As shown in the diagram, threee equal charges are spaced evenly in a row. The magnitude of each charge is +2e, and the distance
between two adjacent charges is 1.50 nm. Then the central charge is displaced 0.350 nm to the right while the other two charges are held in place. After the displacement, what is the magnitude and direction of the net force that the outer two charges exert on the central charge?
4.27*10^-10 to the right
4.27*10^-10 to the left
3.20*10^-10 to the left
3.20*10^-10 to the right
4.27*10^-10 to the right
4.27*10^-10 to the left
3.20*10^-10 to the left
3.20*10^-10 to the right
2 answers:
You might be interested in
The first part of the text is missing, you can find on google:
"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."
Let's now solve the different parts.
(a) 3.03 s
The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by
where
h = 45 m is the initial height
g = 9.8 m/s^2 is the acceleration of gravity
When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:
(b) 18.5 m/s
For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.
The horizontal position is given by
where
is the horizontal speed, which is constant
t is the time
At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:
The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.
(c) 35.0 m/s at 58.1 degrees below the horizontal
At the impact, we know that the horizontal speed is still the same:
we need to find the vertical velocity. This can be done by using the equation
where
is the initial vertical velocity
g is the acceleration of gravity
t is the time
Substituting t = 3.03 s, we find the vertical velocity at the time of impact:
So the magnitude of the velocity at the impact (so, the speed at the impact) is
The angle instead can be found as:
so, 58.1 degrees below the horizontal.