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:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
<h3>PART (a)</h3>
For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
<h3>
PART (b)</h3>
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
<h3>PART(c)</h3>
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
<h3>
PART (d)</h3>
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)
Answer:
Given, a = 3, r = 1/2, n = 10
%r is common ratio
%n is number of terms
%a is the first term of the series
Sum = 0;
a = 3;
r = 1/2;
for i = 0 : 1 : 10;
Sum = Sum + a * r ^ i;
end
Sum
The Travel Speed at which the welding operation can be accomplished is given as 7.147mm/s. See the computation below.
<h3>What is travel speed?</h3>
The travel speed of the welding flame or gun across the workpiece is simply measured in millimeters per minute.
Travel speed, along with voltage and amperage, is one of three factors in arc welding that affect the amount of heat input.
<h3>What is the calculation that justifies the above answer is?</h3>
We are given the following:
Melting temperature = 1650 K
k = (3.33 * 10⁻⁶) * (1,650)²
k = 9.066 J/mm^3
The formula for velocity here is given as:
V = f1 * f2 * Rh / (k*A)
V = 0.9 * 0.6 * 3000 / (9.066*25)
V = 7.147 mm/s
Hence, the Travel Speed at which the welding operation can be accomplished is given as 7.147mm/s.
Learn more about welding operation at;
brainly.com/question/22494632
#SPJ1
Answer:
This is self locking.
Explanation:
Given that
Diameter d= 40 mm
Pitch (P)= 6 mm
Torque T= 1 KN.m
Friction(μ) = 0.1
Thread is double square.
Condition for self locking
L= n x P
N= number of start
Here given that double start so n=2
L=2 x 6 = 12 mm
(μ = 0.1)
So from we can that this is self locking.