2025 Edexcel IGCSE Physics Practice Paper with Answers

A star with a mass much larger than the Sun will expand and cool after the main sequence to become a red supergiant. It will then undergo a massive explosion called a supernova. The remaining core will collapse into either a neutron star or a black hole, depending on its remaining mass. Therefore, the correct evolutionary sequence is shown in option B.

B: 1 mark for identifying the correct pathway of stellar evolution for a high-mass star.

First, convert the time from minutes to seconds: \(t = 4.0 \times 60 = 240\text{ s}\). Use the formula for electrical energy transferred: \(E = V \times I \times t\). Substitute the values: \(E = 6.0\text{ V} \times 0.50\text{ A} \times 240\text{ s} = 720\text{ J}\). This matches option C.

C: 1 mark for the correct calculation of electrical energy transferred.

Use the redshift formula: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\). Calculate the change in wavelength: \(\Delta \lambda = 612\text{ nm} - 600\text{ nm} = 12\text{ nm}\). Substitute the values into the formula: \(\frac{12}{600} = \frac{v}{3.0 \times 10^8}\). This simplifies to \(0.02 = \frac{v}{3.0 \times 10^8}\). Therefore, \(v = 0.02 \times 3.0 \times 10^8 = 6.0 \times 10^6\text{ m/s}\). This corresponds to option B.

B: 1 mark for the correct calculation of the recessional velocity of the galaxy using the redshift equation.

During flight, an aircraft can build up a significant static charge due to friction with the air. When refueling, fuel flowing through the hose can also generate static charge. Connecting an earthing wire provides a low-resistance path for electrons to flow to or from the ground. This safely discharges the aircraft, preventing any electrostatic sparks that could ignite the highly flammable fuel vapor.

MP1: Identify that static charge builds up on the aircraft (due to friction with air during flight). (1) MP2: State that the wire allows charge / electrons to flow to or from the ground (discharging the aircraft). (1) MP3: Explain that this prevents a spark which could cause a fire / explosion / ignite the fuel. (1)

The frequency of a wave is determined solely by the source, so it remains constant when transitioning between mediums. Using the wave equation \(v = f \lambda\), since the frequency \(f\) is constant, a decrease in wave speed \(v\) must result in a proportional decrease in wavelength \(\lambda\).

MP1: State that the frequency remains constant / unchanged. (1) MP2: State that the wavelength decreases. (1) MP3: Explain the change using the wave equation \(v = f \lambda\) (or state that wavelength is directly proportional to speed for a constant frequency). (1)

To find the spring constant, the student should calculate the gradient of the linear (straight-line) section of the force-extension graph, which represents \(k = \text{Force} / \text{extension}\). To identify if the limit of proportionality has been exceeded, the student should locate the point where the graph stops being a straight line and begins to curve, indicating that extension is no longer directly proportional to force.

MP1: State that the spring constant is equal to the gradient/slope of the straight-line/linear portion of the graph. (1) MP2: State that the limit of proportionality is the point where the graph stops being linear / starts to curve. (1) MP3: Explicitly relate gradient calculation to axes, e.g., change in force divided by change in extension. (1)

Light from distant galaxies is shifted towards the red end of the electromagnetic spectrum (longer wavelengths), which shows that the galaxies are moving away from us. Observations show that more distant galaxies have a greater red-shift, meaning they are receding faster. This supports the idea that space itself is expanding, implying that the universe must have originated from a single, extremely dense and hot starting point in the past.

MP1: State that red-shift indicates light is shifted to longer wavelengths because galaxies are moving away (receding) from us. (1) MP2: Explain that more distant galaxies show greater red-shift / are moving away faster. (1) MP3: Conclude that this shows the universe is expanding, supporting the idea of a common origin / single starting point (the Big Bang). (1)

To find the pressure difference, we use the formula: \(p = \rho \times g \times h\). Substituting the given values: \(p = 1000\text{ kg/m}^3 \times 10\text{ m/s}^2 \times 15\text{ m} = 150,000\text{ Pa}\) (or \(1.5 \times 10^5\text{ Pa}\)).

MP1: Recall and use the equation \(p = \rho g h\). (1) MP2: Substitute the values correctly: \(1000 \times 10 \times 15\). (1) MP3: Give the correct value with its unit: \(150,000\text{ Pa}\) / \(150\text{ kPa}\) / \(1.5 \times 10^5\text{ N/m}^2\). (1)

To minimize exposure when handling radioactive materials in a school laboratory, teachers should: 1. Store sources in a secure, lead-lined wooden box when not in use. 2. Use long-handled tongs or forceps to carry the source, increasing the distance from the body. 3. Ensure the source is always pointed away from both the teacher and the students, and keep a safe distance.

Award 1 mark for each valid safety precaution up to a maximum of 3 marks: Use tongs / forceps to handle the sources (do not touch directly) (1); Store sources in lead-lined boxes when not in use (1); Point sources away from students/people (1); Minimize the time the source is out of the container (1); Keep a safe distance from the source (1).

First, calculate the useful work done in lifting the load: \(\text{Work done} = \text{force} \times \text{distance} = 45\text{ N} \times 2.0\text{ m} = 90\text{ J}\). Next, calculate the power output by dividing work done by time: \(\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{90\text{ J}}{3.0\text{ s}} = 30\text{ W}\).

MP1: Calculate the work done / energy transferred: \(45 \times 2 = 90\text{ J}\). (1) MP2: Use \(\text{Power} = \text{work done} / \text{time}\): \(90 / 3\). (1) MP3: Give the correct numerical value with unit: \(30\text{ W}\) (or \(30\text{ J/s}\)). (1)

According to Faraday's law of electromagnetic induction, the size of the induced voltage is proportional to the rate at which magnetic field lines are cut. To increase this voltage, you can: 1. Push the magnet into the coil with a higher speed. 2. Use a stronger magnet (which increases magnetic field density). 3. Increase the number of turns of wire on the coil.

Award 1 mark for each valid method up to a maximum of 3 marks: Move the magnet faster / at higher speed (1); Use a stronger magnet / magnet with a stronger magnetic field (1); Increase the number of turns / loops in the coil (1); Wrap the coil around a soft iron core (1).

1. Find the change in momentum of the toy car: \(\Delta p = m \times v - m \times u = 0.5 \text{ kg} \times 0 \text{ m/s} - 0.5 \text{ kg} \times 4 \text{ m/s} = -2.0 \text{ kg m/s}\). 2. Use the force formula: \(F = \frac{\Delta p}{t} = \frac{-2.0}{0.2} = -10 \text{ N}\). 3. Since we want the magnitude, the average force is 10 N.

- 1 mark: Recall or use of formula \(F = \frac{mv - mu}{t}\) (or \(F = \frac{\Delta p}{t}\)). - 1 mark: Substitution of values: \(F = \frac{0.5 \times 4}{0.2}\). - 1 mark: Correct evaluation of magnitude of force with units: \(10 \text{ N}\).

1. Friction between the flowing fuel and the pipe/tanker walls causes electrostatic charge to build up on the vehicle. 2. A large build-up of static charge could create a high potential difference, leading to a spark. 3. The metal earthing strap provides a low-resistance path for the charge to safely flow to the ground, preventing any electrostatic discharge (spark) that could ignite the fuel vapor.

- 1 mark: Identify that friction (due to fuel flow) causes electrostatic charge to build up. - 1 mark: State that a spark could occur due to this charge build-up. - 1 mark: State that the earthing strap safely conducts the charge to the ground (preventing sparks/explosions).

1. Write down the pressure formula for liquids: \(p = \rho \times g \times h\). 2. Rearrange the formula to solve for depth \(h\): \(h = \frac{p}{\rho \times g}\). 3. Convert pressure from kilopascals to pascals: \(p = 350 \text{ kPa} = 350,000 \text{ Pa}\). 4. Substitute the values: \(h = \frac{350,000}{1000 \times 10} = \frac{350,000}{10,000} = 35 \text{ m}\).

- 1 mark: Recall the formula \(p = \rho g h\) (or rearranged form). - 1 mark: Convert kPa to Pa and substitute values: \(h = \frac{350,000}{1000 \times 10}\). - 1 mark: Correct final answer with unit: \(35 \text{ m}\).

1. Light from distant galaxies is observed to have longer wavelengths than expected, shifted towards the red end of the spectrum (red-shifted). 2. This shift indicates that galaxies are moving away from the observer (receding). 3. More distant galaxies exhibit greater red-shifts, which means they are moving faster. This shows that the entire Universe is expanding, suggesting that if we go back in time, all matter must have originated from a single, hot, dense point (the Big Bang).

- 1 mark: Identify that red-shift indicates galaxies are moving away from us. - 1 mark: State that more distant galaxies have greater red-shift (moving faster). - 1 mark: Conclude that this expansion implies the Universe originated from a single point.

1. When the bicycle wheel turns faster, the magnet inside the dynamo rotates at a higher speed. 2. This increases the rate at which the magnetic field lines cut across the coil of wire. 3. According to Faraday's Law, a faster change in magnetic flux linkage induces a larger voltage (electromotive force) and therefore a larger electric current, making the bulb glow brighter.

- 1 mark: Mention that turning faster increases the rate of cutting of magnetic field lines (or rate of change of magnetic flux). - 1 mark: State that this induces a larger voltage/electromotive force (emf). - 1 mark: Connect the larger voltage/current to the increased brightness of the light.

1. Use the relationship between critical angle \(c\) and refractive index \(n\): \(\sin(c) = \frac{1}{n}\). 2. Substitute \(n = 1.52\): \(\sin(c) = \frac{1}{1.52} \approx 0.6579\). 3. Find \(c\) by taking the inverse sine: \(c = \sin^{-1}(0.6579) \approx 41.14^{\circ}\).

- 1 mark: Recall formula \(\sin(c) = \frac{1}{n}\). - 1 mark: Correct substitution: \(\sin(c) = \frac{1}{1.52}\). - 1 mark: Correct final angle: \(41.1^{\circ}\) (accept \(41^{\circ}\) or \(41.14^{\circ}\)).

1. Both cans transfer energy to their surroundings via radiation, conduction, and convection. 2. Matte black surfaces are excellent emitters of infrared (thermal) radiation, whereas shiny silver surfaces are poor emitters (they are good reflectors of radiation). 3. Therefore, the matte black can will radiate energy to the surroundings at a higher rate, cooling down faster.

- 1 mark: State clearly that the matte black can cools down faster. - 1 mark: Explain that matte black is a better/more efficient emitter of thermal (infrared) radiation. - 1 mark: Contrast with shiny silver (or state that it leads to a higher rate of heat transfer).

1. Determine how many times the activity has halved: - After 1 half-life: \(800 \text{ Bq} \div 2 = 400 \text{ Bq}\) - After 2 half-lives: \(400 \text{ Bq} \div 2 = 200 \text{ Bq}\) - After 3 half-lives: \(200 \text{ Bq} \div 2 = 100 \text{ Bq}\) 2. We see that 3 half-lives have elapsed in a total time of \(15 \text{ days}\). 3. Calculate the duration of one half-life: \(T_{1/2} = \frac{15 \text{ days}}{3} = 5 \text{ days}\).

- 1 mark: Determine that 3 half-lives have elapsed (e.g. by showing halving steps). - 1 mark: Set up equation: \(3 \times \text{half-life} = 15 \text{ days}\). - 1 mark: Correct calculation of half-life with unit: \(5 \text{ days}\).

1. As the fuel flows through the delivery pipe, there is friction between the fuel and the pipe surface.
2. This friction causes the transfer of electrons, leading to a build-up of static electrical charge on both the pipe and the aircraft's fuel tank.
3. If the charge builds up sufficiently, a potential difference is created which can discharge as a spark. This is highly dangerous as the spark can ignite volatile fuel vapours and cause an explosion.

MP1: Friction between fuel and pipe transfers electrons [1 mark]
MP2: Resulting in a build-up of electrostatic charge / voltage [1 mark]
MP3: Spark can discharge and ignite flammable fuel vapours / cause an explosion [1 mark]

1. Recall the wave equation: \( v = f \lambda \).
2. Rearrange the formula to make wavelength the subject: \( \lambda = \frac{v}{f} \).
3. Convert the frequency from megahertz to hertz: \( f = 2.5 \text{ MHz} = 2.5 \times 10^6 \text{ Hz} \).
4. Substitute the values into the equation: \( \lambda = \frac{1540}{2.5 \times 10^6} = 0.000616 \text{ m} \).
5. Round to 2 significant figures: \( 0.00062 \text{ m} \) (or \( 6.2 \times 10^{-4} \text{ m} \) or \( 0.62 \text{ mm} \)).

MP1: Rearrangement of wave equation: \( \lambda = \frac{v}{f} \) [1 mark]
MP2: Correct substitution including frequency conversion to Hz: \( \frac{1540}{2.5 \times 10^6} \) [1 mark]
MP3: Final calculated answer with correct unit and significant figures: \( 0.00062 \text{ m} \) or \( 0.62 \text{ mm} \) [1 mark]

1. Recall the formula for pressure difference in a liquid: \( p = h \times \rho \times g \).
2. Substitute the given values into the equation:
\( p = 120 \text{ m} \times 1030 \text{ kg/m}^3 \times 10 \text{ N/kg} \).
3. Calculate the final pressure:
\( p = 1,236,000 \text{ Pa} \) (or \( 1.24 \times 10^6 \text{ Pa} \)).

MP1: Recall of the liquid pressure formula: \( p = \rho g h \) [1 mark]
MP2: Correct substitution: \( 1030 \times 10 \times 120 \) [1 mark]
MP3: Correct final answer with unit: \( 1,236,000 \text{ Pa} \) or \( 1.24 \times 10^6 \text{ Pa} \) (accept \( 1.21 \times 10^6 \text{ Pa} \) if \( g = 9.8 \text{ m/s}^2 \) is used) [1 mark]

1. Calculate the initial momentum before the collision:
Total initial momentum = \( (m_{\text{car}} \times u_{\text{car}}) + (m_{\text{truck}} \times u_{\text{truck}}) = (0.80 \times 3.0) + (1.2 \times 0) = 2.4 \text{ kg m/s} \).
2. State the conservation of momentum: Total momentum before = Total momentum after.
3. Express the final momentum with the combined mass: \( (m_{\text{car}} + m_{\text{truck}}) \times v = (0.80 + 1.2) \times v = 2.0 \times v \).
4. Set the two equal: \( 2.4 = 2.0 \times v \).
5. Solve for \( v \): \( v = \frac{2.4}{2.0} = 1.2 \text{ m/s} \).

MP1: Use of conservation of momentum: \( m_1 u_1 = (m_1 + m_2) v \) [1 mark]
MP2: Substitution of values: \( 0.80 \times 3.0 = (0.80 + 1.2) \times v \) [1 mark]
MP3: Correct evaluation of final velocity: \( 1.2 \text{ m/s} \) [1 mark]

1. When the hydrogen in the core of a massive star is depleted, the star expands and its outer layers cool, forming a red supergiant.
2. Once nuclear fusion reactions cease, the star undergoes rapid gravitational collapse and explodes in a massive explosion called a supernova.
3. The extremely dense core left behind collapses further to become either a neutron star or, if the mass is high enough, a black hole.

MP1: Expansion and cooling of the star to become a red supergiant [1 mark]
MP2: Violent explosion known as a supernova [1 mark]
MP3: Remaining core collapses to form either a neutron star or a black hole [1 mark]

1. Determine the number of half-lives that have elapsed by halving the initial activity progressively:
- \( 1600 \text{ Bq} \rightarrow 800 \text{ Bq} \) (1 half-life)
- \( 800 \text{ Bq} \rightarrow 400 \text{ Bq} \) (2 half-lives)
- \( 400 \text{ Bq} \rightarrow 200 \text{ Bq} \) (3 half-lives)
2. Since the activity halves 3 times, 3 half-lives have elapsed over the period of \( 24 \text{ hours} \).
3. Calculate the duration of one half-life:
\( T_{1/2} = \frac{24 \text{ hours}}{3} = 8 \text{ hours} \).

MP1: Recognition that the activity has decreased by 3 half-lives (halving 3 times) [1 mark]
MP2: Division of total elapsed time by the number of half-lives: \( \frac{24}{3} \) [1 mark]
MP3: Correct final value of half-life with unit: \( 8 \text{ hours} \) [1 mark]

1. Double-glazed windows feature two panes of glass separated by a narrow gap filled with trapped dry air or a vacuum.
2. Conduction is significantly reduced because air is a very poor thermal conductor, and a vacuum completely prevents conduction as there are no particles to transfer kinetic energy.
3. Convection is also minimized because the gap between the glass panes is too narrow to allow a complete convection current to establish and circulate.

MP1: Air (or vacuum) is trapped in a gap between the two glass panes [1 mark]
MP2: Air is a very poor thermal conductor (or vacuum prevents conduction due to lack of particles) [1 mark]
MP3: The narrow gap prevents the circulation of air, reducing convection currents [1 mark]

1. Recall the orbital speed equation: \( v = \frac{2 \pi r}{T} \).
2. Calculate the orbital period \( T \) in seconds:
\( T = 365 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \)
\( T = 3.1536 \times 10^7 \text{ s} \).
3. Substitute the values into the formula:
\( v = \frac{2 \times \pi \times 1.50 \times 10^{11} \text{ m}}{3.1536 \times 10^7 \text{ s}} \)
\( v \approx 29885.7 \text{ m/s} \).
4. Express to 3 significant figures: \( 2.99 \times 10^4 \text{ m/s} \) (or \( 29,900 \text{ m/s} \)).

MP1: Recall and use of orbital speed equation: \( v = \frac{2 \pi r}{T} \) [1 mark]
MP2: Conversion of orbital period \( T \) into seconds (\( 3.15 \times 10^7 \text{ s} \)) [1 mark]
MP3: Correct evaluation of speed: \( 29,900 \text{ m/s} \) or \( 3.0 \times 10^4 \text{ m/s} \) [1 mark]

1. Calculate the change in momentum of the ball: \(\Delta p = m(v - u)\). Taking the direction to the right as positive: \(u = -15 \text{ m/s}\) and \(v = 25 \text{ m/s}\). \(\Delta p = 0.060 \times (25 - (-15)) = 0.060 \times 40 = 2.4 \text{ kg m/s}\). 2. Use the force-momentum formula: \(F = \frac{\Delta p}{\Delta t}\). 3. Substitute the values: \(F = \frac{2.4}{0.050} = 48 \text{ N}\).

1 mark for calculating the change in momentum (2.4 kg m/s) or showing the calculation \(0.060 \times 40\). 1 mark for the correct formula relating force, change in momentum, and time: \(F = \frac{\Delta p}{t}\). 1 mark for the correct final answer of 48 N (with correct unit).

1. First, calculate the pressure exerted by the water using the liquid pressure formula: \(p = \rho g h\). \(p_{\text{water}} = 1025 \times 10 \times 25 = 256250 \text{ Pa}\) (or \(256.25 \text{ kPa}\)). 2. Find the total pressure by adding the atmospheric pressure to the water pressure: \(p_{\text{total}} = p_{\text{water}} + p_{\text{atm}}\). 3. Convert atmospheric pressure to Pascals: \(101 \text{ kPa} = 101000 \text{ Pa}\). \(p_{\text{total}} = 256250 + 101000 = 357250 \text{ Pa}\) (or \(357 \text{ kPa}\) or \(3.6 \times 10^5 \text{ Pa}\)).

1 mark for calculating the pressure due to water (256250 Pa or 256.25 kPa) using \(p = \rho g h\). 1 mark for adding atmospheric pressure to the liquid pressure (converting units consistently). 1 mark for the correct final total pressure of 357 kPa (or 357250 Pa, or 3.6 x 10^5 Pa) with unit.

1. Use the formula relating refractive index and critical angle: \(\sin(c) = \frac{1}{n}\). 2. Substitute the refractive index of glass: \(\sin(c) = \frac{1}{1.52} \approx 0.6579\). 3. Calculate the inverse sine: \(c = \arcsin(0.6579) \approx 41.1^\circ\) (or \(41^\circ\)).

1 mark for stating the correct equation: \(\sin(c) = \frac{1}{n}\). 1 mark for correct substitution: \(\sin(c) = \frac{1}{1.52}\). 1 mark for calculating the correct critical angle of 41.1 degrees (allow 41 degrees).

1. As the aircraft flies, friction between its metal body and the air molecules (or dust/ice particles) causes the transfer of electrons, making the aircraft electrostatically charged. 2. When the aircraft lands, it remains charged. During refuelling, a potential difference exists between the aircraft and the ground/fuel pipe. 3. This potential difference can cause a spark, which could ignite highly flammable fuel vapours, causing an explosion.

1 mark for stating that friction/rubbing with air/dust causes a transfer of electrons/charge to the aircraft. 1 mark for explaining that a potential difference between the charged aircraft and the fuel pipe/ground can lead to a spark. 1 mark for stating that this spark can ignite the fuel (or fuel vapours) causing a fire/explosion.

First, calculate the total momentum before the collision:
\( p_{\text{initial}} = (m_1 \times u_1) + (m_2 \times u_2) \)
\( p_{\text{initial}} = (0.60 \text{ kg} \times 2.5 \text{ m/s}) + (0.40 \text{ kg} \times 0 \text{ m/s}) = 1.5 \text{ kg m/s} \)

According to the conservation of momentum, the total momentum after the collision must equal the total momentum before the collision:
\( p_{\text{final}} = p_{\text{initial}} = 1.5 \text{ kg m/s} \)

After the collision, the two cars stick together, so their combined mass is:
\( m_{\text{total}} = 0.60 \text{ kg} + 0.40 \text{ kg} = 1.0 \text{ kg} \)

Use the momentum formula to find the common velocity \( v \):
\( p_{\text{final}} = m_{\text{total}} \times v \)
\( 1.5 \text{ kg m/s} = 1.0 \text{ kg} \times v \)
\( v = \frac{1.5}{1.0} = 1.5 \text{ m/s} \)

- **1 mark**: Principle of conservation of momentum expressed in words or formula \( (m_1 u_1 = (m_1 + m_2)v) \)
- **1 mark**: Correct substitution of values into the momentum equation \( (0.60 \times 2.5 = 1.0 \times v) \)
- **1 mark**: Correct calculation of final velocity \( (1.5) \)
- **0.5 marks**: Correct unit (m/s)

First, calculate the gravitational potential energy (GPE) gained by the crate, which is the useful work output:
\( GPE = m \times g \times h \)
\( GPE = 12 \text{ kg} \times 10 \text{ N/kg} \times 5.0 \text{ m} = 600 \text{ J} \)

Next, calculate the useful power output of the motor:
\( P_{\text{useful}} = \frac{\text{Work Done}}{\text{Time}} \)
\( P_{\text{useful}} = \frac{600 \text{ J}}{8.0 \text{ s}} = 75 \text{ W} \)

Finally, use the efficiency formula to calculate the electrical power input:
\( \text{Efficiency} = \frac{\text{Useful Power Output}}{\text{Total Power Input}} \)
\( 0.75 = \frac{75 \text{ W}}{\text{Power Input}} \)
\( \text{Power Input} = \frac{75}{0.75} = 100 \text{ W} \)

- **1 mark**: Calculation of useful work done \( (600 \text{ J}) \) or useful power output \( (75 \text{ W}) \)
- **1 mark**: Use of the efficiency formula rearranged to find input power
- **1 mark**: Correct final value of power \( (100) \)
- **0.5 marks**: Correct unit (W or Watts)

To calculate the energy transferred, use the relationship between energy, charge, and voltage:
\( E = Q \times V \)

Identify the given values from the question:
- Voltage \( V = 12 \text{ V} \)
- Charge \( Q = 450 \text{ C} \)
(Note: The time of 3.0 minutes is extra information and is not needed because the total charge is already provided.)

Substitute the values into the equation:
\( E = 450 \text{ C} \times 12 \text{ V} = 5400 \text{ J} \) (or \( 5.4 \text{ kJ} \))

- **1 mark**: Select and state the correct formula \( E = Q \times V \)
- **1 mark**: Correct substitution of values into the formula \( (450 \times 12) \)
- **1 mark**: Correct numerical calculation \( (5400) \)
- **0.5 marks**: Correct unit (J or Joules)

To calculate the pressure exerted by a column of liquid, use the formula:
\( p = \rho \times g \times h \)

Where:
- \( \rho = 1030 \text{ kg/m}^3 \)
- \( g = 10 \text{ N/kg} \)
- \( h = 25 \text{ m} \)

Substitute the values into the formula:
\( p = 1030 \times 10 \times 25 \)
\( p = 257500 \text{ Pa} \) (or \( 2.575 \times 10^5 \text{ Pa} \) or \( 257.5 \text{ kPa} \))

- **1 mark**: State the correct formula \( p = \rho \times g \times h \)
- **1 mark**: Substitute correct values into the formula \( (1030 \times 10 \times 25) \)
- **1 mark**: Correct calculation of the numerical answer \( (257500) \)
- **0.5 marks**: Correct pressure unit (Pa, Pascal, or \( \text{N/m}^2 \))

First, convert the frequency from megahertz (MHz) to hertz (Hz):
\( f = 98.0 \text{ MHz} = 98.0 \times 10^6 \text{ Hz} \)

Next, use the wave speed equation to find wavelength \( \lambda \):
\( v = f \times \lambda \)
\( \lambda = \frac{v}{f} \)

Substitute the values:
\( \lambda = \frac{3.00 \times 10^8 \text{ m/s}}{98.0 \times 10^6 \text{ Hz}} \)
\( \lambda \approx 3.0612 \text{ m} \)

Rounding to 3 significant figures gives \( 3.06 \text{ m} \).

- **1 mark**: State the correct formula \( v = f \times \lambda \) or rearranged version
- **1 mark**: Convert frequency correctly to Hz \( (98.0 \times 10^6) \)
- **1 mark**: Correct numerical calculation of the wavelength to 3 s.f. \( (3.06) \)
- **0.5 marks**: Correct unit (m or metres)

First, calculate the total radius of the orbit \( r \) by adding the Earth's radius and the altitude:
\( r = 6400 \text{ km} + 400 \text{ km} = 6800 \text{ km} \)

Next, convert the orbital period \( T \) from minutes to seconds:
\( T = 93 \text{ minutes} \times 60 \text{ s/minute} = 5580 \text{ s} \)

Use the orbital speed formula:
\( v = \frac{2 \pi r}{T} \)

Substitute the values:
\( v = \frac{2 \times \pi \times 6800 \text{ km}}{5580 \text{ s}} \)
\( v = \frac{42725.66}{5580} \approx 7.66 \text{ km/s} \)

Rounding to 2 significant figures gives \( 7.7 \text{ km/s} \).

- **1 mark**: Calculate the correct total orbital radius \( (6800 \text{ km}) \)
- **1 mark**: Use of correct orbital speed equation \( v = \frac{2\pi r}{T} \) with time converted to seconds \( (5580 \text{ s}) \)
- **1 mark**: Correct calculation of orbital speed \( (7.7 \text{ or } 7.66) \)
- **0.5 marks**: Correct unit (km/s)

The heat required to change the state of a substance without a change in temperature is given by the formula:
\( Q = m \times L \)

Where:
- \( m = 0.25 \text{ kg} \)
- \( L = 3.3 \times 10^5 \text{ J/kg} \)

Substitute the values into the equation:
\( Q = 0.25 \text{ kg} \times (3.3 \times 10^5 \text{ J/kg}) \)
\( Q = 82500 \text{ J} \) (or \( 82.5 \text{ kJ} \) or \( 8.3 \times 10^4 \text{ J} \))

- **1 mark**: State the correct formula \( Q = m \times L \)
- **1 mark**: Correct substitution of values into the formula \( (0.25 \times 3.3 \times 10^5) \)
- **1 mark**: Correct calculation of energy \( (82500 \text{ or } 83000) \)
- **0.5 marks**: Correct unit (J, Joules, or kJ)

First, find the number of half-lives that have elapsed during the 20 hours:
\( \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{20 \text{ hours}}{4.0 \text{ hours}} = 5 \text{ half-lives} \)

Next, calculate the remaining activity by halving the initial activity 5 times:
- 1 half-life: \( 800 / 2 = 400 \text{ Bq} \)
- 2 half-lives: \( 400 / 2 = 200 \text{ Bq} \)
- 3 half-lives: \( 200 / 2 = 100 \text{ Bq} \)
- 4 half-lives: \( 100 / 2 = 50 \text{ Bq} \)
- 5 half-lives: \( 50 / 2 = 25 \text{ Bq} \)

Alternatively, using the formula:
\( A = A_0 \times \left(\frac{1}{2}\right)^n = 800 \times \left(\frac{1}{2}\right)^5 = 800 \times \frac{1}{32} = 25 \text{ Bq} \)

- **1 mark**: Determine the number of elapsed half-lives \( (5) \)
- **1 mark**: Show a clear method of halving the activity repeatedly or using the formula
- **1 mark**: Correct calculation of final numerical value \( (25) \)
- **0.5 marks**: Correct unit (Bq or Becquerels)

First, calculate the electrical power input to the motor: \(P_{\text{in}} = I \times V\) which gives \(P_{\text{in}} = 2.5 \text{ A} \times 24 \text{ V} = 60 \text{ W}\). Next, calculate the useful power output using the efficiency of the motor: \(\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}}\) so \(0.75 = \frac{P_{\text{out}}}{60 \text{ W}}\) which gives \(P_{\text{out}} = 0.75 \times 60 \text{ W} = 45 \text{ W}\). Since the crate is lifted at a constant speed, the useful power output is related to the work done per second against gravity: \(P_{\text{out}} = \frac{\text{Work Done}}{\text{time}} = \frac{F \times d}{t} = F \times v\). Here, the upward force \(F\) equals the weight of the crate: \(F = m \times g = 45 \text{ kg} \times 10 \text{ N/kg} = 450 \text{ N}\). Now, substitute the values to find the speed \(v\): \(45 \text{ W} = 450 \text{ N} \times v\), so \(v = \frac{45}{450} = 0.10 \text{ m/s}\).

[1 mark] Use of \(P = I \times V\) to find input power of \(60 \text{ W}\). [1 mark] Multiplication by efficiency to find useful power of \(45 \text{ W}\). [1 mark] Use of \(P = F \times v\) (or equivalent work/time formulas) with weight \(F = 450 \text{ N}\). [0.5 marks] Correct final speed of \(0.10 \text{ m/s}\) (accept \(0.1 \text{ m/s}\)) with correct units.

First, determine the total orbital radius \(r\) by adding the radius of the Earth to the altitude of the satellite: \(r = R_{\text{Earth}} + \text{altitude} = 6400 \text{ km} + 600 \text{ km} = 7000 \text{ km}\). Next, convert the orbital period \(T\) from minutes into seconds: \(T = 96 \text{ minutes} = 96 \times 60 \text{ s} = 5760 \text{ s}\). Use the formula for orbital speed: \(v = \frac{2 \pi r}{T}\). Substitute the values into the formula: \(v = \frac{2 \times \pi \times 7000 \text{ km}}{5760 \text{ s}}\), which gives \(v \approx \frac{43982.3}{5760} \approx 7.636 \text{ km/s}\). Rounding to 2 significant figures gives: \(v = 7.6 \text{ km/s}\).

[1 mark] Correct calculation of total orbital radius: \(r = 7000 \text{ km}\). [1 mark] Conversion of time period to seconds: \(T = 5760 \text{ s}\). [1 mark] Use of \(v = \frac{2 \pi r}{T}\) with substitution. [0.5 marks] Correct final answer rounded to 2 significant figures with unit: \(7.6 \text{ km/s}\). (Allow \(7600 \text{ m/s}\) or \(7.6 \text{ km/s}\)).

Define the initial direction of travel of the toy car as the positive direction. Calculate the total momentum before the collision: \(p_{\text{initial}} = m_{\text{car}} u_{\text{car}} + m_{\text{truck}} u_{\text{truck}} = (0.80 \text{ kg} \times 3.0 \text{ m/s}) + (1.2 \text{ kg} \times 0 \text{ m/s}) = 2.4 \text{ kg m/s}\). According to the principle of conservation of momentum, the total momentum after the collision must equal the total momentum before the collision: \(p_{\text{final}} = p_{\text{initial}} = 2.4 \text{ kg m/s}\). Since the toy car bounces backwards, its velocity after the collision is negative: \(v_{\text{car}} = -0.60 \text{ m/s}\). Set up the equation for the total momentum after the collision: \(p_{\text{final}} = m_{\text{car}} v_{\text{car}} + m_{\text{truck}} v_{\text{truck}}\) which gives \(2.4 = (0.80 \text{ kg} \times -0.60 \text{ m/s}) + (1.2 \text{ kg} \times v_{\text{truck}})\). Rearranging this: \(2.4 = -0.48 + 1.2 v_{\text{truck}}\). Rearrange to solve for \(v_{\text{truck}}\): \(1.2 v_{\text{truck}} = 2.4 + 0.48 = 2.88\), so \(v_{\text{truck}} = \frac{2.88}{1.2} = 2.4 \text{ m/s}\). Since the value is positive, the truck moves in the same direction as the initial motion of the car.

