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`        What strength fishing line is needed to stop a 19-lb salmonswimming at 9.2 ft/s in a distance of 4.5 in.?`
2 years ago

Arun
24497 Points
```							Dear Ahmed

Initial speed (vi) of the salmon is, vi = 9.2 ft/s
Final speed (vf) of the salmon is, vf = 0 ft/s
So the average speed of the salmon will be, vav = (vi + vf) /2
= (0 ft/s+9.2 ft/s)/2 =    = 4.6 ft/s
Time required (t) to stop the salmon is equal to the distance travelled (x) by the salmon divided by the average velocity (vav) of the salmon.
So, t = x/ vav
To obtain the time required (t) to stop the salmon, substitute 4.5 in for x and 4.6 ft/s for vav in the equation  t = x/ vav,
t = x/ vav
=(4.5 in)/(4.6 ft/s) =(4.5 in×0.0833 ft/1 in)/(4.6 ft/s) = (0.38 ft) /(4.6 ft/s) = 8.3×10-2 s
The deceleration a of the salmon will be, a = Δv/Δt.
To find out the   deceleration a of the salmon, substitute 9.2 ft/s for Δv and 8.3×10-2 s for Δt in the equation a = Δv/Δt,
a = Δv/Δt
= (9.2 ft/s)/(8.3×10-2 s) =110 ft/s2
To find out the force on the salmon, substitute 19-lb for W, 110 ft/s2 for a, and 32 ft/s2 for g in the equation F = Wa/g,
F = Wa/g
= (19-lb) (110 ft/s2)/ (32 ft/s2) = 65.31 lb
Rounding off to two significant figures, the force will be 65-lb.
From the above observation we conclude that, the force required to stop the 19-lb salmon swimming at 9.2 ft/s in a distance of 4.5 in would be 65-lb.

Regards

```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions