Historical Geology Lab Manual Answers
Appendix 3 Answers to Exercises (3) Answers to Exercises – Physical Geology The following are suggested answers to the exercises embedded in the various chapters of Physical Geology. The answers are in italics. Click on a chapter link to go to the answers for that chapter. (Answers to the chapter-end questions are provided in Appendix 2.) Chapter 1 1.1 Find a piece of granite Responses will vary, but your sample should look something like the one shown below.
Granitic rocks are hard and strong and difficult to break. They are dominated by feldspar (this one has both white plagioclase and pink potassium feldspar), but almost all have some quartz (which looks glassy) and a few per cent of dark minerals, like the black amphibole in this one. An example of a granitic rock SE 1.2 Plate motion during your lifetime – It depends where you live of course, but if you live anywhere in Canada and anywhere in the US east of the San Andreas fault, then you’re on the North America Plate, and that is moving towards the west at 2 to 2.5 cm/year. So if you’re around 20 years old, that plate has moved between 40 and 50 cm to the west in your lifetime.
1.3 Using geological time notation – 2.75 ka is 2,750 years, 0.93 Ga is 930,000,000 years or 930 million years, 14.2 Ma is 14,200,000 years or 14.2 million years. 1.4 Take a trip through geological time – 1) The oxygenation of the atmosphere started at around 2.5 Ga (2500 Ma). It was a catastrophe for many organisms because they could not survive the strong oxidizing effects of free oxygen. 2) We don’t really know the answer to this, but it’s not very long if you include insects, and there is evidence of insect damage to some of the earliest plants.
3) Plants on land allowed for animals on land, so without land plants, we wouldn’t be here. Examples of mechanical weathering SE 5.2 Chemical weathering Chemical change Process Pyrite to hematite oxidation Calcite to calcium and bicarbonate ions dissolution Feldspar to clay hydrolysis Olivine to serpentine hydrolysis Pyroxene to iron oxide oxidation 5.3 Describe the weathering origins of sands Sand description Possible processes Fragments of coral etc.
From a shallow water area near to a reef in Belize Reefs are constantly being eroded by ocean waves, and the fragments are washed inshore by currents and then further eroded by wave action. Angular quartz and rock fragments from a glacial stream deposit near Osoyoos Quartz-bearing rocks have been eroded and transported by a glacier.
The fragments may have been moved a short distance by a stream, but not enough to produce rounding. SE from Earthquakes Canada at. Most of the earthquakes between the Juan de Fuca (JDF) and Explorer Plates are related to transform motion along that plate boundary,.
The string of small earthquakes adjacent to Haida Gwaii are likely aftershocks of the 2012 M7.7 earthquake in that area. Most of the earthquakes around Vancouver Island (V.I.) are related to deformation of the North America Plate continental crust by compression along the subduction zone. Earthquakes that are probably caused by fracking are enclosed within a red circle on the map. 11.2 Moment Magnitude Estimates from Earthquake Parameters Length (km) Width (km) Displacement (m) Comments MW? 60 15 4 The 1946 Vancouver Island earthquake 7.3 0.4 0.2.5 The small Vancouver Island earthquake shown in Figure 11.13 4.0 20 8 4 The 2001 Nisqually earthquake described in Exercise 11.3 6.8 1,100 120 10 The 2004 Indian Ocean earthquake 9.0 30 11 4 The 2010 Haiti earthquake 7.0 The largest recorded earthquake had a magnitude of 9.5.
Could there be a 10? A possible solution is 2500 km long and 300 km wide with 55 m of displacement. (Other solutions are possible.) These are unreasonable numbers because subduction zones don’t tend to fail over that length (typically not much more than 1200 km), rupture zones cannot be that wide because that takes us into the asthenosphere, and displacements are never likely to be that great. 11.3 Estimating Intensity from Personal Observations Building Type Floor Shaking Felt Lasted (seconds) Description of Motion Intensity?
Folded rocks (in yellow) and fold axes (pink) SE 12.2 Types of faults Top left: a normal fault, implying extension Top right: A reverse fault, compression Bottom left: a series of normal faults, extension Bottom right: a right lateral fault (implies that there is shearing, but it is not possible to say if there is extension or compression) 12.3 Putting strike and dip on a map See map below for strike and dip symbols. Relative ages, from youngest to oldest:. dyke (youngest). fault. layer g (although this layer isn’t intersected by the fault or the dyke so it is not possible to know that it is older based on the information available). layer f. layer e.
layer d. layer c. layer b. layer a (oldest). SE. A fine sand grain (0.1 mm) is resting on the bottom of a stream bed. (a) What stream velocity will it take to get that sand grain into suspension?
20 cm/s (b) Once the particle is in suspension, the velocity starts to drop. At what velocity will it finally come back to rest on the stream bed? 1 cm/s. A stream is flowing at 10 cm/s (which means it takes 10 s to go 1 m, and that’s pretty slow).
(a) What size of particles can be eroded at 10 cm/s? No particles, of any size, will be eroded at 10 cm/s, although particles smaller than 1 mm that are already in suspension will stay in suspension. (b) What is the largest particle that, once already in suspension, will remain in suspension at c0 cm/s? A 1 mm diameter particle should remain in suspension at 10 cm/s. 13.4 Determining Stream Gradients. Gradients of Priest Creek (in red) SE The length of the creek between 1,600 m and 1,300 m elevation is 2.4 km, so the gradient is 300/2.4 = 125 m/km.
Use the scale bar to estimate the distance between 1,300 m and 600 m and then calculate that gradient. 5.2 km, with a gradient of 700/5.2 = 134 m km. Estimate the gradient between 600 and 400 m.
Historical Geology Lab Manual Answer Key
3.6 km, with a gradient of 200/3.6 = 56m /km. Estimate the gradient between 400 m on Priest Creek and the point where Mission Creek enters Okanagan Lake. 4 km, with a gradient of 60/4.0 = 15 m/km 13.5 Flood Probability on the Bow River. Calculate the recurrence interval for the second largest flood (1932, 1,520 m 3/s). Ri = 96/2 = 48 years. What is the probability that a flood of 1,520 m 3/s will happen next year?
1/48 = 0.02 or 2%. Examine the 100-year trend for floods on the Bow River. If you ignore the major floods (the labelled ones), what is the general trend of peak discharges over that time? In general the peak discharges are getting lower (from an average of around 400 m 3/s in 1915 to an average of about 300 m 3/s in 2015) Chapter 14 14.1 How Long Will It Take? I = (37-21)/80 = 0.2, V= 0.0002 x 0.2 = 0.00004 m/s. At that rate it will take 2,000,000 s for the groundwater to flow from the gas station to the stream. That is 555 hours, or 23 days.
14.2 Cone of depression The cone of depression increases the gradient of the water table in the area around the well. That should increase the rate at which water flows towards the well. 1952 farmall super c manual.
14.3 What is your water table doing? BC Ministry of the Environment at The water-level for a random observation well in BC is shown above. The water table is slowly rising at this location. Since 2004 the lowest water level has risen from just above 4 m below surface to around 3.6 m above surface and the highest level has risen from around 0.3 m below surface to nearly at surface (0 m). Prior to 2004, where the points are not joined with lines, the trend appears to be similar. 14.4 What goes on at your landfill? – Responses will vary 14.5 Finding a leaking UST in your community – Responses will vary 14.6 Manipulating a Contaminant Plume What could you do at wells A and C to prevent this?
Explain and use the diagram below to illustrate the expected changes to the water table and the movement of the plume. Implications of pumping water from wells B and C and injecting water into well A SE Possible Answer: Injection into well A will cause water table to rise there (like the reverse of a cone of depression), thus reversing flow direction to the right of well A and moving the plume towards Well B.
Extraction from Wells B and C will cause cones of depression and help to reverse the flow and pull the plume back from the stream. Both wells B and C may receive contaminants and so the water from both may need treatment. Chapter 15 15.1 Sand and water – responses will vary 15.2 Classifying Slope Failures. SE 15.3 How Much Does a House Weigh and Can It Contribute to a Slope Failure?
A typical 150 m 2 (approximately 1,600 ft 2) wood-frame house with a basement and a concrete foundation weighs about 145 t (metric tonnes). But most houses are built on foundations that are excavated into the ground. This involves digging a hole and taking some material away, so we need to subtract what that excavated material weighs.
Assuming our 150 m 2 house required an excavation that was 15 m by 11 m by 1 m deep, that’s 165 m 3 of “dirt,” which typically has a density of about 1.6 t per m 3. 165 m 3 of excavated soil x 1.6 t/m 3 = 264 t – thus the excavated material weighs about 1.8 times as much as the house. In this case weight has been removed from the slope by building the house. Chapter 16 16.1 Pleistocene Glacials and Interglacials Describe the nature of temperature change that followed each of these glacial periods. In each case the temperature drops slowly building to a peak of glaciation, and then each of the glacial periods is followed by a very rapid increase in temperature. The current interglacial (Holocene) is marked with an H. Point out the previous five interglacial periods.
The previous 5 interglacials are labelled 1 to 5 on the diagram below. Interglacial 2 had two distinct warm episodes. SE after USGS at Chapter 17 17.1 Wave Height versus Length This table shows the typical amplitudes and wavelengths of waves generated under different wind conditions.
The steepness of a wave can be determined from these numbers and is related to the ratio: amplitude/wavelength. Calculate these ratios for the waves shown. The first one is done for you.
How would these ratios change with increasing distance from the wind that produced the waves? Amplitude Wavelength Ratio m m ampl./length 0.27 8.5 0.03 1.5 33.8 0.04 4.1 76.5 0.05 8.5 136 0.06 14.8 212 0.07 Within increasing distance from the source the wave heights would gradually decrease and so the ratios would decrease. 17.2 Wave Refraction. Possible locations of coastal deposits SE 17.4 A Holocene Uplifted Shore The melting of glacial ice around the world at the end of the last glaciation (between 14 and 8 ka – see Figure 17.25) led to relatively rapid sea-level rise (by a total of approximately 125 m) which resulted in this area being submerged. That was a eustatic process. In response to the loss of ice in this region of coastal British Columbia there was a slower isostatic rebound of the crust, which is why this area is now back up above sea level. 17.5 Crescent Beach Groynes.
Geology Lab 10 Answer Key
SE 18.4 Salt chuck No answer possible 18.5 Understanding the Coriolis effect No answer Chapter 19 19.1 Climate Change at the K-Pg Boundary The short-term climate impact was significant cooling because the dust (and sulphate aerosols) would have blocked incoming sunlight. This effect may have lasted for several years, but its intensity would have decreased over time.
The longer-term impact would have been warming caused by the greenhouse effect of the carbon dioxide. 19.2 Albedo Implications of Forest Harvesting Clear-cutting (or any logging activity) leads to a net increase in albedo, so the albedo-only impact is cooling.
19.3 What Does Radiative Forcing Tell Us? Using the ΔT = ΔF. 0.8 equation the expected temperatures for 2011, 1980 and 1950 compared with the estimated 13.4 C in 1750 should be: 2011 vs 1750 ΔT = 0.8. 2.29 = 1.8 ° C (13.4 + 1.8 = 15.2) 1980 vs 1750 ΔT = 0.8. 1.25 = 1.0 ° C (13.4 + 1.0 = 14.4) 1950 vs 1750 ΔT = 0.8. 0.57 = 0.5 ° C (13.4 + 0.5 = 13.9).
SE using climate data from Environment Canada, and ENSO data from: 19.5 How Can You Reduce Your Impact on the Climate? – responses will vary Chapter 20 20.1 Where does it come from? – responses will vary 20.2 The importance of heat and heat engines Deposit Type Is Heat a Factor? If So, What Is the Role of the Heat? SE after Geological Survey of Canada 21.2 Purcell Rocks Down Under?
The Mesoproterozoic quartzite phyllite schist of Tasmania may correlate with the Purcell rocks of Canada. The main difference is that while the Tasmanian rocks are metamorphosed, the Purcell rocks are generally unmetamorphosed. 21.3 What Is Vancouver Island Made Of?
1) Less than 10% of Vancouver Island is Paleozoic (the Devonian volcanic rocks – Dv) 2) The most common rock type is the Triassic Karmutsen Volcanic rock (basalt – Tv). The most common rocks by age are the Mesozoic rocks (Jurassic volcanic, Jurassic granite and Triassic volcanic). SE 21.4 Dinosaur Country?
This Cretaceous Dinosaur Park Formation sandstone is clearly cross-bedded implying that it was deposited in a stream environment. 21.5 The Volume of the Paskapoo Formation 1) The 60,000 km 2 area of source rock would have to have been eroded to a depth of 750 m to create 45,000 km 3 of sediment 2) 500 m is 500,000 mm so the rate is 500,000 mm/ 4,000,000 years = 0.125 mm/year Chapter 22 (answers provided by Karla Panchuk) 22.1 How do we know what other planets are like inside? KP, after Jenkins, J. Et al, 2015, Discovery and validation of Kepler-452b: a 1.6REarth super Earth exoplanet in the habitable zone of a G2 star, Astronomical Journal, V 150, DOI 10.1088/0004-6256/150/2/56.
Table 22.5 Calculate the radius of star Kepler-452 Sun Kepler-452 Ratio Temperature (degrees Kelvin) 5778 5757 1.0036 Luminosity (x 1026 Watts) 3.846 4.615 1.20 Radius (km) 696,300 768,317. The temperatures of the sun and Kepler-452 are very similar, but the small difference is important. Keep 4 decimal places. Table 22.6 Calculate the radius of planet Kepler-452b Decrease in brightness. Earth radius (km) Kepler-452b radius rplanet (km) Kepler-452b radius/ Earth radius 197 x 10 -6 6378 10,784 1.7. Because we know this is a decrease, you don’t need to keep the negative sign.
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Physical Geology Answer Key
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