ANSWERS TO QUESTIONS ON PAGE 28
Answers: 1. The answer is choice c” or “aphanitic”. “Aphanitic” refers to the fine-grained texture achieved through rapid cooling where large crystals do not have the time to develop. In contrast, in plutonic rocks “phaneritic” or coarse-grained texture, indica- tive of slow-cooling rates, is to be expected.
“Dendritic” commonly refers to a type of geomorphological stream drainage pattern. In a “dendritic” stream pattern there are many contributing streams which join together into the tributaries of the main stream. “Dendritic” patterns are to be expected where the river channel follows the slope of the overall terrain.
“Oolitic” describes a type of texture found in some sedimentary rocks, such as limestone. “Oolites” are small rounded particles that form when layers of typically calcite are deposited around some tiny particles, such as sand grains or fossil fragments, which are rolled back and forth in quiet waters. Some oolitic limestones have proven to be excellent petroleum reservoirs.
2. The answer is choice “a” or “+4.” Each oxygen anion has a valence of -2; thus, four of these result in a total of -8. If the radical of the silicon tetrahedron has an overall value of -4, then the silicon cation must contribute a charge of +4.
3. The answer is choice “b” or “Gutenberg discontinuity”. It occurs at an approximate depth of 2,900 kilometers.
The “Mohorovic discontinuity” refers to the boundary between the Earth’s crust and the Earth’s mantle. It occurs at a depth of 10-12 kilometers under the ocean and 40-50 kilometers under the continent.
The “Bullen discontinuity” or “Lehmann discontinuity” applies to the boundary between the Earth’s molten outer core and its more solid inner core. It occurs at an approximate depth of 5,200 kilometers.
4. The answer is choice b” or [1.32 x 109 dyne cm-2]. The proof follows (please refer to the figure below): “Stress” is force per unit area or: S = F/A or F = SA
(1)
The coefficient of friction is: U = F/W
The area over which “F” acts is: A = aT
Substituting (3) into (1): F = SaT
The weight of Block X is: W = LaTpg
U = F/W = SaT/LaTpg = S/Lpg S = ULpg
(2) (3) (4) (5)
To calculate the stress value that we seek, we substitute (4) and (5) into (2) and solve for “S”:
(6)
Equation (6) gives us the results that we want. Substituting the values given in our example, equation (6) becomes: S = (0.5) (1,000,000 cm) (2.70 gm cm-3) (980 cm sec-2) S = 1.32 x 109 dyne cm-2
(7)
Equation (7) is choice “b” in our example. Please note that the magnitude of “S” in (7) is equivalent to the “unconfined compressive strength” of very hard rocks, such as granite, basalt, quartzite and slate, but that it exceeds their “shear strength” by more than 2.5 times! Thus, it is important to consider potential physical constraints to keep our geological models and related solutions realistic!
5. The answer is choice “c” or L = 3.78 km. The proof follows: Solving equation (6) in question 4 for “L”: S = ULpg
(6) L = S / (Upg) (9) (8)
Substituting actual values: L = (5 x 108 dyne cm-2) / [(0.5) (2.70 gm cm-3) (980 cm sec-2)] L = 3.78 km
Equation (9) corresponds to our choice “c” and takes into account physical constraints in an attempt to make our model more geologically-realistic!
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