Physics 4PH1 | Paper 1 Mock 01 | Written Paper 1
Question 1 Report
A balloon seller uses a portable burner to heat air inside a hot-air balloon before launch. Fig. 1 shows the balloon envelope, basket and open lower mouth. The heated air can move in and out of the balloon, so it is not trapped in a sealed container. The surrounding air is cooler.
(a) State how heating changes the speed of the air molecules inside the balloon. [1]
(b) Explain why the heated air becomes less dense than the cooler air outside. [3]
(c) State the direction of the upthrust on the balloon. [1]
(d) Explain why the balloon rises when the upthrust is greater than its weight. [2]
(e) Describe what the pilot should do to descend slowly. [2]
(f) Calculate the weight of a basket and equipment with mass 180 kg. Take gravitational field strength as 10 N/kg. [2]
(g) State which gas property is reduced when the heated air spreads out: density, mass, or temperature? [1]
(h) Explain why the burner must be switched off before the balloon is packed away. [2]
(a) Heating makes the air molecules move faster. [1]
(b) The molecules gain kinetic energy. They spread further apart and occupy a greater volume. The same mass of air therefore has less mass per unit volume, so its density is lower than that of the surrounding cooler air. [3]
(c) Upthrust acts upwards. [1]
(d) If upthrust is greater than weight, there is a resultant upward force. The balloon therefore accelerates upwards and rises. [2]
(e) The pilot should reduce or turn down the burner. The air then cools, becomes denser, and the upthrust decreases, allowing a slow descent. [2]
(f)
\[W=mg=180\times10=1800\text{ N}\]
Weight = \(1800\text{ N}\). [2]
(g) The property reduced when the heated air spreads out is density. [1]
(h) The burner must be switched off so hot parts can cool, reducing the risk of burns or fire damage to the fabric. [2]
Answer Details
(a) Heating makes the air molecules move faster. [1]
(b) The molecules gain kinetic energy. They spread further apart and occupy a greater volume. The same mass of air therefore has less mass per unit volume, so its density is lower than that of the surrounding cooler air. [3]
(c) Upthrust acts upwards. [1]
(d) If upthrust is greater than weight, there is a resultant upward force. The balloon therefore accelerates upwards and rises. [2]
(e) The pilot should reduce or turn down the burner. The air then cools, becomes denser, and the upthrust decreases, allowing a slow descent. [2]
(f)
\[W=mg=180\times10=1800\text{ N}\]
Weight = \(1800\text{ N}\). [2]
(g) The property reduced when the heated air spreads out is density. [1]
(h) The burner must be switched off so hot parts can cool, reducing the risk of burns or fire damage to the fabric. [2]
Question 2 Report
Fig. 1 shows equipment used to time a metal trolley travelling down a ramp. A card fixed to the trolley passes through two light gates. The data logger measures the time between the gates. The distance between the light gates is 0.80 m. A student adds small masses to the trolley box but keeps the ramp angle unchanged.
(a) State the instrument used by the data logger to measure the time interval. [1]
(b) Describe how the student can calculate the average speed of the trolley between the light gates. [2]
(c) Calculate the average speed when the measured time is 0.40 s. [2]
(d) Explain why adding mass to the trolley does not necessarily make its acceleration down the ramp greater. [3]
(a) The data logger uses light gates to measure the time interval. [1]
(b) Use the known distance between the light gates [1] and divide it by the measured time interval:
\[\text{average speed}=\frac{\text{distance}}{\text{time}}\]
[2]
(c) \[\text{average speed}=\frac{0.80\text{ m}}{0.40\text{ s}}=2.0\text{ m/s}\]
The average speed is 2.0 m/s. [2]
(d) Increasing mass increases the downhill component of the trolley's weight. [1] But it can also increase friction or other resistive forces. [1] As acceleration depends on resultant force divided by mass, \(a=F_{\text{resultant}}/m\), the acceleration may remain the same and need not increase. [1]
Answer Details
(a) The data logger uses light gates to measure the time interval. [1]
(b) Use the known distance between the light gates [1] and divide it by the measured time interval:
\[\text{average speed}=\frac{\text{distance}}{\text{time}}\]
[2]
(c) \[\text{average speed}=\frac{0.80\text{ m}}{0.40\text{ s}}=2.0\text{ m/s}\]
The average speed is 2.0 m/s. [2]
(d) Increasing mass increases the downhill component of the trolley's weight. [1] But it can also increase friction or other resistive forces. [1] As acceleration depends on resultant force divided by mass, \(a=F_{\text{resultant}}/m\), the acceleration may remain the same and need not increase. [1]
Question 3 Report
A railway operator studies the motion of an unpowered maintenance cart on a straight track. Fig. 1 shows the cart before it rolls into a sand-filled stopping box. The cart has a mass of 500 kg and enters the sand at 4.0 m/s. It comes to rest after travelling 1.6 m through the sand. Assume the retarding force is constant.
