Answer:
6.78571 m/s
55.13387 Joules
Explanation:
[tex]v_s[/tex] = Velocity of squid = 2.5 m/s
[tex]v_w[/tex] = Velocity of water
[tex]m_w[/tex] = Mass of water = 1.75 kg
Mass of squid
[tex]m_s=6.5-1.75=4.75\ kg[/tex]
Momentum of squid
[tex]p_s=m_sv_s\\\Rightarrow p_s=4.75\times 2.5\\\Rightarrow p_s=11.875\ kgm/s[/tex]
As the momentum of the system is conserved
[tex]p_s+p_w=0\\\Rightarrow p_s=-p_w\\\Rightarrow 11.875=-p_w\\\Rightarrow 11.875=-m_wv_w\\\Rightarrow v_w=-\frac{11.875}{1.75}\\\Rightarrow v_w=6.78571\ m/s[/tex]
The speed of the water is 6.78571 m/s
Kinetic energy the squid creates is given by
Kinetic energy of the squid + Kinetic energy of water
[tex]K=K_s+K_w\\\Rightarrow K=\frac{1}{2}(m_sv_s^2+m_wv_w^2)\\\Rightarrow K=\frac{1}{2}(4.75\times 2.5^2+1.75\times 6.78571^2)\\\Rightarrow K=55.13387\ J[/tex]
Kinetic energy the squid creates by this maneuver is 55.13387 Joules
(a) The speed of water to escape the predator [tex]V_w=6.758\frac{m}{s}[/tex]=
(b) Kinetic energy = 55.13J
It is given that
Velocity of squid [tex]V_S[/tex]= 2.5 m/s
Velocity of water [tex]V_w=?[/tex]
Mass of water [tex]m_w[/tex] = 1.75 kg
The mass of the squid will be given as
[tex]m_s=6.5-1.75=4.75kg[/tex]
Momentum of squid
[tex]P_s=m_sv_s=4.75\times2.5=11.875\frac{kgm}{s}[/tex]
Since by the conservatiom of momentum
[tex]P_s+P_w=0[/tex]
[tex]P_s=-P_w[/tex]
[tex]11.875=-P_w[/tex]
[tex]-m_w v_w=11.875[/tex]
[tex]V_w=-6.785\frac{m}{s}[/tex]
So the speed of the water is 6.78571 m/s
The kinetic energy created by squid will be
The kinetic energy of the squid + Kinetic energy of water
[tex]K=K_s+K_w[/tex]
[tex]\dfrac{1}{2} (m_sv_s^2+m_wv_w^2)[/tex]
[tex]\dfrac{1}{2} (4.75\times(2.5)^2+1.75+(6.78^2))[/tex]
[tex]KE=55.13J[/tex]
Thus
(a) The speed of water to escape the predator [tex]V_w=6.758\frac{m}{s}[/tex]=
(b) Kinetic energy = 55.13J
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