Which Physics Constants Should You Use in HSC Questions?

A question tells you to use \(g = 10.0\ \mathrm{m\,s^{-2}}\), but your HSC Physics data sheet gives \(g = 9.8\ \mathrm{m\,s^{-2}}\). Which value should you use?

Use the value explicitly supplied in the question. The data sheet is your reference when the question does not provide a more specific value. If the question gives a different value for the same quantity and conditions, follow its instruction rather than replacing it with the value you remember.

However, there is an important catch: sometimes two different values are both correct because they describe different physical situations. Understanding that difference is what stops you from choosing the wrong constant.

01Which value should you use?

For HSC calculations, follow this order:

1. An explicit instruction in the question. If it says to take \(g = 10.0\), use \(10.0\).
2. Values supplied for the particular situation. If a question gives Earth’s mass as \(5.97 \times 10^{24}\ \mathrm{kg}\), use that value rather than a less precise value on your data sheet.
3. The HSC data sheet. If the question does not supply a value, use the relevant value from the sheet provided with your examination.
4. A remembered or independently sourced value. Only use this if neither the question nor the supplied data sheet provides what you need.

This is a priority order, not permission to ignore physical context. A value for Earth’s surface gravitational acceleration cannot automatically be used for a satellite hundreds of kilometres above Earth.

If two values genuinely describe the same quantity under identical conditions but contradict each other, don’t average them or silently choose whichever looks more convenient. Check whether the question specifies which value to use. If it does not, identify the inconsistency and state your assumption.

What does the HSC data sheet actually contain?

The data sheet contains more than just universal physical constants. It includes measured quantities, particle properties, approximate planetary values, and properties of particular materials.

The 2025 HSC Physics data sheet provides the following examples:

These are values from the 2025 examination sheet, included here as an illustration. Always check the actual data sheet for the examination year you are preparing for.

Why can two different values both be correct?

There are four common reasons.

1. Rounding and significant figures

The speed of light in a vacuum is exactly \(299\,792\,458\ \mathrm{m\,s^{-1}}\) under the modern SI system.

However, the HSC data sheet gives:

\[ c = 3.00 \times 10^8\ \mathrm{m\,s^{-1}} \]

These are not contradictory. The second value is simply rounded to three significant figures.

The same applies to Planck’s constant. Its exact SI value is \(6.62607015 \times 10^{-34}\ \mathrm{J\,s}\), while the HSC sheet provides a rounded version.

The International System of Units (SI) fixes the numerical values of several fundamental constants, including \(c\) and \(h\). The number of digits printed on an examination sheet is a separate question of calculation precision.

2. Different physical conditions

Some quantities depend on where or under what conditions they are measured.

Gravitational acceleration is a good example. The value \(9.8\ \mathrm{m\,s^{-2}}\) is an approximation for Earth’s surface. It is not the gravitational acceleration everywhere in the universe.

Similarly, the speed of sound depends on the properties and temperature of the medium through which it travels.

A question might therefore supply \(343\ \mathrm{m\,s^{-1}}\) rather than the data sheet’s \(340\ \mathrm{m\,s^{-1}}\). If it specifies the conditions for that value, use it.

3. Different physical meanings

Consider the charge on an electron.

The data sheet gives:

\[ q_e = -1.602 \times 10^{-19}\ \mathrm{C} \]

But you will sometimes see the elementary charge written as:

\[ e = +1.602 \times 10^{-19}\ \mathrm{C} \]

These values have the same magnitude but opposite signs.

The electron has a negative charge. The elementary charge \(e\) is the positive magnitude of that charge.

They are not competing estimates of the same signed quantity.

4. Different levels of approximation

Earth’s mass on the data sheet is \(6.0 \times 10^{24}\ \mathrm{kg}\).

A question might instead give:

\[ M_E = 5.97 \times 10^{24}\ \mathrm{kg} \]

The second value is more precise. It does not mean the data sheet is wrong. It means the question has supplied a more detailed approximation for its particular calculation.

Use the supplied value.

02Why the speed of light has different published values: an experimental example

The history of measuring the speed of light gives a useful example of why different sources can report different numerical values.

In the nineteenth century, scientists wanted to measure the speed of light using a terrestrial experiment rather than relying on astronomical observations.

The challenge was timing. Light travels so quickly that ordinary human reaction times are nowhere near precise enough.

Fizeau’s rotating toothed wheel experiment

In 1849, French physicist Hippolyte Fizeau developed an ingenious solution.

His apparatus consisted of a light source, a rotating toothed wheel, a distant mirror, and an observer.

The light followed this path:

1. A light source produced a narrow beam.
2. The beam passed through a gap in the rotating toothed wheel.
3. It travelled approximately 8 kilometres to a distant mirror.
4. The mirror reflected the beam back towards the wheel.
5. The returning beam passed through the wheel and reached the observer, provided a tooth did not block it.

What should happen if light travels instantaneously?

The returning beam should pass straight back through the same gap. The wheel would have no time to rotate appreciably during the light’s journey.

But Fizeau observed something different.

At a sufficiently high rotational speed, a tooth moved into the position previously occupied by the gap. The returning light was blocked.

This showed that light took a finite amount of time to travel to the distant mirror and back.

The wheel effectively acted as a very fast clock.

A quick estimate of the time involved

Suppose the mirror is 8.0 km from the wheel.

The total distance travelled by the light is:

\[ d = 2(8.0 \times 10^3) \]

Using the modern approximate speed of light:

\[ t = \frac{d}{c} \]

\[ t = \frac{1.60 \times 10^4}{3.00 \times 10^8} \]

\[ t = 5.33 \times 10^{-5}\ \mathrm{s} \]

That is approximately \(53\) microseconds.

This is an illustrative estimate using the modern value of \(c\), not Fizeau’s recorded experimental result.

The time is far too short for a person to measure with a stopwatch. The rotating wheel made it possible to detect the delay indirectly.

Why did later experiments give different values?

Fizeau obtained approximately:

\[ c = 313\,300\ \mathrm{km\,s^{-1}} \]

This was roughly 5% higher than the modern value.

The experiment was an important breakthrough, but limitations in the measurement of the wheel’s rotation, the tooth geometry, and the light’s travel time affected its precision.

Léon Foucault later improved the approach by replacing the toothed wheel with a rotating mirror.

Instead of relying on a beam being blocked, Foucault measured the small angular displacement of the returning beam caused by the mirror’s rotation.

His 1862 measurement was approximately:

\[ c = 299\,796\ \mathrm{km\,s^{-1}} \]

This was much closer to the modern value.

The important lesson is that different experimental results do not necessarily mean a physical constant has changed. They may reflect improvements in measurement precision and experimental design.

Today, the speed of light in a vacuum is fixed by the definition of the metre. Its exact SI value is \(299\,792\,458\ \mathrm{m\,s^{-1}}\).

The HSC data sheet rounds this value for practical calculations.

03Choose the equation before choosing the constant

Knowing which value to use is only half the problem. You also need the correct equation.

A common mistake is to see a familiar constant in the question and immediately substitute it into an equation.

Instead, identify the physical situation first.

\(g\) or \(G\)? Gravitational acceleration versus gravitational constant

These two symbols are easy to confuse.

For an object close to Earth’s surface, where the change in height is small, use:

\[ F_g = mg \]

Here, \(m\) is the object’s mass and \(g\) is the local gravitational acceleration.

For gravitational attraction between two masses separated by a distance \(r\), use:

\[ F_g = \frac{GMm}{r^2} \]

Here, \(M\) and \(m\) are the interacting masses, and \(r\) is the distance between their centres.

The gravitational acceleration at distance \(r\) from a spherical body is:

\[ g(r) = \frac{GM}{r^2} \]

Notice that \(g\) changes with distance, while \(G\) remains the same.

For a satellite in circular orbit, gravitational force supplies the centripetal force:

\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]

Rearranging:

\[ v = \sqrt{\frac{GM}{r}} \]

This equation is useful when you know the mass of the central body and the satellite’s orbital radius.

It would be incorrect to substitute Earth’s surface value of \(g\) into this equation as though it remained constant throughout the orbit.

For more on the assumptions behind constant acceleration, read Constant Net Force and Acceleration in HSC Physics.

\(c\) or \(v\)? Vacuum speed versus speed in a medium

The symbol \(c\) represents the speed of light in a vacuum.

For electromagnetic waves travelling through a vacuum:

\[ c = f\lambda \]

Here, \(f\) is frequency in hertz (Hz), and \(\lambda\) is wavelength in metres (m).

Inside a material such as glass, light travels more slowly. Its speed is:

\[ v = \frac{c}{n} \]

Here, \(n\) is the refractive index of the material.

The frequency does not change when light crosses the boundary. Its speed and wavelength change instead.

So if a question gives you \(c\) and describes light travelling through glass, don’t automatically substitute \(c\) as the speed inside the glass.

\(h\) or \(c\)? Photon energy versus wave speed

For the energy of a photon, use:

\[ E = hf \]

Here:

• \(E\) is photon energy in joules (J).
• \(h\) is Planck’s constant in joule-seconds (J s).
• \(f\) is frequency in hertz (Hz).

You do not need \(c\) if the frequency is already supplied.

If the question gives wavelength instead, you may first need to determine the frequency using the appropriate wave-speed relationship.

The equation must match the information supplied and the physical situation.

04Worked examples

05Questions and solutions

The following questions test increasingly difficult situations involving supplied constants, different physical conditions, and competing equations.

Question 1

An equipment case has a mass of \(65.0\ \mathrm{kg}\) and rests on Earth’s surface.

A question instructs you to use:

\[ g = 9.81\ \mathrm{m\,s^{-2}} \]

The HSC data sheet gives \(g = 9.8\ \mathrm{m\,s^{-2}}\).

Calculate the weight of the case.

Solution 1

The weight of the case is approximately \(638\ \mathrm{N}\).

Use the value of \(g\) supplied in the question because it is explicitly specified.

Weight is the gravitational force acting on an object:

\[ W = mg \]

Substitute:

\[ \begin{aligned} W &= 65.0 \times 9.81\\ &= 637.65\ \mathrm{N} \end{aligned} \]

To three significant figures:

\[ \boxed{W = 638\ \mathrm{N}} \]

The weight acts vertically downwards.

The data sheet’s \(9.8\) is a rounded approximation. Using the more precise value supplied in the question gives the appropriate result.

Question 2

Use the following values to estimate Earth’s gravitational acceleration at its surface:

\[ G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \]

\[ M_E = 6.0 \times 10^{24}\ \mathrm{kg} \]

\[ r_E = 6.371 \times 10^6\ \mathrm{m} \]

The data sheet separately lists:

\[ g = 9.8\ \mathrm{m\,s^{-2}} \]

(a) Calculate Earth’s gravitational acceleration using the universal gravitational law.

(b) Explain why your calculated result does not need to match \(9.8\) exactly.

(c) If a subsequent question explicitly instructs you to use \(g=9.8\ \mathrm{m\,s^{-2}}\) to calculate an object’s weight near Earth’s surface, which value should you use?

Solution 2

(a) The calculated gravitational acceleration is approximately \(9.86\ \mathrm{m\,s^{-2}}\).

(b) The small difference is caused by the approximations in the supplied planetary values.

(c) Use \(9.8\ \mathrm{m\,s^{-2}}\) in the subsequent calculation because that is the value explicitly instructed.

Step 1: Select the appropriate equation.

We know Earth’s mass, the distance from its centre, and the universal gravitational constant.

Therefore:

\[ g = \frac{GM_E}{r_E^2} \]

This equation describes the gravitational field produced by a spherical body.

Step 2: Substitute the values.

\[ g = \frac{(6.67 \times 10^{-11}) (6.0 \times 10^{24})} {(6.371 \times 10^6)^2} \]

Simplifying:

\[ g \approx 9.86\ \mathrm{m\,s^{-2}} \]

To two significant figures:

\[ \boxed{g \approx 9.9\ \mathrm{m\,s^{-2}}} \]

Step 3: Explain the difference.

The mass of Earth on the data sheet is given to only two significant figures.

This means it is an approximate value. Combining it with the other rounded quantities will not necessarily reproduce the listed gravitational acceleration exactly.

There is no contradiction between the two results.

Step 4: Apply the instruction in part (c).

If the next question explicitly says to use \(g=9.8\), use that value.

You should not substitute \(9.86\) simply because you calculated it in the previous part.

The key distinction is that the first calculation derives an approximate value from Earth’s mass and radius, while the next part specifies the approximation to use.

Question 3

A beam of light travels from air into a glass block.

The light has a frequency of:

\[ f = 5.00 \times 10^{14}\ \mathrm{Hz} \]

The refractive index of the glass is \(n=1.50\).

The question supplies the speed of light in a vacuum as:

\[ c = 2.998 \times 10^8\ \mathrm{m\,s^{-1}} \]

The data sheet gives \(3.00 \times 10^8\ \mathrm{m\,s^{-1}}\).

(a) Calculate the speed of light inside the glass.

(b) Calculate its wavelength inside the glass.

(c) Explain why substituting the vacuum speed directly into \(c=f\lambda\) would produce an incorrect wavelength.

Solution 3

(a) The speed of light inside the glass is \(2.00 \times 10^8\ \mathrm{m\,s^{-1}}\).

(b) Its wavelength is approximately \(400\ \mathrm{nm}\).

(c) The speed of light in glass is lower than its speed in a vacuum, so using \(c\) directly as the speed inside glass gives the wrong wavelength.

Step 1: Identify the physical situation.

The light is travelling through glass, not a vacuum.

We therefore use:

\[ v = \frac{c}{n} \]

Here, \(v\) is the speed inside the glass and \(n\) is its refractive index.

Step 2: Calculate the speed.

Use the value of \(c\) supplied by the question.

\[ \begin{aligned} v &= \frac{2.998 \times 10^8}{1.50}\\ &= 1.9987 \times 10^8\ \mathrm{m\,s^{-1}} \end{aligned} \]

To three significant figures:

\[ \boxed{v = 2.00 \times 10^8\ \mathrm{m\,s^{-1}}} \]

Step 3: Calculate the wavelength.

The wave equation is:

\[ v = f\lambda \]

Rearrange:

\[ \lambda = \frac{v}{f} \]

Substitute:

\[ \begin{aligned} \lambda &= \frac{1.9987 \times 10^8} {5.00 \times 10^{14}}\\ &= 3.9974 \times 10^{-7}\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \boxed{\lambda = 4.00 \times 10^{-7}\ \mathrm{m}} \]

Or:

\[ \boxed{\lambda = 400\ \mathrm{nm}} \]

Step 4: Explain the misconception.

The tempting approach is to use \(c=f\lambda\) because it is the familiar equation for electromagnetic waves.

However, \(c\) represents the speed in a vacuum.

Inside glass, the wave travels at \(v=c/n\). Its frequency remains unchanged, but its wavelength decreases.

If you incorrectly used \(c\) directly, you would obtain approximately \(600\ \mathrm{nm}\), rather than the correct \(400\ \mathrm{nm}\).

The issue is not which numerical value of \(c\) you chose. It is that you used the vacuum speed for light travelling through a different medium.

Question 4

An electron starts from rest and is accelerated through a potential difference of \(150\ \mathrm{V}\).

It then enters a uniform magnetic field perpendicular to its velocity.

The question supplies:

\[ m_e = 9.11 \times 10^{-31}\ \mathrm{kg} \]

\[ q_e = -1.60 \times 10^{-19}\ \mathrm{C} \]

The magnetic field strength is:

\[ B = 2.00 \times 10^{-3}\ \mathrm{T} \]

(a) Calculate the electron’s speed after acceleration.

(b) Calculate the radius of its circular path in the magnetic field.

(c) Explain why the negative sign of the electron’s charge must be retained when determining the direction of its motion, but not when calculating the radius.

Assume negligible energy losses and non-relativistic motion.

Solution 4

(a) The electron reaches a speed of approximately \(7.26 \times 10^6\ \mathrm{m\,s^{-1}}\).

(b) The radius of its circular path is approximately \(2.07 \times 10^{-2}\ \mathrm{m}\).

(c) The charge’s sign determines the direction of the magnetic force, while its magnitude determines the radius.

Step 1: Calculate the kinetic energy gained.

The electron starts from rest and is accelerated through a potential difference.

The magnitude of its kinetic energy gain is:

\[ \Delta K = |q_e|V \]

We use the magnitude of the charge because kinetic energy increases when the electron accelerates.

Substitute:

\[ \begin{aligned} \Delta K &= (1.60 \times 10^{-19})(150)\\ &= 2.40 \times 10^{-17}\ \mathrm{J} \end{aligned} \]

Step 2: Calculate the electron’s speed.

Using conservation of energy:

\[ \Delta K = \frac{1}{2}m_ev^2 \]

Rearrange:

\[ v = \sqrt{\frac{2\Delta K}{m_e}} \]

Substitute:

\[ v = \sqrt{ \frac{2(2.40 \times 10^{-17})} {9.11 \times 10^{-31}} } \]

Therefore:

\[ \boxed{v = 7.26 \times 10^6\ \mathrm{m\,s^{-1}}} \]

Step 3: Select the magnetic force relationship.

The electron enters the magnetic field perpendicular to its velocity.

The magnetic force magnitude is:

\[ F_B = |q_e|vB \]

This force provides the centripetal force:

\[ |q_e|vB = \frac{m_ev^2}{r} \]

Rearrange:

\[ r = \frac{m_ev}{|q_e|B} \]

Step 4: Calculate the radius.

\[ r = \frac{ (9.11 \times 10^{-31}) (7.26 \times 10^6) }{ (1.60 \times 10^{-19}) (2.00 \times 10^{-3}) } \]

\[ r = 0.0207\ \mathrm{m} \]

Therefore:

\[ \boxed{r = 2.07 \times 10^{-2}\ \mathrm{m}} \]

Step 5: Explain the role of the negative charge.

This is where the distinction between signed charge and charge magnitude matters.

The radius is a positive distance, so we use \(|q_e|\) in its calculation.

However, the direction of the magnetic force depends on the sign of the charge.

For a positive charge moving to the right with a magnetic field directed into the page, the right-hand rule gives an upward force.

An electron has a negative charge, so its force acts in the opposite direction: downwards.

The negative sign is essential for predicting the trajectory, even though it does not make the radius negative.

Question 5

A satellite is moving in a stable circular orbit around Earth.

Its altitude above Earth’s surface is \(629\ \mathrm{km}\), and its measured orbital period is:

\[ T = 5.83 \times 10^3\ \mathrm{s} \]

For this question, use:

\[ G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \]

\[ M_E = 5.97 \times 10^{24}\ \mathrm{kg} \]

\[ R_E = 6.371 \times 10^6\ \mathrm{m} \]

The HSC data sheet also lists:

\[ g = 9.8\ \mathrm{m\,s^{-2}} \]

and:

\[ M_E = 6.0 \times 10^{24}\ \mathrm{kg} \]

Two students disagree about the correct gravitational acceleration to use.

• Student A argues that the satellite experiences \(9.8\ \mathrm{m\,s^{-2}}\) because this is the gravitational acceleration given on the data sheet.
• Student B argues that gravitational acceleration must be calculated using \(g(r)=GM_E/r^2\), because the satellite is above Earth’s surface.

(a) Calculate the satellite’s orbital radius from Earth’s centre.

(b) Calculate its orbital speed using its measured period.

(c) Calculate its centripetal acceleration using its orbital speed.

(d) Calculate the gravitational acceleration at the satellite’s orbital radius. Determine whether this agrees with the result in part (c).

(e) Calculate the gravitational acceleration at Earth’s surface using the same values of \(G\), \(M_E\), and \(R_E\). Explain which student’s reasoning is correct and why.

Assume Earth is spherical, the satellite follows a circular orbit, and atmospheric drag and other external influences are negligible.

Solution 5

Student B is correct. The gravitational acceleration at the satellite’s orbital radius is approximately \(8.13\ \mathrm{m\,s^{-2}}\), not \(9.8\ \mathrm{m\,s^{-2}}\).

The important distinction is that \(9.8\ \mathrm{m\,s^{-2}}\) describes the approximate gravitational acceleration near Earth’s surface. The satellite is further from Earth’s centre, so the gravitational field is weaker.

The measured orbital period provides an independent way to check this conclusion.

Step 1: Calculate the orbital radius.

The question gives the satellite’s altitude above Earth’s surface.

However, the gravitational equation requires the distance from Earth’s centre.

Therefore:

\[ r = R_E + h \]

Convert the altitude into metres:

\[ h = 629 \times 10^3\ \mathrm{m} \]

Substitute:

\[ \begin{aligned} r &= 6.371 \times 10^6 + 0.629 \times 10^6\\ &= 7.000 \times 10^6\ \mathrm{m} \end{aligned} \]

Step 2: Calculate orbital speed using the measured period.

For circular motion:

\[ v = \frac{2\pi r}{T} \]

Substitute:

\[ \begin{aligned} v &= \frac{2\pi(7.000 \times 10^6)} {5.83 \times 10^3}\\ &\approx 7.54 \times 10^3\ \mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{v = 7.54 \times 10^3\ \mathrm{m\,s^{-1}}} \]

Step 3: Calculate the centripetal acceleration.

The satellite is travelling in a circular path, so:

\[ a_c = \frac{v^2}{r} \]

Substitute:

\[ a_c = \frac{(7.54 \times 10^3)^2} {7.000 \times 10^6} \]

Therefore:

\[ \boxed{a_c = 8.13\ \mathrm{m\,s^{-2}}} \]

This is the acceleration required to maintain the circular orbit.

Step 4: Calculate gravitational acceleration at the orbital radius.

Now use the universal gravitational equation:

\[ g(r) = \frac{GM_E}{r^2} \]

Substitute:

\[ g(r) = \frac{ (6.67 \times 10^{-11}) (5.97 \times 10^{24}) }{ (7.000 \times 10^6)^2 } \]

Therefore:

\[ \boxed{g(r) = 8.13\ \mathrm{m\,s^{-2}}} \]

This agrees with the centripetal acceleration calculated in part (c), within rounding.

That agreement is important.

Gravity is providing the centripetal acceleration required for the satellite’s circular motion. Both equations describe the same physical interaction from different perspectives.

Step 5: Check gravitational acceleration at Earth’s surface.

Now use Earth’s actual radius rather than the satellite’s orbital radius.

\[ g_{\mathrm{surface}} = \frac{GM_E}{R_E^2} \]

Substitute:

\[ g_{\mathrm{surface}} = \frac{ (6.67 \times 10^{-11}) (5.97 \times 10^{24}) }{ (6.371 \times 10^6)^2 } \]

Therefore:

\[ \boxed{ g_{\mathrm{surface}} \approx 9.81\ \mathrm{m\,s^{-2}} } \]

This is approximately the \(9.8\ \mathrm{m\,s^{-2}}\) listed on the HSC data sheet.

The two gravitational acceleration values are both reasonable:

• Near Earth’s surface: approximately \(9.8\ \mathrm{m\,s^{-2}}\).
• At the satellite’s orbital radius: approximately \(8.13\ \mathrm{m\,s^{-2}}\).

They differ because gravitational acceleration decreases with distance from Earth’s centre.

Step 6: Test Student A’s prediction.

Student A uses the surface value \(g=9.8\) as though it applies throughout the orbit.

This would predict:

\[ v = \sqrt{gr} \]

\[ v = \sqrt{9.8 \times 7.00 \times 10^6} \]

\[ v \approx 8.28 \times 10^3\ \mathrm{m\,s^{-1}} \]

But the measured orbital period gives:

\[ v = 7.54 \times 10^3\ \mathrm{m\,s^{-1}} \]

The predictions do not agree.

Using the surface value of \(g\) at the satellite’s location gives an incorrect orbital speed.

The important lesson is that a supplied constant must describe the physical conditions of the problem. A value can be perfectly correct in one situation and inappropriate in another.

06What to check next

Before substituting any constant into an HSC Physics equation, make these three decisions:

• Value: Has the question supplied a more specific value than the data sheet?
• Meaning: Does the symbol represent the correct physical quantity, including its sign and units?
• Conditions: Does the equation apply to the situation being described?

Once those decisions are made, check that your equation is dimensionally consistent.

A useful next step is How to Check a Physics Equation Using Dimensional Analysis.

Remember: choosing the correct constant does not automatically make an equation correct. You need both the right numerical value and the right physical model.

Syllabus note

This reference uses the Physics Stage 6 Syllabus (2017) context relevant to HSC students through 2026. NESA’s Physics 11-12 Syllabus (2025) begins implementation in 2027, with the first HSC examination under the new syllabus scheduled for 2028.

The examples use values from the 2025 HSC data sheet for illustration. Always practise with the data and formulae sheets relevant to your examination year. NESA also advises students to be familiar with the supplied constants, formulae, and SI units.