A 4.5 kg puffer fish expands to 40% of its mass by taking in water. When the puffer fish is threatened, it releases the water toward the threat to move quickly forward. What is the ratio of the speed of the puffer fish forward to the speed of the expelled water backwards?

2:5

Explanation:

We have the initial mass of the fish = 4.5kg

The mass of water intake = 40%of 4.5kg

= 0.40x4.5kg

= 1.8kg

The formula for conserving momentum

M1V1 = M2V2

M1 = mass of the puffer fish = 4.5kg

M2 = mass of the applied water = 1.8kg

V1 = velocity of the puffer fish

V2 = velocity of water

We are then to find v1/v2

V1/V2 = M2/M1

V1/V2 = 1.8/4.5

= 1/2.5

This is also 2/5

Therefore we have the ratio to be 2:5

it becomes positively charged by charging it

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s