The percentage uncertainty in the diameter \(d\) is calculated as:

$$\% \text{ uncertainty in } d = \frac{0.01}{5.00} \times 100\% = 0.2\%$$

Since sphere volume is \(V = \frac{1}{6}\pi d^3\), the exponent for \(d\) is 3:

$$\% \text{ uncertainty in } V = 3 \times (\% \text{ uncertainty in } d) = 3 \times 0.2\% = 0.6\%$$

Converting megawatts to kilowatts:

$$3.7\text{ MW} = 3.7 \times 10^6\text{ W} = 3.7 \times 10^3\text{ kW}$$

SI prefix rules state that "kilo" must always be written in lowercase (\(\text{k}\)). Therefore, \(\text{kW}\) is correct while \(\text{KW}\) is incorrect.

Potential difference is a measurable physical quantity. In contrast, kelvin and minute are measurement units, and flavour is a non-quantifiable qualitative property.

Applying Ohm's Law \(V = I R\):

$$V = (15\,\mu\text{A}) \times (240\text{ k}\Omega) = (15 \times 10^{-6}\text{ A}) \times (240 \times 10^3\,\Omega) = 3.6\text{ V}$$

Systematic errors shift readings by a constant bias in one direction. Averaging repeated measurements reduces random errors, but has no effect on systematic errors.

Spring constant (\(k = F/x\)) is a physical quantity. Metre is a unit, percentage uncertainty is a measure of precision, and quark flavour is a property classification.

An average apple has a mass of roughly \(0.1\text{ kg}\). Lifting it through a height of \(1\text{ m}\) requires work equal to:

$$W = mgh \approx 0.1\text{ kg} \times 9.81\text{ m s}^{-2} \times 1\text{ m} \approx 1\text{ J} = 1 \times 10^0\text{ J}$$

Breaking down joules into SI base units gives \(\text{J} = \text{kg m}^2 \text{s}^{-2}\). Multiplying by seconds:

$$\text{J s} = (\text{kg m}^2 \text{s}^{-2}) \times \text{s} = \text{kg m}^2 \text{s}^{-1}$$

The central value \(10\text{ m s}^{-2}\) is close to the true value \(9.81\text{ m s}^{-2}\), making it accurate. However, the range \(\pm 2\text{ m s}^{-2}\) is broad, so it is not precise.

Electromotive force is energy per unit charge:

$$\text{e.m.f.} = \frac{\text{Energy}}{\text{Charge}} = \frac{\text{J}}{\text{C}} = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-1}$$

From \(g = \frac{2s}{t^2}\), combining percentage uncertainties yields:

$$\% \text{ uncertainty in } g = (\% \text{ uncertainty in } s) + 2 \times (\% \text{ uncertainty in } t) = 2\% + 2(3\%) = 8\%$$

A vector quantity has both magnitude and direction, whereas a scalar quantity possesses magnitude only.

Extension \(x = 250 - 225 = 25\text{ mm}\). Absolute uncertainties add up during subtraction: \(\Delta x = 2 + 2 = 4\text{ mm}\).

$$\% \text{ uncertainty} = \frac{4}{25} \times 100\% = 16\%$$

Momentum (\(\vec{p} = m\vec{v}\)) depends on velocity direction, making it a vector quantity. Speed, temperature, and Young modulus are scalar quantities.

Drag force follows \(F = \frac{1}{2} C_d \rho A v^2\). Expressing the drag coefficient gives \(C_d = \frac{2F}{v^2 \rho A}\), showing that \(n = 2\).

Every physical quantity measurement must explicitly include a numerical magnitude along with an appropriate unit.

Comparing prefix multipliers: \(\text{kilo} = 10^3\), \(\text{milli} = 10^{-3}\), and \(\text{pico} = 10^{-12}\). This order strictly decreases in size.

Weight is a gravitational force vector pointing toward the gravitational center. Mass, volume, and density have no directional property.

Mean current is \(\frac{3.04 + 3.08}{2} = 3.06\text{ A}\). Fluctuation uncertainty is \(\pm 0.02\text{ A}\). Accuracy uncertainty is \(1\% \times 3.06\text{ A} \approx \pm 0.03\text{ A}\).

$$\text{Total uncertainty} = 0.02 + 0.03 = \pm 0.05\text{ A}$$

Velocity \(v = 100\text{ km h}^{-1} \approx 27.8\text{ m s}^{-1}\).

$$E_k = \frac{1}{2} m v^2 = \frac{1}{2} (600,000) (27.8)^2 \approx 2.3 \times 10^8\text{ J}$$

The order of magnitude is \(10^8\text{ J}\).

Both period (time duration) and potential difference (voltage) are measurable physical quantities. In contrast, units such as kelvin or non-quantifiable properties cannot be listed as physical quantities.

The official SI base unit for thermodynamic temperature is the kelvin, represented by the symbol \(\text{K}\). Celsius is a derived scale, not the base SI unit.

A scalar quantity is completely defined by its numerical magnitude alone and has no spatial directional component associated with it.

Since Work \(W = F \times d\), 1 Joule equals \(1\text{ N m}\). Substituting this yields:

$$\text{J N}^{-1}\text{m}^{-1} = \frac{\text{J}}{\text{N m}} = \frac{\text{J}}{\text{J}} = 1$$

Because it cancels out completely to a pure number, it is not a physical measurement unit.

The recorded values are tightly clustered together within \(\pm 0.1\text{ kg}\), indicating high precision. However, their calculated mean (\(80.2\text{ kg}\)) deviates significantly from the true value (\(75.2\text{ kg}\)), making the measurement not accurate.

The prefix giga (\(\text{G}\)) corresponds to a factor of \(10^9\):

$$128\text{ GB} = 128 \times 10^9\text{ B} = 1.28 \times 10^{11}\text{ B}$$

High precision corresponds to a narrow, sharp distribution peak (Y and Z). High accuracy corresponds to a distribution centered precisely on the true value \(V\) (X and Z).

An expressed physical measurement fundamentally requires both a numerical magnitude and a corresponding unit of measurement.

Impulse is defined as \(\text{Force} \times \text{time}\), which has SI units of \(\text{N s}\). By the impulse-momentum theorem, change in momentum shares this identical unit structure.

Reading from the micrometer scale, the initial reading is \(2.84\text{ mm}\) and the second reading is \(14.68\text{ mm}\). The measured displacement is:

$$\Delta x = 14.68 - 2.84 = 11.84\text{ mm}$$

One complete wave cycle spans \(4\text{ cm}\) on the screen and represents a period of \(T = 500\,\mu\text{s}\). The time-base setting is calculated as:

$$\text{Time-base} = \frac{500\,\mu\text{s}}{4\text{ cm}} = 125\,\mu\text{s cm}^{-1}$$

Converting decijoules (\(\text{dJ}\)) to millijoules (\(\text{mJ}\)):

$$43\text{ dJ} = 4.3\text{ J} = 4300\text{ mJ} = 4.3 \times 10^3\text{ mJ}$$

In classical physics, every physical quantity is formally defined as the product of a numerical value (magnitude) and a unit.

The analogue scale indicates \(7.6\) divisions on a total scale of \(10\) divisions. For a full-scale deflection of \(250\text{ mA}\):

$$\text{Reading} = \frac{7.6}{10} \times 250\text{ mA} = 190\text{ mA}$$

Distance, static pressure, thermodynamic temperature, and time duration are all scalar quantities because none of them possess directional vector components.

The total stack height is \(2.7\text{ cm} = 27\text{ mm}\). Dividing by the total number of paper sheets (\(210\)):

$$\text{Thickness per sheet} = \frac{27\text{ mm}}{210} \approx 0.13\text{ mm}$$

A persistent scale under-reading of 1% is a systematic error. To correct the reading of \(305\text{ mm}\):

$$\text{True width} = \frac{305\text{ mm}}{0.99} \approx 305 \times 1.01 = 308\text{ mm}$$

The measured diameter is \(1.03 - (-0.05) = 1.08\text{ mm}\). When subtracting measured quantities, absolute uncertainties are added together:

$$\text{Total uncertainty} = 0.02 + 0.02 = 0.04\text{ mm}$$

Rearranging \(\mu = \frac{e\tau}{m}\) and inserting base units (\(e \rightarrow \text{A s}\), \(\tau \rightarrow \text{s}\), \(m \rightarrow \text{kg}\)):

$$\text{Base units of } \mu = \frac{(\text{A s}) \cdot \text{s}}{\text{kg}} = \text{A s}^2 \text{kg}^{-1}$$

Area \(A = \frac{\pi d^2}{4} = \frac{\pi (7.0)^2}{4} = 38.48\text{ mm}^2\). The percentage uncertainty in area is twice that of the diameter:

$$\% \Delta A = 2 \times \left(\frac{0.1}{7.0}\right) \times 100\% = 2.857\%$$ $$\Delta A = 38.48 \times 0.02857 \approx 1\text{ mm}^2$$

A typical racehorse has a mass of approximately \(500\text{ kg}\), yielding a weight of \(W = mg \approx 500 \times 9.81 \approx 5000\text{ N}\). Therefore, an estimate of \(6 \times 10^2\text{ N}\) (\(\approx 60\text{ kg}\)) is completely unrealistic.

By vector resolution, the total speed squared is the sum of horizontal and vertical component squares (\(v^2 = v_x^2 + v_y^2\)). Rearranging for horizontal velocity:

$$v_x^2 = v^2 - v_y^2 \implies v_x = (v^2 - v_y^2)^{1/2}$$

One wave cycle occupies \(4\text{ cm}\). With a time-base of \(2.00\text{ ms cm}^{-1}\), period \(T = 4\text{ cm} \times 2.00\text{ ms cm}^{-1} = 8.0\text{ ms} = 8.0 \times 10^{-3}\text{ s}\). The frequency is:

$$f = \frac{1}{T} = \frac{1}{8.0 \times 10^{-3}\text{ s}} = 125\text{ Hz}$$

Parallax variations (1) and friction from a sticking needle (3) cause unpredictable scatter (random errors). A persistent +5% scale offset (2) and zero offset (4) shift all readings constantly in one direction (systematic errors).

Resistance is defined as \(R = \frac{V}{I}\). Substituting base units for potential difference (\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\)) and current (\(\text{A}\)):

$$\text{Base units of } R = \frac{\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}}{\text{A}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}$$

From $g = \frac{4\pi^2 l}{T^2}$, percentage uncertainties combine as:

$$\% \Delta g = \% \Delta l + 2(\% \Delta T) = \frac{0.001}{0.420}\times 100\% + 2\left(\frac{0.1}{1.3}\times 100\%\right)$$ $$\% \Delta g = 0.238\% + 15.38\% \approx 16\%$$

For Screen 1: $T_1 = 3.0\text{ div} \times 0.02\text{ s div}^{-1} = 0.06\text{ s}$. For Screen 2: $T_2 = 1.5\text{ div} \times 0.04\text{ s div}^{-1} = 0.06\text{ s}$. Equal periods imply identical frequencies.

Resolving velocity vectors relative to North ($30^\circ$ East of North):

$$\text{Northerly component} = 75 \cos(30^\circ) = 65\text{ m s}^{-1}$$ $$\text{Easterly component} = 75 \sin(30^\circ) = 38\text{ m s}^{-1}$$

Substituting $\text{N} = \text{kg m s}^{-2}$ into the expression:

$$\text{N s}^2 \text{m}^{-1} = (\text{kg m s}^{-2}) \cdot \text{s}^2 \cdot \text{m}^{-1} = \text{kg}$$

The kilogram is the base unit of Mass.

For $R = \frac{V}{I}$, relative uncertainties sum directly:

$$\% \Delta R = \% \Delta V + \% \Delta I = \frac{0.1}{500.0}\times 100\% + \frac{0.01}{50.00}\times 100\% = 0.02\% + 0.02\% = 0.04\%$$

Reading 1 with wire = $2.58\text{ mm}$. Reading 2 (zero offset) = $+0.13\text{ mm}$. Corrected diameter:

$$\text{Diameter} = 2.58 - 0.13 = 2.45\text{ mm}$$

Initial reading is $0.13\text{ mA}$ and final reading is $0.47\text{ mA}$. Because zero errors affect all readings equally, the net current change is:

$$\Delta I = 0.47 - 0.13 = 0.34\text{ mA}$$

Sphere volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2.05)^3 = 36.09\text{ cm}^3$. Fractional uncertainty in volume is three times that of radius:

$$\% \Delta V = 3 \times \left(\frac{0.01}{2.05}\right) \times 100\% = 1.463\%$$ $$\Delta V = 36.09 \times 0.01463 = \pm 0.5\text{ cm}^3$$

A low-resistance shunt placed in parallel bypasses excess current to extend ammeter range, whereas a high-resistance multiplier placed in series drops excess voltage to extend voltmeter range.

Operating voltage $V = 12\text{ V}$ requires the $0\text{--}20\text{ V}$ range. Operating current $I = \frac{P}{V} = \frac{1.2\text{ W}}{12\text{ V}} = 0.1\text{ A}$, which requires the $0\text{--}0.5\text{ A}$ range.

When multiple independent observers obtain identical, systematically shifted measurement results, it indicates a systematic calibration error in the instrument rather than individual human random error.

From $g = \frac{4\pi^2 l}{T^2}$, combining percentage uncertainties yields:

$$\% \Delta g = \frac{0.2}{87.3}\times 100\% + 2\left(\frac{0.05}{1.9}\times 100\%\right) = 0.229\% + 5.263\% = 5.5\%$$

Force ($\vec{F} = m\vec{a}$) and momentum ($\vec{p} = m\vec{v}$) both require a defined direction for complete description, classifying them strictly as vector quantities.