Question

Answer it please................!!!!!

Answer 1

option 2-
1/2mv^3

Mass of water flowing out per second through the hose is mv.

Therefore, kinetic energy imparted per second to the water is 12×(mv)×v2=12mv3

Answer 2

d/dt (KE) = d/dt [1/2 mv^2] = 1/2 [ dm/dt v^2 + m.2v dv/dt]

d/dt (KE) = 1/2. mvv^2 + 0 (dv/dt = 0)

Here m = a p, dm/dt = a p v = mv

= 1/2 mv^3

so the answer is option 2