To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.
PART A) Normal Force.
Here,
Normal reaction of the ring is N and velocity of the ring is v
PART B) Acceleration
Negative symbol indicates deceleration.
<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>
Answer:
v = 6.45 10⁻³ m / s
Explanation:
Electric force is
F = q E
Where q is the charge and E is the electric field
Let's use Newton's second law to find acceleration
F- W = m a
a = F / m - g
a = q / m E g
Let's calculate
a = -1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8
a = 0.228 10² -9.8
a= 13.0 m / s²
Now we can use kinematics, knowing that the resting parts electrons
v² = v₀² + 2 a y
v =√ (0 + 2 13.0 1.6 10⁻⁶)
v = 6.45 10⁻³ m / s
If both particles have the SAME electrical charge, then they repel.
If they have DIFFERENT electrical charge, then they attract.
Protons have + charge .
Electrons have - charge .
So two protons (A) or two electrons (D) push apart.
One proton and one electron (C) pull together.
I'd say B.) Increasing the voltage of the battery.