Answer:
Use the method on the image and solve it.
Answer:
1) d
2) 5 m/s
3) 100
Explanation:
The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:
(i) 
The equation for velocity v and a constant acceleration a is:
(ii) 
1) Solve equation (ii) for acceleration a and plug the result in equation (i)
(iii) 
(iv) 
Simplify equation (iv) and use the given values v = 0, x₀ = 0:
(v) 
2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:
3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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Answer:
5 Days to Seconds = 432000
Explanation:
Yes I'm pretty sure you can