Explanation:
its either a or d however i say the best choice is d
<u>Answer:
</u>
Cat has 2.02 seconds to right itself.
<u>Explanation:
</u>
Initial height of cat from ground = 20 meter.
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8
, we need to calculate time when s = 20 meter.
Substituting
So, cat has 2.02 seconds to right itself.
To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

= Coulomb's constant
q = Charge
r = Radius
At the same time

The values of variables are the same, then if we replace in a single equation we have this expression,

If we replace the values, we have finally that the charge is,




Therefore the potential energy of the system is 
Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.
We use only one variable at a time to find the accurate result. We want to see how the result of experiment changes everytime with a single variable.