Recent content by tkfn

Well what i did first is find out the units of the constant 1147 in the original equation so the principle of dimensional homogeneity at both sides of the = sign is satisfied: \frac{kcal}{h·m^{2}·ºC}=\frac{kcal·cm^{0.4}·s^{0.6}}{h·m^{2.6}·ºC^{2}}·ºC·\frac{m^{0.6}}{cm^{0.4}·s^{0.6}} Then I...

We are asked to transform this equation: h=1147(1+0.05625T)\frac{V^{0.6}}{D^{0.4}} where the units of each variable h, T, V and D are: h->\frac{kcal}{h·m^{2}·ºC} T->ºC V->\frac{m}{s} D->cm into another equation where the units of each variable h, T, V and D now are...