Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
Answer: 4.7rad
Explanation:
Angular displacement =s/r
Where s=distance traveled
r=radius
Angular displacement =141m/30m
Angular displacement =4.7rad.
Absorption occurs when <span>all of the energy from light waves is transferred to a medium! </span>
Answer:
Therefore, the moment of inertia is:
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
(1)
Where:
- I is the moment of inertia
- M is the mass of the pendulum
- d is the distance from the center of mass to the pivot
- g is the gravity
Let's solve the equation (1) for I
Before find I, we need to remember that
Now, the moment of inertia will be:
Therefore, the moment of inertia is:
I hope it helps you!
