10th Class Physics — MCQs & Practice Questions

1. Uses E = mcΔθ with Δθ = 70 °C88 − 18, not 88[1]
2. E = 1.5 × 4200 × 70 = 4.41 × 10⁵ J[1]
3. Uses t = E/P with P = 2000 Wconverting kW to W[1]
4. t = 441000 / 2000 = 220 saccept 220.5 s[1]
5. Energy is lost to the surroundings / the container is also heatedeither reason accepted[1]

(a) 4.4 × 10⁵ J (b) 220 s (c) heat lost to surroundings

1. Temperature rises from −10 °C to 0 °C[1]
2. Horizontal section at 0 °C while the ice melts[1]
3. Temperature rises from 0 °C to 100 °C[1]
4. Longer horizontal section at 100 °C while the water boilsthe boiling plateau must be longer than the melting one[1]
5. During the flat sections the energy supplied breaks bonds between particles rather than raising kinetic energythe explanation mark[1]

1. Heating makes particles vibrate more and push further apart, so each metal expands[1]
2. The two metals expand by different amounts for the same temperature risethis is the essential point[1]
3. The strip bends toward the metal that expands less[1]
4. Used in a thermostat / fire alarm / oven switchany valid use[1]

Temperature is the average kinetic energy of the particles, measured in °C or K. Heat is the total thermal energy of all the particles, measured in joules, and depends on how many there are.

The two metals expand by different amounts for the same temperature rise, so the strip curves toward the metal that expands less.

The energy supplied is used to break the bonds holding the particles in the lattice, not to increase their kinetic energy. Since temperature measures kinetic energy, it does not change.

1. Recognises two stages: melting, then warminga single-stage answer cannot score more than two[1]
2. Melting: E₁ = mL = 0.20 × 3.34 × 10⁵ = 6.68 × 10⁴ J[1]
3. Warming: E₂ = mcΔθ = 0.20 × 4200 × 20[1]
4. E₂ = 1.68 × 10⁴ J[1]
5. Total = 6.68 × 10⁴ + 1.68 × 10⁴ = 8.36 × 10⁴ Jadding the two stages[1]

8.36 × 10⁴ J

1. A vacuum contains no particles[1]
2. So there is no conduction and no convection through itboth must be named[1]
3. Silvered surfaces are poor emitters of infrared radiation[1]
4. And good reflectors, so radiation is reflected back into the liquid[1]
5. The stopper prevents convection currents / evaporation carrying energy away[1]
6. And is made of an insulating material, so conduction through the top is small[1]

1. Convection[1]
2. Air near the radiator is heated and expands[1]
3. Its density decreases, so it risesthe density step is the mark most often skipped[1]
4. Cooler denser air sinks to replace it, setting up a convection current[1]

1. Metal is a much better conductor than wood[1]
2. It conducts thermal energy away from the hand quickly[1]
3. The rapid energy loss from the skin is what is sensed as cold — the rail is not actually colderthe point of the question[1]

Water heated at the bottom expands, becomes less dense and rises, while cooler denser water sinks to replace it. This convection current heats the whole kettle.

A white, shiny surface is a poor absorber and a good reflector of infrared radiation, so less energy is absorbed and the interior stays cooler.

1. Matt black is the best absorber of infrared radiation[1]
2. So the panel absorbs as much of the Sun's energy as possible[1]
3. Metal is a good conductor[1]
4. So energy passes quickly from the panel surface to the water in the pipes[1]
5. The glass traps a layer of air above the panel[1]
6. Preventing convection currents from carrying energy away[1]
7. Heated water expands, becomes less dense and rises[1]
8. Cooler denser water from the tank sinks to replace it, forming a convection current[1]

Wavelength is the distance between two neighbouring points that are in phase, for example crest to crest. Frequency is the number of complete waves passing a fixed point each second.

In a transverse wave the vibration is perpendicular to the direction of travel, for example light. In a longitudinal wave it is parallel to the direction of travel, for example sound.

The frequency stays the same. The speed decreases and, since v = fλ, the wavelength decreases in proportion.

1. Uses λ = length ÷ number of waves[1]
2. λ = 3.0 / 15 = 0.20 m[1]
3. Uses v = fλ[1]
4. v = 12 × 0.20 = 2.4 m s⁻¹[1]

λ = 0.20 m, v = 2.4 m s⁻¹

1. The waves spread out after passing through the gap[1]
2. This is diffraction[1]
3. The wavelength and frequency are unchanged; only the shape of the wavefront changes[1]
4. The waves spread out more[1]
5. Because the gap width is closer to the wavelengthmaximum spreading when gap ≈ λ[1]
6. Converts 25 mm = 0.025 m[1]
7. Uses v = fλ[1]
8. v = 8.0 × 0.025 = 0.20 m s⁻¹[1]

(c) 0.20 m s⁻¹

1. The waves change direction at the boundary[1]
2. They bend towards the normal[1]
3. The waves travel more slowly in the shallow water[1]
4. The end of each wavefront entering the shallow region slows first[1]
5. So the wavefront pivots, changing the direction of travel[1]
6. The direction does not change[1]
7. Because the whole wavefront slows at the same instant, so there is nothing to pivot aboutthe wavelength still shortens[1]

1. The sound gets quieter and eventually cannot be heard[1]
2. The hammer is still seen to strike, so the bell is still vibratingthis detail is a mark — it rules out the bell simply stopping[1]
3. There are fewer air particles to be compressed and to pass the vibration on[1]
4. Sound requires a medium to travel through[1]
5. It cannot travel through a vacuum[1]

1. Total distance travelled = v × t = 1500 × 0.40 = 600 m[1]
2. Recognises this is the distance down and backthe mark the question exists to test[1]
3. Depth = 600 / 2[1]
4. = 300 m[1]

300 m

1. B is louder than A[1]
2. Because it has a larger amplitude[1]
3. B has a higher pitch than A[1]
4. Because waves closer together means a higher frequency[1]

As the cone moves forward it pushes air particles together, forming a compression. As it moves back the particles spread apart, forming a rarefaction.

The sound reflects from a distant hard surface and travels back to the listener. Because sound travels at a finite speed, the reflected sound arrives later than the direct one.

1. Distance travelled by the first echo = 340 × 0.60 = 204 m[1]
2. Recognises this is there and back[1]
3. Distance to the nearer cliff = 102 m[1]
4. Distance travelled by the second echo = 340 × 1.0 = 340 m[1]
5. Halves it[1]
6. Distance to the further cliff = 170 m[1]
7. Particles in water are much closer together than in air[1]
8. So collisions between them are more frequent[1]
9. And the vibration is passed on more quickly[1]

(a) 102 m (b) 170 m

1. Uses n = sin i / sin r[1]
2. sin r = sin 40° / 1.50 = 0.643 / 1.50 = 0.4285[1]
3. r = 25.4°accept 25°[1]
4. Uses sin c = 1/n = 1/1.50 = 0.667, giving c = 41.8°accept 42°[1]
5. Light must travel from a denser to a less dense medium[1]
6. The angle of incidence must exceed the critical angleboth conditions needed[1]

(a) 25.4° (b) 41.8° (c) denser → less dense, and i > c

1. Light strikes the boundary between the core and the cladding[1]
2. The core is optically denser than the cladding[1]
3. The angle of incidence is greater than the critical angle[1]
4. So total internal reflection occurs, repeatedly, and the light follows the fibre[1]

1. Real and invertedthe object lies between F and 2F[1]
2. It is also magnified[1]
3. A projector (or a film projector / slide projector)[1]
4. At 5 cm the object is inside F, so the image becomes virtual, upright and magnifieda magnifying glass[1]

(a) real, inverted, magnified (b) projector (c) becomes virtual, upright and magnified

A real image is formed where light rays actually meet and can be caught on a screen. A virtual image is formed where rays only appear to come from and cannot be projected.

Light from the submerged part refracts as it leaves the water and enters the air, bending away from the normal. The eye assumes light travels in straight lines, so the straw appears displaced.

1. Uses sin c = 1/n[1]
2. sin c = 1/1.33 = 0.752[1]
3. c = 48.8°accept 49°[1]
4. Total internal reflection occurs[1]
5. Because 60° is greater than the critical angle of 48.8°[1]
6. And the light is travelling from the denser medium to the less dense oneboth conditions[1]
7. Light repeatedly totally internally reflects at the core–cladding boundary, so it follows the fibre even around bends[1]
8. Advantage: much higher data capacity / less signal loss over distance / no electrical interferenceany one[1]
9. Second valid point about the mechanism or the advantage[1]

(a) 48.8° (b) total internal reflection (c) repeated TIR at the core–cladding boundary

Electrons are transferred from the cloth onto the rod. The rod gains electrons and becomes negative; the cloth is left with a positive charge.

A region in which a charge experiences a force. The direction of the field is the direction of the force on a small positive test charge.

The rod induces a separation of charge in the paper, drawing the opposite charge to the near side. That near charge is closer to the rod, so its attraction outweighs the repulsion of the far side.

1. Converts 25 mm = 0.025 m[1]
2. Uses E = V/d = 5000 / 0.025[1]
3. E = 2.0 × 10⁵ V m⁻¹[1]
4. F = QE = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N[1]

E = 2.0 × 10⁵ V m⁻¹, F = 6.4 × 10⁻¹⁴ N

1. Friction between the tanker and the air, and between the fuel and the tank walls, transfers electrons[1]
2. The tanker is insulated by its rubber tyres, so the charge cannot escape and accumulates[1]
3. The charge raises the potential of the tanker[1]
4. Eventually a spark jumps to a nearby earthed object[1]
5. The spark could ignite the fuel vapour and cause an explosion[1]
6. The wire provides a conducting path to earth[1]
7. Electrons flow along it until the tanker is at the same potential as the earth[1]
8. So charge leaks away steadily instead of building to a spark[1]

1. The lines are parallel and equally spaced, showing a uniform field[1]
2. They run from the positive plate to the negative plate[1]
3. The force is upward, towards the positive plate[1]
4. Because the drop is negative, so the force is opposite to the field direction[1]
5. Increase the potential difference across the plates[1]
6. Move the plates closer together, or increase the charge on the dropany one[1]

(b) upward, towards the positive plate

The rate of flow of electric charge, I = Q/t. Its unit is the ampere (A), where one ampere is one coulomb per second.

The current through a metallic conductor is directly proportional to the potential difference across it, provided the temperature stays constant.

The metal ions vibrate more strongly, so the moving electrons collide with them more often. Each collision impedes the flow, so the resistance rises.

1. Converts the area: 0.50 mm² = 0.50 × 10⁻⁶ m²the step most often dropped[1]
2. Uses ρ = RA/L[1]
3. ρ = (0.068 × 0.50 × 10⁻⁶) / 2.0[1]
4. ρ = 1.7 × 10⁻⁸ Ω mcopper[1]
5. Doubling the length doubles the resistance: 0.136 Ω[1]

ρ = 1.7 × 10⁻⁸ Ω m; R = 0.14 Ω

1. The graph passes through the origin[1]
2. It is a straight line at low potential difference[1]
3. It then curves towards the V axis, so the gradient falls[1]
4. A larger current heats the filament[1]
5. The ions vibrate more and the electrons collide with them more often[1]
6. So the resistance increases and the current no longer rises in proportion[1]
7. R = V/I = 6.0 / 0.50 = 12 Ω[1]
8. P = VI = 6.0 × 0.50 = 3.0 W[1]

(c) 12 Ω and 3.0 W

1. Total resistance = 20 + 25 = 45 Ω[1]
2. Uses I = V/R[1]
3. I = 9.0 / 45 = 0.20 A[1]
4. Uses V = IR for the thermistor[1]
5. V = 0.20 × 25 = 5.0 V[1]
6. The potential difference across the thermistor decreases[1]
7. Its resistance falls as it warms, so it takes a smaller share of the supply voltage[1]

(a) 0.20 A (b) 5.0 V (c) it falls

1. Uses V_s/V_p = N_s/N_p[1]
2. V_s = 230 × 5000/200[1]
3. V_s = 5750 V[1]
4. Step-upmore turns on the secondary[1]
5. Direct current produces a constant magnetic field in the core[1]
6. A changing field is required to induce an e.m.f., so nothing is induced in the secondarythe word "changing" is the mark[1]

(a) 5750 V (b) step-up (c) d.c. gives a constant field, and induction needs a changing one

1. The needle deflects to one side while the magnet is moving[1]
2. Move the magnet faster[1]
3. Use a stronger magnet, or a coil with more turnsany second valid change[1]
4. The needle returns to zero — no deflection[1]
5. There is no change in the magnetic field through the coil, and induction requires a change[1]

1. For a given power, a higher voltage means a smaller currentfrom P = VI[1]
2. Power lost as heat in the cables is I²R, so a smaller current wastes much less energythe I² is the key point[1]
3. A step-up transformer raises the voltage at the power station[1]
4. A step-down transformer lowers it again for safe use in homes[1]

Increase the current through the coil, or use a stronger magnetic field. (Also acceptable: increase the number of turns on the coil.)

Soft iron is easily magnetised and demagnetised, so it follows the rapidly alternating field of the primary and carries the changing flux efficiently to the secondary.

1. Uses I = P/V = 20 × 10⁶ / 25 × 10³[1]
2. I = 800 A[1]
3. Uses P = I²R = 800² × 4.0[1]
4. = 2.56 × 10⁶ W = 2.56 MWnearly 13% of the output[1]
5. New current = 20 × 10⁶ / 400 × 10³ = 50 A[1]
6. New loss = 50² × 4.0[1]
7. = 1.0 × 10⁴ W = 10 kW[1]
8. A reduction by a factor of 256, because the current fell by 16 and the loss depends on I²the point of the question[1]
9. Turns ratio N_s : N_p = 400 : 25 = 16 : 1[1]

(a) 800 A (b) 2.56 MW (c) 50 A and 10 kW (d) 16 : 1

1. Uses V_s/V_p = N_s/N_p[1]
2. V_s = 230 × 5000/200[1]
3. V_s = 5750 V[1]
4. Step-upmore turns on the secondary[1]
5. Direct current produces a constant magnetic field in the core[1]
6. A changing field is required to induce an e.m.f., so nothing is induced in the secondarythe word "changing" is the mark[1]

(a) 5750 V (b) step-up (c) d.c. gives a constant field, and induction needs a changing one

1. The needle deflects to one side while the magnet is moving[1]
2. Move the magnet faster[1]
3. Use a stronger magnet, or a coil with more turnsany second valid change[1]
4. The needle returns to zero — no deflection[1]
5. There is no change in the magnetic field through the coil, and induction requires a change[1]

1. For a given power, a higher voltage means a smaller currentfrom P = VI[1]
2. Power lost as heat in the cables is I²R, so a smaller current wastes much less energythe I² is the key point[1]
3. A step-up transformer raises the voltage at the power station[1]
4. A step-down transformer lowers it again for safe use in homes[1]

Increase the current through the coil, or use a stronger magnetic field. (Also acceptable: increase the number of turns on the coil.)

Soft iron is easily magnetised and demagnetised, so it follows the rapidly alternating field of the primary and carries the changing flux efficiently to the secondary.

1. Uses I = P/V = 20 × 10⁶ / 25 × 10³[1]
2. I = 800 A[1]
3. Uses P = I²R = 800² × 4.0[1]
4. = 2.56 × 10⁶ W = 2.56 MWnearly 13% of the output[1]
5. New current = 20 × 10⁶ / 400 × 10³ = 50 A[1]
6. New loss = 50² × 4.0[1]
7. = 1.0 × 10⁴ W = 10 kW[1]
8. A reduction by a factor of 256, because the current fell by 16 and the loss depends on I²the point of the question[1]
9. Turns ratio N_s : N_p = 400 : 25 = 16 : 1[1]

(a) 800 A (b) 2.56 MW (c) 50 A and 10 kW (d) 16 : 1

1. Radio, microwave, infrared, visible, ultraviolet, X-ray, gammatwo marks for the full correct order; one for a mostly correct order[2]
2. 3.0 × 10⁸ m s⁻¹[1]
3. Gamma rays have a much higher frequency, so each photon carries much more energy[1]
4. They are ionising — able to remove electrons from atoms, which can damage or mutate cellsthe word "ionising" is what the mark scheme looks for[1]

Order as listed; 3.0 × 10⁸ m s⁻¹; gamma is ionising because of its far higher photon energy

1. Converts MHz to Hz: 96.0 × 10⁶ = 9.60 × 10⁷ Hzthe prefix is where this question is won or lost[1]
2. Rearranges c = fλ to λ = c/f[1]
3. λ = 3.0 × 10⁸ / 9.60 × 10⁷ = 3.1 m[1]

3.1 m

1. Infrared use: remote controls / thermal imaging / heatersany one[1]
2. Infrared danger: skin burns[1]
3. Ultraviolet use: sterilising / security marking / detecting forgeriesany one[1]
4. Ultraviolet danger: skin cancer / eye damage[1]

They are all transverse waves, and they all travel at 3.0 × 10⁸ m s⁻¹ in a vacuum. (Also acceptable: none requires a medium; all carry energy.)

Microwaves pass through the atmosphere with little absorption, so the signal reaches the satellite and returns. Infrared is strongly absorbed by the atmosphere.

1. X-rays pass through soft tissue but are absorbed by the denser bone[1]
2. Producing a shadow image on the detector[1]
3. Gamma rays are ionising and can kill living cells[1]
4. A focused beam destroys the tumour cells[1]
5. But healthy cells would also be damaged or mutated, so exposure elsewhere must be minimised[1]
6. Ultraviolet is ionising enough to kill bacteria[1]
7. So it sterilises surfaces and instruments[1]
8. Staff must wear eye protection / avoid skin exposure, since UV causes eye damage and skin cancer[1]

1. 1/R = 1/8.0 + 1/8.0 = 0.25, so R = 4.0 Ωone mark for the method, one for inverting correctly[2]
2. Total = 3.0 + 4.0 = 7.0 Ω[1]
3. I = V/R = 9.0 / 7.0 = 1.3 Aaccept 1.29[1]
4. Uses V = IR for the 3.0 Ω resistor[1]
5. V = 1.29 × 3.0 = 3.9 V[1]

(a) 4.0 Ω (b) 7.0 Ω (c) 1.3 A (d) 3.9 V

1. Its resistance decreases[1]
2. The thermistor now takes a smaller share of the supply voltage[1]
3. So a larger share appears across the fixed resistor[1]
4. The output voltage therefore increases[1]

1. Each lamp receives the full mains voltage[1]
2. So each operates at its correct brightness[1]
3. If one lamp fails, the circuit through the others is unbroken[1]
4. So the rest continue to work, and each can be switched independently[1]

1. The total resistance halves[1]
2. Because the second branch gives the current an additional path, so more current flows for the same potential difference[1]

It halves. The second branch gives the current an additional path, so more current flows for the same potential difference.

1. There is only one path for the charge to follow[1]
2. So the same charge passes through each component every second[1]

There is only one path for the charge to follow, so the same charge passes through each component every second.

1. Uses 1/R = 1/R₁ + 1/R₂[1]
2. 1/R = 1/4.0 + 1/12 = 0.25 + 0.0833 = 0.3333[1]
3. R = 3.0 Ωinverting is a separate mark[1]
4. Total resistance = 6.0 + 3.0 = 9.0 Ω[1]
5. Uses I = V/R[1]
6. I = 12 / 9.0 = 1.33 A[1]
7. P.d. across the parallel section = 1.33 × 3.0 = 4.0 Vboth branches share this[1]
8. Uses I = V/R for the 12 Ω branch[1]
9. I = 4.0 / 12 = 0.33 A[1]

(a) 3.0 Ω (b) 1.33 A (c) 0.33 A

Adding a small amount of another element to a pure semiconductor to change its conductivity. It produces n-type, where the carriers are free electrons, and p-type, where they are holes.

In a semiconductor, heat frees more charge carriers, so more current can flow. In a metal the number of carriers is fixed, and heating makes the ions vibrate more so the electrons collide with them more often.

The depletion layer widens and essentially no current flows. This one-way behaviour is used to rectify alternating current into direct current.

