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

Question #114319
Find the conversion factors of:
1. nm/min to μm/h
2. g/cn³ to kg/m³
1
2020-05-11T20:04:21-0400

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."

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