What is unique about the movement of uranus compared to the other outer planets
2 answers:
You might be interested in
Answer:
a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b) The maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) The maximum magnitude of the acceleration of the sphere is = 30.93
The maximum acceleration occurs at maximum excursion from equilibrium.
Explanation:
Given that :
Frequency (f) = 4.80 Hz
Amplitude (A) = 3.40 cm
a)
The total distance traveled by the sphere during one cycle of simple harmonic motion is:
d = 4A (where A is the Amplitude)
d = 4(3.40 cm)
d = 13.60 cm
Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm
b)
As we all know that:
Differentiating the above expression with respect to x ; we have :
Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;
Then:
We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium
substituting for
in the above expression;
Therefore, the maximum speed is = 102.54 cm/s
The maximum speed occurs at maximum excursion from equilibrium.
c) Again;
By differentiation with respect to t;
The maximum acceleration of the sphere is;
where;
where A= 3.40 cm = 0.034 m
The maximum magnitude of the acceleration of the sphere is = 30.93
The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of
a) Zinc (work function: 4.3 eV)
The equation for the photoelectric effect is:
(1)
where
is the energy of the incident photon, with
h = Planck constant
c = speed of light
= wavelength
= work function of the metal
K = maximum kinetic energy of the photoelectrons emitted
The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,
So we can rewrite (1) as
where we have:
V = 1.93 V
e is the electron charge
First of all, let's find the energy of the incident photon:
Converting into electronvolts,
And now we can solve eq.(1) to find the work function of the metal:
so, the metal is most likely zinc, which has a work function of 4.3 eV.
b) The stopping potential is still 1.93 V
Explanation:
The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.
Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.