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
I think your answer is D
because you already said its not a B is a liquid and not a solid so its harder to have a chemical change C is solutions and D is compound and I know for a fact that a compound is a solid and if C is also a solid then that could also be a possible answer choice.
I would say B because it is near the ocean which can cause a tsunami but also because of the wind coming from the ocean (it might cause hurricanes and lots of storms) I’m not sure though but that’s what I think makes sense. Good Luck!
Answer:
The momentum of bath cars is 40000 Ns which make the difficulty to stop each car in aspect of fprce is the same.
Explanation:
Momentum (P) =mass(m) × velocity (v)
For car A,
P = m × v = 1000 × 40 = 40000 Ns
For car B,
P = m × v = 4000 × 10 = 40000 Ns
Force (F) = Momentum change(ΔΡ)/ time taken(t)
F = ΔΡ/t
When stopping the car the momentum changes from 40000 Ns to 0
So momentum change in both cars is the same. So to stop the two cars in a given time (t) you need the same force, which means you will feel same difficulty.
I think its a higher frequency
