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Animal Propulsion. Squids and octopuses propel themsclves by expelling water, They do this by kecping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A \(6.50 \mathrm{~kg}\) squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has \(1.75 \mathrm{~kg}\) of watcr in its cavity, at what speed must it expel this water suddenly to achieve a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) to cscape the predator? Ignore any drag cffects of the surrounding water. (b) How much kinetic energy does the squid create by this mancuver?

Short Answer

Expert verified
The water needs to be expelled at a speed of \(9.29 \mathrm{~m/s}\) for the squid to reach a speed of \(2.50 \mathrm{~m/s}\). The squid creates \(20.31 \mathrm{J}\) of kinetic energy by this maneuver.

Step by step solution

01

Understand the Conservation of Momentum Principle

The conservation of momentum states that the total momentum before and after an event must be the same if no external forces are acting on the system. In this situation, the squid expelling water is an internal force. The total momentum before the squid expels water is zero, because it's at rest. Therefore, the total momentum after the squid expels water must also be zero.
02

Calculate the Speed of the Expelled Water

Let's say \(v\) is the speed of the expelled water. Equating the initial and final momentums, we have: \(0 = 6.50 \mathrm{~kg} \times 2.50 \mathrm{~m} / \mathrm{s} + 1.75 \mathrm{~kg} \times (-v)\). Solving for \(v\), we find \(v = \frac{6.50 \mathrm{~kg} \times 2.50 \mathrm{~m} / \mathrm{s}}{1.75 \mathrm{~kg}}\). Calculating through, \(v = 9.29 \mathrm{~m/s}\).
03

Calculate the Kinetic Energy of the Squid

The kinetic energy (KE) of an object is given by the formula \(KE = 0.5 \times m \times v^2\), where \(m\) is the mass and \(v\) is the speed. Therefore, the kinetic energy of the squid is \(KE = 0.5 \times 6.50 \mathrm{~kg} \times (2.50 \mathrm{~m/s})^2 = 20.31 \mathrm{J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Animal Propulsion Physics
Imagine the remarkable sight of a squid or octopus darting through the water - this is possible thanks to the fascinating principles of animal propulsion physics. In simple terms, these creatures move by exerting a force on the water, ejecting it swiftly out of their body to push themselves in the opposite direction. This is a practical application of Newton’s Third Law of Motion, which states that for every action, there is an equal and opposite reaction. When the squid contracts its cavity to force out water, the water’s expulsion creates a propulsive force that pushes the squid forward.

Understanding this biological adaptation helps us grasp fundamental physics in a living context. It also highlights how efficient and adapted marine organisms are to their liquid environment, utilizing their own body structure to generate motion and survive in the vast ocean.
Momentum Calculation
Momentum is a key concept in physics, defined as the product of an object's mass and velocity, symbolically given as \( p = m \times v \). When dealing with problems such as our squid’s escape maneuver, calculations revolve around the law of conservation of momentum, which tells us that in a closed system with no external forces, the total momentum remains constant.

In the case of the squid, its initial momentum is zero since it's at rest. When it expels the water to escape, the momentum of the squid-plus-water system must remain zero. To calculate the velocity at which the squid ejects the water, we simply equate the momentum of the squid moving in one direction to the momentum of the expelled water in the opposite direction, using the equation \( m_1 \times v_1 = m_2 \times v_2 \), ensuring we preserve the physical law. This results in the understanding that the heavier the squid, or the faster it needs to move, the greater the speed at which it must expel water.
Kinetic Energy in Physics
Kinetic energy represents the energy an object possesses due to its motion. Calculated with the equation \( KE = 0.5 \times m \times v^2 \), it’s directly proportional to the mass of the object and the square of its velocity. This means that even a small increase in velocity results in a much larger increase in kinetic energy. In our exercise involving the squid, the kinetic energy generated during its rapid escape tells us how effective its movement is in relation to the energy expended. It quantifies the energetic cost of the squid's response to a threat - an integral aspect of understanding the dynamics of animal behavior from an energetic standpoint.

Understanding kinetic energy in the context of animal motion sheds light on the evolutionary adaptations of animals and their capacity to perform life-saving maneuvers. The ability to calculate and predict kinetic energy is essential in numerous fields, from biological sciences to engineering and beyond, illustrating just how broadly the concept applies.

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Most popular questions from this chapter

Two figure skaters, one weighing \(625 \mathrm{~N}\) and the other \(725 \mathrm{~N}\) push off against cach other on frictionless ice. (a) If the heavicr skater travels at \(1.50 \mathrm{~m} / \mathrm{s}\), how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this cnergy come from?

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of \(200 \mathrm{~kg}\) and is traveling east with a velocity of magnitude \(5.00 \mathrm{~m} / \mathrm{s}\). Find the final velocity of the car in cach casc, assuming that the handcar docs not leave the tracks. (a) A \(25.0 \mathrm{~kg}\) mass is thrown sideways out of the car with a velocity of magnitude \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the car's initial velocity, (b) \(\mathrm{A} 25.0 \mathrm{~kg}\) mass is thrown backward out of the car with a velocity of \(5.00 \mathrm{~m} / \mathrm{s}\) relative to the initial motion of the car. (c) A \(25.0 \mathrm{~kg}\) mass is thrown into the car with a velocity of \(6.00 \mathrm{~m} / \mathrm{s}\) relative to the ground and opposite in direction to the initial velocity of the car.

An ice hockey forward with mass \(70.0 \mathrm{~kg}\) is skating due north with a speed of \(5.5 \mathrm{~m} / \mathrm{s}\). As the forward approaches the net for a slap shot, a defensive player (muss \(110 \mathrm{~kg}\) ) skates toward him in order to apply a body-check. The defensive player is traveling south at \(4.0 \mathrm{~m} / \mathrm{s}\) just before they collide. If the two players become intertwined and move together after they collide, in what direction and at what speed do they move after the collision? Friction between the two players and the ice can be neglected.

A \(15.0 \mathrm{~kg}\) fish swimming at \(1.10 \mathrm{~m} / \mathrm{s}\) suddenly gobbles up a \(4.50 \mathrm{~kg}\) fish that is initially stationary, Ignore any drag effects of the Water. (a) Find the speed of the large fish just after it eats the small one. (b) How much total mechanical energy was dissipated during this meal?

\(A 12.0 \mathrm{~kg}\) shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of \(150 \mathrm{~m} / \mathrm{s}\). At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was relcased in the explosion?

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