ANSWERS TO “TEST YOUR KNOWLEDGE” ON PAGE 40 Answers:
1. The answer is choice “b” or “sauropod.” Pleurocoelus was a small member of the sauropod family, which includes giants such as brachiosaurus and diplodocus. These lizard-hipped dinosaurs were herbivorous.
2. The answer is choice “a” or “Load casts.” These sedimentary structures form during soft-sediment deformation shortly after burial and prior to lithification. They generate due to the differential loading caused by a denser layer of sediment being deposited on top of a softer, less dense, often saturated material capable of flowage. The “hanging” lobes, knobs, bulges and lumps form as columns of the denser strata sink into the underlying, softer, plastic layer.
“Boudinage” refers to cylindrical, lenticular or rectangular structures that develop from the stretching or extension of firm but flexible strata embedded in a yielding, less competent matrix.
Concretions, common in sedimentary rocks and soil, are hard and compact masses of mineral matter that form as mineral cement precipitates within the spaces between particles or clasts. They may be oval, quasi-spherical or have irregular shapes.
3. The answer is choice “c” or “Decreasing potential volume change.” In general, for clay-rich soils and sediments, as the Atterberg limits and indices increase:
•Potential volume change increases. •Compressibility increases. •Organic content increases. •Strength decreases. •Permeability decreases.
Clays with high values of Atterberg limits and indices are typically of low strength, low permeability and high potential volume change:
4. The answer is choice “b” or “Spodumene” [Li,Al(SiO3)2]. Spodumene belongs to the pyroxene family with a hardness of 6.5-7.0 and perfect prismatic cleavage in two directions. Hiddenite (pale emerald green) and kunzite (pink to lilac) are
gemstone varieties of spodumene. Cinnabar (HgS) is a sulfide or mercury whereas wolframite [(Fe,Mn)WO4] is an oxide of iron, manganese and tungsten.
5. The answer is choice “c” or “A few days.” The calculation follows. Recall that as we drop below ~ 1300°C crystallization will begin in a cooling magma. We are given:
•Um = 50 kgm-1sec-1 (i.e., 50 pascal-second or 5 x 104 centipoise at 1200°C) •Dm = 2,700 kgm-3 (at 1200°C) •Do = 3.8 gcm-3 = 3,800 kgm-3 •do = 0.25 cm = 0.0025 m •g = 9.8 msec-2
The settling velocity (Vset) may be approximated by using Stoke’s Law, where: Vset = (2/9) (g) (do)2 (Do – Dm) / (Um)
Substituting our numbers into (1) we obtain: Vset = (2/9) (9.8 msec-2) (0.0025 m)2 (3,800 kgm-3 – 2700 kgm-3) / (50 kgm-1sec-1) Vset = 2.99 x 10-4 msec-1 = 9,443 myr-1
For our 100-meter thick sill: T = [(100m x 1 yr) / (9,443 m)] x (365 days/yr) = 3.9 days
(1)
(2) (3)
(4)
From (4) and in our much over-simplified model, we see that it would take roughly 4 days for the crystal to settle down to the bottom of the cooling sill.
As stated in the question itself, this exercise is intended to simply provide us with an understanding of the order of magnitude of the geologic process versus what we might have intuitively conjectured. Be aware that the simplifying assumptions that we make can give rise to substantial differences in the potential answer. But, hopefully, this type of exercise gets us within the ballpark of reality and is better than just a simple guess or an arm-waving derived supposition.
42 TPG •
Oct.Nov.Dec 2020
www.aipg.org
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