The initial void ratio is the <em>parameter </em>which is used to show the structural foundations for each <em>specimen of sand </em>so that the method and speed of compression would be <em>measured</em>.
Relative density is the mass per unit volume of each specimen of sand which is <em>measured </em>and it has to do with the<em> relative ratio</em> of the density of the sand.
Unit weight is the the exact weight per cubic foot of the sand which is measured.
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AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.
Explanation:
Answer:
(a) Increases
(b) Increases
(c) Increases
(d) Increases
(e) Decreases
Explanation:
The tensile modulus of a semi-crystalline polymer depends on the given factors as:
(a) Molecular Weight:
It increases with the increase in the molecular weight of the polymer.
(b) Degree of crystallinity:
Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.
(c) Deformation by drawing:
The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.
(d) Annealing of an undeformed material:
This also results in an increase in the tensile strength of the material.
(e) Annealing of a drawn material:
A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.
Answer:
B) 5.05
Explanation:
The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:
Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2
Given that:
Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05
Maximum outer diameter = 35 + 0.05 = 35.05
Minimum inner diameter = 25 - 0.05 = 24.95
Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05
or
Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05
Therefore the LMC wall thickness is 5.05