1. The two components are in series across the supply[1]
2. They share the supply voltage in proportion to their resistances[1]
3. As it gets darker the resistance of the LDR increases[1]
4. So the LDR takes a larger share of the supply voltage[1]
5. The base voltage, taken across the LDR, therefore rises[1]
6. When the base voltage exceeds about 0.6 V the transistor switches on[1]
7. A small base current then allows a much larger collector current[1]
8. That current flows through the lamp, lighting ita relay would be used for a mains lamp[1]

1. An OR gate is required[1]
2. Inputs 0,0 give output 0[1]
3. Inputs 0,1 and 1,0 both give output 1[1]
4. Inputs 1,1 give output 1[1]
5. An AND gate would now be needed[1]
6. The alarm would not sound if only one sensor detected the fire[1]
7. A real fire could therefore be missed, so the design is less safe[1]

(a) OR gate (b) AND gate — but it fails to warn when only one sensor triggers

1. Only the positive halves of each cycle appear[1]
2. The negative halves are missing, leaving gaps along the axis[1]
3. Half-wave rectification[1]
4. Add a capacitor in parallel with the resistor[1]
5. It charges during each pulse and discharges through the gaps, holding the voltage upa four-diode bridge would also give full-wave rectification[1]

1. 92 protonsthe proton number[1]
2. 235 − 92 = 143 neutrons[1]
3. 92 electrons, since the atom is neutral[1]
4. Same number of protons (92); different number of neutrons (146 instead of 143)[1]
5. Chemical behaviour depends on the electrons[1]
6. Both have 92 electrons arranged identically, so they react in the same way[1]

(a) 92 p, 143 n, 92 e (b) same Z, different A (c) identical electron arrangement

1. Most alpha particles passed straight through[1]
2. So the atom is mostly empty space[1]
3. Some were deflected through large angles[1]
4. So there is a concentrated positive charge repelling them[1]
5. A very few were reflected almost straight back[1]
6. So the nucleus is very small and contains most of the atom's mass[1]

1. In that model the positive charge is spread thinly throughout the atom[1]
2. So the repulsive force on an alpha particle anywhere would be small[1]
3. It could not produce the large-angle deflections or backscattering that were observed[1]

The nucleon number A is the total number of protons and neutrons in the nucleus. The proton number Z is the number of protons alone.

They have the same number of protons and therefore the same number of electrons in the same arrangement. Chemical behaviour is decided by the electrons, not by the number of neutrons.

1. Almost all the alpha particles passed straight through[1]
2. A very small fraction were deflected through large angles, some straight back[1]
3. Passing straight through shows the atom is mostly empty space[1]
4. Large deflections show a concentrated region of positive charge[1]
5. Which repels the positive alpha particle[1]
6. Backward scattering shows that region is also very massive, and very small[1]
7. So each alpha particle meets at most one nucleus[1]
8. A thicker foil would cause multiple scattering and the result could not be interpreted[1]

1. 18 hours is 3 half-lives18 ÷ 6[1]
2. Halves three times: 800 → 400 → 200 → 100[1]
3. 100 counts per minute[1]
4. Each half-life removes only half of what remains, so some always remains[1]
5. Temperature or pressure[1]
6. Chemical state or physical form of the sampleany second valid factor[1]

(a) 100 counts/min (b) halving never reaches zero (c) temperature, pressure, chemical state

1. Alpha: A = 218 − 4 = 214[1]
2. Z = 84 − 2 = 82lead[1]
3. Beta: A unchanged at 214[1]
4. Z = 82 + 1 = 83bismuth[1]
5. A neutron changes into a proton and an electron, and the electron is emitted[1]

(a) ²¹⁴₈₂Pb (b) ²¹⁴₈₃ (c) a neutron becomes a proton plus an emitted electron

1. Gamma[1]
2. It is penetrating enough to leave the body and be detected, and least ionising so it does least damageboth halves wanted[1]
3. Short, so the activity falls quickly and the patient is not exposed for long[1]
4. But not so short that it decays away before the scan can be completedthe balance is the point of the question[1]

Handle it with long tongs to increase the distance from the body, and return it to its lead-lined container immediately after use. (Also acceptable: never point it at anyone; minimise exposure time.)

The low level of ionising radiation always present in the environment. Sources include radon gas from rocks, cosmic rays, medical X-rays and food.

