Answer:
Option (c) and option (d)
Explanation:
Eutectic system is one in which a solid and homogeneous mixture of two or more substances resulting in the formation of super lattice is formed which can melt or solidify at a temperature lower than the melting point of any individual metal.
Eutectic alloys are those which have its components mixed in a specific ratio.
It is the composition in an alloy system for which both the liquidus and solidus temperatures are equal.
Eutectic alloys have the composition in which the melting point of the metal is lower than the other alloy composition.
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Answer:
The average thickness of the blubber is<u> 0.077 m</u>
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
Answer:
1
Created on Nov 3, 2018 @author: ASLand
7import atexit
#Read, nanes of both files
Rrintll"Enter tvo files to be compared below
userliamel input ("Enter the nome of the first file: ")
userliame2 input("Enter the name of the second file: ")
ROpen each file
f1 - open(userNamel, r')
@17 f2 = opan(useriame 2, )
tread all the lines into a list
d1 f1.readlines ()
d2 f2.readlines()
re equivalent, print "Yes" else pri
oiterate, and conpare
#11
the
y
if dl == d2:
print("Yes")
atexit
elif for i in range(@, min(len (d1), len(d2))):
if di[i]!=d2[i]:
PCint("No")
print(d1[i])
pcint(d2[])