A) a 5.00-kg squid initially at rest ejects 0.250 kg of fluid with a velocity of 10.0 m/s. what is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00-n frictional force opposing the squid’s movement?
(b) how much energy is lost to work done against friction?

(a) The recoil velocity of the squid is 0.5 m/s

(b) W = 0.25 J

Explanation:

(a)

Momentum of the squid = momentum of the fluid

MV = mv

Where M = mass of the squid, V = recoil velocity of the squid, m = mass of the  fluid, v = recoil velocity of the fluid.

Making V the subject of formula,

V = mv/M

Where m = 0.25 kg, M = 5.00 kg, v = 10 m/s

V = (0.25 × 10)/5

V = 0.5 m/s

The recoil velocity of the squid is 0.5 m/s

(b)

Work done against friction (W) = Frictional force(F) × Distance(d)

W = F ×d Equation 2

Where F = 5.0N, d = distance.

d = velocity × time

Where Velocity = 0.5 m/s, Time = 0.1 s

∴ d = 0.5 × 0.1 = 0.05 m

Substituting these values into equation 2,

W = 5 × 0.05

W = 0.25 J

Therefore 0.25 J of energy is lost to the work done against friction.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

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