Page 663 - 9780077418427.pdf

Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefiles
tiL12214_appd_633-642.indd Page 640 09/10/10 8:36 AM user-f463
tiL12214_appd_633-642.indd Page 640 09/10/10 8:36 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

Using the same method, determine the slope for the remainder of Example 19.4, p. 488
the earthquake position and depth values in the table. Results are Determine the number of magnitude differences, and multiply the
shown below: ground movement change factor by itself for each change in magnitude.
3
__ Richter magnitude 1 = 7.2
−6.3 × 1 0 m

slope =

2
22.1 km
Richter magnitude 2 = 5.2
m _
= −290 surface wave difference = ?
km
3
__ magnitude
−9.8 × 10 m
slope =

3
43.5 km
difference = Richter magnitude 2 − Richter magnitude 1
m _
= −230 = 5.2 − 7.2
km
= −2 magnitudes
4
__
−2.28 × 1 0 m

slope = 91.2 km surface wave
4
difference = (10)(10)
m _
= −250
km = 100
4
−42.0 × 1 0 m
__ The negative magnitude difference means the first earthquake had

slope = 160.3 km approximately 100 times greater ground motion (surface wave ampli-
5
m _ tude) than the second earthquake.
= −262
km
Average the results.
CHAPTER 20
m _
m _
m _
( −240 + −290 + −230
) (
) (
)

km
km
km
Example 20.2, p. 507
m _
m _
) (
(
)
+ −250 + −262 Determine the surface area of the pieces of rock before weathering and
slope average = km km the surface area after weathering. Then subtract the before case from
5
m _ the after case to determine the increase in total surface area caused by
= −250
km physical weathering.
displacement = 0.6 mm displacement
_

CHAPTER 19 age = 67 yr rate = age
Example 19.2, p. 484 rate = ? Convert distance from m to cm.
West Fault West Fault _
100cm

0.6 m ( )
Δy = 500 m _ _ 1 m
Δy
Δy
slope = ∴ Δx = 60 cm

m _ Δx slope
slope = 1.3 m _
60 cm
500 m
Δx = _ rate = 67 yr

m _
Δx = ? 1.3
m _
cm
= 0.9
500 m _ yr
_

= 1.3 m _

m
Example 20.4, p. 509
= 380 m This type of problem is best solved by subdividing it into steps that
consider the volume of sediment eroded in 1 million years and the
East Fault East Fault
thickness of sediment associated with this volume.
Δy = 500 m _ _
Δy
Δy

slope = ∴ Δx =

m _ Δx slope Step 1: Volume Removed
slope = 1.8 m
500 m

Δx = _ 4 Mg m _ m _
_
m _
Δx = ? 1.8 m = 7.06 × 1 0 (1 × 1 0 yr) ρ = ∴ V =
6

m yr V ρ
10
500 m _ = 7.06 × 10 Mg 7.06 × 1 0 Mg
10
_
= 1.8 m _ V = __

Mg
m Mg 1.35 _
_

ρ = 1.35 m 3
3
= 280 m m
10 Mg
7.06 × 1 0 _
The extension is the sum of the horizontal displacement (Δx) of the V = ? = _

Mg
east and west faults: 1.35 _

m 3
Δx w = 380 m extension = Δx w + Δx e 10 3
= 5.2 × 1 0 m
Δx e = 280 m = 380 m + 280 m
extension = ? = 660 m
640 APPENDIX D Solutions for Follow-Up Example Exercises D-8