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# A disk between vertebrae in the spine is subjected to a

## Problem 38PE Chapter 5

College Physics by Urone | 1st Edition

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Problem 38PE

A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have the shear modulus of 1×109 N / m2 . The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter.

Step-by-Step Solution:

Problem 38PE

Solution 38 PE

Step 1:

In this question, we need to find shear deformation of disk between vertebrae in the spine

Data given

Shear force

Shear modulus

Height of disk

Diameter of disk

Step 2 :

We shall find the area of the disk

Substituting values we get

Thus we have area of the cylinder as

Step 3 of 4

Step 4 of 4

##### ISBN: 9781938168000

This textbook survival guide was created for the textbook: College Physics by Urone, edition: 1st. This full solution covers the following key subjects: disk, shear, its, cylinder, equivalent. This expansive textbook survival guide covers 34 chapters, and 3125 solutions. College Physics by Urone was written by and is associated to the ISBN: 9781938168000. Since the solution to 38PE from 5 chapter was answered, more than 1270 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 38PE from chapter: 5 was answered by , our top Physics solution expert on 07/07/17, 04:39PM. The answer to “A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have the shear modulus of 1×109 N / m2 . The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter.” is broken down into a number of easy to follow steps, and 49 words.

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