Aswathi posted an Question
September 03, 2020 • 12:35 pm 30 points
  • IIT JAM
  • Geology (GG)

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

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 kumar 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] = 5m

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    Sajan sarthak best-answer

    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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    thank you sir 😊

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