systems and tenses - scienze.uniroma2.it€¦ · systems&and&tenses& •...

SYSTEMS AND TENSES

English for Scien/sts Maria Cris/na Teodorani

SYSTEMS AND TENSES

•  Scien.fic subjects essen.ally deal with both systems’ descrip.ons and working systems. In order to simply describe systems or state something about them, or tell about the way they usually work we use the simple present tense. But as soon as the system is at work, whatever system, we use the present con3nuous.

SYSTEMS AND TENSES

•  Present simple: statements generally acknowledged, well-‐known behaviours, descrip.ons, shared results in Math models, conclusions and theory modeling.

•  Present con.nuous: system’s behaviour while interac.ng with something as being emphasized at the .me of speaking.

•  Past simple and past con.nuous: same logic

GRAMMAR REVIEW: PRESENT PASSIVE

•  DESCRIBING A SYSTEM (Present simple): •  ACTIVE: The solu.ons of a first order differen.al equa.on (Subject) produce (Verb) a slope field (Object).

•  PASSIVE: A slope field (Subject) is produced (Verb) by the solu.ons of a first order differen.al equa.on (0bject).


•  SYSTEMS AT WORK (Present con.nuous): •  ACTIVE: The solu.ons of a first order differen.al equa.on (Subject) are producing (Verb) a slope field (Object).

•  PASSIVE: A slope field (Subject) is being produced (Verb) by the solu.ons of a first order differen.al equa.on (0bject).


•  General rule for the present passive: •  Present simple: Subject + am/is/are + past par3ciple (3rd column of the verbal paradigms).

•  Present con.nuous: Subject + am/is/are + being + past par3ciple (3rd column of the verbal paradigms).

TEXT BUILDING: REDOX REACTIONS(1)

•  Redox is an acronym that stands for reduc.on/oxida.on. During a chemical reac.on, or equa.on, some reactants are being transformed into some products. We generally associate an oxida.on state to the charge an atom would have if all bonds to atoms of different elements were 100% ionic. Thus the oxida.on number is connected to the charge.

REDOX REACTIONS

•  Let's consider, for instance, the molecule of sodium chloride. We know that sodium is an alkaline metal and that it has one valence electron in group I, while chlorine is an halogen of group VII that just needs 1 electron to have full 8 valence electrons in its shell. Consequently, in the forma.on of NaCl, Na is going to give electrons and Cl is going to get them.

REDOX REACTIONS

•  As a result of this we can write Na+Cl -‐. Here Na+ means +1 charge because the sodium is giving the electron, while Cl-‐ means -‐1 charge because the chlorine is ge>ng it. The bond is ionic. If the bond were covalent, we would focus on par.al posi.ve or nega.ve charges. In the forma.on of the molecule of water, the oxygen is gaining 2 electrons from the 2 hydrogens, which are losing them, the Hs being more electroposi.ve and the O being more electronega.ve.

REDOX REACTIONS

•  Consequently the oxida.on number of hydrogen in H20 is +1, while the oxygen's is -‐1. As a result of this we can say that in a molecule of water the hydrogen are oxidized by the oxygen: the electrons are taken away from them, so that they have a posi.ve charge.

•  Now, let's study the following combus.on:

REDOX REACTIONS

•  CH4 + 2O2 → CO2 + 2H2O Here a molecule of methane is reac3ng with two molecules of oxygen in order to produce a molecule of carbon dioxide plus 2 molecules of water plus some heat (being esothermic, the reac.on produces more heat than you put into it). In CH4 an atom of carbon is bounded with 4 hydrogens.

REDOX REACTIONS

•  While reac3ng, being more electronega.ve, the carbon is taking 4 electrons from the hydrogens, so its charge is going down by four. As a result its oxida.on number is -‐4 , while the hydrogen's is +1. Thus, we can write C-‐4H+1. In CO2, the carbon's oxida.on state is +4, which means that it is giving up 4 electrons, and really it only has 2 electrons to give up, for it has 4 electrons in its valence shell.

