Why does water wet glass
1 answer:
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Answer:
The true stress at true strain 0.05cm/cm is 80MPa
Explanation:
Given that
the strength coefficient is K
true strain is ε
strain hardening exponent is n
initial diameter of bar is d = 1cm, (10mm)
tensile force is F
engineering stress(S) = 120
the engineering stress(S) =
To find force (F) =
120 =
F = 120 * (π/4) * (100)
F = 9425N
Calculate the true strain (ε) = In (l₀ / l₁)
where
l₀ = initial length of the metallic bar = 3cm
l₁ = final length of metallic bar = 3.5cm
ε = In (3.5 / 3)
= In 1.1667
= 0.154cm/cm
Calculate the true stress (σ) at fracture point
=
tensile force is F and final diameter of bar is d₁ (d in the eqn)
Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)
σ =
= 140MPa
To find the strength coefficient (K) of the material bar
K =
K =
= 356.75MPa
To calculate the true stress σ true strain of 0.05cm/cm
K = 356.75MPa
σ =
=
= 80MPa
The true stress at true strain 0.05cm/cm is 80MPa
Answer:
e) v₁ = 29.7 m / s
Explanation:
Let's propose the solution of the problem, let's start at the moment
Initial
p₀ = m v₁
Final
= (m + M) v
The moment is preserved
p₀ =
m v₁ = (m + M) v
v = m / (m + M) v₁ (1)
a) energy is conserved
Let's look for kinetic energy
Initial
K₀ = ½ m v₁²
Final
= ½ (m + M) v²
Let's replace v
= ½ (m + M) [m / (m + M) v₁]²
= ½ m² / (m + M) v₁²
Let's look for the relationship of these energies
Ko / = ½ m v₁² / (½ m² / (m + M) v₁²)
Ko / = (m + M) / m = (1 + M / m)²
The kinetic energy changes therefore it is not conserved in the process, the missing energy is converted into potential heat energy during the crash
b) The impulse is conserved because the system is defined as formed by the two bodies and the externals are of action and reaction, so for the complete system the sum is zero and the moment does not change in value
c) in this case the system is already formed by the two bodies and since there is no rubbing the mechanical energy is conserved, transforming from kinetics to potential
d) when the pendulum oscillates the speed changes from v to zero, so the moment is not conserved, this is because there is an external force acting on the system, the force of gravity
e) For this part let's start at the end of the movement
It is system (bullet + block) moves, energy is conserved
Final. Highest point
= U = (m + M) g h
Initial. Lowest point
Em₀ = K = ½ (m + M) v2
Em₀ =
½ (m + M) v² = (m + M) g h
v = √ 2gh
Let's look for the height (h) by trigonometry
Cos 15 = x / L
h = L-x
h = L - L cos 15
h = L (1- cos 15)
We replace
v = √ (2gL (1- cos 15))
Now we use equation (1) of momentum conservation
v = m / (m + M) v1
v₁ = (m + M) / m v
v1 = (0.1 +2.0) /0.1 RA (2 9.8 3 (1- cos 15))
v₁ = 21 √ (2.00)
v₁ = 29.7 m / s