(a) Calculate the initial momentum of the cart. [2]
(b) Calculate the initial kinetic energy of the cart. [3]
(c) Calculate the magnitude of the deceleration of the cart in the sand. [3]
(d) Calculate the retarding force exerted by the sand. [2]
(e) Explain why using a longer sand box reduces the force on the cart. [3]
(f) State one change to the cart that would increase its momentum at the same speed. [1]
(g) Describe the main energy transfer as the cart stops. [2]
(a) Momentum is \(p=mv\):
\[p=500\times4.0=2000\text{ kg m/s}\]
The initial momentum is 2000 kg m/s. [2]
(b) Use \(E_k=\frac{1}{2}mv^2\):
\[E_k=\frac{1}{2}\times500\times4.0^2=4000\text{ J}\]
The initial kinetic energy is 4000 J. [3]
(c) Use \(v^2=u^2+2as\):
\[0=4.0^2+2\times a\times1.6\]
\[a=-5.0\text{ m/s}^2\]
The deceleration has magnitude 5.0 m/s². [3]
(d) \[F=ma=500\times5.0=2500\text{ N}\]
The retarding force is 2500 N. [2]
(e) A longer sand box gives a greater stopping distance. [1] This increases stopping time or reduces the deceleration. [1] Since \(F=ma\), a smaller deceleration requires a smaller resultant force. [1]
(f) Increase the cart's mass. Since \(p=mv\), this increases momentum at the same speed. [1]
(g) As the cart stops, its kinetic energy is transferred mainly to thermal energy in the sand and cart. [1] Some energy may also be transferred as sound. [1]
Answer Details
(a) Momentum is \(p=mv\):
\[p=500\times4.0=2000\text{ kg m/s}\]
The initial momentum is 2000 kg m/s. [2]
(b) Use \(E_k=\frac{1}{2}mv^2\):
\[E_k=\frac{1}{2}\times500\times4.0^2=4000\text{ J}\]
The initial kinetic energy is 4000 J. [3]
(c) Use \(v^2=u^2+2as\):
\[0=4.0^2+2\times a\times1.6\]
\[a=-5.0\text{ m/s}^2\]
The deceleration has magnitude 5.0 m/s². [3]
(d) \[F=ma=500\times5.0=2500\text{ N}\]
The retarding force is 2500 N. [2]
(e) A longer sand box gives a greater stopping distance. [1] This increases stopping time or reduces the deceleration. [1] Since \(F=ma\), a smaller deceleration requires a smaller resultant force. [1]
(f) Increase the cart's mass. Since \(p=mv\), this increases momentum at the same speed. [1]
(g) As the cart stops, its kinetic energy is transferred mainly to thermal energy in the sand and cart. [1] Some energy may also be transferred as sound. [1]
Question 4 Report
Fig. 1 shows part of the circuit inside a plug fitted to a 2.3 kW kettle. The kettle is connected to a 230 V mains supply. The wires in the flex are insulated and the plug has a fuse in the live wire.
