Answer:
critical stress required for the propagation is 27.396615 ×
N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 ×
N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) = 
.....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 ×
m
so now put value in equation 1 we get
( σc ) =
( σc ) = 
( σc ) = 27.396615 × N/m²
so critical stress required for the propagation is 27.396615 × N/m²
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:
Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.
AC Signal = An electrical signal that alternates between positive and negative voltage.
DC Signal = An electrical signal that only has positive voltage.
Rectify = A fancy word for converting something.
Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.
The load resistor is just there to prevent a short circuit from happening.
Answer:
sum2 = 0
counter = 0
lst = [65, 78, 21, 33]
while counter < len(lst):
sum2 = sum2 + lst[counter]
counter += 1
Explanation:
The counter variable is initialized to control the while loop and access the numbers in <em>lst</em>
While there are numbers in the <em>lst</em>, loop through <em>lst</em>
Add the numbers in <em>lst</em> to the sum2
Increment <em>counter</em> by 1 after each iteration
