a vertebra is subjected to a shearing force of 460 n. the shear modulus for vertebra is 8.0 × 10^{10} n/m^{2}, and the vertebra may be modeled as a cylinder 3.00 cm high and 6.00 cm in diameter. determine the shear deformation δx. δx = m question credit: openstax college physics

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Accepted Answer

# Explanation:
## Step1: Calculate the cross - sectional area of the cylinder
The formula for the area of a circle is $A=\pi r^{2}$. Given the diameter $d = 6.00\ cm=0.06\ m$, the radius $r=\frac{d}{2}=0.03\ m$.
$$
\begin{align*}
A&=\pi r^{2}\
&=\pi\times(0.03)^{2}\
&= 2.827\times10^{-3}\ m^{2}
\end{align*}
$$
## Step2: Use the formula for shear deformation
The formula for shear deformation is $\Delta x=\frac{Fh}{SA}$, where $F = 460\ N$ (shearing force), $h=3.00\ cm = 0.03\ m$ (height of the cylinder), $S = 8.0\times10^{10}\ N/m^{2}$ (shear modulus), and $A$ is the cross - sectional area.
$$
\begin{align*}
\Delta x&=\frac{Fh}{SA}\
&=\frac{460\times0.03}{8.0\times10^{10}\times2.827\times10^{-3}}\
&=\frac{13.8}{2.262\times10^{8}}\
&=6.1\times10^{-8}\ m
\end{align*}
$$

# Answer:
$6.1\times 10^{-8}$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the cross - sectional area of the cylinder

Step2: Use the formula for shear deformation

Answer: