Exercises — SI units — seven base units and all derived units

A base unit is one of the seven fundamental bricks; a derived unit is built from them by an equation.

• $\text{kelvin (K)}$ → base (temperature).
• $\text{mole (mol)}$ → base (amount of substance).
• $\text{candela (cd)}$ → base (luminous intensity).
• $\text{newton (N)}$ → derived, from $F=ma$, so ${\text{kg⋅m⋅s}}^{-2}$.
• $\text{watt (W)}$ → derived, from $P=E_W/t$ (power = work per time), so ${\text{kg⋅m}}^2{\text{⋅s}}^{-3}$.

From work $E_W=Fd$ (work = force along a distance): multiply the newton's bricks by a metre. Here $\text{J}$ is just the name we give the resulting brick-pattern. $[\text{J}]=[\text{N}]\cdot [\text{m}]=({\text{kg⋅m⋅s}}^{-2})\cdot \text{m}={\text{kg⋅m}}^2{\text{⋅s}}^{-2}.$

What: pressure is force spread over an area. Why substitute: the pascal has no independent standard — it is generated by this equation. $[p]=\frac{[F]}{[A]}=\frac{{\text{kg⋅m⋅s}}^{-2}}{{\text{m}}^2}={\text{kg⋅m}}^{-1}{\text{⋅s}}^{-2}.$ The $\text{m}$ in the numerator meets ${\text{m}}^2$ below: ${\text{m}}^{1-2}={\text{m}}^{-1}$.

Step 1 (acceleration): $a=\frac{\Delta v}{\Delta t}=\frac{30-0}{10}=3{\text{m/s}}^2$. Step 2 (force): $F=ma=1200\times 3=3600\text{N}$. Unit check: $[F]=\text{kg}\cdot \frac{\text{m}}{{\text{s}}^2}={\text{kg⋅m⋅s}}^{-2}=\text{N}$, as required. See Kinematics — velocity and acceleration for the $a=\Delta v/\Delta t$ step and Newton's Laws of Motion for $F=ma$.

Charge: $[Q]=[I][t]=\text{A⋅s}=\text{C}$. Voltage: energy per charge, so $[V]=\frac{[E_W]}{[Q]}=\frac{{\text{kg⋅m}}^2{\text{⋅s}}^{-2}}{\text{A⋅s}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}.$ The ${\text{s}}^{-2}$ over ${\text{s}}^{+1}$ gives ${\text{s}}^{-3}$; the $\text{A}$ in the denominator becomes ${\text{A}}^{-1}$.

What we do: divide charge bricks by voltage bricks. Why: capacitance is "charge stored per volt applied", so its recipe is exactly that ratio $Q/V$. $[\mathcal{C}]=\frac{[Q]}{[V]}=\frac{\text{A⋅s}}{{\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}}.$ Dividing means subtract exponents and flip the denominator's signs:

Step 1 — get the ohm's bricks. Why? We can't check $I^{2}R$ until we know what $R$ contributes, and $R=V/I$ tells us: divide the volt's bricks by an ampere. $[R]=\frac{[V]}{[A]}=\frac{{\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}}{\text{A}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-2}.$ (The lone $\text{A}$ below lowers the ampere power by one: ${\text{A}}^{-1-1}={\text{A}}^{-2}$.)

Step 2 — multiply by $I^2$. Why? The power dissipated in a resistor is $P=I^{2}R$, so to test its unit we substitute $[I]=\text{A}$ squared and the ohm bricks we just found. $[I^{2}R]={\text{A}}^2\cdot ({\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-2})={\text{kg⋅m}}^2{\text{⋅s}}^{-3}\underset{={\text{A}}^0=1}{\underbrace{{\text{A}}^{2-2}}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}=\text{W}.$ Why the amperes vanish: the current appears squared ($+2$) and the ohm carries ${\text{A}}^{-2}$, so they add to ${\text{A}}^0=1$, leaving exactly the watt.

We write magnetic flux as $\Phi$ (the Greek capital "phi", a common symbol for flux — a measure of how much magnetic field threads through a loop). Reading the bricks against the table, ${\text{kg⋅m}}^2{\text{⋅s}}^{-2}{\text{⋅A}}^{-1}$ is the weber (Wb), the unit of that flux $\Phi$. It comes from $\Phi =V\cdot t$ (flux = voltage × time), since $[V][t]=({\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1})\cdot \text{s}={\text{kg⋅m}}^2{\text{⋅s}}^{-2}{\text{⋅A}}^{-1}$, matching exactly.

Step 1 — isolate $G$. Why? The unit of $G$ is hidden until $G$ is alone on one side, so we rearrange the law: $G=\frac{F{r}^2}{m_1{m}_2}$. Since $[r]=\text{m}$ we have $[r^2]={\text{m}}^2$. Substitute base bricks: $[G]=\frac{[F][r^2]}{[m{]}^2}=\frac{({\text{kg⋅m⋅s}}^{-2})\cdot {\text{m}}^2}{{\text{kg}}^2}={\text{m}}^3{\text{⋅kg}}^{-1}{\text{⋅s}}^{-2}.$ Step 2 — density. $[\rho ]=\frac{\text{kg}}{{\text{m}}^3}={\text{kg⋅m}}^{-3}$. Step 3 — product $G\rho$. Why multiply? The expression asks for $G\rho$, so we combine bricks by adding exponents: $[G\rho ]=({\text{m}}^3{\text{⋅kg}}^{-1}{\text{⋅s}}^{-2})({\text{kg⋅m}}^{-3})=\underset{1}{\underbrace{{\text{m}}^{3-3}}}\underset{1}{\underbrace{{\text{kg}}^{-1+1}}}{\text{s}}^{-2}={\text{s}}^{-2}.$ Step 4 — square root. Why? The expression is $\sqrt{G\rho }$, and a square root halves every exponent: $[\sqrt{G\rho }]={\text{s}}^{-1}$ = a frequency (hertz). Beautiful: $\sqrt{G\rho }$ is an inverse-time — it sets a natural gravitational timescale.

Why a check and not a solve? We are not finding a number; we test whether the bricks on both sides match. If they don't, the equation is definitely wrong (see Dimensional Analysis). $\left[\frac{GM}{r}\right]=\frac{({\text{m}}^3{\text{⋅kg}}^{-1}{\text{⋅s}}^{-2})\cdot \text{kg}}{\text{m}}={\text{m}}^{3-1}{\text{⋅kg}}^{-1+1}{\text{⋅s}}^{-2}={\text{m}}^2{\text{⋅s}}^{-2}.$ The $2$ is dimensionless and drops. Take the square root: $\sqrt{{\text{m}}^2{\text{⋅s}}^{-2}}={\text{m⋅s}}^{-1}=\text{speed}.$ Consistent. (Note: consistency does not prove the factor $2$ — dimension checks can't catch pure numbers.)

Why this route? The figure builds the watt as newton ÷ second (from $P=E_W/t$). Here we build it a different way — force × velocity — and if physics is consistent the bricks must agree. Substitute ($[F]={\text{kg⋅m⋅s}}^{-2}$, $[v]={\text{m⋅s}}^{-1}$) and add exponents because we multiply: $[Fv]=({\text{kg⋅m⋅s}}^{-2})({\text{m⋅s}}^{-1})={\text{kg⋅m}}^{1+1}{\text{⋅s}}^{-2-1}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}=\text{W}.$ Same box as the figure's watt — both routes agree. See Work, Energy and Power.

Step 1 — isolate $B$. Why? Same reason as before: the field's unit only reveals itself once $B$ is alone, so $B=\frac{F}{qv}$. Step 2 — build the denominator's bricks, exponent by exponent. Why carefully? Stray $\text{s}$ powers hide here, so we write every base brick's exponent explicitly before cancelling. With $[q]=\text{A⋅s}$ and $[v]={\text{m⋅s}}^{-1}$:

What we do: divide Planck's bricks by charge's bricks. Why: $X$ is defined as the ratio $h/e$, so its unit-recipe is exactly $[h]/[e]$.

(a) $a^0=1$ (a pure number), so $[Y]=[m]\cdot [a^0]=\text{kg}\cdot 1=\text{kg}$. Just a mass. (b) Every exponent tells you how many copies of that brick you multiply in. Power $0$ = "zero copies of the brick" = the empty product = the dimensionless $1$. So $[\text{anything}{]}^0$ is dimensionless — it adds nothing to the recipe. Edge case: if the quantity itself is literally zero (like $a=0{\text{m⋅s}}^{-2}$), then $a^0=0^0$ is mathematically undefined. In physics we sidestep this: unit analysis treats $a$ as a symbol carrying bricks, never as the number $0$, so $[a{]}^0=1$ is always safe. The trap is only the numerical $0^0$, not the dimensional one. (c) The radian (from the opening definition) is another dimensionless quantity: $\text{m}/\text{m}=1$. Both $a^0$ and $\text{rad}$ land on the same "no-brick" pattern $1$ — a unit check cannot tell them apart, which is exactly why named-but-dimensionless units exist to remind humans what the ratio meant.

Every rung used the same move: write $[X]$, substitute the seven base bricks, add exponents when multiplying / subtract when dividing, and let dimensionless numbers fall away.