the largest tendon in the body, the achilles tendon, connects the calf muscle to the heel bone of the foot. this tendon is typically 15.0 cm long, 5.40 mm in diameter, and has a youngs modulus of 1.65×10⁹ pa. if an athlete has stretched the tendon to a length of 15.9 cm, what is the tension t, in newtons, in the tendon? t = n

Question

Accepted Answer

# Explanation:
## Step1: Calculate the cross - sectional area $A$
The formula for the area of a circle is $A=\pi r^{2}$. Given the diameter $d = 5.40\space mm=5.40\times10^{- 3}\space m$, so the radius $r=\frac{d}{2}=2.70\times10^{-3}\space m$.
$$
\begin{align*}
A&=\pi r^{2}\
&=\pi\times(2.70\times 10^{-3})^{2}\
&=\pi\times7.29\times10^{-6}\space m^{2}\
&\approx2.3\space\times10^{-5}\space m^{2}
\end{align*}
$$
## Step2: Calculate the change in length $\Delta L$
The original length $L = 15.0\space cm = 0.150\space m$, the new length $L'=15.9\space cm = 0.159\space m$. Then $\Delta L=L'-L$.
$$
\begin{align*}
\Delta L&=0.159 - 0.150\
&=0.009\space m
\end{align*}
$$
## Step3: Use Young's modulus formula $Y=\frac{F/A}{\Delta L/L}$ to find the force $F$ (tension $T$)
Rearrange the formula $Y=\frac{F/A}{\Delta L/L}$ to $F = Y\times A\times\frac{\Delta L}{L}$. Given $Y = 1.65\times10^{9}\space Pa$, $A\approx2.3\times10^{-5}\space m^{2}$, $\Delta L = 0.009\space m$, $L = 0.150\space m$
$$
\begin{align*}
T&=Y\times A\times\frac{\Delta L}{L}\
&=1.65\times 10^{9}\times2.3\times10^{-5}\times\frac{0.009}{0.150}\
&=1.65\times10^{9}\times2.3\times10^{-5}\times0.06\
&=1.65\times2.3\times0.06\times10^{4}\
&=2.28\times10^{3}\space N
\end{align*}
$$

# Answer:
$2280\space N$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the cross - sectional area \(A\)

Step2: Calculate the change in length \(\Delta L\)

Step3: Use Young's modulus formula \(Y=\frac{F/A}{\Delta L/L}\) to find the force \(F\) (tension \(T\))

Answer: