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Squid rely on jet propulsion when a rapid escape is necessary A 1.5 kg squid at rest pulls 0.10 kg of water into its mantle, then ejects this water at a remarkable 45 m/s. Right after this ejection, how fast is the squid moving? Express your answer with the appropriate units.

Respuesta :

skyluke89

Answer:

-3 m/s

Explanation:

The problem can be solved by using the law of conservation of momentum.

In fact, the system squid+water inside is initially at rest, so the total momentum is zero:

[tex]p_i=0[/tex]

while the final momentum is:

[tex]p_f = m_s v_s + m_w v_w[/tex]

where

[tex]m_s=1.5 kg[/tex] is the mass of the squid

[tex]v_s[/tex] is the velocity of the squid

[tex]m_w = 0.10 kg[/tex] is the mass of water

[tex]v_w = 45 m/s[/tex] is the water velocity

Since the total momentum must be conserved,

[tex]p_i = p_f = 0[/tex]

So we have

[tex]m_s v_s + m_w v_w =0\\v_s = -\frac{m_w v_w}{m_s}=-\frac{(0.10 kg)(45 m/s)}{1.5 kg}=-3 m/s[/tex]

where the negative sign means the direction is opposite to that of the water.

onyebuchinnaji

The final velocity of the squid after the ejection of the water is 3 m/s backwards.

The given parameters;

  • mass of the squid, m₁ = 1.5 kg
  • initial velocity of the squid, u₁ = 0
  • mass of water, m₂ = 0.1 kg
  • initial velocity of the water, u₂ = 0
  • final velocity of the ejection, v₂ = 45 m/s

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.5(0) + 0.1(0) = 1.5v₁  +  45(0.1)

0 = 1.5v₁  + 4.5

1.5v₁ = -4.5

[tex]v_1 = \frac{-4.5}{1.5} \\\\v_1 = -3 \ m/s[/tex]

Thus, the final velocity of the squid after the ejection of the water is 3 m/s backwards.

Learn more here:https://brainly.com/question/24424291

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