3 A picture is supported by two vertical strings; if the weighi
1 answer:
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They were going at a velocity 4.07m/s
<u>Explanation:</u>
Distance s =5 m
initial velocity u= 0.8 m/s
Acceleration a =1.6m/s2
We have to calculate the velocity with which they were going afterwards i.e final velocity.
Use the equation of motion
They were going with a velocity 4.07 m/s afterwards.
The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force 
which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
, 
. Re-arranging, we can solve this equation to find 
, the coefficient of dynamic friction:
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
Answer:
x = -3 cm
Explanation:
The electrical potential is the sum of the potentials of each charge
V = k ∑
let's apply this to our case where the potential is V = 0 for x = 0
0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))
in our case
q₁ = + 2.0 10⁻⁶ C
q₂ = - 6.0 10⁻⁶ C
q₃ = + 3.0 10⁻⁶ C
x₁ = -1.0 cm = 1.0 10⁻² m
x₂ = +2.0 cm = 2.0 10⁻² m
we substitute in the equation
0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + 3 10⁻⁶ / x)
3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴
3 10⁻⁶ / x = -1 10⁻⁴
x = - 3 10⁻² m
x = -3 cm