Here is the answer I got from another person but it was incorrect. Maybe taking

Question

Here is the answer I got from another person but it was incorrect. Maybe taking a look at it helps.

apply conservation of momentum

final momentum initial momentum

(1.5 - 0.2)*2.5 + 0.2*v = 1.5*0.4

==> v = -13.25 m/s (velocity of water with respect to a refernce frame at rest)

velocity of water wih respect to squid = -13.25 - 2.5

= -15.75 m/s <<<<<<<<<--------------Answer

Problem 11.46 Squids rely on jet propulsion to move around. A 1.50 kg squid (including the mass of water inside the squid) drifting at 0.40 m/s suddenly ejects 0.20 kg of water to quickly get itself moving at 2.50 m/s. Part A If drag is ignored over the small interval of time needed to expel the water (the impulse approximation), what is the water's ejection speed relative to the squid? Express your answer with the appropriate units. Enter positive value if the direction of the velocity is the same as the direction of the initial velocity of the squid, and negative value if the direction of the velocity is opposite to the direction of the initial velocity of the squid. 15.75 m Submit My Answers Give Up Incorrect, Try Again; 5 attempts remaining Provide Eeedback Continue

Explanation / Answer

1.50 (0.4 ) = 0.20 (v) + 1.30 (2,50)

0.6= 020 (v) + 3.25

v = ( -2.65)/ 0.20 = -13.25 m/s

I thnk the question involves finding water speed relatgive to pre ejectio n speed of squid= -13.25 - 0.4=-13.65 m/s

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