Electromagnets project student guide y’all i need helppppo
1 answer:
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Answer:
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
Explanation:
Given that
Yield strength ,Sy= 240 MPa
Tensile strength = 310 MPa
Elastic modulus ,E= 110 GPa
L=380 mm
ΔL = 1.9 mm
Lets find strain:
Case 1 :
Strain due to elongation (testing)
ε = ΔL/L
ε = 1.9/380
ε = 0.005
Case 2 :
Strain due to yielding
ε '=0.0021
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
For computation of load strain due to testing should be less than the strain due to yielding.
Answer:
Approximately (assuming that the partner is holding the other end of the coil stationary.)
Explanation:
In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.
The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.
There would be a node between:
- the antinode at the end of the spring that is being shaken, and
- the antinode between the two ends of this spring.
Overall, the nodes and antinodes on this spring would be:
- node at the end that is being held still,
- antinode (as mentioned in the question),
- node (inferred, not mentioned in the question), and
- antinode at the end that is being shaken.
The distance between two adjacent nodes is equal to one-half (that is, ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is,
) of the wavelength of the wave.
Thus, if the wavelength of the wave in this question is , the length of this spring would be:
.
The question states that the length of this coiled spring is . In other words,
. The wavelength of this wave would be
.
The frequency of this wave is the number of cycles in unit time:
.
Hence, the speed of this wave would be:
.
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>