REDOX REACTIONS

•  So, what is ge>ng oxidized and what is ge>ng reduced? Let's write down the first half reac.on:

C-‐4 → C+4 + 8 e-‐ Here carbon is going from an oxida.on number of -‐4 on the leh side of this equa.on, to an oxida.on number of +4 on the right side: 8 electrons are being taken away from carbon, so it is being oxidized.

REDOX REACTIONS

•  As for the second half reac.on: 4O + 8 e-‐→ 4O-‐2 we are shown 4 oxygens with a zero oxida.on state (being in the elemental form) turning into 4 oxygens with a -‐2 oxida.on state, so each of these oxygens are taking 4 electrons, the two of them, thus there are 8electrons.

REDOX REACTIONS

•  The oxida.on state, that is, the hypothe.cal charge, is going down, or it is being reduced by carbon, as well as the carbon above is being oxidized by oxygen. Finally, what is the oxidizing agent, what is the thing that is oxidizing? Oxygen, of course. is the oxidizing agent, while carbon is the reducing agent.

REDOX REACTIONS

•  Redox can also be reviewed from a biological point-‐of-‐view. Biologists usually say oxida.on deals with losing hydrogen atoms, while reduc.on deals with gaining hydrogen atoms, though the essen.al meaning stays the same. The reac.ons within cells which result in the ATP (adenosine triphosphate) synthase using energy stored in glucose are referred to as cellular respira.on. It requires oxygen as the final electron acceptor.

REDOX REACTIONS

•  The equa.on for aerobic respira.on is C6H12O6 + 6O2 → 6CO2 + 6 H2O + energy

•  Here we are combining glucose with molecular oxygen so that cellular respira.on is being made for. We end up with 6 carbon dioxides and six molecules of water, while the energy produced is made up of some heat and about 38 ATPs.

REDOX REACTIONS

•  Glucose is completely broken down to CO2 + H2O though, during fermenta.on, it is only par3ally broken down.

•  Let’s take a look at the half reac.ons: H12 → 6 H2 (read: Hydrogen 12 yielding to 6 hydrogens 2) says the hydrogen preserves a +1 oxida.on number (o.n.) on both sides of the equa.on so that nothing is happening with respect to oxida.on and reduc.on,

REDOX REACTIONS

•  while C6 → 6C + 24e-‐ shows the number of electrons lost by carbon in cellular respira.on, the carbon being oxidized by the oxygen. Finally, O6 + 6O2 + 24e-‐ → 6O2 + 6O

emphasizes the fact that these 24 electrons are the same electrons carbon is losing, so that oxygen, which is gaining electrons, is being reduced by carbon.

REDOX REACTIONS

•  Where does the energy come from? It is produced because the electrons are going from a higher energy state, or level, to a lower one (lower orbitals are more stable):

C6H12O6 + 6O2 → 6CO2 + 6 H2O ↕ ↕ ↘ ↙ oxidized reduced oxygens taking e-‐

Therefore carbon is losing hydrogens, while oxygen is gaining hydrogens.

REDOX REACTIONS

•  Use the following redox reac.ons to build a coherent text with appropriate verbs in the present tense (both ac.ve and passive forms).

•  Zn + CuSO4 à ZnSO4 + Cu •  Fe + 2HCl à FeCl2 + H2 •  H3PO3 + MnO4 à HPO4 + Mn •  2H2 + O2 à 2H20

TEXT BUILDING: ENTHALPY(2)

•  We want to figure out and quan3fy the concep.on of enthalpy of forma.on of a substance. To this extent, let’s consider a P(V) diagram, where P is the pressure and V is the volume of the system (3):

ENTHALPY

•  The area under the curve in the clockwise direc.on is the net work done by the system. Here ΔU = 0 -‐that is, there is no change in the internal energy of the system. We know that

ΔU = Q – W = 0 where Q is the heat applied to the system and W is the work done by the system.

