Aswathi Asked a Question
September 3, 2020 7:05 ampts 30 pts
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 displacement is___________ m.
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  • Rahul pandey Best Answer
    Answer: Net slip= 5 meters Net slip = sqrt[(dip-slip)^2 + (strike-slip)^2] = sqrt[(3m)^2 + (4m)^2] = sqrt[9+16] = sqrt [25] ...
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  • Sajan sarthak sahoo thankyou
    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
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    Aswathi
    thank you sir 😊