Chapter 09 Ans
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Chapter 09 Ans

Course Number: GEOL 3041, Fall 2009

College/University: LSU

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Answers for Chapter 9: Trace Elements in Chemical Fractionation V.M. Goldschmidt proposed the following simplistic rules to qualitatively predict element distribution: A. Two ions with the same radius and valence should enter into solid solution in amounts proportional to their concentration. In other words, they should behave about the same. B. If two ions have a similar radius and the same valence, the smaller...

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for Answers Chapter 9: Trace Elements in Chemical Fractionation V.M. Goldschmidt proposed the following simplistic rules to qualitatively predict element distribution: A. Two ions with the same radius and valence should enter into solid solution in amounts proportional to their concentration. In other words, they should behave about the same. B. If two ions have a similar radius and the same valence, the smaller ion is preferentially incorporated into the solid as compared to the liquid. C. If two ions have a similar radius but different valence, the ion with the higher charge is more readily incorporated into the solid than the liquid. 1. Look at the periodic table and use Rules A and B to predict the general behavior of Rb in minerals and melts: to what element will it behave similarly? Will it enter first-forming minerals on Bowens Series, or remain in late melts? In which minerals will it concentrate? Rb will behave similarly to K, concentrating late melts, and in K-feldspar and micas. 1 2. Ni, in contrast to Rb, typically has a valence of +2. To what major elements will it most likely be similar? In which minerals is it likely to concentrate? Will in concentrate in late melts? Ni2+ behaves similarly to Mg2+ and Fe2+ concentrating in early mafics and not in late melts. 3. Use the periodic table and Rule B to predict whether Fe2+ or Mg2+ will concentrate more in liquids than will coexisting mafic minerals. Can you support your contention with a binary phase diagram from Chapter 6? Mg2+ is smaller than Fe2+, so should concentrate more in the solid phase. This is supported by the Fo-Fa diagram (Figure 6.10) in which the solidus is more Mg-rich than the liquidus at any temperature. 4. Use Rule C to predict whether Cr+3 and Ti+4 are preferred in solids as compared to liquids. High valence predicts they incorporate preferentially into solids. A more rigorous and quantitative approach is to actually measure the distribution of elements between minerals and coexisting liquids in melting experiments. From them we can define: i(liquid) i(solid) (9.1) (9.2) Always write the reaction before you analyze one. distribution constant KD = X solid i X liquid i Where X solid is the mole fraction of component i in the solid phase = ni/(ni + nj +nk) i KD = C = D for traces L CS (9.3) distribution coefficient or partition coefficient Where CS is the concentration of component i in the solid. 5. Use Table 9.1 (next page) to determine whether Rb is an incompatible or a compatible element. What about Ba? Ni? Cr? Eu? Rb is highly incompatible (excluded from olivine, pyroxenes, even plagioclase). Ni and Cr are compatible. Eu3+ is incompatible, so is Eu2+ (but it concentrates in plagioclase). 6. Consider a magma with phenocrysts of olivine, orthopyroxene, clinopyroxene, and plagioclase. The olivine phenocrysts contain 1200 ppm Ni. Use Table 9.1 to estimate the concentration of Ni in all the other phases. CL = CS/DOl = 1200/14.0 = 86 ppm Ni. COpx = CL * DOpx = 86 * 5.0 = 430 ppm CCpx = CL * DCpx = 86 * 7.0 = 602 ppm CPl = CL * DPl = 86 * 0.01 = 0.9 ppm 2 Bulk D for a rock: Di = W D A A i (9.4 ) WA = weight fraction of mineral A in the rock = the distribution coefficient for the element i in mineral A. Suppose you have a garnet lherzolite with the following mode (volume %) Olivine Orthopyroxene Magnetite 54% 21% 2% Clinopyroxene Garnet 14% 9% D iA Table 9-1. Partition Coefficients (CS/CL) for Some Commonly Used Trace Elements in Basaltic and Andesitic Rocks Mineral Ol Opx Cpx Garnet Magnetite Density 3.6 3.5 3.4 4.0 5.2 Rb Sr Ba Ni Cr La Ce Nd Sm Eu Dy Er Yb Lu Olivine 0.01 0.014 0.01 14.0 0.7 0.007 0.006 0.006 0.007 0.007 0.013 0.026 0.049 0.045 Opx 0.022 0.04 0.013 5.0 10.0 0.03 0.02 0.03 0.05 0.05 0.15 0.23 0.34 0.42 Cpx Garnet 0.031 0.042 0.06 0.012 0.026 0.023 7.0 0.955 34.0 1.345 0.056 0.001 0.092 0.007 0.23 0.026 0.445 0.102 0.474 0.243 0.582 3.17 0.583 6.56 0.542 11.5 0.506 11.9 Plag Amph Magnetite 0.071 0.29 1.83 0.46 0.23 0.42 0.01 6.8 29. 0.01 2.0 7.4 0.148 0.544 2. 0.082 0.843 2. 0.055 1.34 2. 0.039 1.804 1. 0.1/1.5* 1.557 1. 0.023 2.024 1. 0.02 1.74 1.5 0.023 1.642 1.4 0.019 1.563 * Eu3+/Eu2+ Italics are estimated Data from Rollinson (1993). 7. Use the tables above to calculate the bulk distribution coefficient for Nd and Sm in the rock. Ol Opx Cpx Grt Mt Sum Vol % Density 54 3.6 21 3.5 14 3.4 9 4 2 5.2 mass Mass % 194.4 0.54 73.5 0.20 47.6 0.13 36 0.10 10.4 0.03 361.9 1.00 D(Nd) 0.006 0.03 0.23 0.026 2 WD 0.003 0.006 0.030 0.003 0.057 0.100 D(Sm) 0.007 0.05 0.445 0.102 1 WD 0.004 0.010 0.059 0.010 0.029 0.111 8. According to your calculations above, is Nd or Sm more incompatible? Nd Is the difference great? Not at all Will a partial melt of the lherzolite have more Nd and Sm than the initial rock? More Nd Would you expect the concentration of Nd in the melt to vary with the % melted? Yes. It would be greater for lower % melted (low F). Given the following: 3 Rare Earth Elements CL 1 = C O Di (1 F ) + F Batch melting (9.5) CO is the concentration of the trace element in the original assemblage before melting began F is the weight fraction of melt produced [= melt/(melt + rock)] 9. What is meant by batch melting? Melt stays in contact with the source and remains in equilibrium with the residual unmelted minerals. 10. How can the concept of batch melting be applied to partial melting? Consider the composition of any partial melt as that of an equilibrium melt to some melting fraction, at which point it separates from the solid residuum. This can be done for any melt fraction. 11. What does Equation (9.5) reduce to as F 1? As F 1, the equation reduces to CL/CO = 1 12. In plain English, what does your answer to question 11 tell you about CL at high F (and how might that be applied to the Earth)? At high melt fractions the concentration of a trace element in (a primary) magma approaches the value of that element in the source melted. 13 What does Equation (9.5) reduce to as F 0? As F 0, the equation reduces to CL/CO = 1/D. 14. If we determine CL for a melt representing low F, and we know Di what can we infer about the nature of the source rocks? We can use the melt compositions to estimate the concentration of a number of trace elements in the source, thereby constraining it somewhat. 4 15. Given the plot in Figure 9.2, for what types of elements and for what degrees of partial melting will the concentration of an element vary most extremely? For elements with low D and at very low % melting. 100 10 D = 0.001 D = 0.1 D = 0.5 16. How can the batch melting concept be applied to incremental partial melting in which melts are generated and removed in 10% increments? What variables in Equation (9.5) change as melting proceeds? Consider this as a series of incremental batches each remaining in equilibrium until some percentage melt is reached and then released. Melting then begins anew, so CO changes and, since the remaining mineralogy changes as melt is generated and removed, one must recalculate Di for each new batch. CL/C O 1 D=1 D=2 D=4 D = 10 0.1 0 0.2 0.4 0.6 0.8 1 F Figure 9.2. Variation in the relative concentration of a trace element in a liquid vs. source rock as a fiunction of D and the fraction melted, using equation (9-5) for equilibrium batch melting. 17. Can we simply reverse the melting process, and treat F as the fraction left un-crystallized in a cooling magma chamber? Why does or doesnt this make sense? Yes, as long as equilibrium is maintained. Equilibrium processes are reversible. 18. Rayleigh fractionation treats melting and crystallization as involving the immediate separation of crystals (during crystallization) or melts (during melting) as soon as they are created. How does this model differ from incremental batch melting/crystallization? Its similar, but the batches are tiny, so the values vary continuously. 5 The rare earth elements (REE) compose the lanthanide series. The series is characterized by progressive filling by electrons of the 4f orbital. 19. Are the REE incompatible or compatible? On what do you base your answer? They are incompatible (D << 1). 20. What can you say about the variation in ionic radius as the atomic number increases? To what can you attribute this effect? Looking at the periodic table inside the front cover I note that the ionic radius decreases progressively as Z increases. This is called the lanthanide contraction and results largely from the increasing number of protons in the nucleus attracting electrons (which also increase, but are in the same subshell). 21. Use Goldschmidts rule B to predict how the incompatibility of the REE varies with atomic number. Incompatibility should decrease with increasing Z (smaller ions will be slightly less strongly incorporated in the melt) 22. Compare your result for question 21 to Table 9.1. The D values do indeed rise progressively with increasing Z. The typical petrologic REE diagram is a plot of CSample/CChondrite vs. atomic number, based on the assumption that chondrite meteorites approximate the trace element (and isotopic) composition of primordial mantle from which many common melts are derived: 23. Draw on the graph below the curve representing the REE pattern for a volcanic rock derived by complete melting of primitive chondritic mantle. Explain your answer. If F = 1 then CL = CO and chondritic mantle is being melted, so melt = rock = chondrite and rock/chondrite = 1.0 6 24. Explain the REE diagram on the right. This is a reasonably typical pattern for partial melting at fairly low F of chondritic material. Lighter REE have lower D and are increasingly concentrated in early melts. 25. Use Table 9.1 on page 3 of this handout to explain the accompanying two REE diagrams. What minerals are indicated to be involved? Compare each diagram to that in Question 24. Explain your reasoning. How might you characterize the nature of the mineral involvement? The top pattern resembles that of question 29, so looks like another fairly low-F partial melt of chondritic mantle, but there is a pronounced negative Europium anomaly that typically correlates with plagioclase fractionation (note higher D for Eu2+ and plagioclase in Table 9.1). Either plagioclase was left behind as a residual phase of was removed by fractional crystallization from the liquid. The middle pattern has a much steeper slope for the medium and heavy REE only (its about the same for the LREE). The D values are high for these elements in garnet, so this pattern indicates that garnet was either a residual phase or was removed by fractional crystallization. 7 26. Spider Diagrams are an extension of the REE diagram concept. If the accompanying diagram represents the pattern for a typical partial mantle melt, what can you say about the way the elements are ordered across the abscissa (x-axis)? In general they are arranged in order of increasing D, left to right. The more incompatible elements are on the left. Only Rb, Sm, and Th are a bit out of sequence. Some phase or phases may be sequestering them more than usual or else the authors estimate of D values is not quite correct. 27. Suppose you analyze two rocks and find rock A has a much higher concentration of La than rock B. a. How could a contrast in the nature of the source that melted be responsible? The source of A may have had more rare earth elements in general. b. How could a difference due to melting of the same source be responsible? A may be a result of lower degree of melting (low F) of a similar source. c. responsible? How could a difference due to crystallization of the same melt be A may have been initially identical to B, but fractional crystallization removed some of the more compatible elements (decreasing concentration toward the right). 8 28. How might you use the ratio of La/Ce to distinguish from among at least some of the variables above or to evaluate the extent of partial melting or fractional crystallization? La/Ce would give the slope from among the REE. If it were similar for both A and B, it would argue against hypothesis (a) above, but could not distinguish between (b) and (c). 29. What REE ratio would you judge to be particularly sensitive to the involvement of garnet in petrogenesis? Explain. Garnet affects the HREE mostly, so Id choose Eu, Gd, or Ty over Lu (depending on what elements were actually analyzed). 30. Refer to Table 9.1 on page 3 of this handout and propose a trace element ratio that might be sensitive to variations in crystal fractionation of olivine vs. clinopyroxene during crystallization of a magma. Explain. Id try Ni/Cr. Ni is very high for only olivine and Cr is high for only clinopyroxene. If Ni/Cr were particularly high Id expect clinopyroxene had fractionated without much olivine. If Ni/Cr were very low, Id propose the opposite. 31. How do isotopes of a particular element vary from each other? Isotopes have the same number of protons in the nucleus (same atomic number) but different numbers of neutrons (different mass). 9 32. To what extent would you predict that two isotopes of any particular element would fractionate chemically during melting or crystallization processes? Explain. They shouldnt fractionate chemically at all, because they are the same element. 33. Isotopes of some elements do indeed fractionate during melting or crystallization processes. On what basis would this be possible? Explain. They must fractionate on the basis of mass differences, because that is the only difference remaining (assuming they did indeed fractionate during the process). 34. Is 1H vs. 3H more likely or less likely to fractionate than 206Pb vs. 207Pb during liquid/solid or liquid/vapor transitions? What about 1H vs. 2H? Explain your answer. Much more likely, because the efficiency of mass fractionation is related to mass /total mass, which is 2/3 for 1H/tritium but only 1/207 for Pb. The value is 1/2 for 1H/deuterium, so that should fractionate well, but not as well as tritium. 35. How might we generalize as to what types of elements are most likely to fractionate efficiently? Light elements, particularly if there is a large mass difference (more than one proton). As a general rule, when two isotopes fractionate during melting, crystallization, vaporization, or condensation, the light isotope is preferentially incorporated into the higher entropy phase (vapor > liquid > solid). We can define the extent of fractionation of oxygen isotopes by: ( O/ O) = 18 16 ( 18 O/ 16 O) sample ( 18 O/ 16 O) SMOW ( 18 O/ 16 O) SMOW x 1000 (9.10) 36. If ocean water (SMOW) evaporates to become clouds, what constraints can you place on values of ( 18O/16O) (abbreviated as 18O) for a sample of water vapor from the clouds? 16 O should be favored over 18O in the vapor compared to the liquid (students may wish to expound of the greater velocity of the H2O molecule with 16O relative to 18O and its greater ability to reach escape velocity during evaporation). Clouds should thus have negative values of 18O because 18O/16O in the sample of vapor should be lower than the value for SMOW. 10 37. Suppose you had a liter of seawater, and evaporated it experimentally. How would you expect the value of 18O in the vapor to vary as evaporation progressed? Explain. 0 18O Value for SMOW Without seeing any data Im guessing a bit, but 18O must certainly be negative, and at a minimum at the lowest percent vaporized. It must therefore have a positive slope on the diagram. Id bet on the concave shape for reasons similar to the low D curve in Figure 9.2. Students may pick a more linear trend, but it is very unlikely to be convex upward. 100 0 % evaporated 38. Would you expect the magnitude of 18O for typical cloud vapor to have a large of small 18O value? Why? Id expect a large negative value, as cloud vapor represents a low vapor fraction from the SMOW host. 39. Rain and snow are simply the result of precipitation from clouds. Use what you have discussed above to explain the accompanying 10 diagram: Lower temperatures results in a higher percentage of vapor condensing to liquid. As in the diagram above (for question 37) the percent condensed would drive the 18O value for the condensed liquid toward the value for the vapor being condensed. Thus high degrees of condensation would reflect the original large negative value for the clouds, and lower degrees of condensation would reverse the earlier evaporation process, returning 18O toward the original value of the SMOW evaporated. 0 -10 18O o/oo -20 M et eo tion ipita Prec ric -30 -40 -50 -40 -30 -20 -10 0 10 20 Mean Annual Air Temperature (o C) Figure 9.9. Relationship between (18O/16O) and mean annual temperature for meteoric precipitation, after Dansgaard (1964). 11 40. What is the difference between a radioactive nuclide and a radiogenic one? A radioactive nuclide is an unstable one that will eventually decay to daughter nuclide(s), and some form of radiation. A radiogenic nuclide is a daughter nuclide produced by the decay of a radioactive one. 41. Given the following equation for the decay of 40 K to 40Ar: 40 Ar = 40Aro + e F I K(e GJ HK 40 - t -1) (9.16) where: e is the decay constant for the electron capture process ( e = 0.581 x 10-10 a-1) is the decay constant for the entire decay process ( = 5.543 x 10-10 a-1) 40 40 40 Ar is the total amount of radiogenic 40Ar daughter in the rock Aro is the original 40Ar in the rock at the beginning point of the decay process K is the total amount of radiogenic 40K remaining in the rock Which of the variables in Equation (9.16) cannot be directly determined either empirically in the lab or analytically from a rock sample? For each, suggest why not. 40 Aro is the only one, because some (at least) of the original Ar is quite likely to have been lost, but one cannot tell how much. and e can be determined empirically, but not by analyzing a rock. Propose a geologically plausible solution to permit solving equation (9.16) for the age of a volcanic rock. Ar is a noble gas, so it is quite likely that nearly all would be lost from a lava at the surface of the Earth. Ar isnt likely to be chemically bound, the lava is quite hot and fluid, and the pressure is very low. The gas should thus escape. The assumption that 40Aro = 0 is probably reliable. Why wont such a solution work for other decay schemes, such as Sr Rb, Sm Nd, and U Pb? The amount of any original non-radiogenic nuclides equivalent to the radiogenic ones cannot be known or even approximated from an analysis. For other isotopic systems, we can solve the original-daughter dilemma by analyzing several cogenetic samples with varying amounts of the parent. For the 87Rb 87Sr system, the decay equation can be recast as: 87 Sr/86Sr = (87Sr/86Sr)o + (87Rb/86Sr) t (9.18) where 87Rb is the radioactive parent, 87Sr radiogenic daughter, and 86Sr a stable Sr nuclide (and therefore constant in any particular rock). The subscripted o refers to initial values prior to any 87 Rb decay. 42. Describe Equation (9.18) as a straight line equation of the type y = mx + b. What is represented by y, x, m, and b? What could you derive from the slope? 12 If y = 87Sr/86Sr and x = (87Rb/86Sr) then m (the slope) = t and b (the intercept) = (87Sr/86Sr)o As written above it reads y = b +mx 43. If a mantle source rock is partially melted, how will Rb/Sr differ between the partial melt and the refractory solid residue? Is the fractionation greatest for small % partial melting or large? Explain your answer (can you find a diagram to help you?). In general, Rb/Sr will be higher than in the source. Rb/Sr is highest when F (melt fraction) is lowest (see Figure 9.3) and approaches the value in the mantle as F 1.0 44. Will Rb/Sr vary as the partial melt evolves by crystal If fractionation? so, will the more primitive or the more evolved liquids be richer in Rb? Why? Yes Rb/Sr will generally vary with fractional crystallization. This is close to partial melting in reverse, so as the fraction crystallized approaches 1.0, Rb/Sr will get higher and higher (as in Figure 9.3), so it is high in the evolved liquids. 45. If a mantle source rock is partially melted, how will 87Sr/86Sr differ between the partial melt and the refractory solid residue? Will crystal fractionation affect the ratio? Explain. 87 Sr/86Sr will not vary noticeably as a result of partial melting because heavy isotopes do not fractionate during melting processes. Chemically, both isotopes are strontium. 46. The continental crust has been derived by fractionation from the mantle over geologic time (Chapters 16 - 18). From your answers to Questions 43 - 45, suggest how Rb/Sr of the depleted mantle has changed during that time. The mantle becomes depleted in Rb more than Sr over time because Rb is favored compared to Sr in the partial melts that eventually are lost to generate continental crust. Rb/Sr thus becomes lower as the mantle evolves. 47. How will the effect described in Question 46 affect 87Sr/86Sr in depleted mantle over time? Because Rb/Sr is reduced in depleted mantle, as 87Rb 87Sr by radioactive decay, there is less and less parent compared to daughter, so the addition of 87Sr to the overall Sr population is reduced. 13 48. How will 87Sr/86Sr in enriched continental crust differ from that in the depleted mantle? Explain. Continental crust is the recipient of all that lithophile Rb from the mantle, so Rb/Sr is much higher in it. There is thus much more 87Rb to decay to 87Sr and less initial Sr, so the addition of 87 Sr to the overall Sr population is much more noticeable. 49. The y-intercept of a Rb-Sr isochron is (87Sr/86Sr)o, the initial 87Sr/86Sr ratio of the melt (and hence of the source, because Sr does not mass fractionate during partial melting). What might this value tell us about the nature of the source material for melts such as granites? Explain. The answers to the previous two questions suggest that the depleted mantle should be low in 87 Sr/86Sr and continental crust (and sediment derived from it) should be high in 87Sr/86Sr. If (87Sr/86Sr)o is low, then, it suggests a mantle origin of the melt. If (87Sr/86Sr)o is high, however, the source or the magma may be crustal or contaminated by crustal components. 50. 147Sm 143Nd behaves similarly to the Rb-Sr system. Both Sm and Nd are rare earth elements, and Sm has a larger atomic number than Nd. Which is preferentially fractionated into partial melts, and is the effect as great as for Rb-Sr? Explain. Because of the lanthanide contraction, Nd (the daughter) in this case is larger and preferred in the partial melt. Fractionation should be considerably less than for Rb-Sr, however, because Rb is an alkali metal with a valence of +1 and Sr is an alkaline earth element with a valence of +2. Sm and Nd are both rare earth elements and are much more similar to each other chemically. They should thus fractionate much less than Rb/Sr. 51. Your answers to Question 50 (should) indicate that the daughter element (Nd) is enriched compared to the parent in depleted mantle material. What effect would this have on 143Nd/144Nd (144Nd is non-radiogenic, and therefore constant) for depleted mantle with respect to enriched continental crust over long time spans? Explain. If Sm/Nd is higher in depleted mantle, then more 143Nd will be produced and it will have more of an effect on the relatively lower Nd concentration present. 143Nd/144Nd will thus increase more quickly than in enriched crust and should have a higher value. 14 52. According to Equation 9.3, the half-life (T) can be expressed by: T = 0.693/ . Use this and the following equations to explain the shape of the concordia. 238 235 232 U 234U 206Pb U 207Pb Th 208Pb ( = 1.5512 x 10-10 a-1) ( = 9.8485 x 10-10 a-1) ( = 4.9475 x 10-11 a-1) (9.22) (9.23) (9.24) Pb* indicates radiogenic Pb, and 1.0 - 2.5 indicate time in Ga. The half-life for 206Pb production is 4.468 x 109 yrs The half-life for 207Pb is 7.04 x 108 yrs Because of its shorter half life 207Pb should increase faster than 206Pb, which is true if one observes the axis scales in the figure. Although 207 Pb gets off to a quicker start, 206Pb must catch up, as 235U must deplete more quickly, producing relatively less 207Pb. If one were to concentrate only on Pb, one would expect the concordia to be concave upward, but this ignores the denominators in each axis. The 235U effect begins to predominate over time, as this decreasing number in the denominator causes 207Pb /235U to increase more quickly than 206Pb/238U. The curve is thus concave downward. You can demonstrate this in an Excel spreadsheet if you like. I used the decay constants and assumed 1000 atoms of each 235U and 238U, then calculated the amounts of 206Pb, 207Pb , 238U and 235 U at .05, 1, 1.5, and 2 Ga and then plotted 206Pb vs. 207Pb , 238U vs. 235U and 206Pb/238U vs. 207Pb / 235 U. The first is concave upward and the latter two are concave downward. 53. Pb is an incompatible large ion lithophile element (LIL). LILs are readily scavenged from rocks by water and melts. If a melting or H2O-flushing event occurred at 2.5 Ga in the diagram above, why would the discordia result? Because both 206Pb and 207Pb are lost in proportions equal to their abundances. The lead loss curve thus runs directly from the appropriate point (time) on the concordia toward the origin (zero lead). Of course all of the lead will not be lost, and different rocks or minerals may retain more lead than others. Thus individual samples will lie at various points distributed along the discordia. 15 PROBLEMS 1. Consider a garnet lherzolite with the following mode (volume %): Olivine Orthopyroxene Clinopyroxene Garnet Magnetite 54% 21% 14% 9% 2% Mineral Ol Opx Cpx Garnet Magnetite Density 3.6 3.5 3.4 4.0 5.2 Use Equation (9.4) and Table 9.1 plus the accompanying mineral density table to calculate the bulk distribution coefficient for Nd and Sm in the rock by filling in the table below. See Prob 09-1 Ans.xls for my solution (below) Vol % Ol Opx Cpx Grt Mt Total 54 21 14 9 2 Density 3.6 3.5 3.4 4.0 5.2 Mass 194.4 73.5 47.6 36 10.4 361.9 WD 0.54 0.20 0.13 0.10 0.03 1.00 DMin Nd 0.006 0.03 0.23 0.026 2 DMin Sm 0.007 0.05 0.445 0.102 1 0.111 0.100 Di Which is element more incompatible, Nd or Sm? Is the difference great? Nd has a lower D, so is more incompatible, but only slightly. Will a partial melt of the lherzolite have more Nd and Sm than the initial rock? More Nd and Sm (both are incompatible: D << 1). Would you expect the concentration of Nd in the melt to vary with the % melted? (Hint: consider the concentration if the source rock were completely melted) Yes. The concentration of Nd would be highest with low melt fraction (F) and gradually be diluted with greater F until it reaches the concentration in the source ad F 1 (see Figure 9.1 for D = 0.1). Would variations in % partial melting result in fractionation as great as for Rb/Sr (Figure 9.3)? Explain. No, Rb differs greatly from Sr, whereas Nd and Sm differ only slightly. 2. Use Equation (9.8) and follow the steps below to create a spreadsheet to model Rayleigh crystal fractionation of Rb and Ni as a basaltic magma crystallizes. Suppose that plagioclase, clinopyroxene, and olivine form and are removed from a liquid in the ratio 5:4:1 over a period of time. See Prob 09-2 Ans.xls for steps (a) and (b) below. 16 a) Calculate D Rb and D Ni. Which is more incompatible in this system? On what do you base your answer? Rb is more incompatible (much lower D) b) Use the Raleigh equation in your spreadsheet to calculate CL/Co for F = 0.05, 0.1, 0.15, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, and 0.9. Plot CL/Co vs. F for each, connecting the points with a line. Note: the exponent function on most spreadsheets is the ^ symbol. Thus 105 is expressed as 10^5. c) Which element would provide a more sensitive measure of the progress of fractional crystallization? Why? Over what range of F would this measure be most effective? Explain. Rb would, as it varies much more with F. Most of that variation occurs at values of F between 0 and 0.4, so this is the range of its greatest usefulness. d) How, and to what extent, does the Rb/Ni ratio vary with F? Would this make a better or worse indicator of the extent of crystal fractionation than Rb or Ni alone? Why? Rb/Ni varies even more than the concentration of Rb itself, because Rb has a negative slope whereas Ni has a positive slope. It thus would make a better indicator of the magnitude of F than either alone. 3. Open the spreadsheet REE.XLS (from Worked Example 2, available from http://www.prenhall.com/winter). Print a REE diagram for F = 0.05, 0.1 and 0.2. You can print only the chart if you click the chart before printing. Then modify the spreadsheet so that it uses the Rayleigh crystal fractionation model, instead of batch melting, and print REE diagrams for the same values of F. Compare the two models from your printouts (you might want to trace the curve for one model onto the chart for the other model for the same value of F). In what ways are the models similar? In what ways are they distinguishable? Could you readily recognize a magma that resulted from partial melting vs. fractional crystallization? Explain. See Prob 09-3 Ans.xls. The models produce quite similar results. The Rayleigh model is a bit more extreme (steeper slopes), but I wouldnt be able to distinguish with any confidence the products of Rayleigh vs. Batch melting on the basis of the REE patterns 4. Using the spreadsheet REE.XLS (from Worked Example 2, available from http://www.prenhall.com/winter) and F = 0.1 for a small degree of melting, create a model REE diagram for 100% of each of the minerals listed. Describe how each mineral controls the shape of the REE pattern in the coexisting liquid. This may be important, because some minerals impart a characteristic shape to the REE pattern either to melts from which that phase has been removed, or to melts derived from a rock in which that mineral is a residual phase. Olivine fractionation produces a relatively flat slope with very high concentrations of the REEs. There is a slight Yb dip that may be meaningful or just uncertainty in the D values. Orthopyroxene produces a steeper negative slope that is nearly constant across the diagram. Clinopyroxene produces a steep HREE slope and relatively flat LREE slope. Garnet produces a very steep LREE HREE slope. Plagioclase fractionation produces a positive slope sith a very pronounced negative europium anomaly. Amphibole scavenges the middle REEs with higher La and slightly higher Yb-Lu, resulting in a saucer shape. 5. The file OCEAN.ROC (available from http://www.prenhall.com/winter) contains the major, minor, and trace element data for a typical mid-ocean ridge basalt (+ signs), and an ocean island basalt (triangles). Using IGPET, create an REE and a spider diagram for these two rock types. 17 Both diagrams can be created under the heading Spider. Use the Sun diagram, normalized to chondrite, as your spider diagram. Include both analyses in the diagrams (via the all option). Try to explain the slope of the plots for each rock based on what you know about trace element fractionation associated with partial melting and fractional crystallization processes. Assume as a starting point that both are created by partial (batch) melting of a primordial mantle with chondritic trace element characteristics. In both diagrams, the incompatibility of the elements increases from right to left. You can see these diagrams in the next chapter (Figure 10.14). The OIB patterns are relatively typical for partial melting of a chondritic source as the concentration of a trace element in the melt is greater with increasing incompatibility, as one would expect. For MORB, however, the slope is positive. No model of crystal settling or source retention can explain a partial melt with lower concentrations of more incompatible elements than of more compatible ones. The only solution is that the source must not be chondritic. If we are to stick with our chondritic early Earth model, the mantle source of MORB (the most common volcanic material on Earth) must have been previously depleted in the incompatible elements. In other words the source must have a steeper positive slope than the partial melt. 6. The Early Paleozoic Stowe Formation in Vermont is part of a discontinuous chain of metabasalts and associated ophiolitic rocks that apparently mark a major suture of the Appalachian system. It is of interest to determine whether the volcanics were created as part of a volcanic arc, or are oceanic slices. The analyses of two samples are listed below: #1 SiO2 47.61 TiO2 1.04 Al2O3 16.20 FeO 10.30 MnO 0.06 MgO 8.86 CaO 14.00 Na2O 1.52 K2O 0.28 P2O5 0.04 Zr (ppm) 74 Y (ppm) 23 Nb (ppm) 11 Cr (ppm) 556 #2 49.95 1.07 15.49 9.14 0.08 6.52 13.41 3.75 0.41 0.09 78 28 11 290 Plot the results on each of the following diagrams: a) Ti/100 Zr Y x 3 (all in ppm. Note 1%=10000 ppm) b) TiO2 MnO x 10 - P2O5 x 10 c) Cr - Y This can be done in Excel and compared to Figure 9.8 or students can use IgPet and choose the Diagrams, Basalt Discrimination to plot the diagrams directly. Doing the latter (using the file Stowe.roc) I got: 18 Are the meta-volcanics of the Stowe Formation part of an island arc sequence on the basis of these data? In (a) both analyses plot as ocean floor (MORB) volcanics. In (b) they are on the border between MORB and ocean island tholeiites. In (c) they could be MORB or within-plate basalts. These Stowe rocks are more compatible with an ocean floor or oceanic is land than an island arc setting and may thus represent a sliver of oceanic crust caught up in the orogen. 7. Use Equation (9.12) in a spreadsheet to calculate the concentration of 87Rb and the radiogenic 87 Sr as a function of time. Begin with 100 atoms of Rb and calculate at intervals of 10 Ga spanning 100 Ga. Plot your results for both elements on the same graph (y = abundance, x = time). Discuss the graph in terms of the half-life. Estimate the half-life from the graph, and compare the results to the one you calculate from equation (9.13). See Prob 09-7 Ans.xls. The graph plots in perfect half-life increments of about 50 Ga. After the first half-life there are only 50 atoms of parent Rb and an equivalent amount of daughter Sr. After two half-lives there are only 25 atoms of Rb and 75 atoms of Sr, etc. 19 8. Three whole rock analyses from a modern batholith have provided the following Sr/Rb data (I have converted the data to 10-19 moles, but you can treat the data as though they involved individual atoms): Sample 1 2 3 Rb 25860 59490 82860 87 Sr 8700 8090 7230 87 Sr 12310 11448 10230 86 a) Use your spreadsheet and plot the data on a 87Sr/86Sr vs. 87Rb/86Sr diagram. Draw a line through the data. Use an x scale of 0-10 and a y scale of 0.68 to 0.94. b) For each sample, use the decay equation (9.12) to calculate the amount of 87Rb remaining, the amount of additional 87Sr created, and the location of the three points after 500 million years have passed. Plot them and draw the best line through the data. Are the three points still collinear? Fit a linear regression to the data and extract the slope. Calculate the age from the slope. Does the age and (87Sr/86Sr)o value from your intercept fit within reason the 87Sr/86Sr of the data above? Show your results. c) Repeat the exercise in part (b) as it would appear after 2 billion years. How does the pattern change? All are done on Prob 09-8.xls. All plot on straight lines (as they should) with increasing slope as age increases. Regression of the slope t, and this returns the age within a percent or so uncertainty and an intercept of a constant 0.7067. This reproduces Figure 9.11 perfectly. 9. Calculate the slope, Rb-Sr and Sm-Nd isochron ages and (87Sr/86Sr)o and (143Nd/144Nd)o for the Guichicovi Complex of southern Mexico using Equations (9.18) and (9.19) and the Sm-Nd, Rb-Sr, and U-Pb geoghronology of the Guichicovi accompanying table (Weber and Khler, 1999): Complex, Precambrian Research 96, 245-262 (1999). 87 For example, use a spreadsheet to plot 87Sr/86Sr as the ordinate vs. 87Rb/86Sr as the abscissa. The slope is then t, where t is the age, and (87Sr/86Sr)o is the intercept at 87Rb/86Sr = 0. Using the linearity of the fit as a measure of the accuracy of the results, would you say the age is a good one? How do the Rb-Sr and Sm-Nd ages compare? Which is more likely to be partially reset by later metamorphic events? What can you infer about these rocks from (87Sr/86Sr)o ? See Prob 09-9 Ans.xls. The fit is nicely linear, particularly the Rb-Sr data. Interestingly, the Sm-Nd age is considerably older (6.54 Ga) than the Rb-Sr age (0.14 Ga) for the same rocks. Id guess the Rb-Sr system is upset by a later metamorphic event. (87Sr/86Sr)o = 0.7048, which implies a mantle origin (or at least little enriched crustal input). Sample G08-4 G09-6 G23 G11 Mx05a Mx10-1 G09-4 Mx05-2 Mx09 Mx04a Mx06-2 Mx10-2 G01 G04-3 G18-2 G22 G03-3 G05-2 G21-4 G24 G29 G31 G35 G51 Rb/86Sr 0.3759 0.8837 0.0433 0.1571 0.1771 0.1696 0.1232 0.2420 0.0149 0.0619 0.1272 0.0567 1.1328 1.2365 1.9544 0.2626 0.1038 4.2842 0.2810 1.4007 4.7074 2.1819 2.6330 0.5537 87 Sr/86Sr 0.71058 0.71779 0.71152 0.71128 0.70930 0.70521 0.70875 0.70883 0.70984 0.70433 0.70635 0.70387 0.70818 0.73093 0.73706 0.70858 0.70535 0.76314 0.71338 0.73361 0.80336 0.73851 0.73480 0.71454 147 Sm/144Nd 0.1071 0.1132 0.1086 0.1609 0.1755 0.1817 0.1278 0.1546 0.1229 0.1188 0.1710 0.1310 0.1074 0.0886 0.1486 0.1244 0.1364 0.1388 0.1395 0.1294 0.1112 0.1343 0.1237 0.1328 143 Nd/144Nd 0.51204 0.51208 0.51208 0.51248 0.51263 0.51275 0.51220 0.51248 0.51213 0.51208 0.51252 0.51219 0.51205 0.51183 0.51242 0.51218 0.51220 0.51217 0.51230 0.51208 0.51174 0.51199 0.51204 0.51215 20 10. Being as specific as you can, what single rock would you expect to have the very highest present-day 87Sr/86Sr ratio in the world? Why would you expect it to be so? Explain fully, being clear what factors are involved. What would be the 87Sr/86Sr ratio of a partial melt of this rock? Why? Students readily pick up on the need for high Rb and a long time in order for Rb 87Sr. Because Rb behaves similarly to K, this indicates a very old K-rich rock. Students are less likely to recognize that 87Sr/86Sr is a ratio, and thus also requires low 86Sr for a really high value of 87Sr/86Sr. Because Sr behaves similarly to Ca, this indicates low Ca is also required. The igneous rock classification (Figure 2.2) suggests that an alkali feldspar syenite (or even a leucite-bearing equivalent) of early Archean or Hadean age would be a likely choice. 11. Use equations (9.15), (9.22) and (9.23) to construct a concordia diagram for 3500 Ma of Earth history. Begin with 1000 atoms each of 235U and 238U at an initial time of 0 years, and increment time in 200 Ma increments, calculating N (the number of atoms remaining) and D* (the number of daughter isotopes = No-N) for both decay schemes at each increment. Then plot 207Pb/235U vs. 206 Pb/238U to construct a concordia diagram. Once you have a diagram, add the following points as a discordia: 207 Pb/235U 5 10 12 15 17 19 206 Pb/238U 0.240 0.355 0.400 0.470 0.520 0.560 You can add the points in Excel by copying them, selecting the chart, and using Paste-Special as a new series. Draw a best-fit line through the discordia points and estimate the initial crystallization age and the metamorphic age. See Prob 09-11.xls. I got an initial crystallization age of 3.22 Ga and a metamorphic age of 1.01 Ga. 21

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LSU - GEOL - 3041
Answers for Chapter 8: Chemical Petrology I: Major and Minor Elements.1. What is AA spectroscopy, and in what way(s) does it differ in principle from XRF? AA is atomic absorption spectroscopy. In AA a solution of dissolved sample is aspirated into a flam
LSU - GEOL - 3041
Answers for Chapter 7: Ternary Systems. Ternary systems are, by definition, 3-component systems. The addition of another component to a 2component system requires 3-dimensions to illustrate the T-X variables for an isobaric system, but we cannot further s
LSU - GEOL - 3041
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LSU - GEOL - 3041
Answers for Chapter 5: An Introduction to Thermodynamics.1. The molar Gibbs free energy of formation of quartz is the energy change involved in the reaction of Si metal with O2 gas to form a mole of quartz SiO2: Si (metal) + O2 (gas) = SiO2 (quartz). Wha
LSU - GEOL - 3041
Answers for Chapter 4: Igneous Structures and Field Relationships.1. How does magma viscosity vary with the concentration of a) SiO2 and b) H2O? Explain. Magma viscosity increases with increasing concentration of SiO2 because higher SiO2 generally result
LSU - GEOL - 3041
Answers for Chapter 3: Textures of Igneous Rocks.1. Other than the externally-imposed cooling rate, what are the three principal rates that govern rock texture? Rates of nucleation, growth, and diffusion. 2. What is typically the determining (limiting) r
LSU - GEOL - 3041
Answers for Chapter 2: Classification and Nomenclature of Igneous Rocks.1. What are the three principal categories of igneous rocks? What characterizes each? Plutonic (intrusive) rocks crystallize beneath the surface of the Earth. They tend to have coars
LSU - GEOL - 3041
Answers for Chapter 1: Some Fundamental Concepts.1. In the early 19th century the origin of igneous rocks was hotly debated between the Plutonists, who believed in an igneous origin, and the Neptunists, who believed that the crystalline nature originated
LSU - GEOL - 3032
Part I Origin and Transport of Sedimentary Materials MONDAY SEPTEMBER 21 EXAM 1 (CHAP. 1 AND 2) Part II Physical Properties of Sedimentary Rock WEDNESDAY OCTOBER 7 EXAM 2 (CHAP. 3 AND 4) Part III Composition, Classification, Diagenesis of Sedimentary Rock
LSU - GEOL - 3032
Glacial facies Glacial environments encompass broad areas Deposits on land = Continental Facies Grounded ice facies, Unstratified Stratified Proglacial/periglacial facies glaciofluvial glaciolacustrine Deposits at sea = Glacial-marine Facies Proximal
LSU - GEOL - 3032
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LSU - GEOL - 3032
Tide-dominated estuaries High tidal energy influences upper reach of estuary Well-mixed water Elongate sand bars parallel length of estuary Herringbone cross-bedding (landward and seaward dip directions) Flaser-bedding mud drapes forms during slack water
LSU - GEOL - 3032
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LSU - GEOL - 3032
Delta cycles During active constructional phase, deltaic deposits prograde basinward Progradation generates coarsening-up succession Delta-front buries prodelta delta-front and delta-plain also includes fine-grain sediment from marsh/bays Regression h
LSU - GEOL - 3032
Bounding surfaces in eolian deposits Bedforms rarely preserved Usually preserve lowest part of bedform foreset (cross bedding) 3 types of bounding surfaces Ractiviation surfaces Formed by erosion of lee faceSuperposition surfaces Formed by dune migrat
LSU - GEOL - 3032
Sedimentology Final-Exam Study Guide The final exam (DECEMBER 11 at 3-5 pm) will have three parts. Part 1 will have 11 multi-part questions divided into 4 groups (A-D) corresponding to chapters 8, 9, 10 and 11. You will have to answer two questions from c
LSU - HIST - 2055
T he South-- (4/23/09)Talked about how slavery shaped the south politically Why most southerners were not directly or indirectly involved in slave ownership 6 million whites in the south, 2 million were members of slave-owning families, 400,000 actuall
LSU - HIST - 2055
SocialandEconomicChangesandResponses(cont.)3/31/09(AFTEREXAMII)II.Responses a. Republicanmotherhood b. Revivals c. Reform d. Riots e. Abolitionism f. Womensrights g. Utopiansocieties(F.)Womensrights ideacamearoundfromabolitionism menpartofabolitionist
LSU - HIST - 2055
Slavery(4/16/09)I. II. III. IV. V. VI. VII. VIII. IX. X. XI. NatureofSouthernSlavery MaturationofAntebellumSlavery DemographicsandDistribution DomesticSlaveTrade ControlofSlaves Labor MaterialExistence SlaveFamily SlaveResponse SlaveCulture FreeBlacks
LSU - HIST - 2055
SectionalPolitics(4/23/09)I. II. III. IV. MexicanWar PositionsofSlaveryintheTerritories Compromiseof1850 KansasNebraskaCrisis a. KansasNebraskaAct b. EffectonParties CaningofCharlesSumner 1856Election DredScottDecision JohnBrownsRaid Electionof1860 Seces
LSU - HIST - 2055
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LSU - HIST - 2055
N ullification (4/2/09)I. II. III. IV. John C. Calhoun South Carolina and the 1828 Tariff Theory of Nullification Nullification Crisis a. 1832 Tariff b. South Carolina moves to Nullify c. Jacksons response d. Settlement-AJ is a major player ( I.) John
LSU - HIST - 2055
FINAL 1. describethemajorissuesduringAndrewJacksonspresidencyandexplainhow thehelpeddefinethebasicdifferencebetweenthe2newpartiesofthetime,the WhigsandtheDemocrats. majorissues: o Indianremoval: BestlandisreservedfortheIndians By1820swhitesettlersaremovin
LSU - HIST - 2055
F INAL QUESTIONS:Comprehensive part: we are given two choicesfirst theme: deals with govt in a more abstract way; govt structure andpower, ideas of what it should be doing. Does not deal with political parties or factions (2nd t heme) o 2nd: talk about
LSU - HIST - 2055
Andrew Jackson in Office (3/31/09)I. II. III. Election of 1824 and corrupt bargain Election of 1828 Profile of Jackson a. Background b. Vision of Society and Government Major Policy Issues a. Indian removal b. Government Land Sales c. Internal Improvemen
LSU - HIST - 2055
WHEN THE MISSISSIPPI RAN BACKWARDS 1. Describe the intentions behind New Madrids founding and the wilderness conditions from which the community was carved. Spain feared the USs rapid westward expansion. In order to foment the alienation of western Americ
LSU - HIST - 2055
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LSU - HIST - 2055
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LSU - HIST - 2055
The Constitution - 2/19/09 I. II. Constitutional Convention Constitution a. Power of the new government b. Structure of the new government c. Reasons it satisfies republicanism Ratification a. Opponents b. Supporters Problems with the ConstitutionIII.IV
LSU - HIST - 2055
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LSU - HIST - 2055
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LSU - HIST - 2055
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LSU - HIST - 2055
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LSU - HIST - 2055
Settlement of English Colonies, Part II I.(1/27/09)II.III.Massachusetts a. Puritanism b. Types of settlers c. Puritan society d. Dissenters i. Roger William ii. Anne Hutchinson e. Comparison with Virginia f. Factors leading to change middle colonies a
LSU - HIST - 2055
h ttp:/www.noteswap.com/index.php ^noteswap! Go sign up! Road to Independence and Revolution 2/10/09 I. French and Indian War a. Events of the war b. British unsatisfied with the colonists role Britains relationship with the colonists changes a. Proclamat
LSU - HIST - 2055
Development of Slavery2/3 I. II. origins of racism slavery in Virginia a. gradual shift to slavery b. reasons for shift slavery in other southern colonies a. south Carolina b. Georgia slavery in the middle and northern colonies slave trade a. village raid
UCSD - BIMM - 100
PS 4, 1Problem Set #4 BIMM100 Winter 2010 Short Answers 1. Define a gene. 2. What is the difference between a simple and complex transcription unit? 3. What does it mean when a gene is referred to as solitary? 4. Why is the sequence of duplicated genes l
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PS 5, 1Problem Set #5 BIMM100 Winter 2010 Short Answers 1. Why are ddNTPs required for automated/cycle sequencing reactions but not for pyrosequencing? 2. In automated sequencing, which are fluorescently labeled, dNTPs or ddNTPs? 3. List the reagents/enz
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UCSD - BIBC - 102
UCSD - BIBC - 102
UCSD - BIBC - 102
Handout 8 Fatty Acid Oxidation: Adipocytes: fat cells Perilipin: a protein that coats lipid droplets in adipocytes When glucagon is released from the pancreas to increase blood glucose, glucagon can bind to a receptor on an adipocyte activating PKA. PKA t
UCSD - BIBC - 102
Erin Yee ejyee@ucsd.edu BIBC 102 DI Thurs 5-6 Handout 2 Chymotrypsin: (see mechanism on page 208-209 in text) A serine protease (enzyme that hydrolyzes peptide bonds) o All have serine in active site (R chain CH2-OH) Cleaves bond on the C-terminus side of
UCSD - BIBC - 102
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UCSD - BIBC - 102
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UCSD - BIBC - 102
Handout 7 Glycogen (continued): Strucutre: Right side or reducing end of the chain is attached to an anchor protein called Glycogenin The left side of the chain is the non-reducing end Enzyme Amylo (1,4-1,6) transglycosylase creates branches in glycogen b
UCSD - BIBC - 102
Metabolic Biochemistry Fall 2009 Homework 1 (35 pts.)Name:_ Section:_1) (4 pts.) Does enzymatic catalysis affect the rate constant (k) for the reaction being catalyzed? Explain why or why not in detail.2) (4 pts.) Does enzymatic catalysis affect the eq
UCSD - BIBC - 102
BIBC 102 Fall 2009 (35 Points)KEY Homework 31) (4 pts.) A test electrode contains a solution of ubiquinone, with both the fully oxidized and fully reduced forms, at 1 M concentration each. A circuit is formed from this electrode using a KCl solution sal
UCSD - BIBC - 102
American - PHYSC - 210
Chapter 1 Introduction and Mathematical Concepts1.1 The Nature of PhysicsPhysics has developed out of the efforts of men and women to explain our physical environment. Physics encompasses a remarkable variety of phenomena: planetary orbits radio and TV
American - PHYSC - 210
Chapter 2Kinematics in One DimensionKinematics deals with the concepts that are needed to describe motion. Dynamics deals with the effect that forces have on motion. Together, kinematics and dynamics form the branch of physics known as Mechanics.2.1 Di
American - PHYSC - 210
Chapter 3Kinematics in Two Dimensions3.1 Displacement, Velocity, and Acceleration ro = initial position r = final position r = r ro = displacement3.1 Displacement, Velocity, and AccelerationAverage velocity is the displacement divided by the elapse
American - PHYSC - 210
Chapter 4 Forces and Newtons Laws of Motion4.1 The Concepts of Force and MassA force is a push or a pull. Contact forces arise from physical contact . Action-at-a-distance forces do not require contact and include gravity and electrical forces.4.1 The
American - PHYSC - 210
Chapter 5 Dynamics of Uniform Circular Motion5.1 Uniform Circular MotionDEFINITION OF UNIFORM CIRCULAR MOTION Uniform circular motion is the motion of an object traveling at a constant speed on a circular path.5.1 Uniform Circular MotionLet T be the t
American - PHYSC - 210
Chapter 6 Work and Energy6.1 Work Done by a Constant ForceW = Fs1 N m = 1 joule ( J )6.1 Work Done by a Constant Force6.1 Work Done by a Constant ForceW = ( F cos ) scos 0 = 1 cos 90 = 0 cos180 = 16.1 Work Done by a Constant ForceExample 1 Pullin
American - PHYSC - 210
Chapter 7 Impulse and Momentum7.1 The Impulse-Momentum TheoremThere are many situations when the force on an object is not constant.7.1 The Impulse-Momentum TheoremDEFINITION OF IMPULSE The impulse of a force is the product of the average force and th
American - PHYSC - 210
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American - PHYSC - 210
Chapter 9 Rotational Dynamics9.1 The Action of Forces and Torques on Rigid ObjectsIn pure translational motion, all points on an object travel on parallel paths.The most general motion is a combination of translation and rotation.9.1 The Action of For
American - PHYSC - 210
Chapter 10 Simple Harmonic Motion and Elasticity10.1 The Ideal Spring and Simple Harmonic MotionFApplied x= kxspring constant Units: N/m10.1 The Ideal Spring and Simple Harmonic MotionExample 1 A Tire Pressure Gauge The spring constant of the sprin
American - PHYSC - 210
Chapter 11 Fluids11.1 Mass DensityDEFINITION OF MASS DENSITY The mass density of a substance is the mass of a substance divided by its volume:m = VSI Unit of Mass Density: kg/m311.1 Mass Density11.1 Mass DensityExample 1 Blood as a Fraction of Body