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:
the answer is at the BOTTOM OF THEIR QUESTION
Explanation:
IT IS CORRECT BTW
Answer:
t = 7,8 s
Explanation:
From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .
When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit
Then for the cat arrives at 3,9 m/s nedds
v = vo + a*t vo = 0 then v = a*t
3,9 ( m/s) = 0,5 ( m/s² ) * t
t = 7,8 s
v = 3,9 m/s =
Answer:
Yes.
Explanation:
Newton's first law says that an object in motion stays in motion and an object at rest stays at rest until acted upon by an unbalanced force.
If an object in motion has balanced forces, it will stay in motion. For example, if an object is falling at terminal velocity (for example, a parachuter), then the force of gravity is equal and opposite to the force of air resistance. The forces are balanced, and the object continues to fall at a constant speed.