Well 200 doubled or (x2)=400 if that’s what it means
Answer:
103239.89 days
Explanation:
Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
a³ / T² = 7.496 × 10⁻⁶ (a.u.³/days²)
where,
a is the distance of the semi-major axis in a.u
T is the orbit time in days
Converting the mean distance of the new planet to astronomical unit (a.u.)
1 a.u = 9.296 × 10⁷ miles

Substituting the values into Kepler's third law equation;
(days)²

T = 103239.89 days
An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days
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)