Two fixed charges, q1 = +1.07µC and q2 = -3.28µC, are 61.8cm apart. Where may a third charge be located so that no net force act
2 answers:
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If it takes 
seconds to reach the car, then the distance 
is 
.
The bear's distance from the tourist's starting point is
For maximum, we set the equations equal to each other:
so the distance is
The bear's distance from the tourist's starting point is
For maximum
so the distance is
Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/(year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg