Answer:
W = -0.480 J
Explanation:
given,
q₁ = 4 μC
q₂ = -4.10 μC


b = 0.381
k = 8.99 × 10⁹ Nm²/C²

![W = [-147.436\times (5.88-2.62)\times 10^{-3}]J](https://tex.z-dn.net/?f=W%20%3D%20%5B-147.436%5Ctimes%20%285.88-2.62%29%5Ctimes%2010%5E%7B-3%7D%5DJ)
W = -0.480 J
Work done by the electric force W = -0.480 J
Explanation:
i hope this helps, its not the same person but its the same equation.
I had this question on a quiz, it was X.
Answer:
a)
, b)
, c)
,
, d) 
Explanation:
a) The gravitational force exerted by the Earth on the satellite is:



b) The centripetal acceleration of the satellite is:


c) The speed of the satellite is:



Likewise, the angular speed is:


d) The period of the satellite's rotation around the Earth is:


Explanation:
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