Answer:
a)5.28 Å , b)3.73 Å , c)3.048 Å
Explanation:
the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.
Therefore, a particular unit cell consist only 1/8th part of an atom.
The lattice constant of a simple cubic primitive cell is 5.28 Å
We know formula of distance,
d =
a)(100)
a=5.28 Å
Distance = =5.28 Å
b)(110)
Distance = = 3.73 Å
c)(111)
Distance= = 3.048 Å
Explanation:
Temperature range → 0 to 80'c
respective voltage output → 0.2v to 0.5v
required temperature range 20'c to 40'c
Where T = 20'c respective voltage
Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)
so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v
Therefore, the formula for the circuit will be
The simplest circuit will be a op-amp
NOTE: Refer the figure attached
Vs is sensor output
Vr is the reference volt, Vr = 0.275v
choose R2, R1 such that it will maintain required ratio
The output Vo can be connected to voltage buffer if you required better isolation.
This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
= - ( εˣ / v )
= - ( -0.00088 / 0.34 )
= - ( - 0.002588 )
= 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress = ......................1
put here value and we get
critical stress =
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
