PHYSICS 151 – Notes for Online Lecture 1.1
Mathematical Tools (units, significant figures, order of magnitude estimation, and
strategies for problem solving)
Units -- Whenever we measure a quantity, that measurement is reported as a number
and a unit of measurement. We say such a quantity has dimensions.
Units are a necessity – they are part of any answer and the answer is wrong without
them.
We will preferentially use metric units, although some of the problems in the book use
English units. Conversions are given in the inside front cover of your book.
SI BASE UNITS (Metric System)
Quantity
Length
Mass
Time
Metric
meter (m)
kilogram (kg)
seconds
American
foot (ft)
slug
seconds
Conversion
1 ft = .305 m
1 slug = 14.6 kg
There are other base units that we will get to later in the course. We will also use many
derived units that are combinations of base units such as the unit for energy called a
joule which is a kg-m2/s2.
Dimensional Analysis – Any valid physics formula must be dimensionally consistent
meaning that each term must have the same units.
v = v0 + at
[ L] = [ L] + [ L] T
[ ]
[T ] [T ] [T ]2
[ L] = [ L] + [ L]
[T ] [T ] [T ]
You Try
It!
Is this equation dimensionally consistent?
x = x0 + at
Metric Prefixes - We often use prefixes to simplify the notation. You’re already used
to using prefixes – we use then in talking about ‘millions’ ‘trillions’, etc.
Peta
tera
giga
mega
kilo
centi
milli
micro
nano
pico
femto
P
T
G
M
k
c
m
µ
n
p
f
1015
1012
109
106
103
10-2
10-3
10-6
10-9
10-12
10-15
Can you express 0.024 in terms of a convenient metric prefix?
You Try
It!
Unit Conversion
For example, there are 60 minutes in one hour. If the answer to a problem were 1.65
hours, and I wanted the answer in minutes, you could do the following:
60 minutes = 1 hour
60 minutes 1 hour
=
=1
1 hour
1 hour
60 minutes
=1
1 hour
This means that the quantity 60 minutes/1 hour equals 1.
⎛ 60 minutes ⎞
Answer =1.65 hour ⎜
⎟ = 99 minutes
⎝ 1 hour ⎠
Note that the unwanted units must cancel. There is a ‘hour’ in the numerator of the
answer and an ‘hour’ in the denominator of the conversion factor, so they cancel, which
leaves only the desired minutes.
Note that these factors are reversible.
60 minutes
=1
1 hour
1 hour
=1
60 minutes
so the same factor will take you from minutes to hours.
EXAMPLE: The Eiffel Tower is 301 m high. What is its height in feet?
We can look up the conversion from meters to feet. 1 meter is 3.281 feet.
1 m = 3.281 ft
3.281 ft
1=
1m
⎛ 3.281 ft ⎞
(301 m) ⎜
⎟ = 988 ft
⎝ 1m ⎠
EXAMPLE: Kangaroos have been clocked at speeds of 65 km/hr. What is their speed
in mi/h?
1 mi = 1.609 km
⎛ 65 km ⎞ ⎛ mi ⎞
⎜
⎟⎜
⎟ = 40 mi h
⎝ h ⎠ ⎝ 1.61 km ⎠
EXAMPLE: A field measures 20 km by 30 km. What is the area in m2?
A = (20 km) x (30 km) = 600 km2
2
⎡ km 2 m 2 ⎤
⎛ 1000 m ⎞
8
2
⎟⎟ = 6x10 8 ⎢
600 km = 600 km ⎜⎜
⎥ = 6x10 m
2
1
km
km
⎠
⎝
⎣
⎦
2
2
This example illustrates why it is so important for you to include units when doing your
calculations. If you accidentally use the wrong conversion factor, you should be able to
catch yourself at the end when the units don’t work out correctly.
EXAMPLE : How old are you in seconds?
Solution:
⎛ 365 days⎞ ⎛ 24 hr ⎞ ⎛ 60 min ⎞ ⎛ 60 s ⎞
age in years ⎜
⎟⎜
⎟
⎟⎜
⎟⎜
⎝ 1 yr ⎠ ⎝ 1 day ⎠ ⎝ 1 hr ⎠ ⎝ 1 min ⎠
⎛
s⎞
age in years × ⎜ 3.2 x10 7 ⎟
yr ⎠
⎝
Age in
years
18
19
20
21
22
Age in
seconds
5.8 x 108
6.1 x 108
6.4 x 108
6.7 x 108
7.1 x 108
You Try
It!
A tile store sells tile at the rate of $2.69 per square foot. How much does the
tile cost per square meter?
Measurement and Significant Figures
No measurement is exact. Although we have atomic clocks that are highly
accurate, the accuracy of the watch on your wrist is probably good enough for you to
make it to class on time. The accuracy you need depends on what you’re going to use
the measurements for.
Let’s return to the example of measuring the lengths of the metal rod. The ruler
in question has as its smallest markings a tenth of an inch. You measure the length and
find that it falls about halfway 4.1 and 4.2 cm. You estimate that the length is 4.15 cm,
but the 0.05 in is a guess, so you would report that the length of the rod as 4.15 ± 0.05
mm. The smallest marking on the measuring device represents the precision of your
measurements. The accuracy of a measurement can be determined by repeating it more
than once.
Indicate whether the arrows are accurate and precise or not.
1)
2)
3)
4)
The accuracy of a measurement is reflected in the way the number is written.
When a number is reported, assume that the number of digits reported is the number
known with any certainty. The uncertainty is generally assumed to be one or two units
of the last digits. When counting the number of significant figures:
• All digits 1 through 9 count as significant figures
• Zeroes to the left of all of the other digits are not significant
• Zeroes between digits are significant
• Zeroes to the right of all other digits are significant if after the decimal point and
may or may not be significant if before the decimal point.
For example
Number
1.2
3.61
19.61
0.017
10.25
0.1020
80
Number of Significant
Figures
2
3
4
2
Possible Range of the Real
Measurement
1.1 - 1.3
3.60 –3.62
19.60 - 19.62
.0016 - .0018
Sometimes, the number of significant digits can be unclear. For example, if we write
80
it is not clear whether the zero is significant or not. Is the measurement between 70 and
90 or between 79 and 81?
If you write 80.0, there are 3 significant figures, because zeroes to the right of the
decimal point are significant
Exponential notation removes this ambiguity. We write all of the significant figures out
front and then show what power of ten the significant figures should be multiplied by.
For example (assuming that there is one unit of uncertainty in the last significant figure:
8 x 101 - means 8 (±1) x 101 = measurement is between 70 and 90
1
1
8.0 x 10 means 8.0 (± 0.1) x 10 = measurement is between 79 and 81
Rules for manipulating significant figures
• Addition or subtraction: keep the place of the digit which is the same as the least
significant place of the numbers you are adding/subtracting.
• Multiplication or division: keep the same number of digits as the multiplicand
with the least number of significant figures.
EXAMPLE: A room is measured to be 5.5 feet wide and 6.75 feet long. What are the
area and perimeter of the room?
A = width × length = 5.5 feet × 6.75 feet = 37.125 feet2.
Do we know this to 5 significant figures? Nope.
5.5 feet = 2 s.f.
6.75 feet = 3 s.f.
We keep the same number of significant figures as the multiplicand with the smallest
number of significant figures, so we can only use 2 s.f.
A = width × length = 5.5 feet × 6.75 feet = 37.125 feet2 = 37 feet2.
P = 2(l + w) = 2 (5.5 + 6.75) = 24.50 ft = 24.5 ft
Geometrical Shapes
You Try
It!
Calculate the volume of a long cylinder that has a radius of r = 4.2 cm and a
length of 26.52 cm. After making the calculation convert your volume to cubic inches.
Estimation is a useful technique for getting an approximate answer and for
determining whether the answer you get is reasonable. Estimation is useful when you
don’t have – or need – exact numbers. For instance, you might be painting a room and
all you need to know is how many gallons of paint you need. Since you can’t buy a
quarter of a gallon of paint, you don’t need to know precisely how many tenths of
gallons are needed.
Physicists like to use the phrase “Order of magnitude estimate”. This just means
that the number is correct to within a factor of ten. Here are three examples of
estimation:
EXAMPLE: You are working for a radio station. The general manager wants to do a
promotional stunt: if the Huskers go undefeated, she wants to fill Memorial Stadium
with Oreo cookies. How much would it cost to do this?
Solution: First, estimate the length, width and height of Memorial Stadium. We’ll
approximate Memorial Stadium as a box. The field is 100 yds. long, plus about 20 yd.
for each end zone, and some extra. Let’s say about 150 yd.
L = 1.5 x 102 yards of where they may were
The width is about 100 yards (40 yards for the field, plus sidelines)
W = 1.00 x 102 yards
The height is about 20 yards.
H = 2.0 x 101 yards.
