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Annette [7]
2 years ago
9

Lolliguncula brevis squid use a form of jet propulsion to swim—they eject water out of jets that can point in different directio

ns, allowing them to change direction quickly. When swimming at a speed of 0.15m/s0.15m/s or greater, they can accelerate at 1.2m/s21.2m/s 2 .
(a) Determine the time interval needed for a squid to increase its speed from 0.15m/s0.15m/s to 0.45m/s0.45m/s.
(b) What other questions can you answer using the data?
Physics
1 answer:
romanna [79]2 years ago
6 0

Answer:

a) t = 0.25 s,  b)  x = 0.075 m

Explanation:

a) For this exercise we will use kinematic relationships in one dimension

         v = v₀ + a t

in the problem they indicate the initial velocity v₀ = 0.15 m / s, the final velocity v = 0.45 m / s and the acceleration of the squid a = 1.2 m / s²

          t = \frac{v -v_o}{a}

       

we calculate

         t = \frac{0.45 - 0.15}{1.2}

         t = 0.25 s

b) We can also find the distance traveled during this acceleration

         v² = v₀² + 2a x

         x = \frac{v^2 -v_o^2 }{2a}

let's calculate

         x = \frac{0.45^2 - 0.15^2 }{2 \  1.2}

         x = 0.075 m

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Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

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One of the particle is at the origin

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The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

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2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

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The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:

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This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

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