A normal fault displaces a sandstone bed such that the dip -slip and the strike-slip components are 3 m and 4 m, respectively. the net-slip of the displaceme

Answer: Net slip= 5 meters Net slip = sqrt[(dip-slip)^2 + (strike-slip)^2] = sqrt[(3m)^2 + (4m)^2] = sqrt[9+16] = sqrt [25] ...

This question should be of oblique fault . If it is. a dip slip fault there will be no strike slip component. 5 is the answer