Answer:
Explanation:
Given conditions
1)The stress on the blade is 100 MPa
2)The yield strength of the blade is 175 MPa
3)The Young’s modulus for the blade is 50 GPa
4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.
5)The temperature of the blade is 800°C.
6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)
where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K
Young Modulus, E = Stress,
/Strain, ∈
initial Strain, 


creep rate in the steady state


but Tinitial = 0


solving the above equation,
we get
Tfinal = 2459.82 hr
Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft