PHRASAL’VERBS’ - uniroma2.it · PHRASAL’VERBS’ • If’the’phrasal’verb’has’an ......

PHRASAL VERBS

English for Scien/sts Maria Cris/na Teodorani

PHRASAL VERBS

•  A phrasal verb is a verb followed by a preposi:on working as a par:cle that belongs to the verb, not to the object.

•  The ‘original’ meaning of the verb gets altered •  Take over à gain control of •  Bring about à cause to happen •  Add up à make sense •  Look on à consider

PHRASAL VERBS

•  Spin off à produce an useful result •  Fall through à fail to come to comple:on •  Run into à meet •  Run up against à encounter (a problem) •  Sort out à find a solu:on •  Set out à give in detail in wri:ng •  Step up à increase •  Wear off à lose effect (a chemical compound)

PHRASAL VERBS

•  Try out à test (a machine) •  Take on à acquire a new characteris:c •  Stand for à represent •  End up à finish in a certain way, or place •  Come up to à equal (expecta:ons, standards) •  Come down to à be in the end a maQer of •  Come about à happen •  Hit upon/on à discover by chance

PHRASAL VERBS

•  Lay down à state a rule •  Make for à result in •  Slow down à decelerate •  Put down to à explain the cause of •  Miss out à fail to include •  Bring up à men:on •  Put up with à tolerate •  Bring on à cause the onset of an illness

PHRASAL VERBS

•  Stamp out à eradicate •  S:ck out à protrude •  Take/bring down à deac:vate purposely •  Dilute down à make less strong (weaker) •  Bear out à support or confirm the truth •  Carry out à complete, put into prac:ce •  Carry off à complete something successfully •  Get over with à come to the end of

PHRASAL VERBS

•  Carry along à transport something as it moves along (especially a fluid).

•  Pull off à manage to succeed •  Make up for à compensate •  Make up à invent •  Point out à indicate / draw aQen:on to •  Give off à send off a scent (liquid/gas)

PHRASAL VERBS

•  She has just shut then engine off. •  At room temperature, the polymers act as a gas bumping against the larger par:cles.

•  How much does the curve s/ck out from where it is tensioned on the side?

•  Because of the smell it is giving off I know this is sulfuric acid for sure.

•  Smoking brings on a lot of diseases

PHRASAL VERBS (1)

•  Opposing structural changes in two symmetrical polypep:des bring about opposing changes to the thermal stability of a complex integral membrane protein.

•  Once dormant their metabolism slows down so much that the pineal is virtually switched off.

•  She took the system down to work on that bug in the tape drive.

•  The PC program has spun off unexpected results .

PHRASAL VERBS

•  If the phrasal verb has an object it can be placed between the verb and the par:cle. If the object is a pronoun then it must be so.

•  I turned the light off = I turned off the light •  But you can only say: I turned it off. •  He put the protractor down or he put down the protractor.

•  But only: He put it down.

PHRASAL VERBS

•  The Government took the company over. •  A moving vortex ring carries the spinning fluid along.

•  The experiment was hard but he carried it off. •  He got over with that awful exercise. •  It took her a hour to dilute the acid down. •  Bees carry out pollen-‐collec:ng movements •  It was a tricky problem, but he pulled it off.

PHRASAL VERBS

•  High seed sets seemed to bear out effec:ve wind pollina:on.

•  They borne it out. •  The serpen:ne outcrop is usually taken over by an abundance of perennial grasses.

•  Perennial grasses take the outcrop over. •  The medicine was tested over individuals. •  They tested it over.

PHRASAL VERBS (2)

•  They are leaving the strips in the spurs for one minute to soak all the nectar up.

•  The strips are being le_ in the spurs for one minute to soak up all the nectar.

•  Beetles appear to be taking up nectar from the petal nectaries

•  A general lowering of pH may result in a decrease of the plants capability to take nutrients up.

FROM PHRASAL TO PREPOSITIONAL VERBS

•  Someone tested nectary secre:ons for glucose with diabetes test paper.

•  Nectary secre:ons were tested for glucose with diabetes test paper (passive).

•  We can’t say: They tested them for •  We have to add: They tested them for glucose •  The use of for here implies that the test is searching for specific data.

PREPOSITIONAL VERBS

•  The preposi:on does not work as a par:cle. •  They look a<er her. •  She is looked a_er (passive). •  We can’t say : they look her a_er. •  Popula:ons of Utah junipers sampled over a large geographic area.

•  We can’t say: …sampled a large area over •  The verb is one thing with the preposi:on

PREPOSITIONAL VERBS (3)

•  The flowers fell off at the end of the first day. •  Some anthers have dropped off during processing.

•  Tetrads were abraded off s:gmata during handling.

•  The most common sense of off in the texts analysed is falling or causing something to fall and/or be separated from the plant (shake off, fall off, drop off, cut off, abraded off).

PREPOSITIONAL VERBS (4)

•  Only the median remains visible before fading out.

•  The mucus on the s:gma:c rays gradually dried out.

•  A comple:ve sense (dry out, fade out) may be seen in the examples above. In contrast with up the use of out is preferred when the ac:on is seen as a (long) process leading to a final state. This is some:mes reflected in the verb choice (fade) or in the use of the adverb (gradually).

PREPOSITIONAL VERBS (5) •  Down is not as frequent as the other par:cles examined here. It is basically used to indicate downward movement as in:

•  The petal has completely folded down over the keel.

•  It is layed down into horizontal posi:on. •  The stamen hangs down over the lip. •  Bundles can be easily followed down the flower. •  Crawl down between the staminodes and the style.

PREPOSITIONAL VERBS (6) •  When an external magne:c field is applied, sharp spectral lines like the n=3→ 2 transi:on of hydrogen split into mul:ple closely spaced lines. First observed by Pieter Zeeman, this splijng is a@ributed to the interac:on between the magne:c field and the magne:c dipole momentum associated with the orbital angular momentum. In the absence of the magne:c field, the hydrogen energies depend upon the principal quantum number n only, and the emissions occur at a single wavelength.

TEXT BUILDING: WATER AUTOIONIZATION (7)

•  Say we have the following reac:ons: •  2H2O (aq) ⇄ H3O+ (aq) + OH-‐ (aq) (1) •  H2O (aq) ⇄ H+ (aq) + OH-‐ (aq) (2) •  They both represent the autoioniza:on of water. Just to give a narra:ve around what could happen, we can say that the first reac:on shows up two reac:ng molecules of water yielding to an hydronium ion and a hydroxide.

WATER AUTOIONIZATION

•  Water is a neutral molecule and as soon as the oxygen’s δ-‐ of one molecule gets a@ached to the hydrogen’s δ+ of the other one we are le< with one less proton than electrons in the products. The hydrogen’s nucleus itself, on the le_ side of the equa:on, which is a proton, gets bumped off or scraped off and ends up on the hydronium molecule on the le_ side while the hydroxide has a nega:ve charge because it has lost a proton.


•  This process where water can kind of spontaneously carry out such a behaviour is called autoioniza:on. We can put this down to the fact that first we just have water by itself, and then, because of some random circumstances of some molecules bumping into each other, some subset of the water molecule will ionize like this, so that one part will lose a proton and the other part will gain one.


•  This is an equilibrium reac:on, so now the two molecules might go bumping against two other ones and become water again; maybe they will bump into each other again and become water once more; the process going back and forth, some equilibrium concentra:on is being brought about. The proper equilibrium reac:on is in the form (1).


•  Anyway, equa:ons (1) and (2) are essen:ally equivalent, although the first describes what exactly happens. The second one is crea:ng a sort of picture: we have one water molecule and there is some small probability that one of its hydrogens just pops off, and we’re just le_ with a hydrogen and a hydroxide, which is just an oxygen and a hydrogen atom together, but the reality is that the H+ ca:ons do not exist in water on their own.


•  Whenever they are in water, in an aqueous solu:on, they essen:ally get a ride with another water molecule, and that is what happens with the hydronium ion. But since people care about hydrogen ca:ons –that is, hydrogen protons in a solu:on, we more o_en consider the (2). What is the equilibrium of this reac:on? It turns out that in just regular water, or pure water, at 25°C, which is roughly room temperature we have [H3O+] = [OH-‐] = 10-‐7 M (molars).


•  It means for every 1 liter of water we have 10 to the minus 7 moles of hydronium ca:ons, or hydrogen’s (that is, 10 to the minus 7 :mes Avogadro’s number, which is preQy a good number, for if you do the Math you end up with 60 :mes 10 to the 16th power molecules!). The concentra:on of hydroxide is also 10-‐7 M. Knowing that we can now figure out the equilibrium constant for the autoioniza:on of water in (2):


•  H2O (aq) ⇄ H+ (aq) + OH-‐ (aq) •  KW = [H+] [OH-‐] = 10-‐14 •  Given that, in aqueous solu:on, one molecule of water splits into one hydrogen ca:on and one hydroxide ion, we can calculate K water (or K sub w for water) as the product of the concentra:ons of the two ions (divided by 1), which is equal to 10 (raised) to the nega:ve 14th (power).


•  In a normal situa:on we would divide this by the concentra:on of the reactants, but in this case the reactants are what the solvent is: actually the probability that what is on the le_ side of the equa:on turns into what is on the right side is maximum, it’s equal to 1, so we should divide by 1 –that is, not dividing at all. People chemistry don’t like talking in this way, they like using logarithms!


•  They take the nega:ve logarithm (which is base 10) on both sides of the equa:on (and if the log is in:mida:ng to you remember it’s just an exponent!), so that it turns out that –log10 KW = –log1010-‐14 = 14. This idea right here is called a pKw, any p in chemistry standing for –log10, so that p stands for power. The same conven:on is used in the hydrogen concentra:on. Since [H+] = 10-‐7 M, we end up with a pH = 7 = -‐log [H+].


•  Of course if we add H+ up to get [H+]=10-‐3, the solu:on thus becoming more acid, the new pH will equal 3. We can do the same exercise for the pOH, which turns out to be equal to 7, because water has the same number of hydrogens and hydroxides, for it is dissocia/ng into the two of them.

EXERCISE

•  Calculate, for example, the pH of 10-‐8 M of a HCl solu:on and build up a coherent text using phrasal verb. Then create a :tra:on curve (plojng the pH changes against the volume of a base added to the solu:on) and appropriately arrange the graph as a part of the text. The en:re discourse must be structured according to an introduc:on/data descrip:on/conclusion layout.

TEXT BUILDING IDEAL GAS INTUITION (8)

•  Let’s say we have a container with a lot of gas molecules bumping into each other and onto the container, all with their velocity vectors. Let’s assume that at each bump that occurs there is no loss of energy, no loss of momentum: it is an ideal gas. We can think of pressure as something pushing on an area, so p=F/S (read: pressure is the ra:o of the exerted force to the surface).

IDEAL GAS INTUITION

•  If we consider a side of the container, every :me the molecules bump against they have a change in momentum, so the force exerted on the wall will be given by the change in momentum over the change in :me: F=Δ P/Δ t. Roughly speaking, the product of the pressure exerted over any point of the container and its volume is propor:onal to a constant –that is, the kine:c energy of the molecules bumping around, so PV = K.

IDEAL GAS INTUITION •  If we were able to squeeze the container out (decrease the volume of the container by squeezing it), the molecules would hit (collide with, bump against) the sides more o_en, so that there would be more changes in momentum, each par:cle would exert more force, so that the pressure would be higher. So if volume goes down, then pressure goes up. If I make the volume bigger, then it will take longer for the molecules to bump against larger surfaces. So, similarly, if the volume goes down, then the pressure goes up.

IDEAL GAS INTUITION

•  Let’s say we have a box with an ini:al volume of 50 meters cubed, so V1 = 50 m3, and an ini:al pressure of 500 Pa (Pascal = N/m2), so that P1 = 500 Pa. Then we squeeze the container down to 20 m3. What’s the new pressure? There is no work on the system, no exchange of energy from outside of the system, just the squeezing, so P1 V1 = P2 V2 and 500 x 50 = 20 x P2 => P2 = 1250 Pa.

IDEAL GAS INTUITION

•  As for temperature we can say, oversemplifying things, that it is propor:onal to the ra:o of the average kine:c energy of the system to the number of molecules. T = k E/N, so that E = NT/k. We are essen:ally saying that PV = (1/k) NT. That is to say that P1 V1 /T1 = P2 V2 /T2. If the temperature inside the container steps up, then the par:cles will bump more, so that also the pressure will go up, assuming the volume stays flat.

IDEAL GAS INTUITION

•  If the temperature goes up and the pressure stays flat, the only way to make the molecules bump across is to increase the volume while increasing the temperature. Roughly speaking, we can consider the moles as being the ra:o of the mass in gram to the molar mass and the universal gas constant R=8,31 J/mol L, thus rewri:ng the equa:ons above as Boyle’s law PV = nRT.

EXERCISE

•  Focus on the verbs in bold. Which are just verbs being followed by preposi:ons and which are phrasal verbs? What is the difference? Explain them in the context and find out any synonyms. Then consider any topics in your area and construct a coherent and cohesive text using phrasal/preposi:onal verbs. The layout of the text must contain an introduc:on, a data/graph descrip:on sec:on and a conclusion.

