A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.60 m to the bottom of the incline is 3.80 m/s. What is the speed of the block when it is 4.60 m from the top of the incline

   h ’= 0.51356 m

Explanation:

For this exercise we can use conservation of energy

starting point. Highest point of the trajectory

         Em₀ = m g h

         sin θ = h / L

         h = L sin θ

         Em₀ = m g L sin  θ

final point. Lowest point of the trajectory

         Em_f = K = ½ mv²

as there is no friction, energy is conserved

        Em₀ = Emf

        m g L sin  θ  = ½ m v²

        sin  θ  = ½ v² / gL

        sin  θ  = ½ 3.80² / (9.8  6.60)

        sin  θ  = 0.111626

         tea = 6.41

They ask us for the speed for L ’= 4.60 m

let's find the height

         sin 6.41 = h '/ L'

h ’= L’ sin 6.41

         h ’= 4.60 sin 6.41

         h ’= 0.51356 m

we use conservation of energy

          m g h ’= ½ m v’2

          v ’= √ (2gh’)

          v ’= √ (2 9.8 0.51356)

          v ’= 3.173 m /s²

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.