OpenLLM-France/wikimedia huggingface dataset:
# A-level Mathematics/OCR/D2/Flows in a Network

* Weights on arcs are called capacities.
* Start of network called the source (S); end of network called the sink (T).
* Arcs which are carrying their full capacity are said to be saturated.
* A cut is a continuous line that separates S from T and does not pass through any nodes.
* With the exception of S and T, the sum of flows into a node must equal sum of flows out of node.
* Maximum flow - minimum cut theorem:
** The flow through a network cannot exceed the value of any cut.
** The maximum flow equals the value of the minimum cut.
* Meaning that if you can find, by inspection, a flow and a cut which have the same value, then the maximum flow - minimum cut theorem tells you that you have obtained the maximum flow.
* Labelling procedure finds the max flow by systematically augmenting any initial flow:
* Being with any initial flow or zero flow.
* Replace each arc with two arcs: one showing the amount by which the flow can be increased (its excess capacity) and the other showing the amount by which the flow can be decreased (its potential backflow).
* If S is still connected to T then find a new flow from S to T and alter the excess capacities and potential backflows as necessary.
* Repeat Step 3 until S is disconnected from T.
* If S is disconnected from T then the max flow is the sum of all the flows of Steps 1 and 3.
* If a network has more than one source, create a new supersource; if it has more than one sink, create a new supersink.
* If a node has a restricted capacity then replace it by two unrestricted nodes connected by an arc of the relevant capacity.
* For networks where arcs have both upper and lower capacities, the value of a cut is:
       the sum of the upper capacities that cross the cut line in the direction from S to T minus the sum of the lower capacities that cross the cut line in the direction from T to S.
* For such a network, for the labelling procedure backflows are calculated using the minimum capacities. E.g. for a flow of 5 along an arc with an upper capacity of 8 and a lower capacity of 1, the excess capacity is 3 and the potential backflow is 4.
# Annotations to James Joyce's Ulysses/Scylla and Charybdis/204
## Annotations

Dumas fils (or is it Dumas père) (French) Dumas Junior (or is it Dumas Senior). The reference is to the French writers Alexandre Dumas, fils and his father Alexandre Dumas, père. Joyce requested that fils and père be set in italics in the errata list he compiled for the fourth impression of Ulysses. This errata list was printed and bound with this impression as pages 733-736.
Hamlet père and Hamlet fils (French) Hamlet Senior and Hamlet Junior. The former is the ghost in Shakespeare's play, the latter is the eponymous hero. Stephen has taken up John Eglinton's allusion to the two Dumas.

## References

* ↑ Gifford (1988) 249. Thornton (1968) 212-213.
* ↑ James Joyce, Ulysses, Shakespeare & Company, Paris (1924).
* ↑ Gifford (1988) 250.
# Annotations to James Joyce's Ulysses/Scylla and Charybdis/208
## Annotations

marcato (Italian) with every syllable accented. Marcato is a musical term; see page 201 for some other examples in this episode.

## References

* ↑ Gifford (1988) 255.
# Annotations to James Joyce's Ulysses/Sirens
## Contents

The following list contains links to each of the thirty-five pages of the eleventh episode Sirens. The pagination is that of the original edition of Ulysses, published in Paris on 2 February 1922.
Episode 11: Sirens
* Page 245
* Page 246
* Page 247
* Page 248
* Page 249
* Page 250
* Page 251
* Page 252
* Page 253
* Page 254
* Page 255
* Page 256
* Page 257
* Page 258
* Page 259
* Page 260
* Page 261
* Page 262
* Page 263
* Page 264
* Page 265
* Page 266
* Page 267
* Page 268
* Page 269
* Page 270
* Page 271
* Page 272
* Page 273
* Page 274
* Page 275
* Page 276
* Page 277
* Page 278
* Page 279
# Annotations to James Joyce's Ulysses/Scylla and Charybdis/205
## Annotations

Eureka! (Anglicized Greek) I've got it! Archimedes is alleged to have cried εὕρηκα (I have found) when he discovered the principle that bears his name.
Egomen (Greek) I, on the one hand. In Greek, ἐγώ (egō) means I, while μέν ...δέ (men ... de) are conjunctions that are used to convey the idea on the one hand ... on the other hand. But μέν is sometimes used to emphasize a single word such as ἐγώ. It is also possible that Stephen means ἐγώ μήν (egō mēn), which would mean: I, surely. In either case, the men particle would not be attached to the egō.
Gifford believes that Joyce is also punning on the title of the London literary journal The Egoist, which serialized A Portrait of the Artist as a Young Man and three-and-a-half episodes of Ulysses. By supporting his work, The Egoist helped Joyce to believe in himself. The Other chap, who is undermining Joyce's self-belief, would then be George Roberts, the reluctant publisher of Dubliners.
Other scholars, rejecting the Greek hypotheses, suggest that Egomen may be a typo for the Latin Egomet (I myself). This could mean Stephen, himself; or it could be a reference to the God of Moses, who identifies himself with the words: Ego sum qui sum (I am that I am). This would imply that the Other chap is Satan.
Egomen is a legitimate word of Greek origin, defined by the Oxford English Dictionary as: A monastic functionary in the Greek Church.

## References

* ↑ Gifford (1988) 251. Thornton (1968) 214.
* ↑ Gifford (1988) 251-252.
* ↑ Abbott, Evelyn (1987). A Primer of Greek Grammar. London: Duckworth. p. 200. ISBN 0-7156-1258-1. {{cite book}}: Unknown parameter |coauthors= ignored (|author= suggested) (help)
* ↑ Gert Lernout, A Horrible Example of Free Thought: God in Stephen's Ulysses.
* ↑ Exodus 3:14.
# Annotations to James Joyce's Ulysses/Scylla and Charybdis/203
## Annotations

Que voulez-vous? Moore would say (French) “What do you expect?” Moore would say. The Irish writer and critic George Moore was fond of embellishing his conversation with French phrases, a habit he developed during the seven years he spent studying art in Paris.

## References

* ↑ Gifford (1988) 247-248.
# Annotations to James Joyce's Ulysses/Scylla and Charybdis/206
## Annotations

Summa contra Gentiles (Latin) Comprehensive Treatise Against the Unbelievers. This is the usual title of a work by St Thomas Aquinas also known as Liber de Veritate Catholicae Fidei Contra Errores Infidelium (Book on the Truth of the Catholic Faith Against the Errors of the Infidels). Allegedly written for Raymond of Peñafort, the head of Thomas's religious order the Dominicans, it was an attempt to demonstrate the truth of Catholic theology without having recourse to so-called revealed truths; the Jews and Islamic Moors of Spain were the intended audience.
Stephen quotes from this compendium at 197.08-09.
Notre ami Moore (French) Our friend Moore. Thornton believes this is an allusion to the words Mon ami Moore (My friend Moore) used by the Irish playwright Edward Martyn, a cousin of George Moore's, when rescuing Moore from a potentially compromising situation. The phrase became a standing joke between the two men, and whenever Martyn was exasperated by Moore's excesses, he would shake his head in resignation and mutter, Mon ami Moore, mon ami Moore.
While Gifford agrees that the phrase alludes to this remark of Edward Martyn, he cites another use as being most relevant to this passage: Mon ami Moore yearns to be le génie de l'amitié, but unfortunately he can never be looked upon as a friend. For he suffers from ... a perennial condition of mental diarrhoea.

## References

* ↑ Gifford (1988) 252. Thornton (1968) 215.
* ↑ Gifford (1988). Thornton (1968) 216.
* ↑ George Moore, Hail and Farewell.
# Annotations to James Joyce's Ulysses/Scylla and Charybdis/201
## Annotations

Piano, diminuendo (Italian) Softly, with gradually diminishing volume. These are common musical terms.
A tempo (Italian) at the original tempo (i.e. speed). This musical direction is generally used after a change of tempo (e.g. accelerando) to indicate that the performer should resume the original tempo. Here, however, there has been no change of tempo — in fact, no tempo has been specified.
Stringendo (Italian) with gradually increasing intensity. This musical direction is often loosely equated with accelerando (gradually getting faster), but its meaning is broader. While it is true that the tempo is usually increased, so too are the volume and intensity of expression. Here, Stephen is becoming more and more excited.
# Annotations to James Joyce's Ulysses/Scylla and Charybdis/207
## Annotations

femme de trente ans (French) thirty-year-old woman. Anne Hathaway was thirty in 1585 or 1586. Shakespeare is thought to have left Stratford-upon-Avon for London in 1585.
La Femme de trente ans (A Woman of Thirty) is also the title of a novel by Honoré de Balzac.