[1 mark] Calculation of initial momentum: \(2.4 \text{ kg m/s}\). [1 mark] Use of conservation of momentum equation, taking direction into account (recognising \(-0.48 \text{ kg m/s}\) for the car's final momentum). [1 mark] Calculation of truck's final speed: \(2.4 \text{ m/s}\). [0.5 marks] Correct direction stated (same direction as initial motion of the car).

First, calculate the pressure due to the column of seawater above the submarine using the hydrostatic pressure formula: \(p_{\text{sea}} = \rho \times g \times h\), where \(\rho = 1025 \text{ kg/m}^3\), \(g = 10 \text{ N/kg}\), and \(h = 120 \text{ m}\). This gives \(p_{\text{sea}} = 1025 \times 10 \times 120 = 1,230,000 \text{ Pa}\). Convert this fluid pressure into kilopascals (kPa) to match the unit of atmospheric pressure: \(1,230,000 \text{ Pa} = 1230 \text{ kPa}\). Now, find the total pressure by adding the atmospheric pressure to the fluid pressure: \(p_{\text{total}} = p_{\text{atm}} + p_{\text{sea}}\), so \(p_{\text{total}} = 101 \text{ kPa} + 1230 \text{ kPa} = 1331 \text{ kPa}\). Alternatively, in pascals: \(p_{\text{total}} = 101,000 \text{ Pa} + 1,230,000 \text{ Pa} = 1,331,000 \text{ Pa}\) (or \(1.33 \times 10^6 \text{ Pa}\)).

[1 mark] Use of \(p = \rho g h\) with correct substitution. [1 mark] Correct calculation of sea pressure as \(1,230,000 \text{ Pa}\) (or \(1230 \text{ kPa}\)). [1 mark] Addition of atmospheric pressure with consistent units. [0.5 marks] Correct final answer with units (e.g., \(1331 \text{ kPa}\) or \(1.33 \times 10^6 \text{ Pa}\)).

For nuclear fusion to occur, two positively charged nuclei must get extremely close to each other. Because they both have positive charges, they exert a strong electrostatic repulsive force on each other. At very high temperatures, the nuclei have large amounts of kinetic energy, allowing them to overcome this electrostatic repulsion and reach a distance small enough for the strong nuclear force to bind them. High pressure is also required to keep the nuclei close together, increasing the frequency of collisions and thus the probability of fusion reactions occurring. Therefore, option B is correct.

1 mark for option B. Reject all other options.

This question concerns the Doppler effect. As the source of the sound (the siren) moves towards the stationary observer, the wavefronts in front of the vehicle are compressed. This causes the wavelength of the waves arriving at the observer to be shorter than the wavelength emitted by the source. Since wave speed \(v = f\lambda\) and the speed of sound in air is constant, a shorter wavelength \(\lambda\) results in a higher frequency \(f\) being detected by the observer. Therefore, option D is correct.

1 mark for option D. Reject all other options.

Using the redshift formula: \( \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} \). First, calculate the change in wavelength: \( \Delta \lambda = 658\text{ nm} - 656\text{ nm} = 2\text{ nm} \). Rearrange the formula to solve for velocity: \( v = c \times \frac{\Delta \lambda}{\lambda_0} = 3.0 \times 10^8\text{ m/s} \times \frac{2\text{ nm}}{656\text{ nm}} = 9.15 \times 10^5\text{ m/s} \) (which rounds to \( 9.1 \times 10^5\text{ m/s} \)). Since the observed wavelength has increased (redshifted), the galaxy is moving away from Earth.

1 mark for calculating wavelength change of \( 2\text{ nm} \) or stating the formula \( \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} \); 1 mark for correct substitution and calculation of velocity (\( 9.1 \times 10^5\text{ m/s} \) or \( 9.2 \times 10^5\text{ m/s} \)); 0.5 marks for stating the galaxy is moving away from Earth.

As the bar magnet falls towards the copper ring, the magnetic flux linkage through the ring changes. This change in magnetic flux induces an electromotive force (e.m.f.) and consequently an induced current in the closed copper ring. According to Lenz's law, the direction of this induced current creates a magnetic field that opposes the change that produced it. This produces a magnetic pole at the top of the ring that is the same as the approaching pole of the magnet, resulting in an upward repulsive force that opposes the motion of the magnet.

1 mark for mentioning that the falling magnet changes magnetic flux linkage, inducing an e.m.f./current in the ring; 1 mark for stating that the induced current creates a magnetic field that opposes the motion of the magnet (Lenz's Law); 0.5 marks for explaining this creates a repulsive force (like poles repel), producing an upward resistive force.

Nuclear fusion requires joining two positively charged nuclei (such as hydrogen isotopes). Because both nuclei are positively charged, they experience a strong electrostatic repulsion when brought close together. Extremely high temperatures provide the nuclei with sufficient kinetic energy to overcome this electrostatic repulsion, while high pressures increase the collision frequency. In contrast, nuclear fission is initiated by neutrons, which are neutral particles and experience no electrostatic repulsion, allowing them to easily enter the target nucleus at room temperatures.

1 mark for identifying that fusion involves positive nuclei which experience electrostatic repulsion; 1 mark for explaining that high temperature provides the kinetic energy to overcome this repulsion (and pressure increases collision rate); 0.5 marks for explaining that fission is initiated by neutrons which are uncharged and feel no repulsion.

Calculate the work done (gravitational potential energy gained) by the crane: \( W = m \times g \times h = 450\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 54,000\text{ J} \). Next, calculate the useful power output using \( P = \frac{W}{t} \): \( P = \frac{54,000\text{ J}}{15\text{ s}} = 3600\text{ W} \) (or \( 3.6\text{ kW} \)).

1 mark for calculating work done or GPE gained (\( 54,000\text{ J} \)) using \( mgh \); 1 mark for substituting work and time into power formula \( P = \frac{W}{t} \); 0.5 marks for correct final answer of \( 3600\text{ W} \) or \( 3.6\text{ kW} \) with correct unit.

Determine the number of half-lives that have elapsed by halving the initial activity until reaching the final activity: \( 960\text{ Bq} \rightarrow 480\text{ Bq} \) (1 half-life), \( 480\text{ Bq} \rightarrow 240\text{ Bq} \) (2 half-lives), \( 240\text{ Bq} \rightarrow 120\text{ Bq} \) (3 half-lives). Since 3 half-lives have elapsed, the total time is: \( 3 \times 15\text{ hours} = 45\text{ hours} \).