1. The time taken for the number of undecayed nuclei in the sample to halve[1]
2. Equivalently, the time for the count rate to fall to half its value; the process is random so this is an average[1]
3. 32 days is 4 half-lives[1]
4. Uses 640 ÷ 2⁴[1]
5. = 40 counts per minute[1]
6. The extra 15 counts per minute is background radiation[1]
7. It is measured with the source removed and subtracted from every reading[1]
8. The proton number increases by 1[1]
9. The nucleon number is unchangeda neutron becomes a proton plus an electron[1]

(b) 40 counts per minute

1. The Earth's rotation on its axis, once every 24 hours[1]
2. The tilt of the Earth's axis, about 23.5°[1]
3. A hemisphere tilted toward the Sun receives light at a steeper angle, concentrated over a smaller area[1]
4. And has more hours of daylight, so it is warmer[1]
5. The distance varies by only a few per cent[1]
6. And the Earth is actually closest to the Sun during the northern winter, which is the opposite of what that explanation would predictthe decisive point[1]

(a) rotation (b) axial tilt, giving steeper light and longer days (c) distance varies little and in the wrong direction

1. Converts the period to seconds: 27.3 × 24 × 3600 = 2.36 × 10⁶ sdays → hours → seconds[1]
2. Uses v = 2πr/T[1]
3. v = 2π × 3.8 × 10⁸ / 2.36 × 10⁶[1]
4. v = 1.0 × 10³ m s⁻¹about 1 km/s[1]

1.0 × 10³ m s⁻¹

1. A planet's orbit is nearly circular; a comet's is highly elliptical[1]
2. A comet's distance from the Sun therefore varies enormously[1]
3. Gravitational attraction is much stronger when it is close to the Sun[1]
4. So it moves fastest at its closest approach and slowest when far away[1]

A planet follows a nearly circular orbit and stays at a roughly constant distance from the Sun. A comet follows a highly elliptical orbit, so its distance and speed change greatly.

The astronaut and the spacecraft are both falling freely towards the Earth with the same acceleration. There is no contact force between them, so the astronaut feels no weight.

1. The direction of motion is continually changing[1]
2. So the velocity is changing, because velocity is a vector[1]
3. A changing velocity is an acceleration[1]
4. The resultant force acts towards the centre of the orbit[1]
5. It is the gravitational attraction of the Earth[1]
6. Gravitational field strength is greater closer to the Earth[1]
7. So a larger centripetal force is needed to hold the satellite in that orbit[1]
8. Which requires a greater orbital speed, so the period is shorter[1]

1. Nuclear fusionnot burning[1]
2. Hydrogen nuclei fuse to form helium[1]
3. It expands to become a red giant[1]
4. Then throws off its outer layers as a planetary nebula, leaving a white dwarf[1]
5. A much more massive star becomes a red supergiant and explodes as a supernova[1]
6. Leaving a neutron star or a black hole[1]

Fusion of hydrogen to helium; red giant → planetary nebula → white dwarf; massive stars instead go supergiant → supernova → neutron star or black hole

1. They are moving away from us[1]
2. The further away a galaxy is, the greater its redshift — so the faster it is receding[1]
3. This shows the Universe is expanding[1]
4. So in the past everything was closer together, in a hot dense state[1]
5. The cosmic microwave background radiation[1]

Receding; further means greater redshift; therefore expansion, therefore a hot dense origin; plus the CMB

1. A more massive star has a much stronger inward gravitational force[1]
2. So the core is hotter and denser, and fusion proceeds much faster[1]
3. It therefore uses up its hydrogen far more quickly than the extra supply lasts[1]

Nuclear fusion in the core, in which hydrogen nuclei join to form helium and release energy.

The light from distant galaxies is shifted towards longer wavelengths, which shows they are moving away from us. The further away a galaxy is, the greater its red-shift and so the faster it recedes.

1. Gravity pulls the material of the star inwards[1]
2. The outward pressure from the energy released by fusion pushes outwards[1]
3. The two are balanced, so the star neither collapses nor expands[1]
4. The core contracts and the outer layers expand and cool[1]
5. The Sun becomes a red giant[1]
6. It then sheds its outer layers and the core remains as a white dwarf[1]
7. Light from almost every distant galaxy is red-shifted, so they are all moving away[1]
8. The more distant the galaxy the greater the red-shift, so the universe is expanding[1]
9. Running that expansion backwards means everything began from a single point[1]