## References

* ↑ Gifford (1988) 253.
* ↑ George Brandes, William Shakespeare: A Critical Study 13-15.
* ↑ Honoré de Balzac, A Woman of Thirty.
# A-level Physics (Advancing Physics)/Resistance and Conductance/Worked Solutions

1. The potential difference across a resistor is 4V, and the current is 10A. What is the resistance of the resistor?
$R={\frac {V}{I}}={\frac {4}{10}}=0.4{\mbox{ }}\Omega$
2. What is the conductance of this resistor?
$G={\frac {1}{R}}={\frac {1}{0.4}}=2.5{\mbox{ S}}$
3. A conductor has a conductance of 2S, and the potential difference across it is 0.5V. How much current is flowing through it?
$I=GV=2*0.5=1A$
I = 1A
4. A graph is drawn of potential difference across an Ohmic conductor, and current. For every 3 cm across, the graph rises by 2 cm. What is the conductance of the conductor?
${\mbox{Gradient }}={\frac {2}{3}}={\mbox{ Resistance}}$
$G={\frac {1}{\frac {2}{3}}}={\frac {3}{2}}=1.5{\mbox{ S}}$
5. On another graph of potential difference and current, the graph curves so that the gradient increases as current increases. What can you say about the resistor?
The resistor is not an Ohmic conductor.
6. 3 resistors, wired in series, have resistances of 1kΩ, 5kΩ and 500Ω each. What is the total resistance across all three resistors?
ΣR = (1 x 10$^3$) + (5 x 10$^3$) + 500 = 6500Ω = 6.5kΩ
7. 2 conductors, wired in parallel, have conductances of 10S and 5S. What is the total resistance of both branches of the parallel circuit?
$G=10+5=15{\mbox{ S}}={\frac {1}{R}}$
$R={\frac {1}{15}}\approx 0.0667{\mbox{ }}\Omega$
8. The circuit above is attached in series to 1 10Ω resistor. What is the total conductance of the circuit now?
$\Sigma R=10+{\frac {1}{15}}={\frac {151}{15}}{\mbox{ }}\Omega$
$G={\frac {15}{151}}\approx 0.0993{\mbox{ S}}$
# A-level Physics (Advancing Physics)/Refraction/Worked Solutions

1. A ray of light is reflected from a mirror. Its angle to the normal when it reaches the mirror is 70°. What is its angle of reflection?
The angles of reflection on both sides of the normal are always equal, so the angle of reflection is 70°.
2. The speed of light in diamond is 1.24 x 10$^8$ m/s. What is its refractive index?
$n={\frac {v_{1}}{1.24\times 10^{8}}}$
Assuming the speed of light in the reference medium is 3 x 10$^8$:
$n={\frac {3\times 10^{8}}{1.24\times 10^{8}}}={\frac {3}{1.24}}\approx 2.42$
3. The refractive index of ice is 1.31. What is the speed of light in ice?
$1.31={\frac {3\times 10^{8}}{v_{2}}}$
$v_{2}={\frac {3\times 10^{8}}{1.31}}\approx 2.29\times 10^{8}{\mbox{ ms}}^{-1}$
4. A ray of light passes the boundary between air and a transparent material. The angle of refraction is 20°, and the angle of incidence is 10°. What is the speed of light in this material? Why is it impossible for this material to exist?
${\frac {3\times 10^{8}}{v_{2}}}={\frac {\sin {10}}{\sin {20}}}\approx 0.508$
$v_{2}\approx {\frac {3\times 10^{8}}{0.508}}\approx 5.91\times 10^{8}$
This is greater than the speed of light in a vacuum (3 x 10$^8$), and so is impossible.
5. What is the critical angle of a beam of light leaving a transparent material with a refractive index of 2?
$\sin {C}={\frac {1}{2}}$
C = 30° (using the sin$^{−1}$ button on a scientific calculator - this means 'inverse sine', often denoted arcsine or asin for short.)
# A-level Physics (Advancing Physics)/Resistivity and Conductivity/Worked Solutions