(a) State the function of the fuse in this electric circuit. [2]
(b) Explain why the fuse is placed in the live wire rather than the neutral wire. [2]
(c) Calculate the current taken by the kettle when it is operating normally. [3]
(d) Which fuse rating is most suitable for this kettle: 3 A, 5 A or 13 A? [2]
(a) A fuse melts when the current becomes too large, breaking the circuit and stopping current to the kettle. [2]
(b) If a fuse were in the neutral wire, a fault could leave the appliance connected to the live supply. A fuse in the live wire disconnects the live supply when it melts, making the appliance safer. [2]
(c) Convert \(2.3\text{ kW}\) to \(2300\text{ W}\), then use \(I=P/V\):
\[I=\frac{2300}{230}=10\text{ A}\]
The normal current is \(10\text{ A}\). [3]
(d) A 13 A fuse is most suitable. It is the smallest available rating above the normal operating current of 10 A, so it should not melt in normal use but protects against excessive current. [2]
Answer Details
(a) A fuse melts when the current becomes too large, breaking the circuit and stopping current to the kettle. [2]
(b) If a fuse were in the neutral wire, a fault could leave the appliance connected to the live supply. A fuse in the live wire disconnects the live supply when it melts, making the appliance safer. [2]
(c) Convert \(2.3\text{ kW}\) to \(2300\text{ W}\), then use \(I=P/V\):
\[I=\frac{2300}{230}=10\text{ A}\]
The normal current is \(10\text{ A}\). [3]
(d) A 13 A fuse is most suitable. It is the smallest available rating above the normal operating current of 10 A, so it should not melt in normal use but protects against excessive current. [2]
Question 5 Report
The diagram shows a battery pack powering a model rescue boat motor. Table 1 gives the terminal voltage of the battery pack at different motor currents. The student uses a data logger so that each reading is taken after the motor speed has become steady. The battery has internal resistance, so some energy is transferred thermally inside the battery when current flows.
| motor current / A | terminal voltage / V |
|---|---|
| 0.50 | 6.0 |
| 1.00 | 5.8 |
| 1.50 | 5.6 |
(a) Calculate the resistance of the motor when its current is 1.00 A. [2]
(b) Calculate the electrical energy transferred to the motor in 120 s at 1.00 A. [3]
(c) Explain why the terminal voltage decreases as the motor current increases. [3]
(d) Calculate the internal resistance of the battery pack using the first and last rows of Table 1. [2]
(a) At \(1.00\ \mathrm{A}\), the terminal voltage is \(5.8\ \mathrm{V}\). Therefore:
\[R=\frac{V}{I}=\frac{5.8\ \mathrm{V}}{1.00\ \mathrm{A}}=5.8\ \Omega\]
Motor resistance = \(5.8\ \Omega\) [2].
(b) The power delivered to the motor is:
\[P=VI=5.8\ \mathrm{V}\times1.00\ \mathrm{A}=5.8\ \mathrm{W}\]
\[E=Pt=5.8\ \mathrm{W}\times120\ \mathrm{s}=696\ \mathrm{J}\]
Energy transferred to the motor = \(696\ \mathrm{J}\) [3].
(c) The battery has internal resistance [1]. A larger current produces a larger voltage drop across this internal resistance [1]. More energy is transferred thermally inside the battery, leaving a lower terminal voltage for the motor [1].
(d) From the first and last readings:
\[\Delta V=6.0-5.6=0.4\ \mathrm{V}\]
\[\Delta I=1.50-0.50=1.00\ \mathrm{A}\]
\[r=\frac{\Delta V}{\Delta I}=\frac{0.4}{1.00}=0.40\ \Omega\]
Internal resistance = \(0.40\ \Omega\) [2].
Answer Details
(a) At \(1.00\ \mathrm{A}\), the terminal voltage is \(5.8\ \mathrm{V}\). Therefore:
\[R=\frac{V}{I}=\frac{5.8\ \mathrm{V}}{1.00\ \mathrm{A}}=5.8\ \Omega\]
Motor resistance = \(5.8\ \Omega\) [2].
(b) The power delivered to the motor is:
\[P=VI=5.8\ \mathrm{V}\times1.00\ \mathrm{A}=5.8\ \mathrm{W}\]
\[E=Pt=5.8\ \mathrm{W}\times120\ \mathrm{s}=696\ \mathrm{J}\]
Energy transferred to the motor = \(696\ \mathrm{J}\) [3].
(c) The battery has internal resistance [1]. A larger current produces a larger voltage drop across this internal resistance [1]. More energy is transferred thermally inside the battery, leaving a lower terminal voltage for the motor [1].
(d) From the first and last readings:
\[\Delta V=6.0-5.6=0.4\ \mathrm{V}\]
\[\Delta I=1.50-0.50=1.00\ \mathrm{A}\]
\[r=\frac{\Delta V}{\Delta I}=\frac{0.4}{1.00}=0.40\ \Omega\]
Internal resistance = \(0.40\ \Omega\) [2].
Question 6 Report
A hospital physicist prepares a patient for a scan using technetium-99m. At the time of injection, the activity of the sample is 320 MBq. Its half-life is 6.0 hours. The gamma detector has a circuit in which radiation produces small pulses of current. The physicist uses the smallest suitable activity because energy from ionising radiation can damage atoms and electrons in body tissue.
(a) State what an activity of 320 MBq means. [2]
(b) Calculate the activity remaining after 18 hours. [3]
(c) Explain why the patient is advised to avoid long periods close to other people immediately after the scan. [2]
(d) Which unit is used for absorbed dose, meaning energy transferred to each kilogram of material? [1]
(e) State one difference between absorbed dose and equivalent dose. [1]
(a) \(320\text{ MBq}=320\,000\,000\text{ Bq}\). This means that 320 000 000 nuclear decays occur each second. [2]
(b) \[\frac{18}{6.0}=3\text{ half-lives}\]
\[320\text{ MBq}\rightarrow160\text{ MBq}\rightarrow80\text{ MBq}\rightarrow40\text{ MBq}\]
Activity remaining = \(40\text{ MBq}\). [3]
(c) After injection, the patient emits ionising gamma radiation. Reducing the time spent close to the patient reduces the radiation dose received by other people. [2]
(d) Absorbed dose is measured in grays, \(\text{Gy}\). [1]
(e) Absorbed dose is energy absorbed per unit mass. Equivalent dose also takes account of the radiation type and its biological effect. [1]
Answer Details
(a) \(320\text{ MBq}=320\,000\,000\text{ Bq}\). This means that 320 000 000 nuclear decays occur each second. [2]
(b) \[\frac{18}{6.0}=3\text{ half-lives}\]
\[320\text{ MBq}\rightarrow160\text{ MBq}\rightarrow80\text{ MBq}\rightarrow40\text{ MBq}\]
Activity remaining = \(40\text{ MBq}\). [3]
(c) After injection, the patient emits ionising gamma radiation. Reducing the time spent close to the patient reduces the radiation dose received by other people. [2]
(d) Absorbed dose is measured in grays, \(\text{Gy}\). [1]
(e) Absorbed dose is energy absorbed per unit mass. Equivalent dose also takes account of the radiation type and its biological effect. [1]
Question 7 Report
The diagram shows part of a phone charger used at a community repair workshop. A transformer changes the 230 V alternating voltage from the mains before the charger electronics supply a battery. The primary coil has 1200 turns and the secondary coil has 60 turns. The transformer has an iron core and is inside a plastic box.
(a) State the type of voltage supplied to the primary coil. [1]
(b) Explain why the voltage in the secondary coil is much smaller than the voltage in the primary coil. [2]
(c) Calculate the secondary voltage. [2]
(a) The primary coil is supplied with an alternating voltage, or a.c. [1]
(b) The secondary coil has far fewer turns than the primary coil. [1] Transformer voltage is proportional to number of turns, so this is a step-down transformer and the secondary voltage is much smaller. [1]
(c) Use the transformer equation:
\[\frac{V_s}{V_p}=\frac{N_s}{N_p}\]
\[\frac{V_s}{230}=\frac{60}{1200}\]
\[V_s=230\times\frac{60}{1200}=11.5\text{ V}\]
The secondary voltage is 11.5 V. [2]
Answer Details
(a) The primary coil is supplied with an alternating voltage, or a.c. [1]
(b) The secondary coil has far fewer turns than the primary coil. [1] Transformer voltage is proportional to number of turns, so this is a step-down transformer and the secondary voltage is much smaller. [1]
(c) Use the transformer equation:
\[\frac{V_s}{V_p}=\frac{N_s}{N_p}\]
\[\frac{V_s}{230}=\frac{60}{1200}\]
\[V_s=230\times\frac{60}{1200}=11.5\text{ V}\]
The secondary voltage is 11.5 V. [2]
Question 8 Report
Fig. 1 shows a beta thickness gauge used as paper moves through a printing factory. A beta source is above the paper and a detector is below it. The detector is connected to a control box. When the count is lower than the value set by the operator, the control box moves the rollers closer together. With normal paper the detector count is 800 counts per minute. A test with unusually thick paper gives 600 counts per minute.
