Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂
Answer:
The note in this question is not an instrument that is negotiable under Article 3 of the U.C.C. Furthermore, it is not payable at any given time on demand due to the fact that principal repayment is not covered at a specified period of time. It shows that the acceleration clause is viable for the payment of the amount upon the default of the maker. This is also for an indefinite period of time.
Explanation:
The note in this question is not an instrument that is negotiable under Article 3 of the U.C.C. Furthermore, it is not payable at any given time on demand due to the fact that principal repayment is not covered at a specified period of time. It shows that the acceleration clause is viable for the payment of the amount upon the default of the maker. This is also for an indefinite period of time.
Answer:
T=194.3C
X=1
P=542.5kPa
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties
First we find the specific volume knowing the temperature and quality of state 1
v(p=400kPa, x=0.75)=0.3471m^3/kg
As a constant volume process is presented, it keeps constant in state 2, so we find the temperature and quality with p = 600kPa and v = 0.3471m ^ 3 / kg
T(P=600kPa, v=0.3471m ^ 3 / kg)=194.3C
x(P=600kPa, v=0.3471m ^ 3 / kg)=1
for the second part we find the pressure, with a quality of 1 and a volume of
v= 0.3471m ^ 3 / kg
p=542.5kPa
Answer:
The large percentage of steel includes less than 0.35% carbon
Explanation:
Carbon is perhaps the most important business alloy of steel. Raising carbon material boosts strength and hardness and enhances toughness. However, carbon often increases brittleness.
The large percentage of steel includes less than 0.35% carbon. Any steel with a carbon content range of 0.35% to 1.86% can be considered as hardened.