1. A material has a conductivity of 10$^6$ S m$^{−1}$. What is its resistivity?
$\rho ={\frac {1}{\sigma }}={\frac {1}{10^{6}}}=10^{-6}{\mbox{ }}\Omega {\mbox{ m}}=1{\mbox{ }}\mu \Omega {\mbox{ m}}$
2. A pure copper wire has a radius of 0.5mm, a resistance of 1 MΩ, and is 4680 km long. What is the resistivity of copper?
$\rho ={\frac {10^{6}\times \pi \times (0.5\times 10^{-3})^{2}}{4680\times 10^{3}}}\approx 168\times 10^{-9}{\mbox{ }}\Omega {\mbox{ m}}=168{\mbox{ n}}\Omega {\mbox{ m}}$
3. Gold has a conductivity of 45 MS m$^{−1}$. What is the resistance of a 0.01m across gold connector, 0.05m long?
First, work out resistivity:
$\rho ={\frac {1}{45\times 10^{6}}}=22.{\bar {2}}\times 10^{-9}{\mbox{ }}\Omega {\mbox{m}}$
Then, substitute everything possible into the resistivity formula:
$22.{\bar {2}}\times 10^{-9}={\frac {(\pi \times (0.5\times 0.01)^{2})R}{0.05}}\approx 1.57\times 10^{-3}R$
$R\approx {\frac {22.{\bar {2}}\times 10^{-9}}{1.57\times 10^{-3}}}\approx 14.2\times 10^{-6}{\mbox{ }}\Omega =14.2{\mbox{ }}\mu \Omega {\mbox{ }}$
4. A strand of metal is stretched to twice its original length. What is its new resistance? State your assumptions.
The material does not change, so resistivity is constant. Length doubles, and we know that volume must be constant.
V = AL
A = V/L
$\rho ={\frac {RA}{L}}={\frac {R({\frac {V}{L}})}{L}}={\frac {RV}{L^{2}}}$
$R_{old}={\frac {\rho }{1}}\times {\frac {L^{2}}{V}}={\frac {\rho L^{2}}{V}}$
When we double L, we get:
$R_{new}={\frac {\rho (2L)^{2}}{V}}={\frac {4\rho L^{2}}{V}}=4\times R_{old}$
We are assuming that ρ and V are constant.
5. Which has the greater resistivity: a plank or a piece of sawdust, made from the same wood?
Sawdust and a plank are artefacts, not materials. Hence, they do not have a resistivity. Even if they did, they are made of the same thing, so they would have equal resistivity.
# A-level Physics (Advancing Physics)/Risks, Doses and Dose Equivalents
## Risk

Radioactivity results in risk - this could be a risk of death, or a risk of developing cancer. In physics, risk is what you expect, on average, to happen:
${\mbox{risk}}={\mbox{probability}}\times {\mbox{consequence}}$
So, if there is a 1 in 500 chance that someone gets run over by a car when crossing the road, the risk involved in allowing 500 people to cross the road is one accident.

## Absorbed Dose

Absorbed dose is measured in grays, commonly denoted Gy. One gray is defined as one joule absorbed per kilogram.
The equation for absorbed dose (A.D) is:
    $A.D={\frac {E}{m}}$
where
    E is the energy given off by the source
    m is the mass of the affected thing.
You may be expected to use the equation
    $E=hf={\frac {hc}{\lambda }}$
where
    h is the Planck constant
    f is the frequency of the photon
    c is the speed of light
    λ is the wavelength of the photon
or
    $E=ne$
where
    n is the number of particles
    e is the magnitude of the charge of one electron
to calculate absorbed dose in terms of numbers of particles with a given frequency, wavelength or energy (in eV). If someone is exposed to a certain activity (particles per second) over a period of time, the absorbed dose accumulates.

