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.
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Answer:
B. to lock the tape into place
Explanation:
the button on the front of the housing locks the tape into place when pressed, preventing the tape from being pulled out further it retracting
Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X
, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X
=
................ 2
Given that; V = 240V, f = 50Hz,
=
, I = 15A, so that;
From equation 2,
X
= 
= 16Ω
To determine the inductance of the inductor,
X
= 2
fL
L = 
= 
= 0.05091
The inductance of the inductor is 0.051H.
The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V