Answer:
(A) As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.
Mathematically:
where:
r = distance between the charges
permittivity of free space
By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.
Answer:
The position of the particle is -2.34 m.
Explanation:
Hi there!
The equation of position of a particle moving in a straight line with constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the particle at a time t:
x0 = initial position.
v0 = initial velocity.
t = time
a = acceleration
We have the following information:
x0 = 0.270 m
v0 = 0.140 m/s
a = -0.320 m/s²
t = 4.50 s (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).
Then, we have all the needed data to calculate the position of the particle:
x = x0 + v0 · t + 1/2 · a · t²
x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²
x = -2.34 m
The position of the particle is -2.34 m.
Is there any types of answer to get an idea