## Dose Equivalent

Absorbed dose does not give a full picture of the potential harm radioactivity can do to you. Different types of radiation do different amounts of damage. Absorbed dose equivalent, measured in sieverts (denoted Sv) attempts to compensate. To calculate the dose equivalent, multiply the dose in grays by the quality factor of the particles absorbed. These quality factors are given in the table below.
| Radiation Type | Quality Factor |
| photons and leptons | 1 |
| neutrons with an energy > 10 MeV or < 10keV, protons | 5 |
| neutrons with an energy between 10 and 100 keV, or between 2 and 20 MeV | 10 |
| neutrons with an energy between 100 keV and 2 MeV and atomic nuclei | 20 |
The equation for the effective dose (E.D) received is:
${\mbox{Effective Absorbed}}={\mbox{Absorbed Dose}}\times {\mbox{quality factor}}$

## Questions

1. A mobile phone emits electromagnetic radiation. 1.2 watts of power are absorbed per. kilogram. Assuming that the radiation is absorbed uniformly across a 5 kg head, what dose of radiation would be delivered to the head when making a 10-minute telephone call?
2. What dose equivalent does this correspond to?
3. How many nuclei are there in 1 mg of Americium-241?
4. A ham sandwich becomes contaminated with 1 μg of Americium-241, and is eaten by an 80 kg person. The half-life of Americium-241 is 432 years. Given that Americium-241 gives off 5.638 MeV alpha particles, how long would it be before a dose equivalent of 6 Sv is absorbed, making death certain?
5. What assumptions have you made?
Worked Solutions
# A-level Physics (Advancing Physics)/Rockets, Hoses and Machine Guns/Worked Solutions

1. A machine gun fires 300 5g bullets per. minute at 800ms$^{−1}$. What force is exerted on the gun?
$F=v{\frac {dm}{dt}}=800\times {\frac {300\times 0.005}{60}}=20{\mbox{ N}}$
2. 1 litre of water is pumped out of a tank in 5 seconds through a hose. If a 2N force is exerted on the tank, at what speed does the water leave the hose?
The flow rate is about 0.2 kg s$^{−1}$.
$2=0.2v$
$v={\frac {2}{0.2}}=10{\mbox{ ms}}^{-1}$
3. If the hose were connected to the mains, what problems would there be with the above formula?
The force would not be exerted on the tank, but would instead be referred back through the pipe - it is not gaining velocity at the nozzle.
4. The thrust of the first stage of a Saturn V rocket is 34 MN, using 131000 kg of solid fuel in 168 seconds. At what velocity does the fuel leave the tank?
$34\times 10^{6}=v{\frac {131000}{168}}$
$v={\frac {34\times 10^{6}\times 168}{131000}}=43600{\mbox{ ms}}^{-1}$
5. Escape velocity from the Earth is 11 km$^{−1}$. What is the velocity of the rocket after the first stage is used up, if the total mass of the rocket is 3 x 10$^6$ kg? How does this compare to escape velocity?
$ma=34\times 10^{6}-mg$
$a={\frac {34\times 10^{6}}{3\times 10^{6}}}-9.81=1.52{\mbox{ ms}}^{-2}$
$v=at=168\times 1.52=255.92{\mbox{ ms}}^{-1}\approx 0.256{\mbox{ kms}}^{-1}$
This is 0.002% of escape velocity, hence the need for the other stages of the rocket.
# A-level Physics (Advancing Physics)/Resonance

Resonance occurs when an oscillating system is driven (made to oscillate from an outside source) at a frequency which is the same as its own natural frequency. All oscillating systems require some form of an elastic force and a mass e.g. a mass at the end of a spring. All oscillators have a natural frequency. If you have a mass on a spring, and give it an amplitude, it will resonate at a frequency:
$f={\frac {1}{2\pi }}{\sqrt {\frac {k}{m}}}$
This frequency is independent of the amplitude you give the oscillator to start with. It is the natural frequency of the oscillator. If you keep giving the oscillator amplitude at this frequency, it will not change the frequency of the oscillation. But, you are still doing work. This energy must go somewhere. The only place it can go is into additional kinetic and gravitational potential energy in the oscillation. If you force an oscillation at its resonant frequency, you add significantly to its amplitude.
In simple terms, resonance occurs when the driving frequency of an oscillation matches the natural frequency, giving rise to large amplitudes.
If you were to force an oscillation at a range of frequencies, and measure the amplitude at each, the graph would look something like the following:
There are many types of oscillators, and so practically everything has a resonant frequency. This can be used, or can result in damage if the resonant frequency is not known.

