coulomb's law
f repulsive = q1q2/4 pi epsilon nought r squared
0.1= q1q2/4 pi epsilon nought 0.911 squared
q1+q2= 7.50 µC
0.1= q1(7.5µC-q1)/4 pi epsilon nought 0.911 squared
solve for q1
Force decreases as distance increases
It could be the third and the first one.
Answer:
F = 85696.5 N = 85.69 KN
Explanation:
In this scenario, we apply Newton's Second Law:

where,
F = Upthrust = ?
m = mass of space craft = 5000 kg
g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)
g = 15.0093 m/s²
a = acceleration required = 2.13 m/s²
Therefore,

<u>F = 85696.5 N = 85.69 KN</u>