# E Y N K

```EVERYTHING YOU NEED TO KNOW FOR PHYSICS
(EXCEPT THE PHYSICS)
Part I: Percent Error
Percent error is a way of measuring how good your lab data is. It compares the value you have
measured with the accepted value—the value you'd see in a textbook or the value you think you
ought to be getting.
If you have a precise, highly accurate, widely-accepted textbook value (such as the mass of the
electron, the universal constant of gravity, etc.), then you assume that the textbook value is
correct, and any possible deviation is due to your error. You compare your differences purely
against the textbook ("actual") value,
%=
v actual − v measured
× 100
v actual
If you do not have a textbook value, if you are working with a calculated, theoretical value (such
as the speed of your cart), then you cannot assume that the calculated, theoretical value is 100%
accurate. Differences between the measured value and the calculated value are not necessarily
signs of an egregious boo-boo in the laboratory: you must address the possibility that your
calculated value is somewhat off. You compare your differences not against the calculated
value, but against the average of the measured and calculated values.
%=
v theoretical − v measured
× 100
a
a=
v theoretical + v measured
2
In standard scientific investigations, a percent error of under 5% is acceptable. In a high school
laboratory, conditions are not as ideal as we might like them, and we won't quibble with anything
under 10%. Anything over 20% is definitely iffy, and anything over 30% is begging for a re-do.
(Anyone who beats my personal record of -428.7% error will get a round of applause and should
probably seek assistance before re-doing the experiment.)
Speaking of which, percent error is an absolute value: it is never negative! No one cares whether
you're 67% too high or 67% too low: either way, you're 67% off, your data is lousy, and you
have to re-do the lab.
Always express percent error as a percent (4.3%) never as a decimal (0.043).
At no point in the year should I ever have to answer the question, "What's the equation for
percent error?" You will be expected to have the procedures memorized or look them up.
-1-
Uncertainty
consistent your data is. All data is science comes with some sort of margin of error. The
narrower the margin of error, the more precise your data is.
The graph above charts the comparative brightness of a certain star as it appears in the sky. The
vertical bars represent the uncertainty. As you can see, when the hazy clouds rolled in at 1:30
a.m., the uncertainty became enormous: the computer measured all sorts of numbers not all of
which were accurate or relevant. Uncertainty is often graphed in a box-and-whisker plot like
this.
The mathematical formula for uncertainty is half the difference between the highest and lowest
vhighest − vlowest
u=
2
It is always positive, and it is always expressed with a ±, as shown below.
Example: July 2011 daily high temperatures
Mean T = 89° Fahrenheit
Highest recorded T = high of 99°
99 − 75
u=
Lowest recorded T = high of 75°
2
T = 89 ± 12
When we say that this was the 3rd-hottest summer in Chicago history, this is what we mean!
-2-
1. The acceleration of gravity is 9.81 m/s2 at sea level. You measure it to be 9.63 m/s2. What is
Ans. 1.8%
Ans. GREAT!
2. According to your calculations, the coefficient of friction (traction) for your sneakers is 0.43.
You measure it to be 0.51. What is your percent error?
Ans. 17%
Ans. Not so hot!
3. According to your calculations, the current in your circuit should be 0.38 A. You measure it
to be 0.56 A. What is your percent error?
What should your next step be?
4. Calculate the percent error.
The speed of sound in dry air = 343 m/s (according to the book)
Measured frequency (ν) = 241 s-1
Measured wavelength (λ) = 1.31 m
Calculated speed = s = λν = 316 m/s
What should your next step be?
-3-
Part II: The Metric System
Virtually all of our work will be done in the metric system. (Get used to it.) More helpfully,
we’ll be working in fundamental units and other units derived directly from those units. These
units, for the most part, are not selected for convenience (minutes, hours, etc.) or for familiarity
(pounds, Calories, etc.) but for the way they make the math work out.
Metric Units for Physics:
Fundamental units
Derived units
Measures
Length
Time
Mass
Temperature
Electric current
Unit
meter
second
kilogram
Kelvin
Ampere
Symbol
m
s
kg
K
A
Equivalent To
Volume
Speed
Acceleration
Force
Energy
Power
Electric charge
meter3
meters per second
meters per second2
Newton
Joule
Watt
Coulomb
m3
m/s
m/s2
N
J
W
C
m3
m/s
m/s2
kg⋅m/s2
kg⋅m2/s2
kg⋅m2/s3
A⋅s
Please note that the standard units of mass is NOT the gram, it is the KILOGRAM! This is
different than in chemistry, and it is done because a Newton needs to be 1 kg⋅m/s2, not 1 g⋅m/s2.
You all have seen metric prefixes before. Here they are, just for old times’ sake:
My Table of Metric Prefixes:
femto
pico
nano
micro
milli
centi
deci
f
p
n
µ
m
c
d
10-15
10-12
10-9
10-6
10-3
10-2
10-1
deka
hecto
kilo
mega
giga
tera
peta
da
h
k
M
G
T
P
101
102
103
106
109
1012
1015
-4-
Most common
Most common
You must have these prefixes memorized by no later than Wednesday September 2nd! You will
be using them on a weekly basis, especially the ones I have marked as most common. On tests
and quizzes, you will be on your own—I will not tell you what kilo stands for!
More importantly, you will need to be ready to convert from one unit into another. In general,
you do this by replacing the little letter with its corresponding power of 10.
Example: Corresponding powers of 10
If…
1 ns = 1 × 10-9 s
1 µs = 1 × 10-6 s
1 ms = 1 × 10-3 s
1 ks = 1 × 103 s
1 Ms = 1 × 106 s
Then…
1s=1×
1s=1×
1s=1×
1s=1×
1s=1×
109 ns
106 µs
103 ms
10-3 ks
10-6 Ms
Example: Converting 136.8 µA to A
To convert a prefix-unit into a base unit, multiply by the associated power of 10.
1 µA = 1 × 10-6 A
136.8 µA = 136.8 (×10-6 A)
…see how you replace the µ with ×10-6?
136.8 × 10-6 A = 1.368 × 10-4 A
Check: did we want our answer in pure Amperes? Yes! So we’re done.
136.8 µ A = 1.368 × 10-4 A.
Example: Converting 324 meters to kilometers
To convert a base unit into a prefix-unit, multiply by the opposite power of 10.
If 1 km = 1 × 103 m, then 1 m = 1 × 10-3 km.
324 m = 324 (× 10-3 km)
324 × 10-3 km = 0.324 km
This is the same as saying:
324
m = 0.324 km
10 3
Example: Converting 901 cg to kg
To convert a prefix-unit into another prefix-unit, first convert to a base unit and then proceed.
If 1 cg = 1 × 10-2 g, then 1 g = 1 × 102 cg.
901 cg = 901 (× 10-2 g)
901 × 10-2 g = 9.01 g
9.01 g = 9.01 (× 10-3 kg)
9.01 × 10-3 kg = 0.00901 kg…
…But an answer of 9.01 × 10-3 kg is much preferred! Fewer zeroes to lose track of, you see.
-5-
1. Express the following in meters:
81.1 cm
Ans. 0.811 m
603,428 µm
0.39 mm
2. Express the following in kilograms:
34.0 g
Ans. 0.0340 kg
652.3 mg
Ans. 0.6523 g = 6.523 × 10-4 kg
0.18 Mg
71,300,000 ng
2 Tg
-6-
Part III: Scientific Notation and Significant Figures
Scientific notation you’ve all seen before. It’s when something long and ungainly and full of
zeroes gets shortened using powers of 10. Here is a quick review to brush the dust off.
To write a BIG number in scientific notation: 306000000
Draw a decimal point where it IS NOW: 306000000.