## Reading

Instead of doing questions this time, read the following articles on Wikipedia about these different types:
Tacoma Narrows Bridge
Resonance in Water Molecules (Microwave Ovens)
"No Highway" - a novel with a plot that uses things suspiciously similar to resonance.
Millenium Bridge
# A-level Physics (Advancing Physics)/Risks, Doses and Dose Equivalents/Worked Solutions

1. A mobile phone emits electromagnetic radiation. 1.2 watts of power are absorbed per. kilogram. Assuming that the radiation is absorbed uniformly across a 5 kg head, what dose of radiation would be delivered to the head when making a 10-minute telephone call?
Total energy absorbed = 1.2 x 10 x 60 $dose={\frac {\mbox{energy}}{mass}}$
Stick in the numbers and the dose is equal to 144 Gy. NB: This is wrong, because question says 1.2 Watts/kg already so you don't need to divide by 5kg.
Note to reader: There is currently something wrong with this question, since 144 Gy is enough to kill people. Mobile phones don't do that. I'm working on it. --Sjlegg (talk) 14:30, 6 April 2009 (UTC) Suggestion : A mobile phone does not emit ionizing radiation as the kinetic energy is to low. Therefore, the gray unit cannot be used for this exercise. The unit of measurement for absorption of electromagnetic energy by a body is the SAR.It measures the time rate of absorption of electromagnetic energy by a body. It is measured in W / kg.
The SAR can be determined from the electric field strength E in the body or from the rate of temperature rise (Δt).
2. What dose equivalent does this correspond to?
0 as EM radiatiom from mobile phones isn't ionising.
3. How many nuclei are there in 1 mg of Americium-241?
The mass of 1 Americium-241 nucleus is 241u, which corresponds to 241 x 1.66 x 10$^{−27}$ = 400 x 10$^{−27}$kg.
${\frac {10^{-3}}{400\times 10^{-27}}}=2.5\times 10^{21}$ NB: This calculation is just wrong- ${\frac {10^{-3}}{241}}*Avogadro'snumber=2.5\times 10^{18}$
4. An ham sandwich becomes contaminated with 1 μg of Americium-241, and is eaten by an 80 kg person. The half-life of Americium-241 is 432 years. Given that Americium-241 gives off 5.638 MeV alpha particles, how long would it be before a dose equivalent of 6 Sv is absorbed, making death certain?
First calculate the activity:
$A=\lambda N$
$t_{\frac {1}{2}}={\frac {\ln {2}}{\lambda }}$
$\lambda ={\frac {\ln {2}}{t_{\frac {1}{2}}}}$
$A={\frac {N\ln {2}}{t_{\frac {1}{2}}}}={\frac {2.5\times 10^{21}\times \ln {2}}{432\times 365.24\times 24\times 60\times 60}}=1.27\times 10^{11}{\mbox{ Bq}}$ NB: The wrong N has been used here, questions says micro not milli.
Then calculate the power (energy per. second):
$P=1.27\times 10^{11}\times 5.638\times 10^{6}\times 1.6\times 10^{-19}=0.115{\mbox{ W}}$
Then calculate the dose per. second:
${\frac {0.115}{80}}=1.43{\mbox{ mGys}}^{-1}$
Multiply this by the quality factor for alpha particles (20) to get a dose equivalent per. second of 0.0287 Svs$^{−1}$.
${\frac {6}{0.0287}}=209{\mbox{ s}}=3{\mbox{ min }}29{\mbox{ s}}$
5. What assumptions have you made?
We assumed that the Americium was absorbed uniformly throughout the body, and that the activity remained constant. The latter is acceptable since this period of time is relatively short. The former is not acceptable.