### Video Transcript

Consider the physical quantities π , π£, π, and π‘ with dimensions dimensions of π equals length, dimensions of π£ equals length times inverse time, dimensions of π equals length times inverse time squared, and dimensions of π‘ equals time. Determine whether each of the following equations is dimensionally consistent. Is π£ squared equals two ππ dimensionally consistent? Is π equals π£ times π‘ squared plus 0.5ππ‘ squared dimensionally consistent? Is π£ equals π divided by π‘ dimensionally consistent? Is π equals π£ divided by π‘ dimensionally consistent?

Letβs start with the first question and write out the dimensions of the terms in this equation. First, the dimensions of π£ squared. Weβre told that π£ has dimensions of length per time. We then square that term so it becomes length squared per inverse time squared.

We want to know if these dimensions are consistent with the dimensions of two times π times π . The number two, as with any pure number, has no dimensions; the dimensions of π are given as length per time squared; and the dimensions of π are length. Combining together the dimensions of π and π , we find they equal L squared per time squared. Therefore, the dimensions of this equation are consistent.

Now letβs look at the next equation. π is equal to π£ times π‘ squared plus 0.5ππ‘ squared. The dimensions of π are length L; the dimensions of π£ are length per time; the dimensions of π‘ squared are time squared, which simplifies this term to L times T.

Then we add a term with the dimensions of π times the dimensions of π‘ squared. The dimensions of π are length times inverse time squared, and thatβs multiplied by the dimensions of π‘ squared, which is time squared. This simplifies the final term in our equation to simply L. Because the first term on the right-hand side of the equation does not have dimensions of L, this relationship is not dimensionally consistent.

Now letβs look at the equation π£ is equal to π divided by π‘. The dimensions of π£ are length times inverse time. And we want to know if thatβs equal dimensionally to the dimensions of π , which are length, divided by the dimensions of T, which are time. One over time is equal to time to the negative first. So, dimensionally, this equation is consistent.

Lastly, letβs look at the equation π is equal to π£ divided by π‘. The dimensions of π are length times inverse time squared. And we want to know if these dimensions are consistent with the dimensions of π£ divided by π‘. The dimensions of π£ are given as length per unit time. And one over the dimensions of π‘, time, is equal to time to the negative one.

When we combine the dimensions on the right-hand side of our equation, we get L times T to the negative two. Therefore, dimensionally, this final equation is consistent.