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1 _ 2 change = case 2 − case 1
v f = at and d = at

2
= 12.6 km − 6.3 km
Rearrange v f = at to solve for t. = 6.3 km
Crustal thickening increased elevation by 6.3 km.
1 _ 1 _
v f = at
a
a
19.15. Case 1: Average Crust Thickness
v f _
= t z crust = 35.0 km h = z crust − z crust _
)
a
ρ crust

g
_
1 _ 2 ρ mantle = 3.3 ( ρ mantle
Substitute this relationship for time in d = at . 3
g
2 cm 2.7
_
g
_
3
_
cm
1 _ _ 2 ρ crust = 2.7 ( 3.3 )
v f

d = a ( ) 3 = 35.0 km − 35.0 km

g
2 a cm _
v
1 _ _ 2 f h = ? cm 3
d = a
2 a 2 _
g

v
1 _ _ 2 f 2.7 _ _ 3
cm
(2a)d = (2a) = 35.0 − 35.0 ( ) km _

2 a
g
3.3

2
3
v = 2ad
cm
f

= 35.0 − 35.0(0.82) km
v f = √ 2ad = 35.0 − 28.7 km
m _

2)
√ (
4
= 2 9.8 1.1 × 10 m = 6.3 km

s
4 m _

√

= 2(9.8)(1.1 × 10 ) m Case 2: Thickened, Less Dense Crust
2
s
z crust = 50.0 km _ )
ρ crust

2
√
5 m _
g
_
= 2.2 × 10 ρ mantle = 3.3 h = z crust − z crust

( ρ mantle

2
s
3
g
_
2 m _ cm 2.6
g
_
= 4.7 × 10 ρ crust = 2.6 ( 3.3 )

3
_
cm

s
g
cm 3 = 50.0 km − 50.0 km _
60 s
60 min
(
2 m _ _ _ _
s (
1 km

= 4.7 × 10 1,000 m ) ( 1 h ) h = ? cm 3

) 1 min
g
_
3 km _

= 1.7 × 10 3
2.6 _
cm
h = 50.0 − 50.0 ( ) km _

g
3.3 _

19.14. The position at which crust floats in the mantle can be calculated
3
cm
from the thickness of the crust and the relative densities of crust = 50.0 − 50.0(0.79) km
and mantle by using the formula provided in the question. = 50.0 − 39.5 km
Determine the height of the continental crust above the mantle = 10.5 km
for average crustal thickness and for the case following crustal change = case 2 − case 1
thickening, then subtract the difference. = 10.5 km − 6.3 km
Case 1: Average Crust Thickness = 4.2 km
z crust = 35.0 km h = z crust − z crust _ Crustal thickening and lowered density increased elevation by
)
ρ crust

g
_ ( ρ mantle 4.2 km.
ρ mantle = 3.3
g
cm 3 2.7
( 3.3 )
_

g
_ _ 3
cm
ρ crust = 2.7 = 35.0 km − 35.0 km CHAPTER 20

g
cm 3 _
h = ? cm 3
20.1. Determine the surface area of the pieces of rock before
g
_ weathering and the surface area after weathering. Th en subtract

2.7 _ _ 3
cm

= 35.0 − 35.0 ( ) cm the before case from the after case to determine the increase in
g
3.3 _ total surface area caused by physical weathering.

cm 3
= 35.0 − 35.0(0.82) km Case 1: Before Weathering
= 35.0 − 28.7 km L = 10 m
= 6.3 km W = 10 m
Case 2: Thickened Crust H = 2 m
z crust = 70.0 km _ pieces = 1
)
ρ crust
h = z crust − z crust A surface = ?

g
_ ( ρ mantle
ρ mantle = 3.3 A surface = pieces [2LH + 2WH + 2LW]

g
cm 3 2.7
_
( 3.3 )

g
_ _ 3 = 1[2(10 m)(2 m) + 2(10 m)(2 m) + 2(10 m)(10 m)]
cm

ρ crust = 2.7 = 70.0 km − 70.0 km 2 2 2

g
cm 3 _ = 1[40 m + 40 m + 200 m ]
2
h = ? cm 3 = 1[280 m ]
2
g
_ = 280 m

2.7 _ _ 3
cm

= 70.0 − 70.0 ( ) cm
g
3.3 _

cm 3
= 70.0 − 70.0(0.82) km
= 70.0 − 57.4 km
= 12.6 km
E-43 APPENDIX E Solutions for Group A Parallel Exercises 685