1 mark for showing a clear halving process or determining that exactly 3 half-lives have passed; 1 mark for multiplying the number of half-lives by the half-life period (\( 3 \times 15 \)); 0.5 marks for the correct final answer of \( 45\text{ hours} \) with unit.

To exert the maximum pressure, the block must be placed on its smallest face. Smallest surface area \( A = 0.15\text{ m} \times 0.10\text{ m} = 0.015\text{ m}^2 \). The weight (force) exerted by the block is \( F = m \times g = 1.8\text{ kg} \times 10\text{ N/kg} = 18\text{ N} \). Using the pressure formula: \( P = \frac{F}{A} = \frac{18\text{ N}}{0.015\text{ m}^2} = 1200\text{ Pa} \) (or \( 1.2\text{ kPa} \)).

1 mark for identifying and calculating the minimum surface area (\( 0.015\text{ m}^2 \)) or calculating the weight of the block (\( 18\text{ N} \)); 1 mark for substituting the correct weight and area into the pressure formula \( P = \frac{F}{A} \); 0.5 marks for correct final answer of \( 1200\text{ Pa} \) (or \( 1200\text{ N/m}^2 \) or \( 1.2\text{ kPa} \)) with unit.

Since momentum is a vector quantity, direction is important. Let the initial direction of motion be positive. Initial velocity \( u = +4.5\text{ m/s} \) and final velocity \( v = -1.5\text{ m/s} \). Initial momentum \( p_i = m \times u = 0.80 \times 4.5 = +3.6\text{ kg m/s} \). Final momentum \( p_f = m \times v = 0.80 \times (-1.5) = -1.2\text{ kg m/s} \). The change in momentum is \( \Delta p = p_f - p_i = -1.2\text{ kg m/s} - 3.6\text{ kg m/s} = -4.8\text{ kg m/s} \). The magnitude of the change in momentum is therefore \( 4.8\text{ kg m/s} \).

1 mark for using opposite signs for the initial and final velocities to account for direction; 1 mark for substituting values into the change in momentum formula \( \Delta p = m(v - u) \); 0.5 marks for correct final magnitude of \( 4.8\text{ kg m/s} \) (or \( 4.8\text{ N s} \)) with correct unit.

Once the massive star runs out of hydrogen in its core, fusion slows down, the core contracts, and outer layers expand and cool, forming a red supergiant. The star continues to fuse heavier elements in its core until it can no longer support itself against gravity. It then undergoes a rapid gravitational collapse and explodes violently in a supernova. Depending on the remaining mass of the core, the remnant will collapse further to form either a highly dense neutron star or a black hole.

1 mark for describing the expansion into a red supergiant after leaving the main sequence; 1 mark for describing the supernova explosion; 0.5 marks for identifying the end state as a neutron star or black hole.

1. As the paint droplets leave the nozzle, they all acquire a positive charge. Since like charges repel, the droplets repel one another, spreading out into a very fine and even mist. 2. The charged paint droplets are attracted to the grounded (neutral) car panel by electrostatic induction (inducing an opposite negative charge on the surface of the panel nearest the droplets). 3. This attraction pulls the paint directly towards the panel, coating the back and edges of the metal panel (shadow areas) and reducing the amount of wasted paint.

1 mark: for stating that like charges repel, causing the droplets to spread out evenly / form a fine mist. 1 mark: for explaining that the charged droplets are attracted to the grounded panel (due to induced charge). 0.5 marks: for explaining that this attraction reduces waste / ensures coverage of hard-to-reach areas (such as the back or edges of the panel).

1. An alternating current in the primary coil creates a continuously changing magnetic field in the soft iron core. 2. This changing magnetic field cuts through the turns of the secondary coil. 3. According to Faraday's law of electromagnetic induction, a changing magnetic field (or rate of change of magnetic flux linkage) is required to induce an alternating voltage across the secondary coil. A direct current would create a constant magnetic field, resulting in zero induced voltage after the initial switch-on.

1 mark: for stating that an alternating current produces a continuously changing/varying magnetic field (in the core). 1 mark: for explaining that this changing magnetic field cuts through the secondary coil to induce a voltage. 0.5 marks: for stating that a direct current would create a constant/stationary magnetic field, which does not induce any voltage (or only induces a temporary spike when turned on/off).

1. First, calculate the pressure due to the column of water using the formula: \(p = \rho g h\). \(p_{\text{water}} = 1025\text{ kg/m}^3 \times 10\text{ N/kg} \times 22.0\text{ m} = 225,500\text{ Pa}\) (or \(225.5\text{ kPa}\)). 2. Next, calculate the total pressure by adding the atmospheric pressure: \(p_{\text{total}} = p_{\text{water}} + p_{\text{atm}} = 225,500\text{ Pa} + 101,000\text{ Pa} = 326,500\text{ Pa}\) (or \(3.27 \times 10^5\text{ Pa}\)). 3. The unit of pressure is Pascals (\(\text{Pa}\)) or Newtons per square metre (\(\text{N/m}^2\)).

1 mark: for calculating the pressure of the water column (\(225,500\text{ Pa}\)) using \(p = \rho g h\). 1 mark: for adding atmospheric pressure to find the total pressure (\(326,500\text{ Pa}\) or \(3.27 \times 10^5\text{ Pa}\) or \(327\text{ kPa}\)). 0.5 marks: for giving the correct unit (\(\text{Pa}\), \(\text{N/m}^2\), or \(\text{kPa}\) if matching the value).

1. When a high-mass star runs out of hydrogen in its core for fusion, the core contracts under gravity and heats up, while the outer layers expand and cool to form a red supergiant. 2. Fusion of heavier elements (up to iron) occurs in the core. Once iron is reached, fusion ceases because fusing iron requires more energy than it releases, causing the core to suddenly collapse. 3. This rapid gravitational collapse leads to a massive rebound shockwave, causing the star's outer layers to be violently ejected in a giant explosion known as a supernova.

1 mark: for explaining the formation of a red supergiant (hydrogen depleted, core contracts, outer layers expand and cool). 1 mark: for stating that fusion continues until iron is formed, leading to core collapse. 0.5 marks: for stating that the collapse results in a violent explosion / ejection of outer layers (supernova).

1. When the package hits the ground, the air-filled bubbles compress, which increases the duration of the impact (increases the time \(\Delta t\) taken for the vase to come to a stop). 2. The change in momentum (\(\Delta p\)) of the vase is determined by its initial speed and final speed (which is zero), which remains the same with or without the packaging. 3. Since force is equal to the rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)), increasing the impact time decreases the average force experienced by the vase, making it less likely to shatter.

