Respuesta :
Answer:
The diameter is 7.64 mm.
Explanation:
Given that,
Force = 6640 N
Length = 380 mm
Elongation = 0.50 mm
Modulus of elasticity = 110 GPa
We need to calculate the area
Using formula of stress
[tex]\sigma=\dfrac{F}{A}[/tex]....(I)
[tex]\sigma=\dfrac{E\Delta l}{l}[/tex]....(II)
From equation (I) and (II)
[tex]\dfrac{F}{A}=\dfrac{E\Delta l}{l}[/tex]
[tex]A=\dfrac{Fl}{E\Delta l}[/tex]
Put the value into the formula
[tex]A=\dfrac{6640\times380\times10^{-3}}{110\times10^{9}\times0.50\times10^{-3}}[/tex]
[tex]A=4.587\times10^{-5}\ m^2[/tex]
We need to calculate the diameter
[tex]A=\dfrac{\pi d^2}{4}[/tex]
[tex]4.587\times10^{-5}=\dfrac{\pi d^2}{4}[/tex]
[tex]d=\sqrt{\dfrac{4\times A}{\pi}}[/tex]
[tex]d=\sqrt{\dfrac{4\times4.587\times10^{-5}}{\pi}}[/tex]
[tex]d=7.64\ mm[/tex]
Hence, The diameter is 7.64 mm.
The diameter of the cylindrical rod will be d=7.64 mm.
What will be the diameter of the cylindrical rod?
It is given that,
Force = F= 6640 N
Length = L= 380 mm
Elongation E=l= 0.50 mm
Modulus of elasticity E = 110 GPA
first, we will find out the cross-section area of the road
Using the formula of stress
[tex]\sigma=\dfrac{F}{A}[/tex]
[tex]\sigma=\dfrac{E l }{L}[/tex]
Now from both equations
[tex]\dfrac{F}{A} =\dfrac{El}{L}[/tex]
[tex]A=\dfrac{FL}{El}[/tex]
Now putting all the values in the above equation
[tex]A=\dfrac{6640\times380\times10^{-3} }{110\times10^{9}\times0.50\times10^{-3} }[/tex]
[tex]A=4.587\times10^{-5} m^{2}[/tex]
To find out the area
[tex]A=\dfrac{\pi}{4}\times d^{2}[/tex]
[tex]d=\sqrt{\dfrac{4\times A}{\pi} }[/tex]
[tex]d=\sqrt{\dfrac{4\times 4.587\times10^{-5} }{\pi} }[/tex]
[tex]d=7.64mm[/tex]
Thus the diameter of the cylindrical rod will be d=7.64 mm.
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