EXERCISE

•  Solve the following problems wri:ng a coherent text in English:

•  I’ve got some gas in a balloon at a P1 = 3 atm , V1= 9 liters and T=constant. What will pressure become if V1 goes from 9 liters to V2=3 liters?

•  If I have an ideal gas at a STP, how many moles do I have in 1 liter? –that is, how many liters will one mole take up?

EXERCISE

•  I have a container with p=12 atm, V=300 ml, T=10°C. how much O2 in grams is there?

•  I have 98 ml of an unknown substance, its mass being 0.081 g at STP. What’s the missing substance?

•  Say I have a balloon of He, V=1m^3, P=5Pa, T=20°C. Calculate the number of moles.

TEXT BUILDING: DIFFUSION AND OSMOSIS (9)

•  Let’s say we have a container with a bunch of water molecules bumping to each other. Inside we have also sugar molecules, though we have many more water molecules. We call the thing there is more of the solvent (in this case water). Whatever there is less of it is the solute. So we say that the sugar has been dissolved into the water.

DIFFUSION AND OSMOSIS

•  The combina:on of the two we call it the solu:on. The solvent is the thing doing the dissolving and the thing that is dissolved is the sugar, that is the solute. We want to focus on diffusion. Let’s consider another container with some molecules of gaseous oxygen randomly bumping to each other with a certain amount of kine:c energy. What is it going to come out in this type of container?


•  A_er a certain :me the system reaches its equilibrium. The par:cles, being ini:ally somewhere closed to each other are going to get rela:vely spread out. This is diffusion: essen:ally the spreading of molecules from high concentra:on to low concentra:on areas. We have to bargain for there are some par:cles per unit space –that is, a certain amount of both solvent and solute. How much?


•  Say we have more than one container, for example two, with some water inside and with a hole in the middle; we have to take into account that there are a bunch of water molecules on either sides. Assuming that both sides have the same level of water, the par:cles going from the right to the le_ equal those going the opposite direc:on with the same pressure.


•  Let’s dissolve some solute into it on the le_-‐hand side so that they are small enough to fit through a hole. Over :me some of the sugar molecules will go to the other side so that the concentra:on might be equal: this is an example of diffusion of the solute. We have a shi_ from a hypertonic solu:on to a hypotonic solu:on –that is, from higher concentra:on to a lower concentra:on.


•  What happens if we have a tunnel where the solute is too big to travel, but water is small enough to travel? Let’s say we have an outside environment with a bunch of water around a semipermeable membrane, so that water can go in and out the membrane (permeable to water), but the solute can’t go out of it, its molecules being too big.


•  Zooming the membrane we realize sugar can’t get through the holes, only water can get back and forth. Which side of the membrane has a higher or lower concentra:on of solute? The inside of the membrane is hypertonic, the outside has a lower concentra:on, so it comes out as being hypotonic.


•  If there had been no sugar molecules water would have had equal likelihood of going in both direc:ons, but being there, they prevent some molecules of water from passing the hole in one direc:on, but not in the other. So we have inward flowing of water to try to equilibrate the concentra:on. Therefore we have flowing of the solvent from a hypotonic situa:on to a hypertonic solu:on, so that the membrane stretches out.


•  Diffusing through a semipermeable membrane is called osmosis. Over :me the pure solvent dilutes the solu:on down, so that its level jumps up, while the solvent’s gets down. If we consider a U-‐shaped tube, this process goes on un:l the hydrosta:c pressure exerted on the membrane’s side touching the solu:on balances the osmo:c pressure –that is, the pressure exerted by the solvent molecules approaching the solu:on.


•  The pressure exerted by the solvent is balanced by an equal and opposite pressure that prevents the solvent from passing. We can thus quan:fy the osmo:c pressure as the pressure we have to apply on a solu:on to prevent the solvent from passing into it:

πV = nRT where π is the osmo:c pressure, V is the volume, R is the universal gas constant


(R = .0821 L atm/mol K = 8.3145 m3 Pa/mol K) T is the temperature and n is the number of moles, that is the amount of substance of a system that contains as many par:cles (atoms, molecules, ions, electrons, nuclei, etc.) as there are atoms in 12 grams of 12C, which means that there are 1 mole of 12C atoms in 12 grams of carbon, so that there are 6.02 x 1023 (Avogadro’s number) atoms in 12 g of C.

EXAMPLE

•  If the concentra:on c = n/V then π = cRT •  EXAMPLE: human blood’s plasma has an osmo:c pressure of 7.65 atm at a temperature T=37 °C. How many grams of glucose must be dissolved in water so that the solu:on be isotonic with the blood’s plasma (glucose molar mass = 180.16 g/mol) supposing the final volume being constant?

EXAMPLE •  P=nRT/V = wRT/MV => w = PMV/RT = 7.65 atm x 180,16 (g/mol) / 0.0821(L atm/mol K) x 310K = 54,15 g/l.

•  The injected solu:on must be isotonic with the plasma because if it were more diluted, the water would flow inside the cells where there is a greater concentra:on of solutes, swelling them up un:l break. If the injected solu:on were more concentrated than the cellular liquid then the opposite would happen: the cells would lose water becoming wrinkled.

EXERCISE

•  Find out any other examples of osmosis and describe it carefully conver:ng the numerals into words and using both phrasal and presposi:onal verbs.

•  Apply the argument to related topics such as nuclei, membranes, ribosomes, eukariotes, prokariotes, endoplasmic re:culus, Golgi bodies, etc. and write a coherent text.

TEXT BUILDING: BIOCHEMISTRY OF A NEUROTRANSMITTER (10)

•  Vesicles containing neurotransmiQers are synthesized in the cell body, or “soma” of the neuron and carried down microtubule tracks by motor proteins. For exocytosis to occur, these vesicles must fuse with the membrane of the axon terminal, releasing their contents into the cle_. NTKA is a newly discovered excitatory neurotransmiQer.

BIOCHEMISTRY OF A NEUROTRANSMITTER

•  A_er being released into the synap:c cle_, NTKA is cleaved into two components, NT and KA, by an enzyme known as NTKAse. The KA component diffuses away into the extracellular space, while the NT component is brought back into the presynap:c neuron via specific membrane-‐bound transport proteins. Inside the axon terminal, NT is covalently bonded to another KA group, forming a new NTKA molecule that can be reused as a neurotransmiQer.


•  A new disease, with an autosomal recessive mode of inheritance, is characterized by NTKA deficiency. Autosomal recessive NTKA deficiency is caused by a hyperac:ve NTKAse enzyme, “∆ NTKAse,” which binds NTKA with a much higher affinity than wild-‐type NTKAse does, quickening the deple:on of NTKA from the synap:c cle_.


•  Interes:ngly, ∆ NTKAse and NTKAse do not appear to have different rates of catalysis at satura:ng substrate, kcat .

Three pa:ents volunteer for a study involving NTKA deficiency. Via in vitro gene:c manipula:on, researchers are able to produce a large quan:ty of NTKAse protein from each of these pa:ents and characterize them biochemically.


•  The Figure shows a Lineweaver-‐Burke plot of NTKAse ac:vity for three pa:ents, each of whom has one or more rela:ves with NTKA deficiency.


•  Pa:ents 1 and 2 are of unknown genotype, while pa:ent 3 is homozygous recessive. The NTKAse ac:vity of Pa:ent 3 was measured in the presence of a drug known to limit the ac:vity of ∆ NTKAse.


•  A Lineweaver-‐Burke plot can be used to characterize an enzyme’s ac:vity in different concentra:ons of substrate.


•  The reciprocal of the enzyme’s rate of reac:on, 1V, is plo@ed against the reciprocal of the substrate concentra:on, 1/[S].


•  The enzyme’s maximum rate of reac:on, V�max and the enzyme’s binding affinity for substrate, KM, can be extracted from the plot’s y-‐ and x-‐intercepts, respec:vely. A low KM �� indicates high binding affinity, whereas a high KM indicates low binding affinity.


•  To summarize things up, the figure makes for a Lineweaver-‐Burke plot used to characterize the V�max and KM �� values of NTKAse in three pa:ents. The V�max and KM �� values of each pa:ent can be extracted by taking the inverses of the y-‐ and x-‐intercepts, respec:vely. The third pa:ent’s assay includes an inhibitor of ∆NTKAse.

EXERCISE: READING COMPREHENSION(10)

•  Read the previous text once more and choose the right answer to this ques:on: How would the signaling ac/vity of NTKA change at the post-‐synap/c neuron if a molecule that specifically bound the NTKAse ac/ve site were added to the synap/c cle<?

1) NTKA would compete with this molecule for binding sites on the post-‐synap:c receptor, and fail to excite the post-‐synap:c neuron.

2) This molecule would block NT reuptake channels, leading to an increase in the synap:c concentra:on of NTKA.

3)  NTKA would remain in the synapse for a longer period of :me, leading to increased binding of post-‐synap:c receptors and more excita:on.

4)  This molecule would act as a compe::ve inhibitor of NTKAse, leading to more rapid degrada:on of NTKA in the synapse.

TEXT BUILDING TAKE-‐OFF VELOCITY AND DISPLACEMENT (11)

•  The take-‐off velocity of an A 439 airbus of the American Airlines has a magnitude of 280 Km/h in the direc:on of going down the runaway. From the moment when it leaves to the actual take-‐off it has a constant accelera:on of 1.0 (m/s)/s or 1.0 m/s2 (that is, 1.0 meters per second per second: a_er every second it goes 1 meter per second faster than it was going at the beginning).

TAKE-‐OFF VELOCITY

•  How long does the take-‐off last? We can convert kilometers per hour into meters per second:

•  How many meters per second do 280 kilometers per hour take up? 280 km/hour x 1 hour/3600s x 1000 m/1km = 78 m/s

•  which is preQy fast: we realize that for every second that goes by the airbus is travelling 78 meters, roughly three fourths the length of a football field!

TAKE-‐OFF VELOCITY

•  So how long will it take to go 78 meters per second? Just 78 seconds. And it adds up to our calcula:ons, or it makes sense, for accelera:on = change in velocity / change in :me, so that

•  Δt = Δv/a = 78m/s / 1.0 m/s^2 = 78 s •  Consequently, the airbus takes about seventy-‐eight seconds to take off.

TAKE-‐OFF DISPLACEMENT

•  Now, how long of a runaway does the same airbus need? Given these numbers, what is the minimum length of the runaway? Actually, we want to figure out the displacement (a vector point that represents the difference in the posi:on of two points), or how far does this plane travel as it accelerates 1 m/s2 to 280 km/h, how much length does this thing cover –that is, we are going to calculate the airbus A380 take-‐off distance.


•  If we assume the accelera:on to be constant, then we will end up with something called the average velocity, which is

•  Va = (Vf + Vi) / 2 = 78 + 0 /2 = 39 m/s. •  We can now calculate the space displacement: s = Va*Δt = 39 m/s * 78 s = 3042 m.

•  So we need over three kilometers for one of these stuff to take off (1.8/1.9 miles, 1 kilometer being 0.621371192 miles).


•  If we graph this in a v(t) system of coordinate axes we will find out that the distance travelled is the area under the curve given by the constant accelera:on up to the 78s over the t-‐axis, and if we take the average velocity for the same amount of :me we will get the exact same area under the curve, or we will take the exact same distance.


•  Generally speaking, we may wonder why distance is the area under the velocity/:me line. Well, let’s say I have something moving to the right with a constant velocity (as the seconds s/ck away it does not change) of 5 m/s and let’s plot it against :me. How far does this thing travel a_er 5 seconds? We know that vΔt=Δs so

•  5m/s*5s=25m.


•  This is exactly the area under the rectangle we obtain by ploRng the graph. The area is going to be the distance travelled because displacement is the product of velocity and the change in :me.

•  Let’s assume a = Δv/Δt = 1m/s2 and the magnitude of ini:al velocity ||v|| = 0.


•  A_er 1 second we are going 1 meter per second faster, a_er 2 seconds we are going another meter per second faster than that, again when we go forth in :me, we will go one second faster (and so up) than that: if we remember the defini:on of the slope we can draw the line corresponding to the accelera:on, which is the slope of that line.


•  So if we accelerate that amount for 5 seconds, how far have we travelled? For every second we can split up a rectangle: the smaller the rectangle, the closer we are going to get to the area under the curve, which is the distance travelled, which is luckily going to be the area of a triangle, A=bh/2. So the magnitude of the displacement (distance) ||s|| = ½(5s*5m/s) = 12.5m.

EXERCISE

•  Choose any similar topics and write down a coherent text using appropriate phrasal/preposi:onal verb.

TEXT BUILDING: DIVERGENCE

•  Let’s say we have a vector field represen:ng the velocity of par:cles of fluid in 3D.

•  Then the divergence of the vector is defined as the del vector dot V, that is, the par:al deriva:ves with respect to x, y, z of the three vector’s components, namely:

DIVERGENCE

•  Let’s say we have a vector field represen:ng par:cles of fluid in 2D.

•  = x2y î + 3y ĵ •  Then what does the div( ) turn out to be? •  div( ) = 2xy + 3 •  The divergence is an operator made up of the sum of the deriva:ve with respect to x :mes the unit vector in the x direc:on ( î ) plus …

DIVERGENCE

•  When we deal with the divergence we start with a vector field.

