<span>Assuming that the particle is the 3rd
particle, we know that it’s location must be beyond q2; it cannot be between q1
and q2 since both fields point the similar way in the between region (due to
attraction). Choosing an arbitrary value of 1 for L, we get </span>
<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>
Rearranging to calculate for d:
<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207 </span>
<span>
We pick the value that is > q2 hence,</span>
d = 2.72075922005613*L
<span>d = 2.72*L</span>
Answer:
Y = mx + b
y = how far up
x = how far along
m = Slope or Gradient (how steep the line is)
b = value of y when x=0
Answer:
0
Step-by-step explanation:
-0.1(40)^2+4(40)
-160+160
0
The answer is d 25% of 18,250 = <span>4562.5.
Make me the brainliest person</span>
Answer:
₹20543.75
Hope you got your answer.
