What is the creative process that helps you overcome writer's block called?
1 answer:
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Answer:
Part 1: It would be a straight line, current will be directly proportional to the voltage.
Part 2: The current would taper off and will have negligible increase after the voltage reaches a certain value. Graph attached.
Explanation:
For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.
V=I*R
where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.
In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.
Answer:
For [1 1 0] and [1 0 1] plane, σₓ = 6.05 MPa
For [0 1 1] plane, σ = 0; slip will not occur
Explanation:
compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.
Given;
Stress direction: [1 0 0] ⇒ A
Slip direction: [1 1 1]
Normal to slip direction: [1 1 1] ⇒ B
∅ is the angle between A & B
Step 1: cos∅ = A·B/|A| |B| = ⇒ cos∅ = 1/
σₓ = τ/cos ∅·cosλ
where τ is the critical resolved shear stress given as 2.47MPa
Step 2: Solve for the slip along each plane
(a) [1 1 0]
cosλ = [1 1 0]·[1 0 0]/(·
)
note: cosλ = slip D·stress D/|slip D||stress D|
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa
(b) [0 1 1]
cosλ = [0 1 1]·[1 0 0]/(·
) = 0
∵ σₓ = 2.47MPa/0, which is not defined
Hence, for stress along [1 0 0], slip will not occur along [0 1 1]
(c) [1 0 1]
cosλ = [0 1 1]·[1 0 0]/(·
)
cosλ = 1/
∵ σₓ = τ/ ·
=
* 2.47MPa = 6.05MPa
See attachment for the space diagram
Answer:
attached below
Explanation:
