Physical Quantities And Units All subtopics

a numerical magnitude / value; a unit

a physical quantity requires both a magnitude and a unit; without a unit, the value has no defined physical meaning / size cannot be compared to a standard

approximately 70 kg (accept 50–100 kg)

assume bathtub dimensions of approximately 1.5 m × 0.6 m × 0.4 m; volume ≈ 0.36 m³; accept any value between 0.1 m³ and 0.5 m³ with a stated assumption

the estimate is not reasonable; a typical apple has a mass of approximately 0.1 kg, so the student’s value is roughly ten times too large

Finding the volume $$V=25\times 10\times 1.5$$ $$V=375{\text{ m}}^{3}$$ Finding the mass $$m=\rho V=1000\times 375$$ $$m=3.75\times {10}^{5}\text{ kg}$$ The mass of water in the pool is approximately 3.75 × 10⁵ kg (2 s.f.).

estimation allows physicists to check whether a calculated answer is physically reasonable; it identifies large arithmetic or unit errors quickly; it is essential when exact data are unavailable; strategy — assume a typical breathing rate of approximately 15 breaths per minute; convert one day to 24 × 60 = 1440 minutes; multiply rate by time to give 15 × 1440 ≈ 2.2 × 10⁴ breaths per day; state assumption that breathing rate is constant over the day

kilogram (kg), metre (m), second (s), ampere (A); kelvin (K)

kilogram; second / kelvin (any two correct)

The kilogram is defined as a base unit by the SI system; the prefix is part of the name only and does not mean it is derived from a smaller unit (the gram).

$$W=Fs$$ $$[W]=({\text{kg m s}}^{-2})(\text{m})$$ $$[W]={\text{kg m}}^{2}{\text{ s}}^{-2}$$

$$P=\frac{W}{t}$$ Units of work = kg m² s⁻² $$[P]=\frac{{\text{kg m}}^{2}{\text{ s}}^{-2}}{\text{s}}$$ $$[P]={\text{kg m}}^{2}{\text{ s}}^{-3}$$ This gives kg m² s⁻³ as required.

$$R=\frac{V}{I}$$ Volt = kg m² s⁻³ A⁻¹ $$[R]=\frac{{\text{kg m}}^{2}{\text{ s}}^{-3}{\text{ A}}^{-1}}{\text{A}}$$ $$[R]={\text{kg m}}^{2}{\text{ s}}^{-3}{\text{ A}}^{-2}$$

An equation in which all terms have the same SI base units.

Left side: energy has base units kg m² s⁻² Right side: $$[m{v}^{2}]=(\text{kg})({\text{m s}}^{-1}{)}^{2}$$ $$[m{v}^{2}]={\text{kg m}}^{2}{\text{ s}}^{-2}$$ Both sides have identical base units, so the equation is homogeneous.

Left side: F = kg m s⁻² Right side: $$\left[\frac{mv}{{t}^{2}}\right]=\frac{(\text{kg})({\text{m s}}^{-1})}{{\text{s}}^{2}}$$ $$={\text{kg m s}}^{-3}$$ The base units differ, so the equation is not homogeneous.

Homogeneity only checks that base units match; it cannot detect missing or wrong dimensionless constants such as ½ or π; an equation may be homogeneous but still numerically wrong.

nano = 10⁻⁹; micro = 10⁻⁶; mega = 10⁶; giga = 10⁹

(a) 4.5 × 10⁹ Hz; (b) 2.5 × 10⁻⁷ m; (c) 6.2 × 10⁻⁶ F

$$470 \mu \text{F}=470\times {10}^{-6}\text{ F}$$ $$=4.70\times {10}^{-4}\text{ F}$$