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(A) The speed of the wave in the rope is given by $v = \frac{L}{t} = \frac{43\, m}{1.4\, s} \approx 30.71\, m/s$.The linear mass density $\mu$ of the

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$110$

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(A) The speed of the wave in the rope is given by $v = \frac{L}{t} = \frac{43\, m}{1.4\, s} \approx 30.71\, m/s$.The linear mass density $\mu$ of the rope is $\mu = \frac{m}{L} = \frac{5.0\, kg}{43\, m} \approx 0.116\, kg/m$.The wave speed in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.Rearranging for tension: $T = v^2 \mu = \left(\frac{43}{1.4}ight)^2 	imes \frac{5.0}{43}$.$T = \frac{43^2}{1.96} 	imes \frac{5}{43} = \frac{43 	imes 5}{1.96} = \frac{215}{1.96} \approx 109.69\, N$.Rounding to the nearest whole number,the tension is approximately $110\, N$.

$A$ $43\, m$ long rope of mass $5.0\, kg$ joins two rock climbers. One climber strikes the rope and the second one feels the effect $1.4\, s$ later. What is the tension in the rope in $N$?

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