Metric System
A physical quantity is something that is measured such as mass, length and time. Each of these units is defined by a number, value or its magnitude and a unit. Each physical quantity has its specific unit and these unit help us to analyze what the given or the asked quantity is.
These units are identified through different systems. Using these different systems in one solution may be confusing, thus, scientists around the world use one system as the commonly used system in dealing with measurements. This system of measurement is the International System of Units (SI).
A list of base quantities with the corresponding unit they are measured is given in the table below.
Quantity |
Unit |
Symbol |
Mass |
kilogram |
kg |
Length |
meter |
m |
Time |
second |
s |
Electric current |
ampere |
A |
Temperature |
kelvin |
K |
Amount of substance |
mole |
mol |
Luminous intensity |
candela |
cd |
Prefixes are used to define smaller or larger values or measurement of the same quantities. Below is the list of prefixes and their corresponding factor:
Prefix |
Symbol |
Multiplying Factor |
pico- |
p |
10^{-12} |
nano- |
n |
10^{-9} |
micro- |
µ |
10^{-6} |
milli- |
m |
10^{-3} |
centi- |
c |
10^{-2} |
deci- |
d |
10^{-1} |
kilo- |
k |
10^{3} |
mega- |
M |
10^{6} |
giga- |
G |
10^{9} |
tera- |
T |
10^{12} |
To convert units, the following must be followed:
- Identify the given unit and the unit in which the given unit must be converted.
- Detemine the conversion factor. The unit to be converted must be placed in the position(numerator or denominator) in which it can be cancelled out so that the only unit left is the desired unit.
Example 1.
The length of the pole is 15 meter. What is its length in decimeters?
Given: 15 meters
Desired unit: decimeter
Conversion Factor: 1 dm = 10^{-1} m
Solution: \(15\;m\;\times\;{1\;dm \over 10^{-1}m}=150\;dm\)
Example 2.
The speed of the race car is 140 km/h. Express the race car's speed in m/s.
Given: speed = 140 km/h
Conversion Factors: 1 km = 10^{3} m and 1 hr = 3,600 s
Solution:
\(140\;\frac{km}{h}\times {10^3\;m \over 1\;km}\times {1\;h \over 3,600\;s}=38.89\;m/s\)
Derived Units
Derived units from the term itself is derived from the base units. It is a combination of two or more base units. These base units are multiplied or divided by one another and not added or subtracted. The following are some commonly used derived units:
Quantity |
Unit |
Derived Unit |
velocity |
m/s | m; s^{-1} |
frequency |
herts (Hz) | 1/s or s^{-1} |
energy |
joule (J) | kg m^{2} s^{-2} |
force |
newton (N) | kg m s^{-2} |
electric charge |
coulomb (C) | A s |
acceleration |
m/s^{2} | m s^{-2} |
Example 1.
A 5-kg object is pushed then accelerates at 2 m/s^{2} . How much force is exerted on the object?
Solution:
\(F=ma=(5\;kg)(2\;m/s^2)=10\;kg\;m/s^2=10\;N\)