Answer:
a) The planar defect that exists is twin boundary defect.
b) The planar defect that exists is the stacking fault.
Explanation:
I am using bold and underline instead of a vertical line.
a. A B C A B <u>C</u><u> </u>B A C B A
In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.
b. A B C A <u>B C B C</u> A B C
In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.
Answer:
a) 65.139 MPa
b) 17490 GPa
Explanation:
a) calculate the flexural strength
we now that for a three-point bending test the bending moment will occur at the midsection
Max stress ( flexural strength ) = MY / I
Max stress can be expressed as ; 3/2 * ( F*l / b*d^2 ) ------- ( 1 )
F ( load at fracture ) = 283
l (distance between support point ) = 50
b = 11.6
d = 5.3
substitute given data into equation 1 above
= 3/2 * (( 283 * 50) /( 11.6 * 5.3^2) )
= 3/2 * ( 14150 / 325.844 )
= 3/2 * 43.425 = 65.139 MPa
B ) Determine deflection ( Δ y )
load = 265 N
elastic modulus = 66 GPa
elastic modulus = tensile stress / tensile strain
Tensile stress = Δ y
strain = load = 66 GPa
therefore Δ y = 265 * 66 = 17490 GPa
Answer:
Q= - 7 KJ
Heat is rejected
Explanation:
In first process:
Process is adiabatic so heat transfer will be zero.
Q= 0 For adiabatic process
Now from first law of thermodynamics
Q = ΔU + W
ΔU is the change in internal energy
Q is the heat transfer
W is the work.
So here Q= 0
And work is transfer to the system .It means that work done on the system so it will be taken as negative.
W= - 12 KJ
Q = ΔU + W
0 = ΔU -12
ΔU = 12 KJ
It means that in first process internal energy of system increase.
In second process:
non-adiabatic process and work done by system is 5 KJ.
Q = ΔU + W
U is the point function and it does not depends on the path follows it depends only on final and initial states.
So fro second process
ΔU = - 12 KJ
Q= -12 + 5
Q= - 7 KJ
In magnitude Q= 7 KJ
It mean that heat is rejected from the system because it give negative sign.
Answer:
F = (21.651N, 17.678N, 6.470N)
Explanation:
The components of a vector in R ^ 3 are given according to their cosine directors, therefore, if we name F the force of 25N and Fx, Fy and Fz the components of the force with respect to the X, Y and Z axes , respectively:
Fx = FCos30 = 25 (0.866) = 21.651
Fy = FCos45 = 25 (0.707) = 17.678
Fz = FCos75 = 25 (0.259) = 6.470
F = (21.651N, 17.678N, 6.470N)
Answer:
- Scenario
- Use case
- Scenarios
- Scenarios
- Use case
Explanation:
A <u>scenario</u> is an actual sequence of interactions (i.e., an instance) describing one specific situation; a <u>use case</u> is a general sequence of interactions (i.e., a class) describing all possible <u>scenarios</u> associated with a situation. <u>Scenarios</u> are used as examples and for clarifying details with the client. <u>Use cases</u> are used as complete descriptions to specify a user task or a set of related system features.
