Question:
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.
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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The net-slip on a fault is calculated using the Pythagorean theorem, as the dip-slip and strike-slip components are perpendicular to each other.
Updated On: Dec 11, 2025
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Correct Answer: 5
Solution and Explanation
Step 1: Understanding Net-Slip.
The net-slip on a fault is the vector sum of the dip-slip (vertical displacement) and strike-slip (horizontal displacement) components. Using the Pythagorean theorem: \[ \text{Net-slip} = \sqrt{\text{(Dip-slip)}^2 + \text{(Strike-slip)}^2}. \] Step 2: Calculation of Net-Slip.
Substituting the given values: \[ \text{Net-slip} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m}. \] Step 3: Conclusion.
The net-slip of the displacement is 5 m.
The net-slip on a fault is the vector sum of the dip-slip (vertical displacement) and strike-slip (horizontal displacement) components. Using the Pythagorean theorem: \[ \text{Net-slip} = \sqrt{\text{(Dip-slip)}^2 + \text{(Strike-slip)}^2}. \] Step 2: Calculation of Net-Slip.
Substituting the given values: \[ \text{Net-slip} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m}. \] Step 3: Conclusion.
The net-slip of the displacement is 5 m.
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