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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>
4.0 cm
Hope this helps (:
Answer:
<em>BCEG,ADF</em>
Explanation:
Between 2 waves, if path difference is mλ, m is an integer and they interfere constructively.
Between 2 waves, if path difference is (m+1/2)λ, m is an integer and they interfere destructively.
<em>BCEG,ADF</em>
Answer:
(a) 1.054 m/s²
(b) 1.404 m/s²
Explanation:
0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ)) - μk·m·g·(1 - sin(θ)) = m·a
Which gives;
0.5·g·cos(θ) - μ·g·(1 - sin(θ) = a
Where:
m = Mass of the of the block
μ = Coefficient of friction
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration of the block
θ = Angle of elevation of the block = 20°
Therefore;
0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°) - μk×9.81×(1 - sin(20°) = a
(a) When the static friction μs = 0.610 and the dynamic friction μk = 0.500, we have;
0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°) - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²
(b) When the static friction μs = 0.400 and the dynamic friction μk = 0.300, we have;
0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°) - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².
