The Most Common Fundamental Units

Dimension Unit
Length Meter ( m )
Mass Kilogram ( kg )
Time Seconds ( s )
Temperature Kelvin ( K )

Standard Prefixes

Multiple Prefix

    \[10^{12}\]

Tera, T

    \[10^9\]

Giga, G

    \[10^6\]

Mega, M

    \[10^3\]

Kilo, K

    \[10^2\]

Hecto, k

    \[10^1\]

Deka, da

    \[10^{-1}\]

Deci, d

    \[10^{-2}\]

Cent, c

    \[10^{-3}\]

Milli, m

    \[10^{-6}\]

Micro, (mu)

    \[10^{-9}\]

Nano, n

    \[10^{-12}\]

Pico, p

Example: 1,000,000 Pa = 1 (times 10^6) MPa

Newton’s Second Law

    \[Force (N) = Mass (kg) (\times) Acceleration (m/(s^2))\]

    \[F=ma ((kg \times m/s^2)=N)\]

Sometimes, the questions indicate weight (W)

    \[W=mg((kg\times m/s^2)=N)\]

    \[Density (\rho) =\frac{mass(kg)}{volume(m^3)}\]

    \[Specific Volume (v) = \frac{v}{m}=\frac{1}{\rho}(\frac{m^3}{kg})\]

    \[Specific Weight(\gamma_s)=\rho (\frac{kg}{m^3})g(\frac{m}{s^2})=\rho g (\frac{N}{m^3})\]

Temperature Conversion

    \[T[\degree F]=[\degree C]\times \frac{9}{5}+32\]

    \[T[K]=[\degree C]+273.15\]

    \[T[\degree R]=([\degree C]+273.15)\times \frac{9}{5}\]

Pressure Measurement

    \[P_2=P_{atm}+\rho g h\]

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Screen Shot 2015-07-28 at 11.03.01 AM

Pressure Relation From Point A-B

    \[P_1+\rho_1 g(a+h) - \rho_2 gh - \rho_1 ga = P_2\]

Pressure Difference

    \[\Delta P=P_1-P_2 = (\rho_2-\rho_1)gh\]

Screen Shot 2015-07-28 at 11.07.14 AM
Formula Units
Newton’s Second Law

    \[Force (N) = Mass \times Acceleration\]

    \[F=ma\]

    \[(kg \times \frac{m}{s^2})=N\]

Weight

    \[W=mg\]

    \[(kg times frac{m}{s^2}=N)\]

Density

    \[\rho = \frac{mass(kg)}{volume(m^3)}\]

    \[\frac{kg}{m^3}\]

Specific Volume

    \[v=\frac{v}{m}=\frac{1}{\rho}\]

    \[\frac{m^3}{kg}\]

Specific Weight

    \[\gamma_s=\rho g\]

    \[\frac{N}{m^3}\]

Temperature Conversion

    \[T[\degree F]=[\degree C]\times \frac{9}{5}+32\]

    \[T[K]=[\degree C]+273.15\]

    \[T[\degree R] =([\degree C]+273.15)\times \frac{9}{5}\]

    \[1 bar = 10^5 Pa=0.1 MPa = 100 kPa\]

    \[1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars\]

    \[1 \frac{kgf}{cm^2}=9.807 \frac{N}{cm^2}=9.807 \times 10^4 \frac{N}{m^2}=9.807\times10^4 Pa = 0.9807 bar=0.9679 atm\]