Answer:
Demire karbon ilavesi ne yapar, ne değiştirir, özellikleri nasıl değiştirir, yansımaları nelerdir? Grafiklerde kullanarak detaylı olarak açıklayın. HMK brainly.com/app/profile/7139574 takip et!!!!
Answer:
Explanation:
Using the expression shown below as:
Where,
is the number of vacancies
N is the number of defective sites
k is Boltzmann's constant = 
is the activation energy
T is the temperature
Given that:
N = 10 moles
1 mole = 
So,
N = 
Temperature = 425°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (425 + 273.15) K = 698.15 K
T = 698.15 K
Applying the values as:
![ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}](https://tex.z-dn.net/?f=ln%5B%5Cfrac%20%7B2.3%7D%7B6.023%7D%5Ctimes%2010%5E%7B-11%7D%5D%3D-%5Cfrac%20%7BQ_v%7D%7B1.38%5Ctimes%2010%5E%7B-23%7D%5Ctimes%20698.15%7D)
Answer:
The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region
Explanation:
The shear modulus (G) is the ratio of shear stress to shear strain. Like the modulus of elasticity, the shear modulus is governed by Hooke’s Law: the relationship between shear stress and shear strain is proportional up to the proportional limit of the material. The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region.
Answer:
(d) a and c are correct
Explanation:
METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity
for example : iron, gold ,silver, copper
ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property
for example ; bronze and brass
so from above discussion it is clear that option (d) will be the correct option
Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
