Answer:
a) Q = 251.758 kJ/mol
b) creep rate is 
Explanation:
we know Arrhenius expression is given as

where
Q is activation energy
C is pre- exponential constant
At 700 degree C creep rate is
% per hr
At 800 degree C creep rate is
% per hr
activation energy for creep is
= 
![\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%5C%25%7D%7B5.5%20%5Ctimes%2010%5E%7B-2%7D%5C%25%7D%20%3D%20e%5E%7B%5B%5Cfrac%7B-Q%7D%7BR%28800%2B273%29%7D%5D%20-%5B%5Cfrac%7B-Q%7D%7BR%28800%2B273%29%7D%5D%7D)
![\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%7D%7B5.5%5Ctimes%2010%5E%7B-4%7D%7D%20%3D%20ln%20%5Be%5E%7B%5Cfrac%7BQ%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B1073%7D%20-%20%5Cfrac%7B1%7D%7B973%7D%5D%7D%5D)
solving for Q we get
Q = 251.758 kJ/mol
b) creep rate at 500 degree C
we know





Answer:
L= 50000 lb
D = 5000 lb
Explanation:
To maintain a level flight the lift must equal the weight in magnitude.
We know the weight is of 50000 lb, so the lift must be the same.
L = W = 50000 lb
The L/D ratio is 10 so
10 = L/D
D = L/10
D = 50000/10 = 5000 lb
To maintain steady speed the thrust must equal the drag, so
T = D = 5000 lb
Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:

ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂

θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂

=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain

=3.335 *10^-4

ε(avg) =150 *10^-6
orientation of γmax

θ = 31.71 or -58.29
To determine the direction of γmax

= 1.67 *10^-4
Answer:
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Answer:
(Option B)
Explanation:
The absolute pressure of the air-filled tank is:

