<span>a+b= ?
3i +3j + (3i -3j) = ?
3i + 3j + 3i -3j =?
= 6i + 0j</span>
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:

Explanation:
According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.
Being
and
the masses of pucks a and b respectively, the initial momentum of the system is

Since b is initially at rest

After the collision and being
and
the respective velocities, the total momentum is

Both momentums are equal, thus
Solving for 


The initial kinetic energy can be found as (provided puck b is at rest)


The final kinetic energy is


The change of kinetic energy is

Answer:
-0.0789 m
Explanation:
Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:
y = 0.250 sin(-18.4) ≈ -0.0789 m
Notice it is negative since it is below the x-axis.