Having trouble understanding the relationship between these expressions

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SUMMARY

The discussion centers on the relationship between wave speed and free-fall acceleration in a physics problem involving an astronaut measuring gravitational acceleration on a small planet. The key equation presented is √(MgL/m) = L/t, where M is the mass of the object, g is the gravitational acceleration, L is the length of the wire, and t is the time taken for a pulse to traverse the wire. The participants clarify that the wave speed v can be expressed as both √(T/u) and L/t, emphasizing the equivalence of these expressions in the context of the problem.

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  • Understanding of wave mechanics and wave speed equations
  • Familiarity with gravitational acceleration concepts
  • Basic algebra for rearranging equations
  • Knowledge of tension and mass per unit length in physics
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coldjeanz
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I understand everything up until:

√MgL/m = L/t

Why is that equal to L/t?

I also don't understand this either:

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This is the problem itself however I don't feel it's that important since I am trying to figure out how they are arriving at their conclusions.

An astronaut on a small planet wishes to measure the local value of the free-fall acceleration by timing pulses traveling down a wire that has an object of large mass suspended from it. Assume a wire has a mass of 4.10 g and a length of 1.60 m and that a 3.00 kg object is suspended from it. A pulse requires 43.6 ms to traverse the length of the wire. Calculate g of planet from these data. (You may ignore the mass of the wire when calculating the tension in it.)
 
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They are not saying that the last equality follows from something about \sqrt{MgL/m}, but rather that the wave speed v is equal to both \sqrt{T/u} AND L / t (since speed is just distance divided by time).

The second section is just an algebraic rearrangement of the second equation in the first section.
 
Yeah okay that went through my head for a split second but I couldn't find that in my book so I didn't know where they pulled it from. Thank you!
 

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