Page 711 - 9780077418427.pdf

Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefiles
tiL12214_appe_643-698.indd Page 688 09/10/10 8:37 AM user-f463
tiL12214_appe_643-698.indd Page 688 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

_
20.13. Divide the distance the glacier has retreated by the diff erence of thickness

rate =
the dates of the moraine positions. time
d = 0.96 km Convert thickness from m to mm:
t = 2007 − 1933 = 74 yr _
3
1 × 10 mm
3

1.1 × 10 m ( )
rate of retreat = ? 1 m
6
d _ 1.1 × 10 mm
rate of retreat =
t __
6
1.1 × 10 mm

_ rate = 7
0.96 km

= 6.3 × 10 yr

74 yr
−2
_
−2 km = 1.7 × 10 mm/yr
= 1.3 × 10

yr

21.2. The time break is the difference in age between the oldest strata

20.14. Divide the remaining glacier length by the rate of retreat, then on top of the unconformity and the youngest strata beneath the
add this to the year of the survey to determine the year the unconformity.
glacier will disappear.
8
age of strata above unconformity = 1.76 × 10 yr
d
d = 1.2 km t = __ age of strata beneath unconformity = 2.80 × 10 yr
8

_
−2 km rate of retreat
rate of retreat = 1.25 × 10 time break = ?

yr
1.2 km

= __ time break = (age of strata beneath unconformity)
t = ? −2 km
_
1.25 × 10

yr − (age of strata above unconformity)
8
8
_
1.2
__ km = 2.80 × 10 yr − 1.76 × 10 yr

=

8
km
1.25 × 10 −2 _ = 1.04 × 10 yr

yr
1

= 9.6 × 10 yr 21.3. The time break is the difference in age between the oldest strata
= 96 yr on top of the unconformity and the youngest strata beneath the
2002 + 96 yr = 2098 unconformity. Since only the Devonian rocks are missing, refer
to Figure 21.17 to determine the age of the lower and upper

20.15. Determine the difference in elevation and divide by the
boundaries of the Devonian period.

difference in age.
8
age of strata above unconformity = 3.60 × 10 yr
elevation 1 = 393 m ___ 8
(elevation 1 − elevation 2 )

rate uplift = age of strata beneath unconformity = 4.08 × 10 yr

elevation 2 = 379 m (terrace age 1 − terrace age 2 )
time break = ?
terrace age 1 = 12,500 yr
Convert m to mm time break = (age of strata beneath unconformity)
terrace age 2 = 0 yr − (age of strata above unconformity)
3
1 × 10 mm
rate uplift = ? 393 m ( _ 8 8

)

m =4.08 × 10 yr − 3.60 × 10 yr
7
5
3.93 × 10 mm = 4.8 × 10 yr
3
_
1 × 10 mm

379 m ( 21.4. The principle of cross-cutting relationships determines that the
)

m

5
3.79 × 10 mm flat-lying sedimentary sequence that includes the sandstone is
less than 210 million years old. The principle of superposition

5
5
___
(3.93 × 10 mm − 3.79 × 10 mm)

rate uplift = determines that the flat-lying sedimentary sequence is older

4
(1.25 × 10 yr − 0 yr)
than 189 million years old. Hence, the absolute age of the
5 5
__ _
(3.93 × 10 − 3.79 × 10 ) mm

= sandstone is sometime between 189 and 210 million years old.

yr
4
(1.25 × 10 − 0)

4
__ 21.5. The principle of cross-cutting relationships determines that the
1.40 × 10 mm
=

4
1.25 × 10 yr flat-lying sedimentary sequence that includes the sandstone is

_ less than 56.3 million years old. The principle of superposition
mm
= 1.12
yr determines that the flat-lying sedimentary sequence is older

than 37.3 million years old. Hence, the absolute age of the
sandstone is sometime between 56.3 and 37.3 million years old.
CHAPTER 21
Referring to Figure 21.17, the fossils would date from the
21.1. The sedimentation rate is determined by dividing the thickness Eocene epoch of the Tertiary period.

of the geologic section by the time in which the section was
21.6. The principle of superposition determines the sandstone, shale,

deposited. The section spans the Tertiary period, so the time can
and limestone are younger than the lava fl ow. The principle of

be determined from Figure 21.17 by calculating the diff erence in
cross-cutting relationships determines that the sequence is
age between the beginning and the end of the Tertiary period.
older than the dike. Hence, the absolute age of the limestone
3
thickness = 1.1 × 10 m and shale is sometime between 64.5 and 61.2 million years.
From Figure 21.17: 21.7. Event E cuts across A and D, so it must be younger than these
events. Event A cuts across C and D, so it must be younger than
6
end of Tertiary = 1.6 × 10 yr
6
beginning of Tertiary = 65 × 10 yr these events. Event D overlies B so the principle of
7
6
∴ time = 6.5 × 10 yr − 1.6 × 10 yr superposition dictates it is younger than B. Event B cuts across

7
= 6.3 × 10 yr the youngest layer of C, so B is younger than C. The tilting must

have occurred after the deposition of the layers in C because of
rate = ?
688 APPENDIX E Solutions for Group A Parallel Exercises E-46