ENTHALPY

•  We define enthalpy as H = PV. If we consider a finite change in H we can write

ΔH = ΔU + Δ(PV) = Q – W + Δ(PV) Here ΔH is a state variable because it is the sum of other state variables. If we consider a system with a piston we can write

ΔH = Q – PΔV + ΔPV The change in enthalpy will equal Q if the last two terms cancel out.

ENTHALPY

•  Under what condi.on? If and only if the pressure is constant, then we can factor it out, so we get

ΔH = Q – P Δ V + PΔV = Qp that is, heat at constant pressure. In so doing we are kind of squeezing out the previous diagram, for we are making of the forth path and the return path the same exact path –that is, a horizontal line from A to B.

ENTHALPY

•  As a result, no net work is being added in going from A to B. Most chemical reac.ons are at constant pressure (1 atm). In this case we define enthalpy as the heat content when pressure is constant. Let’s consider the following reac.on as an example:

ENTHALPY

•  C(s) + 2H2 (g) à CH4 + 74 KJ of heat released How much heat is being added to the system? We realize that

H(ini.al) = reactants > H(final) = products and that Qp = -‐ 74 KJ (change in enthalpy). Since this heat is added to the system, what is the change in enthalpy of the reactants’ system rela.ve to the products’ system?

ENTHALPY

•  Hf – Hi = ΔH = -‐74 KJ that is, Hf is lower than Hi by 74 KJ, so we conclude that Hf is at a lower level of energy, or it is more stable and that the reac.on is esothermic, that is, the heat is released by the system. Most chemical reac.ons are at constant pressure (1 atm). We thus define enthalpy of forma.on of a substance as the heat content when pressure is constant.

ENTHALPY

•  Calculate the varia.on of the standard enthalpy of forma.on for the process, at constant pressure, C(diamond) à C(graphite) knowing that

•  Cgraph + O2 (g) à CO2 ΔH°f = -‐393,5 KJ/mole •  Cdiam + O2 (g) à CO2 (g) ΔH°f = -‐395,4 KJ/mole •  You must write a coherent text including numerals and chemical equa.ons.

TEXT BUILDING: MECHANICS APPLIED TO MACHINES (4)

•  Dealing with mechanics applied to machines means dealing with power transmission. Actually, mechanical power is the rate at which work is being provided once the system is in movement. In symbols:

•  P = dW/dt = Fds/dt = Fv •  W =∫Fds =∫F cosθ ds =∫Ft ds

MECHANICS APPLIED TO MACHINES

•  where P is power, W is the work done, F is the exerted force and Ft its tangent component, while s denotes movement in space and v velocity. We define work, generally speaking, as energy transferred by force and, in the same sense, we define energy as the ability to do work. Physically speaking, power is nothing else than a sent unit of work per second.


•  We are going to see how power is being transmiFed as long as a system moves. In systems such as cars and the like, power is transmiFed via fric.on wheels, cogwheels, belts, joints, rod/crank systems, flywheels and similar systems in such a way that a torque is generated between driving and driven, moving and resis.ng structures.


•  A simple example of power transmission between two shahs not too distant from one another is that of fric.on wheels. The figure below shows the scheme of such a transmission.


•  We have two wheels whose diameters are D1 and D2. The first one is placed along the driveshah and has an angular velocity ωm and a torque Mm . The second one belongs to driven shah and has an angular velocity ωu and a torque Mr .


•  If we call I the distance shown in the figure below we get the interaxis I = ½(D1 + D2)


•  If the wheels do not slide, the velocity of the contact point on wheel 1 will equal the velocity of the contact point on wheel 2, so that V1 = V2. Consequently ½(ω1 D1) = ½(ω2 D2). That is to say:

i = ω1 / ω2 = D2 / D1 The transmission ra.o i depends on the diameter of the two wheels. We size the diameters of the two wheels correctly using this last equa.on (together with that of the interaxis I).


•  On a prac.cal level, the uses of fric.on wheels are rather limited, though being silent and having a regular transmission, for we can use them only at low powers. For high powers the force T = f R must be elevated, but since the fric.on coefficient for commonly used materials such as steel or cast iron is rather low (f= .10 -‐ .15), there should be very high pushing forces R, so that shahs, pins, bearings, etc., would be strongly stressed.


•  Since fric.on wheels undergo a huge radial stresses in order to ensure their adherence they do not provide high power transmissions. Anyway, star.ng from two ideal fric.on wheels we can obtain a series of cogs on their external surfaces – that is, a series of projec.ons on the edge of a wheel transferring mo.on by engaging with another series and alterna.ng with empty spaces.


•  Once in mo.on these cogs are being easily interpenetrated; in this case, power transmission is no longer due to fric.on but to the “pushing” force that each cog of the driving wheel is exer.ng on those of the driven wheel. In this way, provided that the built cogs are strong enough, we can transmit high powers.


•  It is possible to convert rota.onal mo.on into linear (transla.onal) mo.on using the pinion/rack mechanism, where the pinion’s rota.onal mo.on is being converted into a transla.onal mo.on by the rack. Given a gear, we define the pinion as the cogwheel with the smallest diameter and the wheel as that with the largest diameter. The interaxes is the distance between the axis of the two wheels.


•  If ω1 is the angular velocity of the pinion and ω2 the angular velocity of the wheel, then we define the transmission ra.o as i = ω1 / ω2


•  An example of this is the car’s steering-‐wheel mechanism. While driving, the steering’s rota.on is being converted into the transla.on of the elements ac.ng upon the wheels.


•  As shown in the figure below, the force transmiFed from the driving wheel to the driven one is the tangent component FT .


•  It is Ft = Fcosθ = Cm / R1 = Cr / R2, where Cm is the machine torque and Cr is the resis.ng torque. Of course the radial component Fr is not responsible for mo.on and cons3tutes a solicita.on all over the shah on which the wheels are keyed. Its module is Fr = Fsinθ. This suggests we ought to render the pressure angle θ very small in order to increase the value of FT .


•  The value of the angle of pressure affects the minimum number of cogs that a wheel can have. In prac.ce, we assign the number of cogs as a func.on of the pressure angle and of the transmission ra.o using the following formula:

Zmin = 2 / [i2 + (1+2i)sin2θ -‐ i]½ •  As for the minimum cogs’ number (Zmin) as a func.on of θ and i we have the following table:


i 1 2 3 4 5 6 7 8 9 10

θ=15° 21 25 26 27 28 28 29 29 29 29

θ=20° 13 15 15 16 16 16 17 17 17 17

θ=25° 9 10 10 11 11 11 11 11 11 11

EXERCISES

•  Write a text describing the previous table. •  Describe how the pinion/rank system works with respect to your own car in mo.on: what happens if you have to turn right, leh, or get straight on?

•  Write a coherent text in your area using appropriate tenses.

DIFFERENTIAL EQUATIONS (DE) (5)

•  Differen.al equa.ons are useful for modeling and simula.ng phenomena and understanding how they operate. They are ‘available’ in different nota.ons:

•  y’’ + 2y’ = 3y •  (read: the second deriva.ve of y (or “ y prime prime”) plus twice the first deriva.ve of y is equal to three y)

DIFFERENTIAL EQUATIONS (DE)

•  Func.on nota.on: •  f’’(x) + 2f’(x) = 3f(x) •  (read: the second deriva.ve of the func.on with respect to x plus….)

•  Leibnitz ‘s nota.on: •  d2y/dx2 + 2 dy/dx = 3y •  (read: the second deriva.ve of y with respect to x twice plus….).

DIFFERENTIAL EQUATIONS (DE)

•  The solu.on of a DE is a func.on or class of func.ons. One solu.on is y1(x) = e-‐3x : it sa3sfies the equa.on y’’ + 2y’ = 3y. In fact since the first deriva.ve is y1’(x) = -‐3e-‐3x (read: y subset 1 prime of x equals nega.ve 3 .mes e (raised) to the nega.ve 3x) and the second deriva.ve is y1’’(x) = 9e-‐3x (read: y subset 1 prime prime of x…), then subs.tu.ng into the equa.on we realize we get an iden.ty, so that it works. No.ce that y1 is a solu.on, but not the only one : for example y2(x) = ex is another solu.on.

PARTICULAR LINEAR SOLUTIONS TO

DIFFERENTIAL EQUATIONS.(6) •  Let’s consider dy/dx = -‐2x + 3y -‐5. A solu.on is

a linear func.on in the form y=mx + b. Therefore we have to find out the m and b that make this linear func.on sa.sfy the DE. In order for y=mx + b to sa.sfy the DE this has to be true for all x in the linear equa.on. So dy/dx = m = -‐2x + 3y -‐5, so that m = -‐2x + 3(mx + b) -‐5 and m = (3m-‐2)x + 3b -‐5.

PARTICULAR LINEAR SOLUTIONS TO DIFFERENTIAL EQUATIONS.

•  This has to be true for all x, so that 3m-‐2=0 and m=2/3=3b-‐5. Consequently b=17/9 and the solu.on is y = (2/3)x + 17/9.

•  EXERCISE. Find out any similar DE and write down a coherent text describing it.

SLOPE FIELDS (7)

•  We use slope fields to visualize solu.ons of differen.al equa.ons.

•  In other words, the solu.ons of a first order DE produce a slope field –that is, their graphical representa.on.

•  Given dy/dx = -‐x/y let’s say we do not know the solu.on and want to give a general sense of what a solu.on might look like.

SLOPE FIELDS

•  To this extent we create a table labeling its columns x, y and dy/dx. When x=0 and y=1, then dy/dx = 0. It means if the solu.on goes through the point (0,1), then the slope is going to be 0 (marked as a short horizontal line parallel to the x-‐axis). At the point (1,1) (read: “1 comma 1” or “ordered pair 1,1”) the slope is -‐1 (a nega.ve downward slope);

SLOPE FIELDS

•  at (1,0) the slope is undefined; at (-‐1,-‐1) the value of the slope is again nega.ve 1, while if the solu.on goes through (1,-‐1), we will have a posi.ve upward slope of 1. When plo�ng the table onto a graph a slope field is being produced. The solu.ons, corresponding to some condi.ons, look like circles.

SLOPE FIELDS (8)

•  EXERCISE . The graph below shows the slope field of dy/dx=x2-‐x-‐2, with the blue, red, and turquoise lines being (x3/3)-‐(x2/2)-‐2x+4, (x3/3)-‐(x2/2)-‐2x, and (x3/3)-‐(x2/2)-‐2x-‐4 respec.vely. Write a coherent introduc.on/data descrip.on/conclusion text including the graph and the Math equa.ons converted into words. Use the present passive in both the system’s descrip.on and func.oning.

SLOPE FIELDS (8)

SEPARABLE DIFFERENTIAL EQUATIONS(9)

•  Given the DE dy/dx = -‐x / yex^2 , we want to find the par.cular solu.on that is going through the point (0,1). This ordered pair is being given as our ini.al condi.on. Separa.ng the variables means get the dy and dx on separate sides. Therefore mul.plying by ydx both sides of the equa.on we are being leH with ydy = -‐xe-‐x^2 dx, so that, integra.ng both sides, we get

SEPARABLE DIFFERENTIAL EQUATIONS

•  ∫ydy = -‐ ½∫2xe-‐x^2 •  (read: “the integral of y (.mes) dy is equal to nega.ve one half .mes the integral of 2x (.mes) e to the nega.ve x squared).

•  As a result we get y2/2 + C1 = ½ e-‐x^2 + C2, where C1 and C2 are two constants.

•  (read: y squared over 2 plus C1 is equal to one half .mes e to the nega.ve x squared, which is the an.-‐deriva.ve, plus C2 ).


•  At the point (0,1) we get ½= ½ + C, where C = C2 – C1 , so that y = e-‐x^2/2, which is the par.cular solu.on that sa3sfies our ini.al condi.on.


•  EXERCISE. Find out any separable DE and build a coherent text. As usual, numerals and/or diagrams must be a part of the text. Equa.ons must be wri�en in words (just once for each new equa.on and only if quan..es are different). The text must be structured according to an introduc.on/data descrip.on/conclusion layout. Support your text with samples of separable DE applica.ons to any phenomena, using appropriate connec.ves when paragraphing.

MODELING WITH DIFFERENTIAL

EQUATIONS (10) •  Let’s use DE for modeling popula.on. If P

stands for popula.on and t stands for .me, say, in days, then dP/dt stands for the rate of change of popula.on with respect to .me.

•  If we consider the rate of change as being propor.onal to the actual popula.on, then we can write dP/dt = KP, which is a reasonable statement since the larger the popula.on the larger the rate of growth at any given .me.

MODELING WITH DIFFERENTIAL EQUATIONS

•  Integra.ng both sides of the equa.on we get ∫dP/dt = ∫kP, or ln|P|= Kt + C1.

•  (read: the logarithm of the module of P (or the absolute value of P) is equal to Kt plus C subset 1).

•  Therefore |P| = eKt+C1 = eKt eC1 = C eKt .

MODELING WITH DIFFERENTIAL EQUATIONS

•  |P| = eKt+C1 = eKt eC1 = C eKt •  (read: the module of P (or the absolute value of P) is equal to e to the kt plus C subset 1, which is equal to e to the kt power .mes e to C subset 1, which is in turn equal to a constant C .mes e to the kt).

•  If P>0, then P= C eKt . •  EXERCISE. Model anything with DE and write down a coherent text.

HOMOGENEOUS DIFFERENTIAL EQUATIONS (11)

•  Given dy/dx = f(x,y), if we algebraically manipulate this statement so that dy/dx = F(y/x), then we can make a variable subs.tu.on that makes it separable. So for example if dy/dx = (x+y)/x then dy/dx = 1 + y/x. By se>ng y/x = V, then we get y=xV, so that dy/dx = V + x dV/dx, or dV= (1/x)dx, from which V = ln|x|+ C. Unsubs3tu3ng back we get y/x = ln|x|+ C or y = xln|x|+ xC. To figure out C we need some ini.al condi.ons.

SECOND ORDER DIFFERENTIAL EQUATIONS (12)

•  A second order DE is in the form a(x)y’’ + b(x)y’ + c(x)y = d(x) •  (read: a of x .mes the second deriva.ve of y with respect to x (or y “prime prime”) plus…)

•  where coefficients are func.ons of x. The associated linear homogenous equa.on is in the form

•  Ay’’ + By’ + Cy = 0, where A, B and C are constants.

SECOND ORDER DIFFERENTIAL EQUATIONS

•  Let’s consider the following example: •  y’’ + 5y’ + 6y = 0 •  In order to find out a solu.on we have to ask ourselves if there is any func.ons that when taking its 1st, 2nd, 3rd, 4th, 5th, … nth deriva.ve it essen.ally becomes the same func.on. The func.on we are searching for is ex .

•  Let’s set y = erx .

SECOND ORDER DIFFERENTIAL EQUATIONS

•  Then •  (erx)’’ + (erx)’ + 6 erx = 0 or •  erx (r2 + 5r + 6) = 0 •  where r2 + 5r + 6 = 0 is the characteris.c equa.on, from which we get r1 = -‐2 and r2 = -‐3, so that y1 = e -‐2x and y2 = e-‐3x .

•  EXERCISE. Write a coherent text with an appropriate layout about any second order DE and their applica.ons.

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. •  h�ps://www.khanacademy.org/science/chemistry/oxida.on-‐reduc.on/redox-‐oxida.on-‐reduc.on/v/balance-‐and-‐

redox-‐reac.ons1 •  h�ps://www.khanacademy.org/science/chemistry/oxida.on-‐reduc.on/redox-‐oxida.on-‐reduc.on/v/prac.ce-‐

determining-‐oxida.on-‐states •  h�ps://www.khanacademy.org/science/physics/thermodynamics/v/heat-‐of-‐forma.on •  h�ps://www.khanacademy.org/test-‐prep/mcat/biomolecules/principles-‐of-‐bioenerge.cs/v/enthalpy-‐1 •  h�ps://www.khanacademy.org/science/physics/torque-‐angular-‐momentum/torque-‐tutorial/v/

rela.onship-‐between-‐angular-‐velocity-‐and-‐speed •  h�ps://www.khanacademy.org/science/physics/torque-‐angular-‐momentum/torque-‐tutorial/v/

rela.onship-‐between-‐angular-‐velocity-‐and-‐speed •  h�ps://www.khanacademy.org/science/physics/torque-‐angular-‐momentum/torque-‐tutorial/v/constant-‐

angular-‐momentum-‐when-‐no-‐net-‐torque •  h�ps://www.khanacademy.org/science/physics/torque-‐angular-‐momentum/torque-‐tutorial/v/cross-‐

product-‐and-‐torque •  h�ps://www.khanacademy.org/math/differen.al-‐equa.ons/second-‐order-‐differen.al-‐equa.ons/

complex-‐roots-‐characteris.c-‐equa.on/v/complex-‐roots-‐of-‐the-‐characteris.c-‐equa.ons-‐1 •  h�ps://www.khanacademy.org/math/differen.al-‐equa.ons/second-‐order-‐differen.al-‐equa.ons/

undetermined-‐coefficients/v/undetermined-‐coefficients-‐1

REFERENCES •  1) Listening exercise at h�ps://www.khanacademy.org/science/chemistry/oxida.on-‐reduc.on/

redox-‐oxida.on-‐reduc.on/v/introduc.on-‐to-‐oxida.on-‐and-‐reduc.on All Khan Academy content is available for free at www.khanacademy.org.

•  2) Listening exercise: h�ps://www.khanacademy.org/science/physics/thermodynamics/v/enthalpy •  3) Graph at h�p://en.wikipedia.org/wiki/Pressure_volume_diagram •  4) Adapted from www.is.tutopesen..it/dipar.men./meccanica/meccanica.pdf •  5) Listening exercise: h�ps://www.khanacademy.org/math/differen.al-‐equa.ons/first-‐order-‐

differen.al-‐equa.ons/differen.al-‐equa.ons-‐intro/v/differen.al-‐equa.on-‐introduc.on •  6) Listening exercise: h�ps://www.khanacademy.org/math/differen.al-‐equa.ons/first-‐order-‐

differen.al-‐equa.ons/differen.al-‐equa.ons-‐intro/v/finding-‐par.cular-‐linear-‐solu.on-‐to-‐differen.al-‐equa.on

•  7) Listening exercise: h�ps://www.khanacademy.org/math/differen.al-‐equa.ons/first-‐order-‐differen.al-‐equa.ons/differen.al-‐equa.ons-‐intro/v/crea.ng-‐a-‐slope-‐field

•  8) From h�p://en.wikipedia.org/wiki/Slope_field •  9) Listening exercise: separable-‐differen.al-‐equa.ons-‐introduc.on •  10) Listening at h�ps://www.khanacademy.org/math/differen.al-‐equa.ons/first-‐order-‐

differen.al-‐equa.ons/modeling-‐with-‐differen.al-‐equa.ons/v/modeling-‐popula.on-‐with-‐simple-‐differen.al-‐equa.on

•  11) Listening. h�ps://www.khanacademy.org/math/differen.al-‐equa.ons/first-‐order-‐differen.al-‐equa.ons/homogeneous-‐equa.ons/v/first-‐order-‐homegenous-‐equa.ons

•  12) Listening exercise: 2nd-‐order-‐linear-‐homogeneous-‐differen.al-‐equa.ons-‐2

systems and tenses - scienze.uniroma2.it€¦ · systems&and&tenses& •...

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