(a) State why beta radiation, rather than alpha radiation, is used in this gauge. [1]
(b) Explain why beta radiation is more suitable than gamma radiation for this use. [3]
(c) Describe how the control box responds when paper becomes too thick. [2]
(d) Calculate the percentage decrease in detector count for the thick paper. [2]
(e) Explain why thicker paper gives a lower count. [2]
(f) State two precautions used to reduce radiation exposure from the source. [2]
(g) Complete the statement: the activity of a source is the number of nuclear decays per _____. [1]
(a) Beta is used because it can pass through paper, whereas alpha radiation is stopped by paper or by air. [1]
(b) Beta is partly absorbed by paper, so a change in thickness produces a measurable change in detector count. Gamma is too penetrating, so the same thickness change would cause a much smaller count change. [3]
(c) When paper becomes too thick, the detector count falls below the set value. The control box moves the rollers closer together, making the paper thinner. [2]
(d)\[\text{decrease}=800-600=200\text{ counts min}^{-1}\]
\[\text{percentage decrease}=\frac{200}{800}\times100=25\%\]
The percentage decrease is 25%. [2]
(e) Thicker paper contains more material. More beta particles are absorbed or stopped before reaching the detector, giving a lower count. [2]
(f) Any two precautions are acceptable: shield the source when not in use, use a sealed source, minimise time nearby, keep workers at a distance, or use remote operation. [2]
(g) Activity is the number of nuclear decays per second. [1]
Answer Details
(a) Beta is used because it can pass through paper, whereas alpha radiation is stopped by paper or by air. [1]
(b) Beta is partly absorbed by paper, so a change in thickness produces a measurable change in detector count. Gamma is too penetrating, so the same thickness change would cause a much smaller count change. [3]
(c) When paper becomes too thick, the detector count falls below the set value. The control box moves the rollers closer together, making the paper thinner. [2]
(d)\[\text{decrease}=800-600=200\text{ counts min}^{-1}\]
\[\text{percentage decrease}=\frac{200}{800}\times100=25\%\]
The percentage decrease is 25%. [2]
(e) Thicker paper contains more material. More beta particles are absorbed or stopped before reaching the detector, giving a lower count. [2]
(f) Any two precautions are acceptable: shield the source when not in use, use a sealed source, minimise time nearby, keep workers at a distance, or use remote operation. [2]
(g) Activity is the number of nuclear decays per second. [1]
Question 9 Report
A mechanic checks the stopping force of a small car on a test track. Fig. 1 shows the car travelling at 18 m/s before braking. Its mass is 950 kg. During a braking time of 3.0 s, its velocity decreases uniformly to 6.0 m/s.
(a) Calculate the acceleration of the car while braking. [2]
(b) Calculate the resultant force on the car. [2]
(c) State the direction of this resultant force. [1]
(d) Calculate the distance moved by the car during the 3.0 s braking time. [3]
(e) Explain why a passenger moves forward relative to the car when it starts to brake. [2]
(a) Acceleration is change in velocity divided by time:
\[a=\frac{v-u}{t}=\frac{6.0-18}{3.0}=-4.0\text{ m/s}^2\]
The acceleration is \(-4.0\text{ m/s}^2\). The negative sign shows that the acceleration is opposite to the original direction of motion. [2]
(b) Apply Newton's second law, \(F=ma\):
\[F=950\times(-4.0)=-3800\text{ N}\]
The resultant force is \(-3800\text{ N}\), with magnitude 3800 N. [2]
(c) The resultant force acts opposite to the direction of motion, so it acts backwards. [1]
(d) With uniform slowing, mean speed is:
\[\text{mean speed}=\frac{18+6.0}{2}=12\text{ m/s}\]
\[s=\text{mean speed}\times t=12\times3.0=36\text{ m}\]
The car moves 36 m. [3]
(e) The passenger has inertia, meaning that they tend to continue at their original velocity. [1] When the car slows, the passenger therefore continues forwards relative to the car until a force, such as from the seat belt, slows them. [1]
Answer Details
(a) Acceleration is change in velocity divided by time:
\[a=\frac{v-u}{t}=\frac{6.0-18}{3.0}=-4.0\text{ m/s}^2\]
The acceleration is \(-4.0\text{ m/s}^2\). The negative sign shows that the acceleration is opposite to the original direction of motion. [2]
(b) Apply Newton's second law, \(F=ma\):
\[F=950\times(-4.0)=-3800\text{ N}\]
The resultant force is \(-3800\text{ N}\), with magnitude 3800 N. [2]
(c) The resultant force acts opposite to the direction of motion, so it acts backwards. [1]
(d) With uniform slowing, mean speed is:
\[\text{mean speed}=\frac{18+6.0}{2}=12\text{ m/s}\]
\[s=\text{mean speed}\times t=12\times3.0=36\text{ m}\]
The car moves 36 m. [3]
(e) The passenger has inertia, meaning that they tend to continue at their original velocity. [1] When the car slows, the passenger therefore continues forwards relative to the car until a force, such as from the seat belt, slows them. [1]
Question 10 Report
Fig. 1 shows an ear defender used by a ground crew member near an aircraft. A microphone records the sound outside and inside the defender. The sound level outside is 98 dB and the sound level inside is 76 dB.
