<h2>
FAULT</h2>
The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts. The fault labeled 'E' cuts through all three sedimentary rock layers (A, B, and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>
Answer:
x₁ = 15 m, x₂ = 12 m
, x_total = 27m, v₁ = 5 m / s
, v₂ = - 3 m / s
Explanation:
In this exercise we will use the kinematics of uniform motion
v = d / t
let's apply this equation for the first move
v₁ = Δx / t = (x₂-x₀) / t
v₁ = (12- (-3)) / 3
v₁ = 5 m / s
the distance traveled is x₁ = 15 m
Now let's analyze the return movement
v₂ = Δx / dt
v₂ = (0 - 12) / 4
v₂ = - 3 m / s
The negative sign indicates that the vehicle is moving to the left
the distance traveled is x₂ = 12 m
The total dystonia is
x_total = x₁ + x₂
x_total = 15 +12
x_total = 27m
In the attached we have the graphics of the movement
Answer:
The (minimum) thickness of the oil slick is 325 nm.
Explanation:
Let the (minimum) thickness of the oil slick = 
Therefore:
2 = (
+
)*λ
/
(1)
similarly,
2=(
)*λ
/ (2)
Thus, equation 1 = equation 2
(+
*λ
= (
*λ
Where:
λ
λ
Therefore:
=650/390=5/3
This shows that,
=2
=1
Thus, using equation 2
=
Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
Mechanical energy
I think
