Answer: D) Earthquake
<em>I hope this helps, and Happy Holidays! :)</em>
<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116
So coefficient of static friction = 0.6116</span>
you can collect water and shine a light though it and record your finings
To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have
Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.
As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.
