Question

how do you use dimensional analysis to convert between different metric units?

Answer 1

ok lets try to explain this with basic physiscs: distance is measured in meters and time in seconds so when you divide those two you get speed which has unit of meters per second now lets put that in formulas:
\[s= distance [m] , t = time [s] \rightarrow \]
\[v = (speed/velocity) = s/t = distance/time [m/s = ms ^{-1}]\]
\[a = v/t = velocity/time = m/s/s = m/s ^{2} = ms ^{-2}\]
now when that is clear lets use little recalculations:
lets say you pass 10 m in 5 seconds so your speed/velocity is:
\[s=10m , t = 5 s \rightarrow v=s/t= 10m/5s = 2m/s =2ms ^{-1}\]
now lets put that in km per second and km per hour:
\[1 km = 1000m \rightarrow 1m = 0,001 km or 1m = 10^{-3}m\]
\[2 m/s = 2\times10^{-3} km/s = 0,002 km/s (km s ^{-1})\]
\[1 h (hour)=3600 s \rightarrow 1s = 3,6\times10^{-3}\]
\[v=2m/s \rightarrow v= 2 *(10^{-3}/3,6\times10^{-3})=2*3,6 = 7,2 km/h (kmh ^{-1})\]

Answer 2

i could do that with pressure if you like cause pressure is force on surface, force is mass times acceleration in units newton or N, and surface is in square meters, so pressure is equal to:
\[F = a \times m= kgms ^{-2} (or N), P (surface) = l \times d=m \times m = m ^{2}\]
\[p(pressure) = F/P = force/surface=N/m ^{2} (kgms ^{-2}/m ^{2}=kgm ^{-1}s ^{-2})\]