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Unformatted text preview: Dimensions and Units Every expression you write down that tries to describe something about the real world has to be dimensionally consistent. After all, what does the expression 5 mi hr + 3 ft mean? Or an equation such as 6 sec- 2 km = 4 kg? Or the statement that the distance from here to downtown Miami is 8? Answer: Nothing! (Well, maybe in the last case there is a context, and you know that the speaker meant eight miles , but often you don’t know this, and it means there’s a mistake.) You can figure out correct units for something just by solving a little algebra. In the equation x = vt , if x = 50 meters and t = 2 seconds, you can solve for the units of v by straight algebra: v = x t = 50 meters 2 seconds = 25 meters second . If, a few lines later, you see an expression such as x = vt 2 / 2, where x , v , and t are supposed to mean the same thing as before, you can immediately see that this makes no sense. The units would be x = 1 2 v t 2 = ⇒ meter = meter second · second 2 , or meter = meter · second , and that’s nonsense. Notice the the factor of 1/2 didn’t play any role in this; it has no units, so (for this purpose) you can ignore it. In the equation, p + ρv 2 / 2 + ρgh = C , I tell you some of the units: ρ : kg / m 3 , v : m / s , h : m , What are all the rest? (You don’t even need to know what all the terms in this expression mean. You’ll encounter it in a much later chapter.) All the terms must have the same units, so just write them down: p + kg m 3 · m s 2 + kg m 3 · g · m = C. p , and g , and C don’t have any units specified, so multiply all the units to get p + kg m · s 2 + kg m 2 · g + C. This means that p and C both must have units of kg (m s 2 ). I can solve for the units of g by g = kg (ms 2 ) kg m 2 = kg · m 2 kg · m s 2 = m s 2 ....
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This note was uploaded on 01/08/2012 for the course PHYSICS 205 taught by Professor Galeazzi during the Fall '11 term at University of Miami.
- Fall '11