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How do physicists describe quantities precisely using units, scalars and vectors, and how do they handle measurement uncertainty?

Basic physics: SI base and derived units, homogeneity of equations, scalars and vectors, resolving and adding vectors, and the treatment of measurement uncertainty.

A focused answer to the Eduqas A-Level Physics Component 1 basic physics content, covering SI base and derived units, checking the homogeneity of equations by units, distinguishing scalars from vectors, resolving and adding vectors, and the treatment of measurement uncertainty.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Eduqas wants you to use the seven SI base units and build derived units from them, check whether an equation is homogeneous (balanced) in its units, distinguish scalar from vector quantities, resolve a vector into perpendicular components and add vectors, and express a measured value with a sensible uncertainty.

The answer

SI base and derived units

A common Eduqas task is to find the base units of an unfamiliar quantity by rearranging the defining equation and substituting the base units of each term. Prefixes scale units in powers of ten: pico (101210^{-12}), nano (10910^{-9}), micro (10610^{-6}), milli (10310^{-3}), kilo (10310^{3}), mega (10610^{6}) and giga (10910^{9}).

Homogeneity of equations

Scalars and vectors

Resolving and adding vectors

For two perpendicular vectors this reduces to Pythagoras for the magnitude and a single inverse-tangent for the direction. Resolving is the standard route into inclined-plane, projectile and equilibrium problems later in the course.

Measurement uncertainty

Examples in context

Dimensional checking catches algebra slips before they cost marks: if a derived expression for a period came out with units of metres, you have made an error. Engineers carry uncertainties through every calculation so they can quote a tolerance, for example the load a cable can safely bear. GPS positioning resolves a satellite signal into components to locate a receiver, and surveyors combine vector displacements to map terrain.

Try this

Q1. State the SI base units of the joule. [2 marks]

  • Cue. From W=FsW = Fs and F=maF = ma, 1 J=1 kgm2s21\ \text{J} = 1\ \text{kg}\,\text{m}^2\,\text{s}^{-2}.

Q2. A force of 12 N12\ \text{N} acts at 3030^\circ above the horizontal. Find its horizontal and vertical components. [2 marks]

  • Cue. Fx=12cos30=10.4 NF_x = 12\cos 30^\circ = 10.4\ \text{N}; Fy=12sin30=6.0 NF_y = 12\sin 30^\circ = 6.0\ \text{N}.

Q3. A length is measured as (5.0±0.1) cm(5.0 \pm 0.1)\ \text{cm}. State its percentage uncertainty. [1 mark]

  • Cue. 0.15.0×100=2%\dfrac{0.1}{5.0} \times 100 = 2\%.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20193 marksThe drag force on a sphere falling through a fluid is given by F=6πηrvF = 6\pi\eta r v, where rr is the radius and vv the speed. Use base units to determine the base units of the viscosity η\eta.
Show worked answer →

Rearrange for η\eta: η=F6πrv\eta = \dfrac{F}{6\pi r v}, so the units are [F][r][v]\dfrac{[F]}{[r][v]}.

Force has base units kgms2\text{kg}\,\text{m}\,\text{s}^{-2}, radius m\text{m}, speed ms1\text{m}\,\text{s}^{-1}.

So [η]=kgms2m×ms1=kgm1s1[\eta] = \dfrac{\text{kg}\,\text{m}\,\text{s}^{-2}}{\text{m} \times \text{m}\,\text{s}^{-1}} = \text{kg}\,\text{m}^{-1}\,\text{s}^{-1}.

Markers reward rearranging for η\eta, substituting the base units of force, radius and speed, and the correct answer kgm1s1\text{kg}\,\text{m}^{-1}\,\text{s}^{-1}.

Eduqas 20214 marksTwo forces of 6.0 N6.0\ \text{N} and 8.0 N8.0\ \text{N} act on a point at right angles to each other. Calculate the magnitude and direction of the resultant force.
Show worked answer →

The forces are perpendicular, so combine them with Pythagoras: F=6.02+8.02=36+64=100=10 NF = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \text{N}.

The direction relative to the 8.0 N8.0\ \text{N} force: θ=tan1 ⁣(6.08.0)=tan1(0.75)=37\theta = \tan^{-1}\!\left(\dfrac{6.0}{8.0}\right) = \tan^{-1}(0.75) = 37^\circ.

Markers reward using Pythagoras for perpendicular vectors, the resultant 10 N10\ \text{N}, and an angle of 3737^\circ from the larger force (or 5353^\circ from the smaller, if clearly stated).

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