(a) Calculate the reduction in sound level caused by the ear defender. [1]
(b) State which sound wave property is mainly reduced by the ear defender. [1]
(c) Explain why long exposure to very loud sound can damage hearing. [3]
(a) \[98\text{ dB}-76\text{ dB}=22\text{ dB}\] The ear defender reduces the sound level by \(22\text{ dB}\). [1]
(b) The wave property mainly reduced is amplitude. [1]
(c) Very loud sound produces large vibrations in the ear. These can damage the hair cells, or sensory cells, in the cochlea. These cells do not repair easily, so the damage can cause hearing loss. [3]
Answer Details
(a) \[98\text{ dB}-76\text{ dB}=22\text{ dB}\] The ear defender reduces the sound level by \(22\text{ dB}\). [1]
(b) The wave property mainly reduced is amplitude. [1]
(c) Very loud sound produces large vibrations in the ear. These can damage the hair cells, or sensory cells, in the cochlea. These cells do not repair easily, so the damage can cause hearing loss. [3]
Question 11 Report
Fig. 1 shows the current-voltage graph for a filament lamp used in a theatre prop. The student obtained each point by changing the supply voltage and waiting briefly before reading the current. Unlike a fixed resistor at constant temperature, the graph becomes less steep as the voltage rises because the filament becomes very hot. At 6.0 V, the current is 0.50 A.
(a) State how the current changes as the voltage increases from 0 V to 8 V. [1]
(b) Explain why the graph becomes less steep at higher voltages. [3]
(c) Calculate the power of the lamp at 6.0 V. [2]
(d) Calculate the energy transferred by the lamp operating at 6.0 V for 4.0 minutes. [3]
(e) Describe how an ammeter and a voltmeter are connected when obtaining this graph. [2]
(a) As voltage increases from \(0\ \mathrm{V}\) to \(8\ \mathrm{V}\), the current increases [1].
(b) At a higher voltage there is a larger current and more heating of the filament [1]. The filament temperature rises [1], increasing its resistance. Therefore each further increase in voltage produces a smaller increase in current, so the graph becomes less steep [1].
(c) At \(6.0\ \mathrm{V}\), the current is \(0.50\ \mathrm{A}\):
\[P=VI=6.0\ \mathrm{V}\times0.50\ \mathrm{A}=3.0\ \mathrm{W}\]
Power = \(3.0\ \mathrm{W}\) [2].
(d) Convert minutes to seconds:
\[t=4.0\times60=240\ \mathrm{s}\]
\[E=Pt=3.0\ \mathrm{W}\times240\ \mathrm{s}=720\ \mathrm{J}\]
Energy transferred = \(720\ \mathrm{J}\) [3].
(e) The ammeter is connected in series with the lamp [1]. The voltmeter is connected in parallel across the lamp [1].
Answer Details
(a) As voltage increases from \(0\ \mathrm{V}\) to \(8\ \mathrm{V}\), the current increases [1].
(b) At a higher voltage there is a larger current and more heating of the filament [1]. The filament temperature rises [1], increasing its resistance. Therefore each further increase in voltage produces a smaller increase in current, so the graph becomes less steep [1].
(c) At \(6.0\ \mathrm{V}\), the current is \(0.50\ \mathrm{A}\):
\[P=VI=6.0\ \mathrm{V}\times0.50\ \mathrm{A}=3.0\ \mathrm{W}\]
Power = \(3.0\ \mathrm{W}\) [2].
(d) Convert minutes to seconds:
\[t=4.0\times60=240\ \mathrm{s}\]
\[E=Pt=3.0\ \mathrm{W}\times240\ \mathrm{s}=720\ \mathrm{J}\]
Energy transferred = \(720\ \mathrm{J}\) [3].
(e) The ammeter is connected in series with the lamp [1]. The voltmeter is connected in parallel across the lamp [1].
