Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that 
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
answer is
a)
3/4
Explanation:
In the last 5 meters of braking, you lose 3/4 of your speed.
Answer:
a) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached
Answer:
When the imposter is sus : O
Explanation:
