Recent content by norcal36

I would express the total length by L0+x...? Then with that I could use V=√T/m/L equation which then would give me V=√2g(L0+x)...This sound correct?

This is the work I've done so far... I=p^2/2dv...I=(10)^2/2(1.2)(343)=.12112 I(A)=P...(.12112)(4pi(500)^2)=380522.366 I=P/A...380522.366/4pi(4000)^2=.0018925 10log(I/I0)...10log(.0018925/10^-12)=92.77db 92.77-7(4)=64.77db I am winging it and that's the closest I can get to the right answer...

You are right the equation I used is for another situation. However if I use the conservation of energy I get mgx=1/2kx^2...mg=1/2kx...kx=2mg...kx=T=2mg. Again I am assuming tension will be the equivalent to kx. The x would be the displacement of the "spring" which equals length. However I am...

Hello again, I did a sum of the forces at the top and bottom of the motion and got: Top...mg=ma Bottom...T-mg=ma Got T=2mg when I substitute the mg in for ma at bottom.Then I assumed kx=T=2mg so x(length of cord) would equal 2mg/k Then V of wave would be ...sq.rt2mg/k Is it that simple? Feel...

Thanks!

Also I apologize for my inexperience with formatting . I don't know what LaTeX is and only way I could show my work was to use Word Possessor.

Thank you for that insight. With that I do get the correct answer with replacing the mg with T1. However, is there a way to manipulate the equation without just replacing that mg with T1? Again thank you for the insight!

Hi new to the physics world and the symbiology is hard to understand completely. Attached is the work I've done to a final solution but I can't seem to get the answer in terms of only density of water and T1 and T2. Thank you for any assistance. [ Mentor Note -- Word file replaced with a...