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mixer [17]
3 years ago
5

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

ng the cavity to force out the water through an opening. A 6.50 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. Part A If the squid has 1.55 kg of water in its cavity, at what speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator
Physics
1 answer:
Zigmanuir [339]3 years ago
4 0

Answer:

10.1 m/s

Explanation:

By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.

And by the law of conservation of momentum,

initial momentum of squid + expelled water = final momentum of squid + expelled water.

Now, the initial momentum of the system is zero.

So, 0 = final momentum of squid + expelled water

0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled

So, MV + mv = 0

MV = -mv

v = -MV/m

= -6.50 kg × 2.40 m/s ÷ 1.55 kg

= -15.6 kgm/s ÷ 1.55 kg

= -10.1 m/s

So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s

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A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude
kondaur [170]

Answer:

81.8 m/s

Explanation:

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v_0=115 m/s (toward east)

So, decomposing along the x- and y- directions:

v_{x0} = 115 m/s\\v_{y0} = 0

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

a=2.88 m/s^2 (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2

So the two components of the velocity after a time t = 25.0 s will be

v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s

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3 years ago
in three situations, a briefly applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice.
victus00 [196]

The work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

F= \frac{mv}{t}

The force applied to an object increases with increases in the velocity of the object.

In the given diagram, the resultant velocity of the puck is calculated as follows;

Figure a:

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Figure b:

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Figure c:

\Delta v = 4 - (-2)\\\\\Delta v = 6 \ m/s

Thus, the work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

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