•  When we take the divergence we have a value for each point x, y, …

•  Taking the divergence makes for point values. •  Make for à result in •  The divergence tells about an actual scalar number at any point in the field.

DIVERGENCE

•  Let’s say the velocity of the fluid is •  = ½ x î + 0 ĵ = ½ x î (only in the x direc:on) •  When x=1 the magnitude of the vector is ½ •  x=2 à v=1 (twice as big) •  x=3 à v=3/2 •  A par:cle right, say, at v=1 in the fluid means its velocity in the x direc:on is going to be 1 m/s to the right

DIVERGENCE

•  For every v the magnitude of the vector does not change (when we move up).

•  It is only dependent on x (it depends on x). •  No upward movement in the x-‐y plane •  The further we go to the right, the faster the par:cles move towards the right.

•  As we increase in the x direc:on the field gets stronger and stronger, so that the par:cle goes faster and faster.

DIVERGENCE

•  The divergence of this vector field at any point is ½ .

•  The divergence is posi:ve •  What do these statements mean? •  We are essen:ally focusing on how much the magnitude of the field increases in the x direc:on.

DIVERGENCE

•  The posi:ve divergence tells us that if we take an arbitrarily infinitely small circle on the right-‐hand side we have par:cles exi:ng really fast. In a given amount of :me, say 1 second, par:cles are moving fast out of the right side. In the same amount of :me we will have some par:cles coming through the le_-‐hand side, but it is a fewer number of par:cles.

DIVERGENCE

•  Since there are more par:cles leaving than coming, the space inside the circle is going to become less dense.

•  The divergence is telling us that at any point the field is less dense: there is more flowing out at any point than flowing in.

•  A posi:ve divergence says the slope of how much our x component is increasing.

DIVERGENCE

•  If we draw circles anywhere all we are going to see in more exi/ng through the right than entering through the le_ and so we are going to get less dense at any given points. Therefore any given point is a source of the field, or a source of par:cles.

•  If the divergenge is nega:ve then the space inside the circle will get denser and denser.

DIVERGENCE

•  A nega:ve divergence means there are more par:cles going in to the right than coming out of to the le_ side.

•  A posi:ve divergence means the par:cle of the field is diver/ng out of a point.

•  A nega:ve divergence means the par:cle is converging to a point.

EXERCISE

•  Build a coherent text en:tled “Divergence” using the notes of these slides. Imagine to write a text to be published in a scien:fic review whose aim is giving a general intui:on of this topic. Write an introduc:on including all the points you are going to treat. Then describe the argument using our examples, including graphs and equa:on as a part of the text. Complete it with an appropriate conclusion.

TEXT BUILDING PERIOD MATRICES

•  To treat a general Riemann surface, choose a homology basis < α1, . . . , αg, β1, . . . , βg > with intersec:on numbers αi・ βj = δij . For ease of visualiza:on, we can take each pair (αi, βi) as simple loops mee:ng just once, in disjoint handles as i varies. The intersec:on number means we can think of αi poin/ng along the posi:ve real axis and βi along the posi:ve imaginary axis at the point of intersec:on.

PERIOD MATRICES

•  Let ai ,bi be the corresponding periods of ω. Cut X along the curves (αi, βi)g1 to obtain a surface with boundary U. Then ω has no periods on Y so it integrates to a func:on f : Y → C. The key point is to no:ce, as in the case of a torus, that if p, p* ∈ ∂U are two points that get glued together onto αi , then f(p*) − f(p) = bi, the period along bi. Similarly, if p, p* end up on bi, then

f(p*) − f(p) = −ai

PERIOD MATRICES

•  Now we use the fact that d(z dz’) = dz dz’. Thus by Stokes’ theorem we have

•  ∫X ω ∧ ω’ = ∫dU f ∧ dω’. (read ∧ cross) •  The curve αi, for example, occurs twice along ∂Y , giving a contribu:on of ∫ai (f(p) − f(p*)) ∧ ω’ = − bi ai’. Similarly from the curves βi we get ∫bi (f(p) − f(p*)) ∧ ω’ = ai bi’. So altogether we have the famous formula:

∫X ω ∧ ω’ = ai bi’ − bi ai’

EXERCISE

•  Find out any rigorous excerpts like this and check all uses of phrasal/preposi:onal verbs.

•  Build up any rigorous scien:fic discourse, for example using your notes, using appropriate phrasal/preposi:onal verbs, especially with references to curves in space.

•  Repeat the same exercise by applying the same Math to any scien:fic subjects.

