Something that there is only so many of or not enough of
Answer:
Work= -7.68×10⁻¹⁴J
Explanation:
Given data
q₁=q₂=1.6×10⁻¹⁹C
r₁=2.00×10⁻¹⁰m
r₂=3.00×10⁻¹⁵m
To find
Work
Solution
The work done on the charge is equal to difference in potential energy
W=ΔU
![Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J](https://tex.z-dn.net/?f=Work%3DU_%7B1%7D-U_%7B2%7D%5C%5C%20Work%3D-kq_%7B1%7Dq_%7B2%7D%5B%5Cfrac%7B1%7D%7Br_%7B2%7D%7D-%5Cfrac%7B1%7D%7Br_%7B1%7D%7D%20%5D%5C%5CWork%3D%28-9%2A10%5E%7B9%7D%29%2A%281.6%2A10%5E%7B-19%7D%20%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7B3.0%2A10%5E%7B-15%7D%20%7D-%5Cfrac%7B1%7D%7B2%2A10%5E%7B-10%7D%20%7D%20%5D%5C%5C%20%20Work%3D-7.68%2A10%5E%7B-14%7DJ)
The equivalent of the Newton's second law for rotational motions is:

where

is the net torque acting on the object

is its moment of inertia

is the angular acceleration of the object.
Re-arranging the formula, we get

and since we know the net torque acting on the (vase+potter's wheel) system,

, and its angular acceleration,

, we can calculate the moment of inertia of the system:
Answer:
1 through 92 occur naturally on Earth