Answer:
Salt can dissolve in water making it an excellent soluble for water
To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.
The angular velocity can be described as

Where,
Final Angular Velocity
Initial Angular velocity
Angular acceleration
t = time
The relation between the tangential acceleration is given as,

where,
r = radius.
PART A ) Using our values and replacing at the previous equation we have that



Replacing the previous equation with our values we have,




The tangential velocity then would be,



Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

Replacing with our values and re-arrange to find 



That is equal in revolution to

The linear displacement of the system is,



Electrons: negative charge
Protons: positive charge
Neutrons: negative charge
The atom would have to have more electrons than protons
Hope this helps :)
This can be solve using the formula P = I^2 * Rwhere P is the powerI is the CurrentR is the resistanceP = I^2 * R
1/4 Watt = I^2 * 100 ohm solve for II^2 = 1/400 I = 0.05 amps then using the formula to solve for the voltage:V = I * RV = 0.05 amps * 100 ohms V = 5 volts