Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
<u>Calculate the value of the M2 </u>
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
<em>Note: Calculation for T2 is attached below</em>
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:
IA = 80/3 kgm^2
Explanation:
Given:-
- The mass of rod AB, m1 = 20 kg
- The length of rod AB, L1 = 2m
- The mass of rod CD, m2 = 10 kg
- The length of rod CD, L2 = 1m
Find:-
What is the moment of inertia about A for member AB?
Solution:-
- The moment of inertia About point "O" the center of rod AB is given as:
IG = 1/12*m1*L^2
- To shift the axis of moment of inertia for any object at a distance "d" from the center of mass of that particular object we apply the parallel axis theorem. The new moment of inertia about any arbitrary point, which in our case A end of rod AB is:
IA = IG + m1*d^2
- Where the distance "d" from center of rod AB to its ends is 1/2*L1 = 1 m.
So the moment of inertia for rod AB at point A would be:
IA = 1/12*m1*L^2 + m1*0.5*L1^2
IA = 1/3 * m1*L1^2
IA = 1/3*20*2^2
IA = 80/3 kgm^2