Name: MSCI 304-Marine Geology Homework - In answering questions, SHOW ALL YOUR W

Question

Name: MSCI 304-Marine Geology Homework - In answering questions, SHOW ALL YOUR WORK depth 3100 m depth 4100 mLundge. In a particular ocean basin a spreading axis (ridge) runs north-south, fracture zone or transform fault but it is offset by an east-west transform fault the north of the fracture zone, the depth to other side of the fracture zone, the depth is 4100 m. extends is 310 tht which extends beyond the ridge offset as a fracture zone. To the seafloor is 3100 m and immediately south of here, on the Answer the following: a. Calculate the age of the seafloor at 3100 m and 4100 m depth. D - 2500+350 SQRT(T). (Where: crust in meters; SQRT-square root function; and T = time in millions of years). b. On the diagram above, which way should the ridge to the south of the transform fault be offset- to the left or right of the ridge indicated to the north of the transform fault? Draw and label the ridge south of the transform fault on the diagram. c. If the point to the north is 500 km away from the ridge, calculate the spreading rate as cm/y (assume spreading is even in both directions away from the ridge). The position of the ridge south of the transform fault has not been surveyed. Assuming that the spreading rate has remained constant; calculate the amount of offset (as km) between the position of the ridge to the north asking zone) d. and to the south resulting from the transform fault? (Note: the question is for the offiet (difference in) distance between the ridge to the north & south of the fracture

Explanation / Answer

a) The age of seafloor at 3100m depth ,

D=2500+350*SQRT(T)

Therefore, T= {(D-2500)/350}^2= {(3100-2500)/350}^2=2.94million years

Similarly, for D=4100

T={(4100-2500)/350}^2= 20.90million years

b) Since the rock on the south is older than the adjacent rock on the north of fracture zone, so the ridge to the south of the fracture zone must have moved East ( to the left with respect to the ridge on the north as seen by standing over it ).

c)The point at 3100m depth on the north side of the fracture zone is 500km away from the ridge and its age is determined to be 2.94 million years. So, the spreading rate of the ridge will be distance/time on one side and equal in the other side(as mentioned in the question). So, the spreading rate will be (distance/age)×2 = (500×10^5cm/2.94×10^6years)×2

= 34.01cm/yr

d) The adjacent rocks on the two sides of the fracture zone differ in age by 20.90-2.94=17.96 million years

The spreading rate of the ridge is found to be 34.01cm/yr.

So, the distance of separation between the two rocks of same age on two sides of the fracture zone is (age difference of adjacent rocks×spreading rate) = 17.96×10^6years×34.01cm/year=6.1082×10^8cm = 6108.2 km

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