Respuesta :
Answer:E
Explanation:
Given
mass of object [tex]m=1000 kg[/tex]
Length of steel rod [tex]L=5 m[/tex]
diameter of rod [tex]d=0.8 cm[/tex]
Young modulus [tex]E=210,000 MN/m^2[/tex]
And we know [tex]E=\frac{stress}{strain}[/tex]
[tex]stress=\frac{Load}{Area}[/tex]
[tex]Strain=\frac{\Delta L}{L}[/tex]
Therefore [tex]\Delta =\frac{FL}{AE}[/tex]
[tex]\Delta =\frac{1000\times 9.8\times 5}{\frac{\pi (0.8\times 10^{-2})^2}{4}\times 210000\times 10^6}[/tex]
[tex]\Delta =\frac{20\times 9.8}{\pi \times 0.64\times 210\times 100}[/tex]
[tex]\Delta =4.64\times 10^{-3} m[/tex]
[tex]\Delta =0.464 cm[/tex]
Answer:
[tex]\Delta L=4.642\,mm[/tex]
Explanation:
Given:
mass of the hanging object, m= 1000 kg
original length of the steel rod, L=5 m
diameter of the rod, d= 0.80 cm
Young's modulus for the rod, E = [tex]210,000 MPa[/tex]
Firstly we find the value of stress:
[tex]\sigma = \frac{Force}{area}[/tex]
[tex]\sigma = \frac{1000\times 9.8}{\pi\times 4^2}[/tex]
[tex]\sigma = 194.9648 MPa[/tex]
Now strain,
[tex]\epsilon= \frac{\sigma}{E}[/tex]
[tex]\epsilon= \frac{194.9648 }{210000}[/tex]
[tex]\epsilon= 0.00093[/tex]
We know,
[tex]\epsilon= \frac{\Delta L}{L}[/tex]
[tex]0.00093= \frac{\Delta L}{5000}[/tex]
[tex]\Delta L=4.642\,mm[/tex]