MOVE the decimal RIGHT, counting how many spaces you must move it:
3.06000000 = 3.06 _ 108
To write a small number in scientific notation: 0.00000000098
Draw a decimal point where it SHOULD BE: 0.0000000009.8
MOVE the decimal LEFT, counting how many spaces you must move it:
0.0000000009.8 = 9.8_ 10-10
Why? Because all numbers written in scientific notation have the proper number of significant
figures. Scientific notation is the easiest way to keep track of significant figures—it is the only
way to express numbers like 400 with two significant figures.* And significant figures in physics
can make the difference between an answer that looks correct and an answer that looks incorrect.
So let’s talk significant figures.
RULE 1: ALL COUNTING NUMBERS 1-9 ARE SIGNIFICANT.
This means that 3.1416 has five significant figures and 6.11 has three significant figures. Count
them!
RULE 2: ALL PLACE-HOLDING ZEROES ARE NOT SIGNIFICANT.
• Any trailing zero that is NOT followed by a decimal point is defined as placeholding.
Examples:
The zeroes in 300 and 21,000 and 7,500 and 8,890 are all place-holders.
This means that 300 has one significant figure, 21,000 has two, and 8,890 has three. The
place-holders don’t count!
Counterexamples:
The zeroes in 300. and 21,004 and 7,500.0 and 8,890.0000 are all significant. The
presence of a decimal point or a final integer makes all the difference!
This means that 300. has three significant figures, 21,004 has five, and 8,890.0000 has
eight.
A Theory: any zero after the decimal point is significant.
This is true EXCEPT IN CASE OF…
How, you ask? 400 with two significant figures is 4.0 × 102. This is markedly different from 4 × 102 and 4.00 ×
102.
*
-7-
• Any zero in a pure decimal that comes in a long starting line of zeroes is defined as
place-holding.
Examples:
The zeroes in 0.03 and 0.00096 and 0.00000131 are all place-holders.
This means that 0.03 has one significant figure, 0.00096 has two, and 0.00000131 has
three. The place-holders don’t count!
Counterexamples:
The trailing zeroes in 0.030 and 0.000960 and 0.00130100 are all significant. Any zero
that comes after a decimal point AND after an integer counts!
This means that 0.030 has two sig figs, 0.000960 has three, and 0.00130100 has six.
When taking data in the lab, be sure to take at least two significant digits, if not three.
Always specify whether your zeroes are trailing or not—write 50. rather than 50, if you can
say for sure that your number is more precise than “fiftyish.” I will always try to give you at
least three significant figures.
There are different rules for significant figures math depending on whether you are
uniformity in data. Multiply/divide math is designed for scaling up or scaling down. Learn
and/or memorize these steps!
• Perform the addition or subtraction.
• Take the number with the fewest decimal places, and truncate your answer to that number of
decimal places.
Example: 30.6 kg + 29.9991 kg
30.6 kg: one decimal place
29.9991 kg: two decimal places
30.6 kg + 29.9991 kg = 60.5991 kg ≈ 60.6 kg
Final answer: 60.6 kg with one decimal place
To Multiply Or Divide:
• Perform the multiplication or division.
• Take the number with the fewest significant figures, and truncate your answer to that
number of significant figures.
Example: 5.412 N _ 30. m
30. m: two significant figures
5.412 N: four significant figures
(5.412 N)(30. m) = 162.36 N•m ≈ 160 N•m
Final answer: 160 Newton-meters with only two significant figures
Moral of the story: hope you’re given more than two significant figures!
-8-
1. How many significant figures in…
3,100 N
Ans. 2
501.0 N
930.0 N
Ans. 4
930 N
0.00375 N
0.000375 N
0.0003750 N
Ans. 4
2. A tank is 13.0 m long, 0.14 m deep, and 3.7776 m wide. Calculate the volume of fluid in the
tank.
Ans. 6.9 m3
3. A torque is a twisting force (as opposed to a pulled force). If one balance beam experiences
four separate torques…
τ1 = 17.23 N×m
τ2 = -13.966 N×m
τ3 = -13.3400 N×m
τ4 = 10 N×m
…what is the total torque on the balance beam? (Hint: “total” means add.)
Extra credit: what does this mean about the balance beam?
4. You calculate the temperature of Hell to be 104.1º C. Your measured value is 1.0 × 102 º C.
-9-
Part IV: Unit Analysis
Unit analysis is the science of units. It has two primary uses:
I) The answer to the question, "Mrs. Eliaser, what units is the answer supposed to be in?"
II) A foolproof method of checking your equations without an answer key
To try some unit analysis, take the equation you're working on, and replace every letter with its
associated unit. Then look and see if the statement you have written is true! If it is true, then
you have a correct equation; if it is false, then you have some checkin' to do!
Unit analysis only makes sense when you write your units as FRACTIONS! Get into the
habit, and remember that PER means DIVIDE!
   
mph
Example: F = ma
m
m/h
h
m
h
m = 14 kg
a = 2.0 m/s2
F = ma = (14)(2.0) = 28 …
But 28 whatses?
Well, if you'd have put the units next to the numbers, you'd see 28 whatses! Watch!
ma = (14 kg)(2.0 m/s2) …now group your units together like this…
ma = (14)(2.0)(kg)(m/s2) …or better still, like this…
ma = (14)(2.0)(kg)( sm )
2
Then just multiply the units together and see!
F = 28 kg ⋅ sm
2
The unit of F—the kilogram-meter-per-second-squared—is named after Sir Isaac Newton, such
2
that 1 Newton = 1 kg ⋅ m/s .
Example: F =
Gm1m2
d2
You know that F is measured in Newtons (N), m is measured in kilograms (kg),
and d is measured in meters (m). What are the units of G?
Simply replace each letter with its component unit, then rearrange. Don’t forget little things
like squares!
F=
Gm1 m2
d2
N=
G(kg)(kg)
N=
(m) 2
Now multiply both sides by m2: N ⋅ m 2 = G ⋅ kg 2
And divide by kg2:
N ⋅ m2
=G
kg 2
-10-
G ⋅ kg 2
m2
2
The units for G are N ⋅ m
kg 2
.
Example: distance = speed × time
d = st
s = 35 mph
t = 15 seconds
d = st = (35 mph)(15 sec) = 525 miles
After 15 seconds, you have gone 525 miles. Right? Um?
d = distance = miles
s = speed = miles per hour
t = time = seconds
distance = (speed)(time)
miles = (mph)(seconds)
⎛ mi ⎞
(sec )
mi =
⎝ hr ⎠
Now, think. Is this a true statement? Of course not!
An example of a true statement would be:
mi =
⎛ mi ⎞
(hr )
⎝ hr ⎠
Conclusion: to use the equation d = st with the data above, you must first convert t in to
hours. Otherwise, the equation will be wrong!
A Note Regarding Fundamental Units:
We already mentioned fundamental units in the context of the metric system. They are defined
by standard weights and measures in France: there is a standard “kilogram” of a platinumiridium alloy, etc. Other units, including kph, Newtons, degrees Celsius, etc. are derived relative
to these fundamental units. In unit analysis, it is usually preferable to use the fundamental units,
since more things will cancel out that way. More importantly, all of your calculations must
be done in fundamental units or in ways that directly relate to them. (That is to say, metersper-second instead of miles-per-hour.) Otherwise, your answers will be inaccurate!
-11-
1. Using the equation p = mv, where m is measured in kilograms (kg) and v is measured in
meters per second (m/s), find the units of p.
Ans. p is measured in kg·m/s
Extra credit: what’s p in physics?
2. Using the equation F =
kQ1Q2
, where F is measured in Newtons (N), Q is measured in
d2
Coulombs (C), and d is measured in meters (m), find the units of k.
Extra credit: what’s the name of this equation and what topic is it describing?
3. Area = length × width. Find the basic metric unit for area using unit analysis. (Careful!
This is not the unit you are used to seeing in math!)
Ans. A is measured in m2
force
to find the basic metric units of pressure. (Careful! This
area
is not the unit you are used to seeing in chemistry! Need the metric unit for force? I
recommend page 3.)
4. Use the equation pressure =
-12-
```