LISTENING EXERCISES •  Listen to the following videos, take notes of the key-‐words while listening

and write down a coherent text using appropriate phrasal/preposi:onal verbs and logical connec:ves.

•  hQps://www.khanacademy.org/science/biology/cellular-‐molecular-‐biology/cell-‐division/v/nuclei-‐membranes-‐ribosomes-‐eukaryotes-‐and-‐prokaryotes. (All Khan Academy content is available for free at www.khanacademy.org).

•  hQps://www.khanacademy.org/science/biology/cellular-‐molecular-‐biology/mitosis/v/mitosis •  hQps://www.khanacademy.org/science/biology/cellular-‐molecular-‐biology/cellular-‐respira:on/v/oxida:on-‐and-‐reduc:on-‐

from-‐biological-‐view •  hQps://www.khanacademy.org/science/biology/cellular-‐molecular-‐biology/cellular-‐respira:on/v/glycolysis •  hQps://www.khanacademy.org/science/physics/one-‐dimensional-‐mo:on/kinema:c_formulas/v/average-‐velocity-‐for-‐

constant-‐accelera:on •  hQps://www.khanacademy.org/science/physics/one-‐dimensional-‐mo:on/kinema:c_formulas/v/deriving-‐displacement-‐as-‐

a-‐func:on-‐of-‐:me-‐accelera:on-‐and-‐ini:al-‐velocity •  hQps://www.khanacademy.org/science/physics/one-‐dimensional-‐mo:on/kinema:c_formulas/v/plojng-‐projec:le-‐

displacement-‐accelera:on-‐and-‐velocity •  hQps://www.khanacademy.org/science/physics/one-‐dimensional-‐mo:on/kinema:c_formulas/v/plojng-‐projec:le-‐

displacement-‐accelera:on-‐and-‐velocity •  hQps://www.khanacademy.org/math/mul:variable-‐calculus/par:al_deriva:ves_topic/curl/v/curl-‐1 •  hQps://www.khanacademy.org/math/mul:variable-‐calculus/par:al_deriva:ves_topic/curl/v/curl-‐2 •  hQps://www.khanacademy.org/math/mul:variable-‐calculus/par:al_deriva:ves_topic/curl/v/curl-‐3

REFERENCES •  1) Adapted from “Phrasal and preposi:onal verbs in specialized texts: a crea:ve device” Mari Carmen Campoy

Cubillo Universitat Jaume available at hQp://www.aelfe.org/documents/text4-‐Campoy.pdf •  2-‐5) Ibidem •  6) Adapted from hQp://hyperphysics.phy-‐astr.gsu.edu/hbase/quantum/zeeman.html •  7) Listening exercise from khanAcademy at hQps://www.youtube.com/watch?v=2q4vSKwaBtw (All Khan Academy

content is available for free at www.khanacademy.org). •  8) Listening: hQps://www.khanacademy.org/science/physics/thermodynamics/v/thermodynamics-‐part-‐1. All

Khan Academy content is available for free at www.khanacademy.org. Listen also to parts 2,3,4,5 and focus on phrasal/preposi:onal verbs while wri:ng the texts down.

•  9) Listening exercise: hQps://www.khanacademy.org/science/biology/cellular-‐molecular-‐biology/cell-‐division/v/diffusion-‐and-‐osmosis. All Khan Academy content is available for free at www.khanacademy.org.

•  10) Adapted from hQps://www.khanacademy.org/test-‐prep/mcat/biological-‐sciences-‐prac:ce/biological-‐sciences-‐prac:ce-‐tut/e/-‐biochemistry-‐of-‐a-‐newly-‐discovered-‐-‐pretend-‐-‐neurotransmiQer

•  11) Listening exercise: hQps://www.khanacademy.org/science/physics/one-‐dimensional-‐mo:on/accelera:on_tutorial/v/airbus-‐a380-‐take-‐off-‐:me. All Khan Academy content is available for free at www.khanacademy.org.

•  12) Listening exercise: hQps://www.khanacademy.org/math/mul:variable-‐calculus/par:al_deriva:ves_topic/divergence/v/divergence-‐1. Listen also to parts 2,3.

•  13) Adapted from hQp://www.math.harvard.edu/~ctm/sem/home/notes/99/course.pdf.

PHRASAL’VERBS’ - uniroma2.it · PHRASAL’VERBS’ • If’the’phrasal’verb’has’an ......

Documents

Transcript of PHRASAL’VERBS’ - uniroma2.it · PHRASAL’VERBS’ • If’the’phrasal’verb’has’an ......