A 57 kg water skier is being pulled by a nylon tow rope that is attached to a bo

Question

A 57 kg water skier is being pulled by a nylon tow rope that is attached to a boat. The unstretched length of the rope is 13 m and its cross-sectional area is 2.0x10^-5 m2. As the skier moves, a resistive force (due to the water) of magnitude 120 N acts on her; this force is directed opposite to her motion. What is the change in the length of the rope when the skier has an acceleration whose magnitude is 0.90 m/s^2? I don't know how to work this problem, Can you please show me and give me the answers? Thank you! Need to find m (meters)____________

Explanation / Answer

F=.9*57+120=171.3 Y (nilon)=3*10^9 Pa so Y=(F/A)/(dL/L) so dL=FL/(AY)=171.3*13/(2*10^(-5)*3*10^9)=.037 m(ans)

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