Measure physical quantities is a foundational skill for science laboratory technologists working in Kenya. Accurate measurement underpins reliable experimental results, quality control, and compliance with regulatory standards in laboratories across hospitals, universities, and industrial research facilities. This chapter focuses on understanding the distinction between basic and derived physical quantities, essential for precise data collection and interpretation in physics-related laboratory techniques.

1.1 Basic and derived physical quantities

Basic and derived physical quantities form the language of measurement in physics. Basic quantities are fundamental and independent, while derived quantities result from combinations of basic quantities according to physical laws. Mastery of these concepts enables laboratory technologists to correctly interpret measurements and convert between units during experiments and data analysis.

1.1.1 Basic physical quantities and their SI units

Basic physical quantities are the fundamental measurable properties that cannot be broken down into simpler quantities. The International System of Units (SI) defines seven basic quantities, each with a unique base unit. These units form the basis for all scientific measurements in Kenyan laboratories, ensuring standardization and reproducibility.

The seven SI basic physical quantities are:

• Length (L): The measure of distance between two points, with the SI unit metre (m). Length measurement is critical in calibrating laboratory apparatus and setting up experiments.

• Mass (M): The amount of matter in an object, measured in kilograms (kg). Mass measurement is essential for preparing chemical solutions and reagents.

• Time (T): The duration of events, measured in seconds (s). Accurate time measurement is necessary for kinetics studies and timed reactions.

• Electric current (I): The flow of electric charge, measured in amperes (A). This is important in experiments involving electrical circuits or sensors.

• Temperature (Θ): The degree of hotness or coldness, measured in kelvin (K). Temperature control and measurement influence reaction rates and material properties.

• Amount of substance (N): The number of elementary entities, measured in moles (mol). Used extensively in chemical quantification and stoichiometry.

• Luminous intensity (J): The perceived power of light, measured in candelas (cd). Relevant in photometry and optical experiments.

Worked Examples

Example 1: A laboratory technician measures the length of a metal rod as 0.75 m. Express this length in centimetres.

Given: Length \(L = 0.75\, m\)

Formula: \(L_{cm} = L_{m} \times 100\)

Substitution: \(L_{cm} = 0.75 \times 100\)

Calculation: \(L_{cm} = 75\, cm\)

Answer: 75 cm

Example 2: A chemical sample has a mass of 250 grams. Convert this mass into kilograms.

Given: Mass \(m = 250\, g\)

Formula: \(m_{kg} = \frac{m_{g}}{1000}\)

Substitution: \(m_{kg} = \frac{250}{1000}\)

Calculation: \(m_{kg} = 0.25\, kg\)

Answer: 0.25 kg

Example 3: A stopwatch records a time interval of 3 minutes 45 seconds. Convert this to seconds.

Given: Time \(t = 3\, min\, 45\, s\)

Formula: \(t_{s} = (minutes \times 60) + seconds\)

Substitution: \(t_{s} = (3 \times 60) + 45\)

Calculation: \(t_{s} = 180 + 45 = 225\, s\)

Answer: 225 seconds

Example 4: A laboratory experiment requires a temperature of 300 K. Convert this temperature to degrees Celsius.

Given: Temperature \(T = 300\, K\)

Formula: \(T_{^\circ C} = T_{K} - 273\)

Substitution: \(T_{^\circ C} = 300 - 273\)

Calculation: \(T_{^\circ C} = 27^\circ C\)

Answer: 27 degrees Celsius

Example 5: An electric current of 5 amperes flows through a circuit for 10 seconds. How much electric charge has passed?

Given: Current \(I = 5\, A\), Time \(t = 10\, s\)

Formula: \(Q = I \times t\)

Substitution: \(Q = 5 \times 10\)

Calculation: \(Q = 50\, C\)

Answer: 50 coulombs

1.1.2 Derived physical quantities and their expressions

Derived physical quantities result from algebraic combinations of basic quantities through multiplication or division, often reflecting physical laws. Common derived quantities include velocity, acceleration, force, and energy. Understanding their formulae and units is vital for laboratory technologists to correctly calculate and interpret experimental data.

Derived quantities are expressed as functions of basic quantities, for example:

• Velocity (v): Rate of change of displacement with time, \(v = \frac{L}{T}\), units metres per second (m/s).

• Acceleration (a): Rate of change of velocity with time, \(a = \frac{v}{T} = \frac{L}{T^2}\), units metres per second squared (m/s²).

• Force (F): Product of mass and acceleration, \(F = M \times a\), units newtons (N).

• Pressure (P): Force per unit area, \(P = \frac{F}{A}\), units pascals (Pa).

• Energy (E): Product of force and displacement, \(E = F \times L\), units joules (J).

Worked Examples

Example 1: A particle moves 100 metres in 20 seconds. Calculate its velocity.

Given: Displacement \(L = 100\, m\), Time \(T = 20\, s\)

Formula: \(v = \frac{L}{T}\)

Substitution: \(v = \frac{100}{20}\)

Calculation: \(v = 5\, m/s\)

Answer: 5 m/s

Example 2: A car accelerates from rest to 30 m/s in 10 seconds. Find its acceleration.

Given: Initial velocity \(u = 0\, m/s\), Final velocity \(v = 30\, m/s\), Time \(t = 10\, s\)

Formula: \(a = \frac{v, u}{t}\)

Substitution: \(a = \frac{30 - 0}{10}\)

Calculation: \(a = 3\, m/s^2\)

Answer: 3 m/s²

Example 3: A mass of 2 kg is accelerated at 4 m/s². Find the force applied.

Given: Mass \(M = 2\, kg\), Acceleration \(a = 4\, m/s^2\)

Formula: \(F = M \times a\)

Substitution: \(F = 2 \times 4\)

Calculation: \(F = 8\, N\)

Answer: 8 newtons

Example 4: A force of 10 N is applied over an area of 2 m². Calculate the pressure exerted.

Given: Force \(F = 10\, N\), Area \(A = 2\, m^2\)

Formula: \(P = \frac{F}{A}\)

Substitution: \(P = \frac{10}{2}\)

Calculation: \(P = 5\, Pa\)

Answer: 5 pascals

Example 5: A force of 15 N moves an object 3 m. Calculate the work done (energy transferred).

Given: Force \(F = 15\, N\), Displacement \(L = 3\, m\)

Formula: \(E = F \times L\)

Substitution: \(E = 15 \times 3\)

Calculation: \(E = 45\, J\)

Answer: 45 joules

1.1.3 SI unit system and unit conversions in laboratory measurements

The SI unit system is the internationally accepted standard for scientific measurements, used universally in Kenyan laboratories to ensure consistency and accuracy. Laboratory technologists must be proficient in converting between SI units and other common units, particularly when interpreting data from equipment calibrated in alternative systems or when reporting results.

Key unit conversions include:

• Length: metres to centimetres, millimetres, kilometres

• Mass: kilograms to grams, milligrams

• Time: seconds to minutes, hours

• Volume: cubic metres to litres, millilitres

• Temperature: kelvin to degrees Celsius

Mastery of conversion factors and dimensional analysis is essential for error-free data handling.

Worked Examples

Example 1: Convert 5000 millimetres to metres.

Given: Length \(L = 5000\, mm\)

Formula: \(L_{m} = \frac{L_{mm}}{1000}\)

Substitution: \(L_{m} = \frac{5000}{1000}\)

Calculation: \(L_{m} = 5\, m\)

Answer: 5 metres

Example 2: Convert 3.5 kilograms to grams.

Given: Mass \(m = 3.5\, kg\)

Formula: \(m_{g} = m_{kg} \times 1000\)

Substitution: \(m_{g} = 3.5 \times 1000\)

Calculation: \(m_{g} = 3500\, g\)

Answer: 3500 grams

Example 3: Convert 2 hours 15 minutes to seconds.

Given: Time \(t = 2\, hr\, 15\, min\)

Formula: \(t_{s} = (hours \times 3600) + (minutes \times 60)\)

Substitution: \(t_{s} = (2 \times 3600) + (15 \times 60)\)

Calculation: \(t_{s} = 7200 + 900 = 8100\, s\)

Answer: 8100 seconds

Example 4: Convert 2500 millilitres to cubic metres.

Given: Volume \(V = 2500\, ml\)

Formula: \(V_{m^3} = \frac{V_{ml}}{1,000,000}\)

Substitution: \(V_{m^3} = \frac{2500}{1,000,000}\)

Calculation: \(V_{m^3} = 0.0025\, m^3\)

Answer: 0.0025 cubic metres

Example 5: Convert 25°C to kelvin.

Given: Temperature \(T = 25^\circ C\)

Formula: \(T_{K} = T_{^\circ C} + 273\)

Substitution: \(T_{K} = 25 + 273\)

Calculation: \(T_{K} = 298\, K\)

Answer: 298 kelvin

1.1.4 Dimensional analysis and its application in verifying physical equations

Dimensional analysis is a technique used to verify the correctness of physical equations by comparing the dimensions of terms on both sides. It ensures that equations are dimensionally consistent, a necessary condition for physical validity. Laboratory technologists use dimensional analysis to check formulas used in experiments and calculations, reducing errors and improving data reliability.

Each physical quantity can be expressed in terms of fundamental dimensions: mass \([M]\), length \([L]\), time \([T]\), electric current \([I]\), temperature \([\Theta]\), amount of substance \([N]\), and luminous intensity \([J]\).

Worked Examples

Example 1: Verify the dimensional consistency of velocity \(v = \frac{L}{T}\).

Given: Length dimension = \([L]\), Time dimension = \([T]\)

Left side dimension: \([v]\)

Right side dimension: \(\frac{[L]}{[T]} = [L][T]^{-1}\)

Since both sides represent velocity dimension \([L][T]^{-1}\), the equation is dimensionally consistent.

Example 2: Check dimensional consistency of force \(F = M \times a\), where acceleration \(a = \frac{L}{T^2}\).

Given: Mass \([M]\), Length \([L]\), Time \([T]\)

Right side dimension: \([M] \times \frac{[L]}{[T]^2} = [M][L][T]^{-2}\)

Force dimension is \([M][L][T]^{-2}\), so the equation is dimensionally consistent.

Example 3: Verify if the equation \(P = \frac{F}{A}\) is dimensionally correct, where pressure \(P\), force \(F\), area \(A\).

Given: Force dimension \([M][L][T]^{-2}\), Area dimension \([L]^2\)

Right side dimension: \(\frac{[M][L][T]^{-2}}{[L]^2} = [M][L]^{-1}[T]^{-2}\)

Pressure dimension is \([M][L]^{-1}[T]^{-2}\), confirming dimensional consistency.

Example 4: Check dimensional correctness of kinetic energy \(E = \frac{1}{2} M v^2\).

Given: Mass \([M]\), Velocity \([L][T]^{-1}\)

Right side dimension: \([M] \times ([L][T]^{-1})^2 = [M][L]^2[T]^{-2}\)

Energy dimension is \([M][L]^2[T]^{-2}\), confirming the formula is dimensionally consistent.

Example 5: Verify the equation for frequency \(f = \frac{1}{T}\), where \(T\) is period.

Given: Time dimension \([T]\)

Right side dimension: \(\frac{1}{[T]} = [T]^{-1}\)

Frequency dimension is \([T]^{-1}\), so the equation is dimensionally consistent.

Chapter Summary

This chapter introduced the distinction between basic and derived physical quantities, explaining how fundamental measurements combine to form more complex ones. It detailed the International System of Units (SI) as the standard framework for expressing these quantities consistently. Methods for converting units within and between measurement systems were outlined to ensure accuracy in calculations. Various measuring instruments were described, emphasizing their appropriate use in obtaining precise physical measurements. The chapter explored practical techniques for measuring different physical quantities, highlighting the importance of correct procedures. Archimedes' principle was presented as a fundamental concept explaining buoyant forces in fluids. The concept of upthrust was examined to understand the force exerted by a fluid on submerged objects. Finally, the law of floatation was discussed alongside the determination of density and relative density, providing essential tools for analyzing the behavior of materials in fluids.