Answer:
95.3 N
Explanation:
The tension in the cable is found by the equation:
[tex]T=\mu v^2[/tex]
Where [tex]\mu[/tex] is the mass density, and [tex]v[/tex] is velocity.
First we find mass density:
[tex]\mu =\frac{m}{L}[/tex] --> [tex]m[/tex] is mass: [tex]m=0.21kg[/tex] and [tex]L[/tex] is length of the cable: [tex]L=4.15m[/tex], so:
[tex]\mu=\frac{0.21kg}{4.15m} =0.0506kg/m[/tex]
And the velocity:
[tex]v=\frac{distance}{time} =\frac{d}{t}[/tex]
the time is [tex]t=0.765s[/tex] and in that time the pulse went down and back along the cable 4 times, if one time down and back is:
2*4.15m=8.3m,
four times this path is:
4*8.3m=33.2m
thus, the velocity is:
[tex]v=\frac{distance}{time} =\frac{33.2m}{0.765s}=43.4m/s[/tex]
And with this data we can now calculate the tension:
[tex]T=(0.0506kg/m)(43.4m/s)^2\\T=95.3N[/tex]
The tension is 95.3N
