena occur.  Scientists assume that the universe is orderly and can be understood. Section 1.2
  • 7.  Scientific Method – general methods of observations, rules for reasoning, and making predictions  Can be broken down into: ◦ Observations & Measurements ◦ Hypothesis ◦ Experiments ◦ Theory ◦ Law Section 1.2
  • 8.  Quantitative data are gathered Section 1.2
  • 9.  Hypothesis – a possible explanation for the observations  Example: Matter consists of small particles (atoms) that simply rearrange themselves  A tentative answer or educated guess  New experiments are designed to test the validity of the hypothesis.  The Hypothesis is supported if it correctly predicts the experimental results Section 1.2
  • 10.  The testing, under controlled conditions, to determine if the results support or confirm the hypothesis  Experimental results can be duplicated by other researchers  No concept or model of nature is valid unless the predictions are in agreement with experimental results. Section 1.2
  • 11.  Theory – tested explanation for a broad segment of basic natural phenomena  Example: Atomic Theory – This theory has withstood testing for 200+ years.  Depending on continued experimentation, theories may be accepted, modified, or rejected. Section 1.2
  • 12.  Scientific Law – after a series of experiments a concise statement (words/math) describes a fundamental relationship of nature  Example – Law of Conservation of Mass (no gain or loss during chemical reaction)  The law simply states the finding, but does not explain the behavior. Section 1.2
  • 14.  Sight, Hearing, Smell, Taste, Touch  Sight and Hearing provide the most information to our brains about our environment.  Sensory Limitations – can be reduced by using measuring devices  Instruments extend our ability to measure and learn about our environment.  Our senses can also be deceived -> Section 1.3
  • 15. Lines “a” and “b” are equal in length! Section 1.3
  • 16. The lines are all horizontal! Section 1.3
  • 19.  Expressed in magnitude and units  Fundamental quantities – length, mass, & time  The study of Force and Motion requires only these three quantities.  Standard Unit – fixed and reproducible value to take accurate measurements Section 1.4
  • 20.  Two major systems of units  British (English) system – only used widely in the United States (miles, inches, pounds, seconds, etc.)  Metric system – used throughout most of the world (kilometers, meters, grams, etc.)  The U.S. “officially” adopted the metric system in 1893, but continues to use the British system. Section 1.4
  • 22.  The measurement of space in any direction  Space has three dimensions – length, width, and height.  Metric Standard Unit = Meter (m), originally defined as 1/10,000,000 of distance from equator to north pole  British Standard Unit = Foot, originally referenced to the human foot. Section 1.4
  • 23. Originally defined as a physical quantity of nature. 1/10,000,000 of the distance from the equator to the pole. Section 1.4
  • 24. The meter is now defined by the distance light travels in a vacuum/time. Section 1.4
  • 25.  The amount of matter an object contains  An object’s mass is always constant  Mass is a fundamental unit that will remain constant throughout the universe.  Metric Standard Unit = Kilogram (kg) – originally defined as the amount of water in a 0.1m cube. Now referenced to a cylinder in Paris Section 1.4
  • 26. U.S. Prototype #20 Kilogram, at NIST in Washington, D.C. Actually – 0.999 999 961 kg of “the” standard in Paris Section 1.4
  • 27.  British Standard Unit = Slug (rarely used)  We use the Pound (lb.)  The pound is actually not a unit of mass, but rather of weight, related to gravitational attraction (depends on where the object is!)  Object: Earth = 1lb.  Moon = 1/6lb.  In fact, the weight of an object will vary slightly depending on where it is on earth (higher altitude less weight) Section 1.4
  • 29.  Time - the continuous, forward flowing of events  Time has only one direction  forward  Second (s) – the standard unit in both the metric and British systems  Originally 1/86,400 of a solar day  Now based on the vibration of the Cs133 atom (Atomic Clock) Section 1.4
  • 30. Originally defined as a fraction of the average solar day. Section 1.4
  • 31. Defined by the radiation frequency of the Cs133 atom Section 1.4
  • 32.  Uses acronym “mks system” from standard units of length, mass, and time – meter, kilogram, second  It is a decimal (base-10) system – this is much better than the British system  Administered by -- Bureau International des Poids et Mesures (BIPM) in Paris  International System of Units (SI)  Contains seven base units Section 1.4
  • 33.  The fundamental units are a choice of seven well- defined units which by convention are regarded as dimensionally independent: ◦ meter, m (length) ◦ kilogram, kg (mass) ◦ second, s (time) ◦ ampere, A (electrical current) ◦ kelvin, K (temperature) ◦ mole, mol (amount of a substance) ◦ candela, cd (luminous intensity) Section 1.5
  • 34.  Easy expression and conversion  Metric examples vs. British examples ◦ 1 kilometer = 1000 meters ◦ 1 mile = 5280 feet ◦ 1 meter = 100 centimeters ◦ 1 yard = 3 feet or 36 inches ◦ 1 liter = 1000 milliliters ◦ 1 quart = 32 ounces or 2 pints ◦ 1 gallon = 128 ounces Section 1.5
  • 35.  Mega, M – 106 – 1,000,000 times the base  Kilo, k – 103 – 1,000 times the base  Centi, c – 10-2 – 1/100th of the base  Milli, m – 10-3 – 1/1000th of the base  See Appendix 1 for complete listing Section 1.5
  • 36.  Liter – volume of liquid in a 0.1m (10 cm) cube (10cm x 10cm x 10cm = 1000 cm3 )  A liter of pure water has a mass of 1 kg or 1000 grams.  Therefore, 1 cubic cm (cc) of water has a mass of 1 gram.  By definition 1 liter = 1000 milliliters (ml)  So, 1 ml = 1 cc = 1 g of pure water.  1 ml = 1 cc for all liquids, but other liquids do not have a mass of 1 g Section 1.5
  • 37.  A Liter is slightly more than a quart. – 1 quart = .946 liter – 1 liter = 1.06 quart Section 1.5
  • 38.  (1 kg = 2.2046 lb on earth)  The amount of water in a 0.10m (10 cm) cube (0.10m3 ) Section 1.5
  • 39.  Metric ton -- mass of 1 cubic meter (1 m3 ) of water  1 m = 100 cm  (100cm)3 = 1,000,000 cm3  Remember that 1000 cm3 = liter  Therefore, there are 1000 liters in 1 m3 of water.  Each liter has a mass of 1 kg.  1 kg x 1000 = 1 metric ton Section 1.5
  • 40.  It is difficult to make all measurements with only the 7 fundamental units.  Derived units are therefore used, these are multiples/combinations of fundamental units.  We’ve already used derived units Volume  length3 , m3 , cm3  Area  length2 , m2 , ft2 , etc.  Speed  length/time, m/s, miles/hour, etc. Section 1.6
  • 41.  Density (r) = mass per unit volume  r=m/v [or m/length3 (since v = length3 )]  How “compact” a substance is  Typical Units used – g/cm3 , kg/m3  Al = 2.7 g/cm3 ,Fe = 7.8 g/cm3 , Au = 19.3 g/cm3  Average for solid earth = 5.5 g/cm3 Section 1.6
  • 42.  Hydrometer – a weighted glass bulb  The higher the hydrometer floats the greater the density of the liquid  Pure water = 1g/cm3  Seawater = 1.025 g/cm3  Urine = 1.015 to 1.030 g/cm3  Hydrometers are used to ‘test’ antifreeze in car radiators – actually measuring the density of the liquid Section 1.6
  • 43.  The denser the liquid the higher the hydrometer floats. Section 1.5
  • 44.  When a combination of units becomes complex and frequently used –  It is given a name ◦ newton (N) = kg x m/s2 ◦ joule (J) = kg x m2 /s2 ◦ watt (W) = kg x m2 /s3 Section 1.6
  • 45.  Relates one unit to another unit  Convert British to Metric (1in  cm)  Convert units within system (1kg  g)  We use “conversion factors” – many are listed on inside back cover of book  1 inch is equivalent to 2.54 centimeters  Therefore “1 in = 2.54 cm” is our conversion factor for inches & centimeters Section 1.6
  • 46.  Question: How many centimeters are there in 65 inches? Section 1.6 2.54 cm = 11 inch • Since 1 in = 2.54 cm  1 in 2.54 cm = 1• Or 2.54 cm 1 in • 65 in. x = 165 cm (the inches cancel out!!)
  • 47.  Step 1 - Choose/Use a Conversion Factor, generally can be looked up.  Step 2 – Arrange the Conversion Factor into the appropriate form, so that unwanted units cancel out.  Step 3 – Multiply or Divide to calculate answer.  Use common sense – anticipate answer! cm inch 54.2 1 or inch cm 1 54.2 for example Section 1.6
  • 48.  How fast in mi/h is 50 km/h?  Conversion Factor is 1km/h=0.621mi/h 50 km/h Starting Value hkm hmi /1 /621.0 x Conversion Factor = 31.05 mi/h Result Section 1.6
  • 49. Section 1.6 50 km/h Starting Value hkm hmi /1 /621.0 x Conversion Factor = 31.05 mi/h Result
  • 50.  Either Conversion Factor can be used:  1km/h = 0.621mi/h or 1mi/h = 1.61km/h  How fast in km/h is 50 mi/h? hmi hkm /621.0 /1 50 mi/h x = 80.5 km/h Starting Value Conversion Factor Same Result 50 mi/h x hmi hkm /1 /61.1 = 80.5 km/h Section 1.6
  • 51. Section 1.6 hmi hkm /621.0 /1 50 mi/h x = 80.5 km/h Starting Value Conversion Factor Same Result 50 mi/h x hmi hkm /1 /61.1 = 80.5 km/h
  • 52.  22 inches = ?? Meters  Inches  centimeters  meters 22 in Starting Value in cm 1 54.2 x Conv. Factor #1 in  cm cm m 100 1 x Conv. Factor #2 cm  m = 0.56 m Result Section 1.6
  • 53. Section 1.6 22 in Starting Value in cm 1 54.2 x Conv. Factor #1 in  cm cm m 100 1 x Conv. Factor #2 cm  m = 0.56 m Result
  • 54.  How would one express “First and 10” in meters?  Conversion Factor is 1yd = 0.914 m 10 yd Starting Value Section 1.6 x Conversion Factor yd m 1 914.0 = ?? Result
  • 55. 10 yd Starting Value Section 1.6 x Conversion Factor yd m 1 914.0 = 9.14 m Result
  • 56. Photo Source: Copyright © Bobby H. Bammel. All rights reserved. Section 1.6
  • 57. 4843 m x m ft 1 28.3 = 15,885 feet above SL (1775 feet higher than the top of Pikes Peak!) Section 1.6
  • 58.  Significant figures (“SF”) – a method of expressing measured numbers properly  A mathematical operation, such as multiplication, division, addition, or subtraction cannot give you more significant figures than you start with.  For example, 6.8 has two SF and 1.67 has three SF. Section 1.7
  • 59.  When we use hand calculators we may end up with results like: 6.8/1.67 = 4.0718563  Are all these numbers “significant?” Section 1.7
  • 60.  General Rule: Report only as many significant figures in the result as there are in the quantity with the least.  6.8 cm/1.67 cm = 4.1(round off 4.0718563) ◦ 6.8 is the limiting term with two SF  5.687 + 11.11 = 16.80 (round up 16.797) ◦ 11.11 is the limiting term with four SF Section 1.7
  • 61.  All non-zero digits are significant ◦ Both 23.4 and 234 have 3 SF  Zeros are significant if between two non-zero digits (‘captive’) – 20.05 has 4 SF, 407 has 3 SF  Zeros are not significant to the left of non-zero digits – used to locate a decimal point (leading zeros) – 0.0000035 has 2 SF  To the right of all non-zero digits (trailing zeros), must be determined from context – 45.0 has 3 SF but 4500 probably only has 2 SF Section 1.7
  • 62.  Exact Numbers – numbers of people, items, etc. are assumed to have an unlimited number of SF  In the process of determining the allowed number of significant figures, we must generally also ‘round off’ the numbers. Section 1.7
  • 63.  If the first digit to be dropped is less than 5, leave the preceding digit unchanged. ◦ Round off to 3 SF: 26.142  26.1  If the first digit to be dropped is 5 or greater, increase the preceding digit by one. ◦ Round off to 3 SF: 10.063  10.1 Section 1.7
  • 64.  Round off 0.0997 to two SF  0.0997  0.10  What about this? 5.0 x 356 = 1780  Round off 1780 to 2 SF  1780  1800 Section 1.7
  • 65.  Many numbers are very large or very small – it is more convenient to express them in ‘powers-of-10’ notation  1,000,000 = 10x10x10x10x10x10 = 106 Section 1.7 000,000,1 1 6 10 1 = = 0.000001 = 10-6
  • 66. Section 1.7 Standard HMCO copyright line
  • 67.  The distance to the sun can be expressed many ways: ◦ 93,000,000 miles ◦ 93 x 106 miles ◦ 9.3 x 107 miles ◦ 0.93 x 108 miles  All four are correct, but 9.3 x 107 miles is the preferred format. Section 1.7
  • 68.  The exponent, or power-of-10, is increased by one for every place the decimal point is shifted to the left. ◦ 360,000 = 3.6 x 105  The exponent, or power-of-10, is decreased by one for every place the decimal point is shifted to the right. ◦ 0.0694 = 6.94 x 10-2 Section 1.7 Rules for Scientific Notation
  • 69.  5.6256 x 0.0012 = 0.0067507   round to 2 SF  0.0067507 rounds to 0.0068   change to scientific notation  0.0068 = 6.8 x 10-3 Section 1.7 Example Rounding/Scientific Notation
  • 70.  0.0024/8.05 = 0.0002981   round to 2 SF  0.0002981 rounds to 0.00030   change to scientific notation  0.00030 = 3.0 x 10-4  **Note that the “trailing zero” is significant** Section 1.7 Example Rounding/Scientific Notation
  • 71.  Read the problem, and identify the chapter principle that applies to it. Write down the given quantities w/ units. Make a sketch.  Determine what is wanted – write it down.  Check the units, and make conversions if necessary.  Survey equations – use appropriate one.  Do the math, using appropriate units, round off, and adjust number of significant figures. Section 1.7 Problem Solving
  • 72.  The earth goes around the sun in a nearly circular orbit with a radius of 93 million miles. How many miles does Earth travel in making one revolution about the sun? Section 1.7 Problem Solving
  • 73.  The earth goes around the sun in a nearly circular orbit with a radius of 93 million miles. How many miles does Earth travel in making one revolution about the sun?  Determine what parts of the question are important and how to attack the problem. Section 1.7 Problem Example
  • 74.  The earth goes around the sun in a nearly circular orbit with a radius of 93 million miles. How many miles does Earth travel in making one revolution about the sun?  In order to solve this problem notice that you need an equation for a circular orbit (circumference)  The radius of 93,000,000 miles is given  Our answer also needs to be in miles (convenient!)  Equation: c = 2pr (p = 3.14159…) Section 1.7 Problem Example
  • 75.  Circumference = c = 2pr (p = 3.14159…)  c = 2 x 3.14159 x 93,000,000 miles  or  c = 2 x 3.14159 x 9.3 x 107 miles  c = 58.433574 x 107 miles  round off and adjust to two SF  c = 5.8 x 108 miles  5.8 x 108 miles = distance that the earth travels in one revolution around the sun Section 1.7 Problem Solving
  • 76.  Density: r = m/V Section 1.7