Answer:
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Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Empirical formula of compound is XF3
Compound consist of 65% F
In 100g of compound there is 65 g of F
= 65 / 19 moles of Fluorine = 3.421 moles
So moles of X = 3.421 / 3 = 1.140 moles
And in 100 g X
consist of 35 g
So the molar mass of X = 35 / 1.140 = 30.71 g = 31
approximately
And it is the mass of phosphorus
So the empirical formula for the compound is PX3
Answer:
But there are ways to harness kinetic energy to either generate useful mechanical work or electricity. This is what many have tried to do to make use of energy that would be otherwise wasted. One way to harness kinetic energy that has popped up many times in recent years has to do with roads and speed bumps
Explanation: