Physical Quantities

From $E=hf$, $h=\dfrac Ef$.

Dimension of energy: $[E]=\text{ML}^2\text{T}^{-2}$; of frequency: $[f]=\text{T}^{-1}$. So

$[h]=\frac{\text{ML}^2\text{T}^{-2}}{\text{T}^{-1}}=\text{ML}^2\text{T}^{-1}$

SI unit: $\text{J s}$ (equivalently $\text{kg m}^2\text{s}^{-1}$), the same as angular momentum. Its value is $6.63\times10^{-34}$ J s.

(d) The value is precise but not accurate.

The true value of $g$ is about $9.8\ \text{m/s}^2$, far outside the quoted range $18$ to $22\ \text{m/s}^2$, so the result is not accurate. The uncertainty of $\pm2$ is small compared with 20 (about 10%), so the value is fairly precise.

(d) $[ML^2T^{-3}]$

Power $=\dfrac{\text{work}}{\text{time}}=\dfrac{[ML^2T^{-2}]}{[T]}=[ML^2T^{-3}]$.

(b) 0.200 m

The reading is 20 cm to the nearest millimetre, i.e. $20.0\ \text{cm} \pm 0.1\ \text{cm}$. Converting to metres gives $0.200\ \text{m}$, and the result must be written with three decimal places so that the last digit (the millimetre place) is retained and the precision of the measurement is not lost.

No. A dimensionally correct equation need not be physically correct. Dimensional analysis cannot detect dimensionless constants (such as $\tfrac12$ or $2\pi$), and it cannot distinguish between quantities that have the same dimensions.

Example: the equation $s=ut+at^2$ is dimensionally consistent, since each term has dimension $[L]$. But the correct relation is $s=ut+\tfrac12at^2$, so the first is wrong. Similarly, $K=mv^2$ is dimensionally consistent but incorrect, as the true kinetic energy is $\tfrac12mv^2$.

Dimensional correctness is only a necessary condition, not a sufficient one.