"If a ball of mass M is dropped from a height h onto a spring with spring constant k (whose equilibrium positions is at height 0), compresses the spring an additional distance (L/2), and then rebounds, what height will the ball reach? Express your answer symbolically."

So I was thinking something along the lines of h= M*k*(L/2) but I'm not sure :/ Any guidiance?

 h ’= k  (h +L/2)²/ (2 M g )

Explanation:

For this exercise, one of the best methods to solve it is with energy conservation.

Starting point. Lower, spring with maximum compression

              Em₀ = Ke = ½ k x²

Final point. Higher after bounce

              = U = M g h ’

             Em₀ = Em_{f}

             ½ k x² = M g h’

the compressed distance is

          x = h+ L / 2

where h is the distance that compresses the spring by the height where it comes from and L/2 the additional compression

        h ’= ½ k x² / M g

we calculate

       h ’= k  (h +L/2)²/ (2 M g )

       h' = (k/2Mg) h2 (1 + L/2h)2

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6