The answer is A
Hope this helps 8)
Answer: you can watch a video on how to solve this question on you tube
Answer:
Explanation:
Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)
fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)
vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)
BFFS = 50+5, BFFS =55 (given) fLW= 6.6
TLC=6+3=9 fLC= 0.65
fM= 0.0
fA= 1.0
FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)
Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)
After: fA= 3.0
FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)
Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln
Answer:
e = 3.97*10^-4
Explanation:
1.8 mm = 0.0018 m
2.6*10^4 mm = 26 m
Elongation is The ratio between the stretched length and the original length.
e = L/L0
This is calculated with Hooke's law:
e = σ/E
Where
σ: normal stress
E: elastic constant
σ = P/A
Where
P: normal load
A: cross section
A = π/4 * d^2
Therefore:
e = P / (A * E)
e = 4 * P / (π * d^2 * E)
e = 4 * 290 / (π * 0.0018^2 * 207*10^9) = 3.97*10^-4
Answer:
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Explanation:
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