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Annette [7]
3 years ago
8

For each description, choose the fault it describes. Fault types may be used more than once.

Physics
2 answers:
MArishka [77]3 years ago
3 0

Answer;

  • Rock layers are placed under compressional stress:  Reverse Fault
  • Rock layers are placed under tensional stress:  Normal Fault
  • Hanging wall has dropped relative to footwall:  Normal Fault
  • Hanging wall has risen relative to footwall: Reverse faults

Explanation;

A normal fault occurs because two tectonic plates are pulling apart from each other, causing tensional stress. The stress allows gravity to pull the hanging wall downward along the footwall, making a cliff-like formation called a scarp. Thus, if you see the hanging wall has dropped relative to the footwall, it is a normal fault. Normal faults occur at divergent boundaries.

A reverse fault happens at convergent boundaries, where two plates colliding produce compressional stress. As the two plates come together, the pressure causes the hanging wall to move upward along the footwall. These breaks are reverse faults because the hanging wall moves in an opposite, or reverse, direction when compared to a normal fault.

A strike-slip fault occurs when tectonic forces cause rock layers to break apart and move in opposite, parallel directions, causing shear stress. The rock layers in a strike-slip fault move sideways past one another.


Harlamova29_29 [7]3 years ago
3 0
Tensional stress creates normal faults. 

<span>Shear stress creates strike-slip faults </span>

<span>If the hanging wall has dropped it is a normal fault </span>

<span>If the hanging wall has risen then it's a reverse fault. 

</span>
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Who’s going faster in the attached image?
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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction tim
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Answer:

14.0 m

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Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

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Distance = Speed × Time

\text{Distance}=20\times 0.5=10\ m

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m

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\text{Distance}=u\times 0.5=0.5u\ m

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u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s

Maximum speed of the car by which it will not hit the deer is 25.1 m/s

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