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Annette [7]
3 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]3 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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A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th
inn [45]

Answer:

|\vec r|=339.82\ m

\theta=-6.67^o

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

\vec r=\vec r_1+\vec r_2

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

x=rcos\theta

y=rsin\theta

And the vector is expressed as

\vec z==

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

\vec r_1==\ km=\ m

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

\vec r_2==\ m

The total displacement is

\vec r=\ m+\ m

\vec r=\ m

In (magnitude,angle) form:

|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m

\boxed{|\vec r|=339.82\ m}

\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169

\boxed{\theta=-6.67^o}

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