Answer:
Equation for SHM can be written
V = w A cos w t where w is the angular frequency and the velocity is a maximum at t = 0
V1 = w1 A cos w1 t
V2 = w2 A cos w2 t
V2 / V1 = w2 / w1 since cos X t = 1 if t = zero
V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2
If the velocity is twice as large the period will be 1/2 long
Kinetic energy increases, potencial energy decreases,
kinetic energy + potential energy = energy, energy can not be destroyed, just transformed
Bumper of a stationary bumper car. The momentum of the
stationary car increases. Which happens to the momentum of the moving bumper
car? It decreases. It stays the same. It is converted to inertia.
Bumper of a stationary bumper car. The momentum of the
stationary car increases. The momentum of the moving bumper car It is converted
to inertia.
Answer:
Explanation:
E=(σ/ε0)
As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."
Answer:

Explanation:
<u>Motion in The Plane</u>
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of
and
as the initial speed and angle, then we have




If we want to know the maximum height reached by the object, we find the value of t when
becomes zero, because the object stops going up and starts going down

Solving for t

Then we replace that value into y, to find the maximum height

Operating and simplifying

We have

The maximum height is

