Two positive charges are labeled q Subscript 1 baseline and q Subscript 2 baseline are 0.1 m apart. Two positively charged parti
cles are separated on an axis by 0.1 meters, and q1 has a 6 µC charge and q2 has a 2 µC charge. Recall that k = 8.99 × 109 N • meters squared over Coulombs squared.. What is the force applied between q1 and q2 in N? In which direction does particle q2 want to go?
Taking square roots and simplifying, x =s /[1+(q2/q1)^0.5]
Assuming an identical distance, the rigidity of Q on 2Q is equivalent in value to the rigidity of 2Q on Q. for that reason, had the area R been stored an identical, the two forces could be equivalent. inspite of the shown fact that, via fact the area is being decreased, we could constantly consult with the equation we use to calculate those forces: F = ok(Q1xQ2)/(R^2) because R is squared and is being halved, the final result's that's it being divided by potential of a million/4. for that reason, the rigidity would be expanded by potential of four, and be 4F.
Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.
Try to chase it or try to put up missing posters if you cant him/her, i never had a dog but i would do that and ask neighbors if they know where the dog is at.
