Answer:
27009.56 mm
Explanation:
Given:
Diameter of the aluminium alloy bar, d = 12.5 mm
Length of the bar, L = 27 m = 27 × 10³ mm
Tensile force, P = 3 KN = 3 × 10³ N
Elastic modulus of the bar, E = 69 GPa = 69 × 10³ N/mm²
Now,
for the uniaxial loading, the elongation or the change in length (δ) due to the applied load is given as:
![\delta=\frac{PL}{AE}](https://tex.z-dn.net/?f=%5Cdelta%3D%5Cfrac%7BPL%7D%7BAE%7D)
where, A is the area of the cross-section
![A=\frac{\pi d^2}{4}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D)
or
![A=\frac{\pi\times12.5^2}{4}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%5Ctimes12.5%5E2%7D%7B4%7D)
or
A = 122.718 mm²
on substituting the respective values in the formula, we get
![\delta=\frac{3\times10^3\times27\times10^3}{122.718\times69\times10^3}](https://tex.z-dn.net/?f=%5Cdelta%3D%5Cfrac%7B3%5Ctimes10%5E3%5Ctimes27%5Ctimes10%5E3%7D%7B122.718%5Ctimes69%5Ctimes10%5E3%7D)
or
δ = 9.56 mm
Hence, the length after the force is applied = L + δ = 27000 + 9.56
= 27009.56 mm
Answer:
Teller, Loan Officer, and Tax Preparer
Explanation:
Answer:
cross-weight is used to tighten it up.
Explanation:
and yes this is important because Cross-weight percentage compares the diagonal weight totals to the car's total weight.
hope this help
(mark this answer as an brainliest answer)
Answer:
a)-True
Hope this helps even tho theres no school right now