Units let us know what property is being measured, and they tell us the magnitude scale.
Metric units follow powers of ten. There are 100 cm in a meter, and 1000 mm in a meter. English units have less predictable relationships. There are 3 feet in a yard, and 12 inches in a foot, and 36 inches in a yard.
Metric Units
The metric system was designed to be a unified and rational system of measures. The metric system improves calculation, communication, and conversion. Amazingly, it has been adopted by almost every country on Earth.
Annoyingly, United States have chosen to not fully adopt the metric system, but the American sciences do use metric. We use metric units on this site.
The metric system defines all units from seven base units. The base units come from measuring physical constants. For example, the meter is defined as the distance light travels in (1 / 299 792 458) seconds.
Quantity | Name | Symbol |
---|---|---|
time | second | s |
length | meter | m |
mass | kilogram | kg |
current | ampere | A |
temperature | kelvin | K |
amount | mole | mol |
light intensity | candela | cd |
Derived units are combinations of these base units. For example, velocity is combination of length divided by time (m/s).
Quantity | Name | Symbols |
---|---|---|
area | meters squared | m² |
volume | meters cubed | m³ |
velocity | meters per seconds | m/s |
acceleration | meters per seconds squared | m/s² |
momentum | kilogram meters per seconds | kg m/s |
Some derived units get a special abbreviation normally written as a capital letter. For example, the unit of force is N, but it stands for kg m/s².
Quantity | Name | Abbreviation | Symbols |
---|---|---|---|
force | newton | N | kg m/s² |
energy | joule | J | kg m²/s² |
power | watt | W | kg m²/s³ |
frequency | hertz | Hz | 1/s |
volume | liter | L | 10⁻³ m³ |
Metric Prefixes
We use metric prefixes to indicate multiplication or division by powers of ten. For example we can replace 1000 with the letter "k".
$$ 1000 \, \mathrm{m} = 1 \, \mathrm{km} $$ $$ 22\,500 \, \mathrm{m} = 22.5 \, \mathrm{km} $$Name | Symbol | Power |
---|---|---|
giga | G | 109 |
mega | M | 106 |
kilo | k | 103 |
centi | c | 10-2 |
milli | m | 10-3 |
micro | μ | 10-6 |
nano | n | 10-9 |
Converting 10 km into meters means multiplying by 1000.
$$10 \, \mathrm{km} = 10\left({\color{DeepPink}1000}\right) \mathrm{m} = 10\,000 \, \mathrm{m}$$ $$10\,000 \, \mathrm{m} = 10\,({\color{DeepPink}1000}) \, \mathrm{m} = 10 \, \mathrm{km}$$Converting 10 ms into meters means multiplying by 0.001.
$$10 \, \mathrm{ms} = 10\left({\color{DodgerBlue}0.001 }\right) \mathrm{s} = 0.01\, \mathrm{s}$$ $$0.01 \, \mathrm{s} = 10\left({\color{DodgerBlue} 0.001 }\right) \mathrm{s} = 10\, \mathrm{ms}$$solution
$$ \mathrm{k} = 1000 $$ $$120\, \mathrm{km} = 120(1000)\, \mathrm{m} = 120\,000\, \mathrm{m}$$solution
$$ \mathrm{n} = 10^{-9} $$ $$450\,\mathrm{nm} = 450\left( 10^{-9}\right) \mathrm{m} = 0.000\,000\,45 \, \mathrm{m}$$In practice, conversions are just moving the decimal to the left for negative exponents and to the right for positive.
$$10 \, \mathrm{km} = 10.\overrightarrow{\undergroup{0}\undergroup{0}\undergroup{0}}{\color{DeepPink}.} \, \mathrm{m} = 10\,000 \, \mathrm{m}$$ $$10\,000 \, \mathrm{m} = 10{\color{DeepPink}.} \overleftarrow{\undergroup{0}\undergroup{0}\undergroup{0}}. \, \mathrm{m} = 10 \, \mathrm{km}$$$$10 \, \mathrm{ms} = 0{\color{DodgerBlue}.}\overleftarrow{\undergroup{0}\undergroup{1}\undergroup{0}}.0 \, \mathrm{s}= 0.01 \, \mathrm{s}$$ $$0.01 \, \mathrm{s} = 0.\overrightarrow{\undergroup{0}\undergroup{1}\undergroup{0}}{\color{DodgerBlue}.}0 \, \mathrm{s}= 10 \, \mathrm{ms}$$
solution
$$10 \, \mathrm{Mm} = 10.\overrightarrow{\undergroup{0}\undergroup{0}\undergroup{0}\undergroup{0}\undergroup{0}\undergroup{0}}{\color{DeepPink}.} \, \mathrm{m} = 10\,000\,000 \, \mathrm{m}$$solution
$$10 \, \mathrm{\mu s} = 0{\color{DodgerBlue}.}\overleftarrow{\undergroup{0}\undergroup{0}\undergroup{0}\undergroup{0}\undergroup{1}\undergroup{0}}.0\, \mathrm{s}= 0.000\,01 \, \mathrm{s}$$Nonmetric Units
Converting outside the metric system is more complex. Other unit systems don't always use powers of ten, so we can't simply move the decimal left or right. To convert we need to multiply by a conversion fraction.
- Build a conversion fraction.
- Put the old unit on the bottom of the fraction.
- Put the new unit on top of the fraction.
- Find out how two numbers are equal. Add numbers to the top and bottom of the fraction so that they are equal.
- Multiply the number you are converting by the conversion fraction. The old unit should cancel to leave just the new unit.
Let's convert 50 minutes into seconds.
$$ 50 \,\mathrm{min} \to \mathrm{s}$$In order to cancel minutes we want to build a fraction with minutes on top and the seconds on the bottom.
$$ 50 \,\mathrm{min} \left( \mathrm{\frac{s}{min}} \right)$$Our fraction can't change the actual value, so it must be equal to one. The top and bottom of the fraction must equal each other.
$$\text{{\color{DeepPink}1 minute} = \color{DodgerBlue}60 seconds}$$ $$ 50 \, \mathrm{min} \left( \frac{{\color{DodgerBlue}60\, \mathrm{s}} }{{\color{DeepPink}1\,\mathrm{ min}}} \right)$$ $$ 50 \, \cancel{\mathrm{min}} \left( \frac{60\, \mathrm{s}}{1 \,\cancel{\mathrm{min}}} \right)$$ $$ 50 \left( \frac{60\, \mathrm{s}}{1} \right)$$ $$3000\, \mathrm{s}$$solution
$$ 50 \,\mathrm{min} \to \mathrm{hour}$$ $$50\,\mathrm{\color{DeepPink}min} \left( \frac{1\,\mathrm{hour}}{60\,\mathrm{\color{DeepPink}min}} \right)$$ $$\frac{50}{60}\,\mathrm{hour}$$ $$0.8\overline{33}\,\mathrm{hour}$$(1 mile = 1.6 kilometers)
solution
$$26.2 \, \textcolor{DeepPink}{ \mathrm{mile}} \left( \frac{1.6 \, \mathrm{km}}{1 \,\textcolor{DeepPink}{ \mathrm{mile}}} \right) = 41.92\, \mathrm{km}$$solution
$$\mathrm{ft \to inches}$$ $$6\, \mathrm{ft} + 1\, \mathrm{in}$$ $$6\, \textcolor{DeepPink}{\mathrm{ft}}\left( \frac{12\,\mathrm{in}}{1\, \textcolor{DeepPink}{\mathrm{ft}}}\right) + 1\, \mathrm{in}$$ $$72\, \mathrm{in} + 1\, \mathrm{in}$$ $$73\, \mathrm{in}$$$$\mathrm{inches \to meters}$$ $$\text {a google search returns: } 1 \, \mathrm{m} = 39.37 \, \mathrm{in}$$ $$73\,\textcolor{DeepPink}{ \mathrm{in}} \left( \frac{1\, \mathrm{m}}{39.37 \, \textcolor{DeepPink}{ \mathrm{in}}}\right) $$ $$73 \left( \frac{1\, \mathrm{m}}{39.37}\right) $$ $$1.85\, \mathrm{m}$$
Space and Time
Our universe has three dimensions of space and one dimension of time. The three different spacial dimensions make communicating information about space ambiguous.
Information about the three dimensions of space needs a direction in addition to the magnitude. The pairing of direction and magnitude is called a vector.
If I say "bike 10 km to get to the store", you still can't find the store because you don't know the direction. I should say "bike 10 km west".
Scalars
A scalar is a variable that has a magnitude, but not a direction in space. Temperature is a scalar. You couldn't say it is 50°C to the left.
Variable | Magnitude |
---|---|
air pressure | 101.3 kPa |
temperature | 21° C |
price | $50 |
speed | 10 m/s |
distance | 3000 m |
Speed and distance seem like they could have a direction, but they are defined as only the magnitude of velocity and displacement vectors.
Vectors →
A vector is a variable that has a magnitude and direction in space. Velocity is a vector. You could say a velocity is 10 m/s in the west direction.
Variable | Magnitude | Direction |
---|---|---|
displacement | 10 m | west |
displacement | 5.5 m | up |
velocity | 20 m/s | 20° above the x-axis |
velocity | 3 m/s | 10 left |
acceleration | 9.8 m/s² | ↓ |
Think of a vector like the hypotenuse of a right triangle. This makes the other two sides of the triangle the components of the vector. Often the components are horizontal and vertical, but they don't have to be.
We can use the pythagorean theorem to solve for the magnitude of the sides.
$$a^{2} + b^{2} = c^{2}$$ Example: You walk 3 miles north and then 4 miles west. How far away are you from your starting location?solution
$$a^{2}+b^{2} = c^{2}$$ $$3^{2}+4^{2} = c^{2}$$ $$25 = c^{2}$$ $$5 \, \mathrm{miles} = c$$Vectors and Angles
We can use trig functions (SOH-CAH-TOA) to find how the vector components relate to the angle of the vector.
For a velocity vector the hypotenuse is v. The adjacent and opposite sides of the triangle are the x and y part of v.
For displacement the equations are the same
solution
$$ v_{y} = (v) \sin \theta $$ $$ v_{y} = (250 \, \mathrm{\tfrac{m}{s}}) \sin(14 \degree) $$ $$ v_{y} = 60 \mathrm{\tfrac{m}{s}} $$solution
Northeast is an angle of 45 degrees to the north.
$$ d_\mathrm{north} = d \cos \theta $$ $$ d_\mathrm{north} = (75 \, \mathrm{m}) \cos(45 \degree) $$ $$ d_\mathrm{north} = 53 \, \mathrm{m} $$