Answer to Question #114319 in Mechanics | Relativity for Mateus Michael Ndinomwene

Question

Find the conversion factors of:
1. nm/min to μm/h
2. g/cn³ to kg/m³

Answer 1

1) We know that $1\,\mu m = 10^{-6}\, m, \; \; 1\,nm = 10^{-9}\,m,$ so $1\,nm = \dfrac{10^{-9}}{10^{-6}}\,\mu m = 10^{-3}\, \mu m.$

Next, we know that 1 hour is equal to 60 min. Therefore,

$1\dfrac{nm}{min} = 1 \dfrac{nm}{\dfrac{1}{60}h} = 60 \dfrac{nm}{h}.$ Then, $60 \dfrac{nm}{h} = 60 \dfrac{10^{-3}\,\mu m}{h} = 6\cdot10^{-2}\,\dfrac{\mu m}{h}.$

If we have, for example, $v = 5 \, nm/min$, it'll be equal to $5\cdot 6\cdot10^{-2}\,\mu m/h = 0.3\, \mu m/h.$

2) We know that $1\,g = 10^{-3}\, kg$ and $1\,cm^{3} = (10^{-2}\,m)^3 = 10^{-6}\,m^3.$ Therefore,

$1\dfrac{g}{cm^3} = 1\dfrac{10^{-3}\,kg}{10^{-6}\,m^3} = 10^3\dfrac{kg}{m^3}.$

If we have, for example, $\rho = 4\,g/cm^3,$ then $\rho = 4\cdot10^3\,kg/m^3 = 4000\,kg/m^3.$