Answer:
The solution for this question is attached below:
The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely w
1 answer:
Answer:
1064.8 kg/m³
Explanation:
Weight of the hydrometer = ρghA where ρ is the density, g is acceleration due to gravity, h is the submerged height and A is the cross sectional area.
W in water = ρwghwA
W in liquid = (ρliq)g hliq A where the cross sectional area is constant
W in water = W in liquid
(ρw)ghwA = (ρliq)g hliq A where ρw is density of water, ρliq is the density of liquid and hw and hliq are the heights of the liquid and that water. g acceleration due to gravity cancel on both sides as well as the constant A
pliq = × 1000 kg /m³ ( density of water) =(
) × 1000 = 1064.8 kg/m³
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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
