One student did an experiment on the rock cycle.
1 answer:
First, when the student added the layers of wax over each other, this became a representation of sedimentary rocks.
Then the student folded his/her palm and squeezed the layers of wax. This means that the student applied heat and pressure on the wax (sedimentary rocks)
Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.
Based on the above, the best choice would be:
<span>d. Heat and pressure can change sedimentary rocks into metamorphic rocks.</span>
Then the student folded his/her palm and squeezed the layers of wax. This means that the student applied heat and pressure on the wax (sedimentary rocks)
Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.
Based on the above, the best choice would be:
<span>d. Heat and pressure can change sedimentary rocks into metamorphic rocks.</span>
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Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
