Answer:
C. strike-slip fault
Explanation:
The scientist must have observed a strike- slip fault.
A fault is an evidence of brittle deformation of the crust in the presence of applied stress on earth materials. Here, the earth material is the rock subjected to tension.
Where a fault occurs, there must have been movement between two blocks of rocks. The direction of movement helps us to delineate the fault type.
- When two blocks moves past each other horizontally, it is a strike-slip fault like rubbing your palms together.
- When a block moves in the direction of the dip, it forms a dip-slip fault which results in a fault-block mountain characterized by graben and horst systems.
Option A, Plateau is a table landform usually a mountain with flat peak.
Option B is a bowl shaped stratigraphic pattern in which the youngest sequence is at the core of the strata or a fold.
So, the most fitting option is C, a strike-slip fault.
Answer:
Sliding friction is the force that sliding objects experience
Explanation:
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
