Exercise 5 Page 310

When we are working with real-world problems we need to consider units of measure. The process of operating with these units is called dimensional analysis. This can guide us to a certain point. For example, let's say that we are given the formula for the volume of a sphere. V = 4/3 π r^3Then we are asked to find the radius of a sphere with a volume of 300 m^3. For this problem, we solve for the radius and obtain a relation. r = sqrt(3/4V) If we are not sure if we isolated correctly, we can check by analyzing the units on both sides of the equation. These must be the of the same dimension. We know the radius has dimensions of length; in this case, m and the volume have dimensions of length cubed m^3, while the constant 34 is adimensional. r = sqrt(3/4V) → m ≠ sqrt(m^3) As their dimensions do not match, this equation is dimensionally incorrect and does not make sense. This suggests we did a mistake while isolating r.