I need question 4B answered please and thank you! Spring 2017 3. Geological inve
ID: 107606 • Letter: I
Question
I need question 4B answered please and thank you! Spring 2017 3. Geological investigations, ground motion recordings, and aftershock locations indicate that the 1994 Northridge Earthquake had a moment magnitude of 6.7, slip of 1.25 m, a shear modulus A-3.00 x 1010 N/m2, average km. and a fault width (depth 20.0 What was the along strike length of the fault in this event? Remember a Newton (N) is the same thing as (kg a. Make sure that you convert all km to m before doing your calculations. You will need your seismic moment to be in units of N for this problem to work. (20 pts) Given Mw) fverage (s) 1.2 sm 00 X LO Iwidth of faut (b) 20km length fault at strike be L. Vog 12 IS log Lo MD La o q Is Mo I 833.83 m Page 2 of 5Explanation / Answer
Answer:4b
According to the given condition in question all the values are same and we have to calculate for average slip, therefore we can assume the average slip by letter S and put in the following equation to ge the value;
1678567.21= 1021.1/ (3*1010 * 20* 103 *S)
1678567.21= 1021.1/ 60*1013 *S
1678567.21= 108.1/ 60 * S
S= 1678567.21*60 /108.1= 100714032.6 /108.1 =0.8 m
Average slip would be 0.8m.
No, because the average slip is about just 0.8 m in this case whereas japan's earthquake were about 50m.