So the volume of Memorial Stadium is:
V = L × W × H =1.5 x 102 yd × 1.0 x 102 yd × 2.0 x 101 yd
= (1.5 × 1.0 ×2.0) x 10(2+2+1) (yd ×yd×yd)
= 3.0 x 105 yd3
Oreo cookies are about 2” in diameter and ½” thick.
1 ft ⎞ 1 ⎛ 1 yd ⎞ 1
yd
⎟=
⎟ = ft⎜
⎝ 12 in ⎠ 12 ⎝ 3 ft ⎠ 36
r = 1” = 1 in⎛⎜
h=
1 ⎛ 1 ft ⎞ 1 ⎛ 1 yd ⎞ 1
yd
in ⎜
ft⎜
⎟=
⎟=
2 ⎝ 12 in ⎠ 24 ⎝ 3 ft ⎠ 72
2
1
1
3
(yd × yd × yd)
Voreo = πr h = 3⎛⎜ yd ⎞⎟ ⎛⎜ yd ⎞⎟ =
⎝ 36 ⎠ ⎝ 72 ⎠ 36 × 36 × 72
1
1
yd 3 =
yd 3
12 × 36 × 72
10 × 40 × 70
2
1
yd 3
1x10 × 4 x101 × 7 x101
1
yd 3
28 x10 3
1
yd 3
4
2.8 x10
1
VOreo = x10 − 4 yd 3
3
1
Now: how many cookies fit in the stadium?
Vstadium 3.0x10 5 yd 3
= 1
# cookies =
3
−4
VOreo
3 x10 yd
=
3.0
1
3
x10 5− ( −4 )
= 9x10 9
= 1x1010 cookies
About 10 billion Oreos could fit inside the stadium. If you figure an average bag has
about 100 Oreos for $3.00, this stunt would cost.
$3.00
⎛
⎞
8
1x1010 cookies⎜
⎟ = 3x10 dollars!
2
⎝ 1x10 cookies⎠
EXAMPLE: What is the total weight of all the people in Lincoln?
Lincoln has approximately 200,000 people. Given the distribution of men, women and
children, we might estimate that the average weight of a person is 130 lbs.
Total weight = 130
lbs
x 200,000people
person
⎡ lbs
⎤
people⎥
⎣ person
⎦
= 1.3x10 2 × 2.0x10 5 ⎢
. × 2.0) x10 ( 2 +5) [ lbs]
= (13
= 2.6x10 ( 7) [ lbs] – or, about 26 million pounds.
EXAMPLE: Approximately what fraction of the area of the United States is covered by
automobiles?
We can approximate the Area of the US as a rectangle about 3000 miles wide
and 1500 miles tall.
AreaUS = 4,500,000 mi2
We can estimate the Area of a car as a rectangle 20 feet long and 8 feet
wide for an area of 160 m2.
2
⎛ 1 mi ⎞
2
−5
Area = 160 m 2 ⎜
⎟ = 6.2 × 10 mi
⎝ 1609 m ⎠
And assume 1 car for every other person of the 300,000,000 people in the US
for a total area for cars of:
Areacars = ( 6.2 × 10−5 mi 2 ) (150, 000, 000 ) = 9, 270 mi 2
Thus, the fraction of the area covered by cars is:
A r e a ca rs
9, 270 m i2
=
≈ 0 .0 0 2
A r e aU S
4, 500, 000 m i2
You Try
It!
How many times does a person’s heart beat in their lifetime?
So far we have applied this technique to problems resembling games, like the number of
gumballs in the gumball machine. However, it is much more than that as it can be
applied to realistic physical systems. For example, lets look at collisions in astronomy.
Our solar system is one of approximately 100,000,000,000 in the Milky Way Galaxy.
Our Milky Way Galaxy is one of approximately 100,000,000,000 in the Universe. Do
stars in these galaxies every collide? Do galaxies every collide?
One way to look at this problem is to calculate the ratio of the size of a star to its
separation from the nearest star to it. This is a simplification of true problem which
would need to involve other variables such as the velocities of the stars.
Star Ratio =
Typical Star Separation
= 1.4 ×109 m
Typical Star Diameter
Galaxy Ratio =
⎛ 9.46 ×1015 m ⎞
⎟
1ly
⎝
⎠ = 2.9 ×107
9
1.4 ×10 m
( 4.3ly ) ⎜
Typical Galaxy Separation 0.66 Mpc
=
= 22
Typical Galaxy Diameter
30kpc
Thus, this calculation suggests that Galaxies do collide and astronomers can actually
take pictures of this happening.