Question

A 15kg tuna fish moving horizontally to the Right at 5m/s Swallows a 2 kg dalagang bukid that is swimming to the left at 7.5m/s .what is the speed of the tuna fish immediately after, if the forces exerated on the fish by the water can be neglected

Answer 1

Answer:

7.5m/ ???????

Explanation:

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Answer 2

The Speed of tuna fish after eating dalagang is 3.53 m/s towards right.

Taking speed of right as positive and left as negative.

Given :

Speed of 15 kg tuna fish , u = 5 m/s.

Speed of 2 kg dalagang , v = -7.5 m/s.

Let , speed of tuna fish be V.

Mass of tuna fish after eating dalagang, m = 15+2 kg = 17 kg.

It is given that no external force is applied .

So, Momentum is conserve and

Initial momentum = Final momentum

And momentum , P = mv ( mass and velocity )

Therefore ,

$(15\times 5) - (2\times7.5)= (15+2)\times V$

So, V = 3.53 m/s.

Hence , this is the required solution.

Learn More :

Momentum

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