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First, find the total orbital radius, \( r \):
\( r = \text{radius of planet} + \text{altitude} = 4800\text{ km} + 1200\text{ km} = 6000\text{ km} = 6.0 \times 10^6\text{ m} \).

Next, convert the time period, \( T \), to seconds:
\( T = 3.0\text{ hours} = 3.0 \times 3600\text{ s} = 10800\text{ s} \).

Use the orbital speed formula:
\( v = \frac{2 \pi r}{T} \)
\( v = \frac{2 \times \pi \times 6000\text{ km}}{10800\text{ s}} \approx 3.49\text{ km/s} \approx 3.5\text{ km/s} \).

1 mark for correct answer C.
- Reject other options because they represent incorrect orbital radii or incorrect conversions of time.

First, find the total resistance of the series circuit:
\( R_{\text{total}} = 4.0\ \Omega + 6.0\ \Omega = 10.0\ \Omega \).

Calculate the current in the circuit:
\( I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{10.0\ \Omega} = 1.2\text{ A} \).

Now, calculate the power dissipated in the \( 4.0\ \Omega \) resistor:
\( P = I^2 \times R = (1.2\text{ A})^2 \times 4.0\ \Omega = 1.44 \times 4.0 = 5.76\text{ W} \approx 5.8\text{ W} \).

1 mark for correct answer B.
- Reject option A (uses only current squared).
- Reject option C (dissipated power of the other resistor).
- Reject option D (incorrectly assumes full battery voltage across the single resistor).

The total pressure at depth \( h \) is given by:
\( p = p_{\text{atmospheric}} + \rho g h \)

Substitute the given values into the equation:
\( 2.5 \times 10^5\text{ Pa} = 1.0 \times 10^5\text{ Pa} + (1000\text{ kg/m}^3 \times 10\text{ N/kg} \times h) \)

Subtract atmospheric pressure from both sides to find the pressure due to the water column (hydrostatic pressure):
\( 1.5 \times 10^5\text{ Pa} = 10000 \times h \)

Solve for depth \( h \):
\( h = \frac{1.5 \times 10^5}{10000} = 15\text{ m} \).

1 mark for correct answer B.
- Reject option C (which ignores atmospheric pressure, giving \( 25\text{ m} \)).

Use the principle of conservation of momentum:
\( \text{total momentum before} = \text{total momentum after} \)

Calculate the initial total momentum:
\( p_{\text{initial}} = (m_1 \times u_1) + (m_2 \times u_2) \)
\( p_{\text{initial}} = (0.20\text{ kg} \times 1.5\text{ m/s}) + (0.30\text{ kg} \times 0\text{ m/s}) = 0.30\text{ kg m/s} \)

The two cars stick together, so their combined mass is:
\( m_{\text{total}} = m_1 + m_2 = 0.20\text{ kg} + 0.30\text{ kg} = 0.50\text{ kg} \)

Calculate the common final velocity \( v \):
\( p_{\text{final}} = m_{\text{total}} \times v \)
\( 0.30\text{ kg m/s} = 0.50\text{ kg} \times v \)
\( v = \frac{0.30}{0.50} = 0.60\text{ m/s} \).

1 mark for correct answer A.
- Reject option B (simple average of velocities).
- Reject options C and D (which ignore correct mass conservation).

First, calculate the energy supplied by the heater in seconds:
\( t = 4.0 \times 60\text{ s} = 240\text{ s} \)
\( E = P \times t = 50\text{ W} \times 240\text{ s} = 12000\text{ J} \)

Use the equation for specific latent heat of fusion:
\( E = m \times L \)
\( 12000\text{ J} = m \times (3.3 \times 10^5\text{ J/kg}) \)

Solve for mass \( m \):
\( m = \frac{12000}{3.3 \times 10^5} \approx 0.03636\text{ kg} \)

Convert kilograms to grams:
\( m = 0.03636 \times 1000\text{ g} \approx 36\text{ g} \).

1 mark for correct answer C.
- Reject option A (fails to convert minutes to seconds).
- Reject options B and D (power of ten errors).

First, convert time from minutes to seconds: \(2 \text{ minutes} = 120 \text{ s}\). Use the formula for electrical energy transferred: \(E = I \times V \times t\). Substitute the values: \(E = 1.5 \text{ A} \times 12 \text{ V} \times 120 \text{ s} = 2160 \text{ J}\).

1 mark for correct substitution with converted time: \(1.5 \times 12 \times 120\). 1 mark for correct numerical value and unit (2160 J or 2.16 kJ).

Use the momentum formula: \(p = m \times v\). Substituting the given values: \(p = 800 \text{ kg} \times 12 \text{ m/s} = 9600 \text{ kg m/s}\).

1 mark for correct calculation of momentum: \(800 \times 12 = 9600\). 1 mark for correct unit: \(\text{kg m/s}\) or \(\text{N s}\).

Use the formula for change of state: \(Q = m \times L\). Substitute the given values: \(Q = 0.25 \text{ kg} \times 3.3 \times 10^5 \text{ J/kg} = 82500 \text{ J}\).

1 mark for correct substitution into the formula: \(0.25 \times 3.3 \times 10^5\). 1 mark for correct final answer with units: \(82500 \text{ J}\) (or \(82.5 \text{ kJ}\)).

When the wire moves between the magnet poles, it cuts through the magnetic field lines. This change in magnetic flux linkage induces a voltage (or potential difference) across the ends of the wire.

1 mark for stating that the wire cuts the magnetic field lines. 1 mark for stating that this induces a voltage / potential difference.

Use the efficiency formula: \(\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}\). Substitute the values: \(\text{efficiency} = \frac{160}{400} = 0.4\) or \(40\%\).

1 mark for correct calculation setup: \(160 / 400\). 1 mark for correct answer: \(0.4\) or \(40\%\) (accept either, no units required).

First, determine the number of half-lives that have elapsed: \(800 \rightarrow 400 \rightarrow 200 \rightarrow 100\), which is 3 half-lives. Divide the total time by the number of half-lives: \(\text{half-life} = \frac{15 \text{ days}}{3} = 5 \text{ days}\).

1 mark for identifying that 3 half-lives have elapsed. 1 mark for the correct calculation and units (5 days).

Use Snell's law: \(n = \frac{\sin(i)}{\sin(r)}\). Substitute the given values: \(n = \frac{\sin(40^\circ)}{\sin(25^\circ)} = \frac{0.6428}{0.4226} \approx 1.52\).

1 mark for correct substitution: \(\frac{\sin(40)}{\sin(25)}\). 1 mark for correct evaluation to 2 decimal places (1.52).

Use the density formula: \(\rho = \frac{m}{V}\). Substitute the values: \(\rho = \frac{16 \text{ kg}}{0.020 \text{ m}^3} = 800 \text{ kg/m}^3\).

1 mark for correct density calculation: 800. 1 mark for correct unit: \(\text{kg/m}^3\).

First, convert the time from minutes to seconds: \(t = 3.0 \text{ minutes} = 3.0 \times 60 = 180 \text{ s}\). Next, use the formula for electrical energy: \(E = I \times V \times t\). Substitute the given values: \(E = 9.5 \text{ A} \times 230 \text{ V} \times 180 \text{ s} = 393 300 \text{ J}\). To two significant figures, this is \(390 000 \text{ J}\) (or \(390 \text{ kJ}\)).

1 mark for converting time to seconds and substituting into the energy formula, e.g., \(9.5 \times 230 \times 180\). 1 mark for the correct final energy value with unit (accept \(393 300 \text{ J}\), \(390 000 \text{ J}\) or \(390 \text{ kJ}\)).

Use the formula for change of state: \(Q = m L\), where \(Q\) is the thermal energy, \(m\) is the mass, and \(L\) is the specific latent heat of fusion. Substitute the given values into the equation: \(Q = 0.080 \text{ kg} \times 80\,000 \text{ J/kg} = 6400 \text{ J}\).

1 mark for correct substitution into the formula \(Q = m L\) (e.g., \(0.080 \times 80\,000\)). 1 mark for the correct final answer with a unit (e.g., \(6400 \text{ J}\) or \(6.4 \text{ kJ}\)).

Use the formula for acceleration: \(a = \frac{v - u}{t}\). Since the skateboarder starts from rest, the initial velocity \(u = 0\). Rearranging the formula for time gives: \(t = \frac{v - u}{a} = \frac{5.4 - 0}{1.8} = 3.0 \text{ s}\).

1 mark for correct rearrangement of the acceleration formula or substitution of the given values (e.g., \(1.8 = \frac{5.4}{t}\)). 1 mark for the correct final time value with unit (e.g., \(3.0 \text{ s}\)).

Use the pressure formula for a liquid column: \(p = \rho g h\), where \(\rho\) is the density, \(g\) is the gravitational field strength, and \(h\) is the depth. Substituting the values: \(p = 1000 \text{ kg/m}^3 \times 10 \text{ N/kg} \times 25 \text{ m} = 250\,000 \text{ Pa}\).

1 mark for correct substitution of values into the pressure formula (e.g., \(1000 \times 10 \times 25\)). 1 mark for correct final answer with unit (e.g., \(250\,000 \text{ Pa}\) or \(250 \text{ kPa}\)).

Use the formula for orbital speed: \(v = \frac{2 \pi r}{T}\). Substituting the given values: \(v = \frac{2 \times \pi \times 5.9 \times 10^{12} \text{ m}}{7.8 \times 10^9 \text{ s}}\). Performing the calculation: \(v \approx 4753 \text{ m/s}\), which rounds to \(4800 \text{ m/s}\) (to two significant figures).

1 mark for correct substitution of values into the orbital speed formula (e.g., \(\frac{2 \times \pi \times 5.9 \times 10^{12}}{7.8 \times 10^9}\)). 1 mark for the correct final speed value with unit (accept in the range \(4700 \text{ m/s}\) to \(4800 \text{ m/s}\)).

First, calculate the useful gravitational potential energy gained by the crate using \(GPE = m g h\): \(GPE = 80 \text{ kg} \times 10 \text{ N/kg} \times 6.0 \text{ m} = 4800 \text{ J}\). Next, calculate efficiency using \(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} = \frac{4800 \text{ J}}{6400 \text{ J}} = 0.75\) (or \(75\%\)).

1 mark for calculation of useful work done as \(4800 \text{ J}\) or substitution into efficiency equation. 1 mark for correct final efficiency as a decimal or percentage (e.g., \(0.75\) or \(75\%\)).

Use the formula relating force, change in momentum, and time: \(F = \frac{m(v - u)}{t}\). Here, the initial velocity is \(4.0 \text{ m/s}\) and the final velocity is \(0 \text{ m/s}\). Substitute the values to find the magnitude of the force: \(F = \frac{0.50 \text{ kg} \times 4.0 \text{ m/s}}{0.20 \text{ s}} = 10 \text{ N}\).

1 mark for calculating the momentum change as \(2.0 \text{ kg m/s}\) or substitution into the force formula (e.g., \(\frac{0.50 \times 4.0}{0.20}\)). 1 mark for the correct force magnitude with unit (e.g., \(10 \text{ N}\)).

By conservation of momentum, the total momentum before the collision must equal the total momentum after the collision in a closed system. Initial momentum: \(p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (0.25\text{ kg} \times 3.6\text{ m/s}) + (0.50\text{ kg} \times 0\text{ m/s}) = 0.90\text{ kg m/s}\). Final momentum: since the car rebounds, its final velocity has an opposite sign: \(v_1 = -0.40\text{ m/s}\). Therefore, \(p_{\text{final}} = m_1 v_1 + m_2 v_2 = (0.25\text{ kg} \times -0.40\text{ m/s}) + (0.50\text{ kg} \times v_2) = -0.10\text{ kg m/s} + 0.50 v_2\). Setting initial momentum equal to final momentum: \(0.90 = -0.10 + 0.50 v_2\). Rearranging: \(1.00 = 0.50 v_2\), which gives \(v_2 = 2.0\text{ m/s}\).

1 mark: State conservation of momentum (total momentum before equals total momentum after in a closed system). 1 mark: Correct calculation of total initial momentum as \(0.90\text{ kg m/s}\). 1 mark: Correct use of negative sign for rebounding velocity leading to expression \(-0.10 + 0.50 v_2\). 1 mark: Correct algebraic rearrangement leading to \(1.00 = 0.50 v_2\). 1 mark: Final value \(2.0\) with correct units \(\text{m/s}\).

In a series circuit, the supply voltage is shared: \(V_{\text{resistor}} = V_{\text{total}} - V_{\text{thermistor}} = 12\text{ V} - 4.5\text{ V} = 7.5\text{ V}\). Using Ohm's law for the fixed resistor to find the circuit current: \(I = V / R = 7.5\text{ V} / 15\ \Omega = 0.5\text{ A}\). Since the components are in series, the same current flows through the thermistor. Resistance of the thermistor is calculated using: \(R_{\text{thermistor}} = V_{\text{thermistor}} / I = 4.5\text{ V} / 0.5\text{ A} = 9.0\ \Omega\).

1 mark: Subtract voltages to find potential difference across fixed resistor: \(12 - 4.5 = 7.5\text{ V}\). 1 mark: Recall and use Ohm's law \(I = V / R\) to calculate current. 1 mark: Correct current value of \(0.5\text{ A}\) (accept with units). 1 mark: Use \(R = V / I\) with thermistor's voltage and calculated current. 1 mark: Correct thermistor resistance of \(9.0\ \Omega\) with standard units of ohms.

Total electrical energy supplied: \(E = P \times t = 2400\text{ W} \times 150\text{ s} = 360,000\text{ J}\). Energy required to heat the water to boiling: \(Q_1 = m c \Delta T = 0.50\text{ kg} \times 4200\text{ J/kg}^\circ\text{C} \times (100 - 20)^\circ\text{C} = 168,000\text{ J}\). Energy remaining to turn water into steam: \(Q_2 = E - Q_1 = 360,000\text{ J} - 168,000\text{ J} = 192,000\text{ J}\). Mass of water boiled away: \(m = Q_2 / L = 192,000\text{ J} / (2.26 \times 10^6\text{ J/kg}) \approx 0.08496\text{ kg}\), which rounds to \(0.085\text{ kg}\) (or \(85\text{ g}\)).

1 mark: Calculate total input energy: \(2400\text{ W} \times 150\text{ s} = 360,000\text{ J}\). 1 mark: Calculate energy to heat water: \(0.50 \times 4200 \times 80 = 168,000\text{ J}\). 1 mark: Calculate remaining energy for vaporisation: \(360,000 - 168,000 = 192,000\text{ J}\). 1 mark: Correct substitution into \(Q = mL\) rearranged as \(m = Q/L\). 1 mark: Correct final mass of \(0.085\text{ kg}\) (or \(85\text{ g}\)) with appropriate units.

First, determine how many half-lives have elapsed as activity drops from \(800\text{ Bq}\) to \(100\text{ Bq}\): \(800 \to 400 \to 200 \to 100\), which corresponds to exactly \(3\) half-lives. Since \(3\) half-lives equal \(15\text{ hours}\), the length of one half-life is \(15 / 3 = 5\text{ hours}\). A total time of \(25\text{ hours}\) corresponds to \(25 / 5 = 5\) half-lives. The final activity is therefore: \(800 / 2^5 = 800 / 32 = 25\text{ Bq}\).

1 mark: Identify that 3 half-lives have elapsed for activity to drop to 100 Bq. 1 mark: Calculate the half-life as \(5\text{ hours}\). 1 mark: Determine that 25 hours is equivalent to 5 half-lives. 1 mark: Halve the initial activity 5 times or divide by 32 to get 25. 1 mark: State the correct final value \(25\) with correct unit \(\text{Bq}\) (or Becquerels).

Using the turns ratio equation: \(V_s / V_p = N_s / N_p\), which gives \(V_s = 12\text{ V} \times (3000 / 150) = 12 \times 20 = 240\text{ V}\). Assuming the transformer is 100% efficient, power input equals power output: \(P_{\text{in}} = P_{\text{out}} \implies V_p \times I_p = V_s \times I_s\). Substituting the known values: \(12\text{ V} \times 4.0\text{ A} = 240\text{ V} \times I_s\). This simplifies to \(48\text{ W} = 240 \times I_s\), giving \(I_s = 48 / 240 = 0.20\text{ A}\).

1 mark: Use turns ratio formula to find secondary voltage. 1 mark: Correct secondary voltage calculation: \(240\text{ V}\). 1 mark: State assumption of 100% efficiency or use equation \(V_p I_p = V_s I_s\). 1 mark: Correct secondary current calculation: \(0.20\text{ A}\). 1 mark: State correct units for both calculated values (volts/V and amperes/A).

First, calculate the volume of the cylinder: \(V = \pi r^2 h = 3.14 \times (0.050\text{ m})^2 \times 0.20\text{ m} = 3.14 \times 0.0025 \times 0.20 = 0.00157\text{ m}^3\). Calculate density: \(\text{Density} = m / V = 12.56\text{ kg} / 0.00157\text{ m}^3 = 8000\text{ kg/m}^3\). Next, calculate the base area of the cylinder: \(A = \pi r^2 = 3.14 \times (0.050\text{ m})^2 = 0.00785\text{ m}^2\). Calculate the force exerted on the table (its weight): \(F = m \times g = 12.56\text{ kg} \times 10\text{ N/kg} = 125.6\text{ N}\). Calculate pressure: \(P = F / A = 125.6\text{ N} / 0.00785\text{ m}^2 = 16,000\text{ Pa}\) (or \(\text{N/m}^2\)).

1 mark: Correct calculation of volume: \(0.00157\text{ m}^3\). 1 mark: Correct calculation of density with units: \(8000\text{ kg/m}^3\). 1 mark: Correct calculation of force (weight) as \(125.6\text{ N}\). 1 mark: Correct calculation of pressure as \(16,000\). 1 mark: Correct unit of pressure (Pa or \(\text{N/m}^2\)) given.

When light from distant galaxies is analyzed, its wavelength is found to be increased, which shifts it toward the red end of the spectrum (redshift). This occurs because the galaxies are moving away from us (the Doppler effect). Observations show that more distant galaxies have greater redshifts, meaning they are moving away faster. This uniform recessional movement indicates that space itself is expanding. If the universe is currently expanding, tracing this process back in time suggests that all matter must have originated from a single, extremely dense and hot starting point (the Big Bang).

1 mark: Explain redshift as the increase in wavelength of light from distant galaxies. 1 mark: Connect redshift to the Doppler effect, indicating galaxies are moving away from us. 1 mark: State that more distant galaxies show greater redshift / are moving faster. 1 mark: Explain that this uniform expansion indicates space itself is expanding. 1 mark: Conclude that tracing the expansion backward in time suggests a single dense, hot origin point (Big Bang).

The nuclear equation is written as: \(^{235}_{92}\text{U} + ^{1}_{0}\text{n} \to ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}\). To control the reaction: 1. A moderator (such as water or graphite) slows down the fast-moving neutrons produced during fission. Slow (thermal) neutrons are much more easily captured by uranium nuclei to maintain the fission process. 2. Control rods (made of boron or cadmium) absorb excess neutrons. By adjusting their depth in the reactor core, operators can limit the number of neutrons available to trigger further fissions, preventing a dangerous runaway chain reaction.

1 mark: Correctly balanced reactants side of equation: \(^{235}_{92}\text{U} + ^{1}_{0}\text{n}\). 1 mark: Correctly balanced products side showing Barium, Krypton, and 3 neutrons: \(^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}\). 1 mark: Explain that the moderator slows down fast-moving neutrons. 1 mark: Explain that slow neutrons are required because they are more likely to cause further fission. 1 mark: Explain that control rods absorb neutrons to regulate/limit the rate of the chain reaction.

First, calculate the total initial momentum of the system. Since the toy truck is stationary: \(p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (0.80 \times 3.5) + (1.20 \times 0) = 2.8\text{ kg m/s}\) (to the right). After the collision, the toy car rebounds to the left, so its velocity is \(-0.70\text{ m/s}\). Total final momentum: \(p_{\text{final}} = m_1 v_1 + m_2 v_2 = (0.80 \times -0.70) + (1.20 \times v_2) = -0.56 + 1.20 v_2\). According to the principle of conservation of momentum: \(p_{\text{initial}} = p_{\text{final}}\), so \(2.8 = -0.56 + 1.20 v_2\). Rearranging gives: \(3.36 = 1.20 v_2\), which yields \(v_2 = 2.8\text{ m/s}\). Since the value is positive, the direction of motion is to the right.

- Use of momentum formula \(p = m \times v\) (1 mark)
- Calculation of initial momentum as \(2.8\text{ kg m/s}\) (1 mark)
- Substitution of correct negative velocity for rebound to get \(-0.56\text{ kg m/s}\) (1 mark)
- Rearranging conservation of momentum equation to find magnitude of \(v_2\) as \(2.8\text{ m/s}\) (1 mark)
- Stating correct direction (to the right) with correct unit (1 mark)

First, calculate the current flowing through the known \(15\,\Omega\) resistor using Ohm's law. Since the resistors are in parallel, each experiences the full potential difference of \(9.0\text{ V}\). \(I_1 = \frac{V}{R_1} = \frac{9.0}{15} = 0.6\text{ A}\). Next, find the current flowing through the unknown resistor \(R\) using the junction rule where the total current is the sum of branch currents: \(I_{\text{total}} = I_1 + I_2 \implies 1.2 = 0.6 + I_2 \implies I_2 = 0.6\text{ A}\). Finally, calculate the value of the unknown resistance: \(R = \frac{V}{I_2} = \frac{9.0}{0.6} = 15\,\Omega\).

- Recall and use of \(V = I \times R\) (1 mark)
- Calculate the current through the \(15\,\Omega\) resistor as \(0.6\text{ A}\) (1 mark)
- State or use that the sum of currents in parallel branches equals total current (1 mark)
- Calculate current in the unknown branch as \(0.6\text{ A}\) (1 mark)
- Calculate unknown resistance \(R = 15\,\Omega\) (1 mark)

First, calculate the thermal energy required to melt the ice using the latent heat equation: \(Q = m \times L = 0.20 \times (3.3 \times 10^5) = 66,000\text{ J}\). Next, relate this energy to power and time: \(E = P \times t\). Rearranging for time: \(t = \frac{E}{P} = \frac{66,000}{80} = 825\text{ s}\).

- Recall and use of \(Q = m \times L\) (1 mark)
- Calculate thermal energy required as \(66,000\text{ J}\) (1 mark)
- Recall and use of \(P = \frac{E}{t}\) or \(E = P \times t\) (1 mark)
- Correct substitution of values into time equation (1 mark)
- Correct calculation of \(825\) with unit of seconds (1 mark)

For part (a), use the transformer relation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearranging for \(N_s\): \(N_s = \frac{N_p \times V_s}{V_p} = \frac{1150 \times 12}{230} = 60\) turns. For part (b), assuming 100% efficiency, electrical power input equals electrical power output: \(V_p \times I_p = V_s \times I_s\). Rearranging for \(I_p\): \(I_p = \frac{V_s \times I_s}{V_p} = \frac{12 \times 4.0}{230} = \frac{48}{230} \approx 0.21\text{ A}\).

- Recall and use of \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) (1 mark)
- Correct calculation of secondary turns as \(60\) (1 mark)
- Recall and use of \(V_p \times I_p = V_s \times I_s\) (1 mark)
- Correct substitution of values into current equation (1 mark)
- Correct calculation of primary current as \(0.21\text{ A}\) (accept \(0.209\text{ A}\)) with unit (1 mark)

First, calculate the total radius of the orbit \(r\) from the center of the Earth: \(r = R_{\text{Earth}} + \text{altitude} = 6400\text{ km} + 600\text{ km} = 7000\text{ km}\). Next, convert the orbital period \(T\) from minutes to seconds: \(T = 97 \times 60 = 5820\text{ s}\). Use the orbital speed formula: \(v = \frac{2 \pi r}{T}\). Substitute the values: \(v = \frac{2 \times \pi \times 7000}{5820} \approx \frac{43982.3}{5820} \approx 7.56\text{ km/s}\). Rounded to two significant figures, this is \(7.6\text{ km/s}\).

- Calculate total orbital radius \(r = 7000\text{ km}\) (1 mark)
- Convert orbital period to seconds \(T = 5820\text{ s}\) (1 mark)
- Recall and use orbital speed formula \(v = \frac{2 \pi r}{T}\) (1 mark)
- Correct substitution of values (1 mark)
- Correct calculation of speed as \(7.6\text{ km/s}\) (accept \(7.56\text{ km/s}\)) with unit (1 mark)

First, calculate the useful work done in lifting the crate (which equals the increase in gravitational potential energy): \(\text{GPE} = m \times g \times h = 45 \times 10 \times 12 = 5400\text{ J}\). Next, calculate the electrical energy supplied to the motor: \(E_{\text{input}} = P_{\text{input}} \times t = 480 \times 15 = 7200\text{ J}\). Alternatively, calculate the useful power output: \(P_{\text{output}} = \frac{\text{GPE}}{t} = \frac{5400}{15} = 360\text{ W}\). Then, calculate the efficiency using either energies or powers: \(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} = \frac{5400}{7200} = 0.75\) (or \(75\%\)).

- Recall and use of \(\text{GPE} = mgh\) (1 mark)
- Calculate useful energy output as \(5400\text{ J}\) (1 mark)
- Recall and use of \(E = P \times t\) or \(P = \frac{E}{t}\) (1 mark)
- Calculate total energy input as \(7200\text{ J}\) OR useful power output as \(360\text{ W}\) (1 mark)
- Calculate final efficiency as \(0.75\) or \(75\%\) (1 mark)

First, find the pressure due solely to the liquid (hydrostatic pressure). \(P_{\text{liquid}} = P_{\text{total}} - P_{\text{atmospheric}} = (4.50 \times 10^5) - (1.01 \times 10^5) = 3.49 \times 10^5\text{ Pa}\). Next, use the liquid pressure formula: \(P = h \times \rho \times g\). Rearranging for depth \(h\): \(h = \frac{P_{\text{liquid}}}{\rho \times g}\). Substitute the values: \(h = \frac{3.49 \times 10^5}{1030 \times 10} = \frac{349,000}{10,300} \approx 33.9\text{ m}\) (or \(34\text{ m}\) to two significant figures).

- Subtract atmospheric pressure from total pressure (1 mark)
- Calculate pressure due to seawater as \(3.49 \times 10^5\text{ Pa}\) (1 mark)
- Recall and use formula \(P = h \rho g\) (1 mark)
- Rearrange equation to make \(h\) the subject (1 mark)
- Calculate depth \(33.9\text{ m}\) (accept \(34\text{ m}\)) with unit (1 mark)

For nuclear fusion to occur, two positively charged nuclei must come close enough together for the short-range strong nuclear force to bind them. Because both nuclei are positively charged, they experience a strong electrostatic force of repulsion. High temperatures provide the nuclei with very high kinetic energies, allowing them to travel fast enough to overcome this repulsion. High pressures increase the concentration of nuclei, making collisions much more frequent.

1 mark for selecting option A. Reject all other options.

The magnitude of the induced voltage is proportional to the rate of cutting of magnetic flux. To increase the peak voltage, the student can use a stronger magnet, move the magnet faster, or use a coil with more turns of wire. Therefore, moving a stronger magnet into the coil at a faster speed will increase the induced voltage.

1 mark for selecting option D. Reject all other options.

Using the formula: \(W = Q \times V\)
Substitute the given values:
\(W = 15 \text{ C} \times 6.0 \text{ V}\)
\(W = 90 \text{ J}\)

1 mark for the correct formula: \(W = Q \times V\). 1 mark for correct substitution: \(15 \times 6.0\). 0.5 mark for correct answer with appropriate unit (90 J).

Nuclei are positively charged and experience electrostatic repulsion. High temperatures provide the high kinetic energy needed to overcome this repulsion and bring the nuclei close enough to fuse. High pressure increases the collision rate of the nuclei.

1 mark for identifying that nuclei repel each other due to positive charges. 1 mark for explaining that high temperature provides enough kinetic energy to overcome this repulsion. 0.5 mark for explaining that high pressure increases the frequency of collisions / brings them close together.

Using the formula: \(v = \frac{2 \pi r}{T}\)
Substitute the values:
\(v = \frac{2 \times \pi \times 7.0 \times 10^6}{5.6 \times 10^3}\)
\(v \approx 7854 \text{ m/s}\) (or \(7900 \text{ m/s}\) to 2 significant figures).

1 mark for selecting the correct formula: \(v = \frac{2 \pi r}{T}\). 1 mark for correct substitution. 0.5 mark for correct final answer with unit (e.g. \(7900 \text{ m/s}\) or \(7.9 \text{ km/s}\)).

Using the formula: \(Q = m L_f\)
Substitute the values:
\(Q = 0.25 \text{ kg} \times 3.3 \times 10^5 \text{ J/kg}\)
\(Q = 82500 \text{ J}\) (or \(83000 \text{ J}\) to 2 significant figures).

1 mark for selecting the correct formula: \(Q = m L\). 1 mark for correct substitution. 0.5 mark for correct final answer with unit (82500 J, 83000 J, or 83 kJ).

Using the formula for pressure difference in a liquid: \(p = \rho \times g \times h\)
Substitute the values:
\(p = 1000 \text{ kg/m}^3 \times 10 \text{ N/kg} \times 15 \text{ m}\)
\(p = 150000 \text{ Pa}\) (or \(150 \text{ kPa}\)).

1 mark for selecting the correct formula: \(p = \rho g h\). 1 mark for correct substitution. 0.5 mark for correct final answer with unit (150000 Pa or 150 kPa).

Let the initial velocity be positive: \(u = +12 \text{ m/s}\). The final velocity is negative because it rebounds in the opposite direction: \(v = -8.0 \text{ m/s}\).
Change in momentum \(\Delta p = m(v - u)\)
\(\Delta p = 0.15 \times (-8.0 - 12)\)
\(\Delta p = 0.15 \times (-20) = -3.0 \text{ kg m/s}\).
The magnitude of the change in momentum is \(3.0 \text{ kg m/s}\).

1 mark for understanding that momentum is a vector and calculating the velocity difference including sign change: \(\Delta v = 12 - (-8) = 20 \text{ m/s}\). 1 mark for calculation of momentum: \(0.15 \times 20\). 0.5 mark for correct answer with unit (3.0 kg m/s or 3.0 N s).

A step-up transformer has more turns of wire on its secondary coil than on its primary coil. The alternating current in the primary coil creates a continuously changing magnetic field in the soft iron core. This changing magnetic field cuts through the secondary coil, inducing a larger alternating voltage across its ends due to the higher number of turns.

1 mark for stating that there are more turns on the secondary coil than on the primary coil. 1 mark for explaining that the alternating current in the primary coil creates a changing magnetic field in the core. 0.5 mark for stating that this changing magnetic field cuts the secondary coil to induce a larger voltage.

Using the formula: \(\sin(c) = \frac{1}{n}\)
Substitute the value of \(n\):
\(\sin(c) = \frac{1}{1.6} = 0.625\)
\(c = \arcsin(0.625) \approx 38.7^\circ\) (which rounds to \(39^\circ\)).

1 mark for using the correct formula: \(\sin(c) = \frac{1}{n}\). 1 mark for calculating \(\sin(c) = 0.625\). 0.5 mark for correct final angle (39 degrees or 38.7 degrees).

1. Convert the initial and final temperatures from Celsius to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 127 + 273 = 400\text{ K}\). 2. Use the pressure law for constant volume: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). 3. Rearrange the formula to solve for the final pressure: \(P_2 = P_1 \times \frac{T_2}{T_1}\). 4. Substitute the values: \(P_2 = 1.5 \times 10^5\text{ Pa} \times \frac{400\text{ K}}{300\text{ K}} = 2.0 \times 10^5\text{ Pa}\).

[1 mark] Conversion of both temperatures to Kelvin (300 K and 400 K). [1 mark] Correct recall and rearrangement of the pressure law formula. [0.5 marks] Correct final pressure value with unit (2.0 x 10^5 Pa or 200,000 Pa).

1. Find the change in wavelength: \(\Delta \lambda = 612\text{ nm} - 600\text{ nm} = 12\text{ nm}\). 2. State the redshift equation: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\). 3. Rearrange the equation to find velocity: \(v = c \times \frac{\Delta \lambda}{\lambda_0}\). 4. Substitute the values: \(v = (3.0 \times 10^8\text{ m/s}) \times \frac{12\text{ nm}}{600\text{ nm}} = 3.0 \times 10^8 \times 0.02 = 6.0 \times 10^6\text{ m/s}\).

[1 mark] Calculation of the change in wavelength (12 nm) or correct substitution of change in wavelength into equation. [1 mark] Correct rearrangement of redshift formula to make velocity (v) the subject. [0.5 marks] Correct calculation of the recessional velocity with unit (6.0 x 10^6 m/s or 6,000,000 m/s).

1. Identify the output power (secondary power, \(P_s\)):
\(P_s = 36\text{ W}\)

2. Use the transformer efficiency formula to find the input power (primary power, \(P_p\)):
\(\text{Efficiency} = \frac{P_s}{P_p}\)
\(0.92 = \frac{36}{P_p}\)
\(P_p = \frac{36}{0.92} \approx 39.13\text{ W}\)

3. Use the relationship between power, current, and voltage in the primary coil:
\(P_p = I_p \times V_p\)
\(39.13 = I_p \times 240\)
\(I_p = \frac{39.13}{240} \approx 0.163\text{ A}\)

Rounding to two significant figures gives \(0.16\text{ A}\).

• Recall and rearrangement of the efficiency equation to find input power (1 mark)
• Calculation of input power as 39.13 W (1 mark)
• Use of \(P = I \times V\) rearranged to make \(I_p\) the subject (1 mark)
• Correct substitution and calculation of primary current to 2 sig figs (1.8 marks)

1. Calculate the voltage across the fixed resistor (\(V_{\text{fixed}}\)):
\(V_{\text{fixed}} = V_{\text{total}} - V_{\text{LDR}}\)
\(V_{\text{fixed}} = 9.0\text{ V} - 6.0\text{ V} = 3.0\text{ V}\)

2. Calculate the current (\(I\)) flowing through the circuit using Ohm's law on the fixed resistor:
\(I = \frac{V_{\text{fixed}}}{R_{\text{fixed}}}\)
\(R_{\text{fixed}} = 1.5\text{ k}\Omega = 1500\ \Omega\)
\(I = \frac{3.0\text{ V}}{1500\ \Omega} = 0.0020\text{ A}\) (or \(2.0\text{ mA}\))

3. Since the components are in series, the same current flows through the LDR. Calculate the resistance of the LDR (\(R_{\text{LDR}}\)):
\(R_{\text{LDR}} = \frac{V_{\text{LDR}}}{I} = \frac{6.0\text{ V}}{0.0020\text{ A}} = 3000\ \Omega\) (or \(3.0\text{ k}\Omega\))

• Calculation of the potential difference across the fixed resistor: 3.0 V (1 mark)
• Recall of \(V = IR\) and calculation of current: 0.002 A or 2.0 mA (1 mark)
• Correct calculation of the LDR resistance: 3000 ̳ (1.8 marks)
• Correct unit (\(\Omega\) or \(\text{k}\Omega\)) consistent with their working (1 mark)

1. Write the balanced nuclear equation:
\(^{2}_{1}\text{H} + {^{3}_{1}\text{H}} \rightarrow {^{4}_{2}\text{He}} + {^{1}_{0}\text{n}}\)
Both the atomic numbers (1 + 1 = 2 + 0) and the mass numbers (2 + 3 = 4 + 1) balance on both sides.

2. Explanation of conditions:
- Hydrogen nuclei are positively charged (protons) and therefore strongly repel each other electrostatically.
- Extremely high temperatures are needed to give the nuclei enough kinetic energy to overcome this electrostatic repulsion and get close enough for the strong nuclear force to bind them together.
- Extremely high pressure (and density) is required to increase the frequency of collisions between the nuclei, ensuring that a sufficient rate of fusion reactions occurs.

• Correct reactants and products in a balanced nuclear equation (2 marks)
• Mention of electrostatic repulsion between positively charged nuclei (1 mark)
• Explanation that high temperature provides high kinetic energy to overcome repulsion (1 mark)
• Explanation that high pressure increases collision frequency/chance of fusion (0.8 marks)

1. Calculate total energy supplied by the kettle:
\(E = \text{Power} \times \text{time} = 2200\text{ W} \times 120\text{ s} = 264000\text{ J}\)

2. Calculate energy required to heat the water to boiling point (\(100\ ^\circ\text{C}\)):
\(Q_1 = m \times c \times \Delta \theta\)
\(\Delta \theta = 100\ ^\circ\text{C} - 20\ ^\circ\text{C} = 80\ ^\circ\text{C}\)
\(Q_1 = 0.50\text{ kg} \times 4200\text{ J/kg}^\circ\text{C} \times 80\ ^\circ\text{C} = 168000\text{ J}\)

3. Calculate remaining energy available for boiling/vaporisation:
\(Q_2 = E - Q_1 = 264000\text{ J} - 168000\text{ J} = 96000\text{ J}\)

4. Calculate the mass of water turned into steam:
\(Q_2 = m_{\text{steam}} \times L\)
\(96000\text{ J} = m_{\text{steam}} \times 2.26 \times 10^6\text{ J/kg}\)
\(m_{\text{steam}} = \frac{96000}{2.26 \times 10^6} \approx 0.04248\text{ kg}\)

5. Calculate mass of liquid remaining in the kettle:
\(m_{\text{remaining}} = m_{\text{initial}} - m_{\text{steam}} = 0.50\text{ kg} - 0.04248\text{ kg} \approx 0.4575\text{ kg}\)

Rounding to two significant figures gives \(0.46\text{ kg}\).

• Calculation of total energy supplied: 264,000 J (1 mark)
• Calculation of energy needed to reach boiling point: 168,000 J (1 mark)
• Subtraction to find energy available for change of state: 96,000 J (1 mark)
• Calculation of the mass of steam produced: 0.042 kg (0.8 marks)
• Correct final remaining mass of water with units: 0.46 kg (1 mark)

1. Use conservation of momentum to find the final velocity (\(v\)) after collision:
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
\(0.25 \times 3.2 + 0.55 \times 0 = (0.25 + 0.55) v\)
\(0.80 = 0.80 v\)
\(v = 1.0\text{ m/s}\)

2. Calculate the total initial kinetic energy (\(KE_i\)):
\(KE_i = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 0.25 \times (3.2)^2 = 0.125 \times 10.24 = 1.28\text{ J}\)

3. Calculate the total final kinetic energy (\(KE_f\)):
\(KE_f = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 0.80 \times (1.0)^2 = 0.40\text{ J}\)

4. Calculate the loss in kinetic energy:
\(\Delta KE = KE_i - KE_f = 1.28\text{ J} - 0.40\text{ J} = 0.88\text{ J}\)

• Use of conservation of momentum to calculate final velocity of 1.0 m/s (1 mark)
• Correct calculation of initial kinetic energy: 1.28 J (1 mark)
• Correct calculation of final kinetic energy: 0.40 J (1 mark)
• Calculation of the difference to find kinetic energy lost: 0.88 J (1.8 marks)

1. Recall the redshift formula:
\(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\)

2. Find the change in wavelength (\(\Delta \lambda\)):
\(\Delta \lambda = \lambda - \lambda_0 = 671.1\text{ nm} - 656.3\text{ nm} = 14.8\text{ nm}\)

3. Substitute the values into the formula:
\(\frac{14.8\text{ nm}}{656.3\text{ nm}} = \frac{v}{3.0 \times 10^8\text{ m/s}}\)

4. Calculate the velocity (\(v\)):
\(v = \frac{14.8}{656.3} \times 3.0 \times 10^8\text{ m/s}\)
\(v \approx 0.02255 \times 3.0 \times 10^8\text{ m/s} \approx 6.765 \times 10^6\text{ m/s}\)

Rounding to two significant figures gives \(6.8 \times 10^6\text{ m/s}\).

• Recall of the redshift formula (1 mark)
• Calculation of change in wavelength \(\Delta \lambda = 14.8\text{ nm}\) (1 mark)
• Substitution of values into formula (1 mark)
• Correct final value and unit to 2 or 3 sig figs: \(6.8 \times 10^6\text{ m/s}\) or \(6.77 \times 10^6\text{ m/s}\) (1.8 marks)

1. Deduct background radiation to find the corrected count rates for the source:
- Corrected rate at 0 hours: \(404\text{ cpm} - 20\text{ cpm} = 384\text{ cpm}\)
- Corrected rate at 3 hours: \(212\text{ cpm} - 20\text{ cpm} = 192\text{ cpm}\)
- Corrected rate at 6 hours: \(116\text{ cpm} - 20\text{ cpm} = 96\text{ cpm}\)

2. Examine how the corrected counts decrease over time:
- From 0 to 3 hours (an interval of 3 hours), the corrected count rate halves: \(\frac{384}{2} = 192\text{ cpm}\)
- From 3 to 6 hours (another interval of 3 hours), the corrected count rate halves again: \(\frac{192}{2} = 96\text{ cpm}\)

3. Since the corrected activity halves exactly every 3 hours, the half-life of the isotope is 3 hours.

• Stating that background radiation must be subtracted to find corrected count rate (1 mark)
• Calculating corrected counts: 384, 192, and 96 cpm (1.8 marks)
• Explaining that the corrected count rate halves over an interval of 3 hours (1 mark)
• Final concluding statement showing half-life is 3 hours (1 mark)

1. Use the critical angle formula to calculate the refractive index (\(n\)) of the glass:
\(\sin(c) = \frac{1}{n}\)
\(n = \frac{1}{\sin(41.8^\circ)}\)
\(\sin(41.8^\circ) \approx 0.6665\)
\(n = \frac{1}{0.6665} \approx 1.50\)

2. Use Snell's Law for the ray entering the glass from air:
\(n = \frac{\sin(i)}{\sin(r)}\)
\(1.50 = \frac{\sin(35.0^\circ)}{\sin(r)}\)
\(\sin(35.0^\circ) \approx 0.5736\)
\(\sin(r) = \frac{0.5736}{1.50} \approx 0.3824\)
\(r = \sin^{-1}(0.3824) \approx 22.5^\circ\)

• Recall of critical angle equation \(n = 1/\sin(c)\) (1 mark)
• Correct calculation of refractive index as 1.50 (1 mark)
• Recall of Snell's Law \(n = \sin(i)/\sin(r)\) and correct rearrangement (1 mark)
• Correct substitution and calculation of the angle of refraction as \(22.5^\circ\) (1.8 marks)

Step 1: Convert the initial temperature from Celsius to Kelvin. \(T_1 = 17 + 273 = 290\text{ K}\). Step 2: Use the pressure law formula: \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\). Step 3: Rearrange the formula to solve for the final temperature \(T_2 = \frac{p_2 \times T_1}{p_1}\). Step 4: Substitute the values: \(T_2 = \frac{1.25 \times 10^5\text{ Pa} \times 290\text{ K}}{1.0 \times 10^5\text{ Pa}} = 362.5\text{ K}\). Step 5: Convert back to Celsius: \(T_2 = 362.5 - 273 = 89.5\text{ }^\circ\text{C}\). Step 6: Identify the key assumption, which is that the volume of the container remains constant.

M1: Convert temperature to Kelvin: \(17 + 273 = 290\text{ K}\) (1). M2: Recall and use \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\) (1). M3: Calculate final temperature in Kelvin: \(362.5\text{ K}\) (1). A1: Correct conversion to Celsius: \(89.5\text{ }^\circ\text{C}\) (accept rounded value of \(90\text{ }^\circ\text{C}\)) (1). B1: State that the volume of the gas/container remains constant (1).