Respuesta :
Answer:
A) The speed of the water must be 8.30 m/s.
B) Total kinetic energy created by this maneuver is 70.12 Joules.
Explanation:
A) Mass of squid with water = 6.50 kg
Mass of water in squid cavuty = 1.55 kg
Mass of squid = [tex]m_1=6.50 kg- 1.55 kg=4.95 kg[/tex]
Velocity achieved by squid = [tex]v_1=2.60 m/s[/tex]
Momentum gained by squid = [tex]P=m_1v_1[/tex]
Mass of water = [tex]m_2=1.55 kg[/tex]
Velocity by which water was released by squid = [tex]v_2[/tex]
Momentum gained by water but in opposite direction = [tex]P'=m_2v_2[/tex]
P = P'
[tex]m_1v_1=m_2v_2[/tex]
[tex]v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s[/tex]
B) Kinetic energy does the squid create by this maneuver:
Kinetic energy of squid = K.E =[tex]\frac{1}{2}m_1v_1^{2}[/tex]
Kinetic energy of water = K.E' = [tex]\frac{1}{2}m_2v_2^{2}[/tex]
Total kinetic energy created by this maneuver:
[tex]K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}[/tex]
[tex]=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules[/tex]
