Answer:
B. Maximum velocity of ejected electrons.
Explanation:
The ejection of electrons form a metal surface when the metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength or higher frequency (or equivalently, above a threshold frequency), which leads to the enough energy of the wave to incident and get absorbed to the exposed surface emits electrons. This phenomenon is known as the photoelectric effect or photo-emission.
The minimum amount of energy required by a metal surface to eject an electron from its surface is called work function of metal surface.
The electrons thus emitted are called photo-electrons.
The current produced as a result is called photo electricity.
Energy of photon is given by:
where:
h = Planck's constant
frequency of the incident radiation.
Answer:
If you use P = 2 * pi * (L / g)^1/2 for the period of the simple pendulum
g = 4 * pi^2 * 1.2 / 2.8^2 = 6.04 m/s2
Note: omega = 2 pi * f = 2 pi / P and omega = (g / L)^1/2
Answer:
Em₀ = U = m g h
, Em_{f} = K = ½ m v²
Explanation:
When a car is on a ramp it has a certain amount of mechanical energy. At the highest point of the ramp the mechanical energy is fully potential given by
Em₀ = U = m g h
As part of this energy descends down the ramp, part of this energy is transformed into kinetic energy and has one part of each, even though the sum remains the initial energy
Em = K + U = ½ m v² + mg y
y <h
when it reaches the bottom of the ramp it has no height therefore there is no potential energy, all of it has been transformed into kinetic energy
Em_{f} = K = ½ m v²
This energy transformation is in the case that the friction force is zero.
If there is a friction force, it performs work against the low car, it is reflected in an increase in the internal energy (temperature) of the car. In this case the energy in the lower part is less than the initial one by a factor
= - fr L
therefore the numeraire values of the velocity are lower, due to the energy lost by friction.
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
where
Now we find the moment of inertia by integrating from
to
The moment of inertia is
(from (-a/2) to
(a/2))
It can be observed without changing the identity of the object.
