ANSWERS TO QUESTIONS ON PAGE 26
Answers: 1. The answer is choice “c” or “gaspeite”, a nickel, iron, magnesium carbonate.
“Malachite” [Cu2 (CO3) (OH)2] is a green, monoclinic copper carbonate hydroxide mineral with hardness of 3.5 to 4. Its color is typically darker than that of “gaspeite”.
“Pyrope” [Mg3Al2(SiO4)3] is a red to purplish-black variety of garnet. The mineral is a cubic “nesosilicate” with a hardness of 7-7.5.
2. The answer is choice “a” or “pliosaurus”.
Diatryma describes an extinct, giant flightless bird of the Eocene Epoch. It grew to a height of about 7 feet. It had massive legs and was probably a strong, fast runner. The head was large and supported a powerful beak.
Ichthyosaurus was a Mesozoic reptilian counterpart of toothed whales. These animals had a fish-like tail, dorsal fin and paddle limbs. The pointed head allowed this carnivore to cut rapidly through the water. They averaged six to eight feet in length, but some probably grew much larger.
3. The answer is choice c” or “solifluction”.
Mudflows through “liquefaction” occur as loosely-packed sediment behaves in fluid fashion in response to ground vibra- tions or through the sudden addition of water, bringing about the collapse of the sediment’s structure.
“Wetlands” are land areas which becomes saturated with water, either permanently or seasonally. “Wetlands” consist of saturated soils, supporting distinct ecosystems and aquatic vegetation.
4. The answer is choice “b” or t = [(P1 - P3)/2] * (sin 2). The proof follows:
We are given: P3 sin – P sin + t cos = 0 P1 cos – P cos – t sin = 0
(1) (2)
To obtain the wanted equation for “t”, we can solve (1) and (2) for “P” and equate both expressions. Thus: P sin = P3 sin + t cos P cos = P1 cos - t sin P = P3 + t cot P = P1 – t tan
P1 – t tan = P3 + t cot P1 – P3 = t tan + t cot P1 – P3 = t (tan + cot )
P1 – P3 = t [(sin / cos ) + (cos / sin )] P1 – P3 = t [(sin2 + cos2 ) / (sin * cos )]
Recalling trigonometric identities: sin2 + cos2 = 1 sin 2 = 2 sin cos sin 2 = sin cos
Substituting (13) and (14) into (11), we obtain: P1 – P3 = t [1 / ( sin 2)] t = [(P1-P3) / 2] * (sin 2 )
(3) (4) (5) (6) (7) (8) (9)
(10) (11)
(12) (13) (14)
(15) (16)
Equation (16) is the answer that we seek and corresponds to our choice “b”. Simple stuff, isn’t it? Oh, well….
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