Respuesta :
Explanation:
Expression for Young's modulus is represented as follows.
Y = [tex]\frac{\text{shear stress}}{\text{shear strain}}[/tex]
= [tex]\frac{\frac{F}{A}}{\frac{\Delta L}{L_{o}}}[/tex]
where, Y = Young's modulus
F = force which acts perpendicular to the surface
A = area of the surface
[tex]L_{o}[/tex] = original length of the rope
[tex]\Delta L[/tex] = elongation in the rope
The given data is as follows.
r = 0.008 m, mass (m) = 65 kg, g = 9.8 m/s
S = [tex]5 \times 10^{9} N/m^{2}[/tex]
[tex]L_{o}[/tex] = 35 m
Force perpendicular to the surface is calculated as follows.
F = mg
= [tex]65 kg \times 9.8 m/s[/tex]
= 637 N
and, Area = [tex]\pi r^{2}[/tex]
= [tex]3.14 \times (0.008)^{2}[/tex]
= [tex]20.096 \times 10^{-5}[/tex]
Putting the given values into the above formula as follows.
[tex]\Delta L = \frac{1}{Y} \times \frac{F}{A} \times L_{o}[/tex]
= [tex]\frac{1}{5 \times 10^{9}} \times \frac{637 N}{20.096 \times 10^{-5}} \times 35 m[/tex]
= [tex]221.879 \times 10^{-4}[/tex]
= [tex]2.21 \times 10^{-2} m[/tex]
Thus, we can conclude that by [tex]2.21 \times 10^{-2} m[/tex] does a 65.0 kg mountain climber stretch her 0.800 cm diameter nylon rope.
