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:
Explanation:
Question is incomplete
Assuming the question you have asked is
You are driving home from school steadily at 95 km/h for 180 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h.
given,
speed of 95 km/h for 180 km
due to rain
speed is reduced to 65 km/h
distance traveled in 4.5 hour
time taken to travel 180 km
d = s x t

t = 1.9 hr
distance traveled in time, t' = 4.5-1.9 = 2.6 hr
Speed of vehicle = 65 Km/h
d' = s x t'
d' = 65 x 2.6
d'= 169 Km
total distance your hometown from school
D = d + d'
D = 180 + 169
D = 349 Km
Answer:
<h2>480</h2>
Explanation:
<h2>R=120÷0.25</h2><h2>R=480 ohms </h2>
because the unit for resistance is in ohms
D. physical property
the bonds between molecules of mercury are breaking so it's physical and it's not changing the chemical composition of the substance
Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC