ANSWERS TO QUESTIONS ON PAGE 28
Answers: 1. The answer is choice a” or “metasomatism.”
“Diagenesis” is the process that involves physical and/or chemical changes of sediment, after deposition, and its transformation into solid bedrock. Consolidation and compaction, cementation and recrystallization are examples of “diagenetic” changes.
“Applanation” describes the physiographic processes which result in the reduction of topographic relief. 2. The answer is choice “c” or “tectonic deformation.”
In soil mechanics, “cation-exchange capacity” describes the number of exchangeable “cations” (positively-charged ions) per dry weight that a soil can hold at a given pH value, and which are available for exchange with the soil-water solution.
“Diatomaceous earth” is a sedimentary deposit made up of nearly-pure silica formed by the accumulation of fossilized “diatom” shells. “Diatoms” are microscopic, single-celled plant organisms that secrete siliceous tests and which constitute an important component of phytoplankton.
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3. The answer is choice “a” or “A type of small to microscopic crustacean.” “Charophytes” relate to fresh-water algae as given in choice “b.” Sorry, no mollusks here!
4. The answer is choice “b” or “Me = 5.98 x 1024 kg. ” The proof follows: Consider a mass “m” near the surface of the Earth. Recall equations (1) through (5):
F = G m1 m2 / r2 F = mg
re = 6,400 km g = 9.8 msec-2
G = 6.67 x 10-11 newton-m2 kg-2
(1) (2) (3) (4) (5)
For (1), let m1 = m, m2 = Me (the Earth’s mass) and r = re (the average Earth’s radius). Thus, (1) becomes: F = G m Me / re 2
(6)
Equating (2) and (6): mg = G m Me /re2
g = G Me / re2
Solving (8) for Me we obtain: Me = g re2 / G
Me = 5.98 x 1024 kg
(7) (8)
(9)
Equation (9) allows us to calculate the answer that we seek. Substituting (3), (4) and (5) into (9): Me = [(9.8 msec-2) (6.4 x 106 m)2] / (6.67 x 10-11 newton-m2 kg-2)
(10) (11)
Equation (11) corresponds to our choice “b” and depicts the numerical value of the Earth’s mass. Wow! That’s heavy!
5. (Submitted by David Abbott, CPG-4570) – The answer is choice “b,” or “~3.3 miles.” The calculation follows: 1 gallon = 0.005 cubic yards/gallon
(1)
Assume mine openings are 2 yards high × 1.3 yards wide ×1 yard long. Then: Mine opening = 2.6 cubic yards
3 million gallons ÷ 0.005 cubic yards/gallon = 15,000 cubic yards 15,000 cubic yards ÷ 2.6 cubic yards/lineal yard = 5,769 lineal yards 5,769 lineal yards × 3 feet/yard = 17,307 feet 17,307 feet ÷ 5,280 feet/mile = 3.3 miles Equation (6) is the answer that we seek.
(2) (3) (4) (5) (6)
In fact, much of the volume will be in stopes, the voids left after ore extraction, which are very irregularly shaped. Plate 8 from USGS Professional Paper 535 (available online from
usgs.gov), shows both map and sectional views of the mine workings of the Gold King Mine, the American Tunnel, and the various mines of the Sunnyside Mine complex. Most of the water is believed to have come from the Sunnyside Mine because the Gold King Mine’s mineralization doesn’t include base metals and the Sunnyside Mine complexes workings do. The American Tunnel is the lowest working in the interconnected complex of underground workings, around 2,000 feet below the Terry Tunnel, which also drains parts of the complex. The American Tunnel is about 2 miles long.
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