Final answer:
The ratio of the speed of the puffer fish forward to the speed of the expelled water backward is 0.4:1, derived from the conservation of momentum principle in physics.
Step-by-step explanation:
The question involves the conservation of momentum principle in physics, where the mass of a puffer fish is ejected to propel itself in the opposite direction. When a 4.5 kg puffer fish expands to 40% of its mass by taking in water, it means it has increased its mass by an additional 1.8 kg (which is 40% of 4.5 kg), making its total mass 6.3 kg when fully expanded. Upon releasing this water, the fish's mass returns to 4.5 kg. According to the conservation of momentum mv (mass × velocity) of the system (puffer fish plus water) is constant before and after the water is expelled, assuming there are no external forces involved and the water expulsion happens in a vacuum or similar idealized environment.
After the water is expelled, the momentum of the fish and the water must be equal and opposite since their initial momentum was zero (they were not moving relative to one another before the water was expelled). Let's assign v_f as the final velocity of the fish and v_w as the final velocity of the water. Since the mass of the expelled water is 1.8 kg, the conservation of momentum can be written as:
4.5 kg × v_f = 1.8 kg × v_w
Solving for the ratio of the velocities:
v_f/v_w = (1.8 kg) / (4.5 kg) = 0.4
So the ratio of the speed of the puffer fish forward to the speed of the expelled water backward is 0.4:1, making the correct answer b) 0.4:1.