1 mark: for stating that the compression of the bubble wrap increases the time taken for the collision / time to come to a stop. 1 mark: for stating that the force is equal to the rate of change of momentum (or quoting \(F = \frac{\Delta p}{\Delta t}\) with constant change in momentum). 0.5 marks: for concluding that the reduced force prevents the vase from breaking.

1. Nuclear fusion involves joining two positively charged nuclei (such as isotopes of hydrogen) together. 2. Since both nuclei have positive charges, they experience a very strong electrostatic force of repulsion when they are close. 3. High temperatures mean the nuclei have extremely high kinetic energies, allowing them to move fast enough to overcome this repulsion. High pressures force the nuclei close enough together so that the attractive strong nuclear force can take over and fuse them.

1 mark: for identifying that the nuclei are both positively charged and experience electrostatic repulsion. 1 mark: for explaining that high temperature provides high kinetic energy to overcome this repulsion. 0.5 marks: for explaining that high pressure increases the collision rate / forces the nuclei close enough for the strong nuclear force to bind them.

1. Identify the wave equation: \(v = f \lambda\), where \(v\) is speed, \(f\) is frequency, and \(\lambda\) is wavelength. 2. Convert the frequency from kilohertz to hertz: \(f = 80\text{ kHz} = 80,000\text{ Hz}\). 3. Rearrange the formula to solve for wavelength: \(\lambda = \frac{v}{f} = \frac{5900\text{ m/s}}{80,000\text{ Hz}} = 0.07375\text{ m}\) (or \(7.38\text{ cm}\) / \(7.4\text{ cm}\)). 4. The correct unit is metres (\(\text{m}\)) or centimetres (\(\text{cm}\)).

1 mark: for using the formula \(v = f \lambda\) (or rearranged correctly: \(\lambda = v/f\)). 1 mark: for converting \(80\text{ kHz}\) to \(80,000\text{ Hz}\) and calculating the value \(0.0738\text{ m}\) (or \(7.38\text{ cm}\)). 0.5 marks: for stating the correct unit of length (\(\text{m}\) or \(\text{cm}\)).

First, calculate the change in momentum of the car. Taking the initial direction as positive: initial momentum = mass * initial velocity = 0.85 kg * 12 m/s = 10.2 kg m/s. Since the car rebounds, the final velocity is -4.0 m/s. Final momentum = mass * final velocity = 0.85 kg * (-4.0 m/s) = -3.4 kg m/s. The change in momentum is final momentum - initial momentum = -3.4 kg m/s - 10.2 kg m/s = -13.6 kg m/s. The magnitude of change in momentum is 13.6 kg m/s. Force is calculated using the formula: force = change in momentum / time. Force = 13.6 / 0.080 = 170 N.

1 mark for calculating the change in momentum (13.6 kg m/s). 1 mark for recalling and using the formula force = change in momentum / time. 1 mark for the correct final answer of 170 N. 0.25 marks for providing the correct unit (N).

The formula for pressure difference in a liquid is: pressure = density * g * depth. Substitute the given values into the equation: pressure = 1025 kg/m^3 * 10 N/kg * 120 m. This gives pressure = 1,230,000 Pa (or 1.23 MPa).

1 mark for recalling the correct formula (p = rho * g * h). 1 mark for correct substitution of values (1025 * 10 * 120). 1 mark for the correct calculation (1230000). 0.25 marks for the correct unit (Pa or N/m^2).

First, calculate the useful gravitational potential energy gained by the crate: GPE = mass * g * height = 45 kg * 10 N/kg * 8.0 m = 3600 J. Next, use the efficiency formula: efficiency = useful energy output / total energy input. Rewrite the formula to find total energy input: total energy input = useful energy output / efficiency = 3600 J / 0.75 = 4800 J.

1 mark for calculating the useful work output (3600 J). 1 mark for recalling and rearranging the efficiency formula. 1 mark for the correct total energy input (4800 J). 0.25 marks for the correct unit (J).

First, convert the time from minutes to seconds: time = 4.0 * 60 = 240 s. Next, calculate the electrical charge that passed through: charge = current * time = 2.5 A * 240 s = 600 C. Then, calculate the electrical energy transferred using energy = charge * voltage = 600 C * 9.0 V = 5400 J.

1 mark for converting time to seconds (240 s). 1 mark for calculating charge (600 C) or using energy = current * voltage * time. 1 mark for the correct final answer (5400 J). 0.25 marks for the correct unit (J).

First, convert the orbital radius into meters: radius = \(2.4 \times 10^5\) km = \(2.4 \times 10^8\) m. Convert the orbital period into seconds: period = 4.0 * 86400 = 345,600 s. Use the orbital speed formula: orbital speed = 2 * pi * radius / period. Orbital speed = 2 * 3.14159 * \(2.4 \times 10^8\) / 345,600 = 1,507,964,473 / 345,600 = 4363 m/s. Rounding to 2 significant figures gives 4400 m/s.

1 mark for converting radius to meters and time to seconds. 1 mark for substituting values into orbital speed formula. 1 mark for the correct calculation to 2 significant figures (4400 m/s). 0.25 marks for correct unit (m/s).

Use the transformer equation: primary voltage / secondary voltage = primary turns / secondary turns. Substitute the given values: 230 / 12 = 1150 / secondary turns. Rearranging the equation gives: secondary turns = 1150 * 12 / 230 = 5 * 12 = 60 turns.

1 mark for recalling the transformer equation. 1 mark for correct rearrangement of the formula. 1 mark for the correct calculation of turns (60). 0.25 marks for specifying the unit as turns.

First, convert the frequency from MHz to Hz: frequency = 98.0 MHz = \(98.0 \times 10^6\) Hz. Use the wave speed equation: wave speed = frequency * wavelength. Rearrange the equation to find wavelength: wavelength = wave speed / frequency = \(3.00 \times 10^8\) / \(98.0 \times 10^6\) = 3.06 m.

1 mark for converting frequency to Hz. 1 mark for rearranging the wave speed equation to make wavelength the subject. 1 mark for the correct calculation (3.06 m). 0.25 marks for the correct unit (m).

First, find the number of half-lives that have passed: number of half-lives = 60 hours / 15 hours = 4. Since the activity halves during each half-life: after 1 half-life, activity = 400 Bq; after 2 half-lives, activity = 200 Bq; after 3 half-lives, activity = 100 Bq; after 4 half-lives, activity = 50 Bq.

1 mark for calculating the number of half-lives (4). 1 mark for showing a halving process. 1 mark for the correct final activity of 50 Bq. 0.25 marks for the correct unit (Bq).