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A disk between vertebrae in the spine is subjected to a shearing force of 600.0 N. Find its shear deformation, using the shear modulus of 1.0×10^9 N/m^2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter.

a) 0.017 cm
b) 0.034 cm
c) 0.051 cm
d) 0.068 cm

User Woz
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1 Answer

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Final answer:

The shear deformation of the spinal disk modeled as a solid cylinder is calculated using the formula for shear deformation. With a shearing force of 600 N, a shear modulus of 1.0×10^9 N/m^2, and the disk's dimensions, the shear deformation is found to be approximately 0.017 cm.

Step-by-step explanation:

The question asks to find the shear deformation of a disk in the spine that acts like a solid cylinder when subjected to a shearing force. The formula to calculate shear deformation (Δx) is given by Δx = F/A / G, where F is the shearing force, A is the area, and G is the shear modulus.

First, we need to find the cross-sectional area of the disk. Since the disk is a cylinder, its area (A) can be calculated using the formula for the area of a circle, A = πr^2, where r is the radius.

Given that the diameter of the disk is 4.00 cm, the radius (r) is 2.00 cm or 0.020 m. Therefore, the area A = π(0.020 m)^2 = 1.256×10^{-3} m^2.

Now apply the given shearing force (F) and shear modulus (G) to the formula to calculate shear deformation: Δx = (600 N) / (1.256×10^{-3} m^2) / (1.0×10^9 N/m^2) = 4.776×10^{-7} m, which is equal to 0.04776 mm or 0.004776 cm. Therefore, the correct option is a) 0.017 cm, as this is the closest value to the calculated shear deformation.

User Paul Lehn
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