Answer:

Explanation:
From conservation of energy states that

Scientists conduct experiments in order to prove a theory or a prediction they have or contradict it, so that then they can write down their results to study them.
Answer:
v =
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:


Answer:
h=2.86m
Explanation:
In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

There is no kinetic energy in the initial state, nor potential energy in the end,

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

The inertia of the bodies is given by the equation,



On the other hand the angular velocity is given by

Replacing these values in the equation,

Solving for h,
