Answer:
speed of ejected out water will be 0.102 m/s
Explanation:
Initially the jellyfish is given at rest
so here since there is no external force on the system of jelly fish and the water in it so the momentum of the whole system will remains conserved
so here we have
[tex]P_i = 0[/tex] initial momentum is zero
now final momentum of both will also be zero
so here we have
[tex]m_1v_1 + m_2v_2 = 0[/tex]
[tex]m_1= 1.3 g[/tex]
[tex]m_2 = (3.2 - 1.3) = 1.9 g[/tex]
[tex]v_2 = 0.070 m/s[/tex]
so we have
[tex](1.3g)(v_1) + (1.9 g)(0.070) = 0[/tex]
[tex]v_1 = 0.102 m/s[/tex]
so speed of ejected out water will be 0.102 m/s