It’s true only if \(a = b\).
It’s obviously false, unless \(b=0\).
It’s true (conditionally on the given supposition that \(a - b\) makes physical sense).
question id: dimension-brackets-1
\[ \newcommand{\dnorm}{\text{dnorm}} \newcommand{\pnorm}{\text{pnorm}} \newcommand{\recip}{\text{recip}} \]
Reading question 15.1 Give three examples of fundamental dimensions. Provide at least two units that are suitable for measuring each of the dimensions.
In this book, we denote the dimension of a quantity \(a\) as \([a]\). This is a special use of the square brackets and has nothing to do with arithmetic. (For computer programmers: the dimension interpretation of \([\,]\) is also unrelated to indexing arrays.)
Reading question 15.2
It’s true only if \(a = b\).
It’s obviously false, unless \(b=0\).
It’s true (conditionally on the given supposition that \(a - b\) makes physical sense).
question id: dimension-brackets-1
It’s the same as \([a - b]\).
It’s zero.
It is without meaning.
It’s the same as \([a]\).
question id: dimension-brackets-2
Never.
The right hand side is meaningless. You can’t do arithmetic on dimensions.
Sometimes, so long as \([a]\) and \([b]\) are different.
Always, regardless of what \([a]\) and \([b]\) are.
question id: dimension-brackets-3
It’s nonsense. Obviously \(17 \neq 1\).
Trick question. There’s no such thing as \([17]\).
All pure numbers are dimensionless, so both 17 and 1 have the same null dimension.
question id: dimension-brackets-4
No information is given about \([b]\).
It must be that \([b] = 1\).
\(b\) can have any dimension whatsoever.
question id: dimension-brackets-5
Reading question 15.3
To change the units of a quantity.
To change the dimension of a quantity.
To confuse students.
Quantities don’t have flavors, and 1 is not a flavor either.
question id: flavor-of-1-2
L
L2
Angle is another fundamental dimension, just as M, L, and T are fundamental dimensions.
Angles are dimensionless.
question id: flavor-of-1-3