Calculating Tension in a Stretched String Using Wave Speed

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To calculate the tension in a stretched string, the wave speed formula v = √(F/(m/L)) is used, where v is the wave speed, F is the tension, and m/L is the linear density. Given a linear density of 0.00500 kg/m and a wave speed of 85 m/s, the equation can be rearranged to solve for tension F. It is suggested to treat the linear density as a separate variable, such as μ or λ, to simplify calculations. Proper algebraic manipulation is necessary to isolate F in the equation. Understanding the relationship between wave speed, tension, and linear density is crucial for solving this problem.
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Homework Statement



A string whose linear density is 0.00500 kg/m is stretched to produce a wave speed of 85 m/s. What tension was applied to the string.

Homework Equations



v=√(F/(m/L))


The Attempt at a Solution



v=0.00500 kg/m m/L= 85 m/s

I can't seem to re-arrange the equation properly or I am just using the wrong one to begin with.
 
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Before plugging in any numbers, use algebra to solve the v=√(F/(m/L)) formula for the tension F. [And it might help to treat the linear density (m/L) as its own variable. I'd call it μ, or maybe λ.]
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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