To solve this problem it is necessary to apply the concepts related to the capacitance in the disks, the difference of the potential and the load in the disc.
The capacitance can be expressed in terms of the Area, the permeability constant and the diameter:
Where,
= Permeability constant
A = Cross-sectional Area
d = Diameter
Potential difference between the two disks,
V = Ed
Where,
E = Electric field
d = diameter
Q = Charge on the disk equal to 
Through the value found and the expression given for capacitance and potential, we can define the electric charge as
Re-arranging the equation to find the diameter of the disks, the equation will be:
Replacing,
Therefore the diameter of the disks is 0.03m
You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.
T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N
Answer:Arctic circle
Explanation:
Arctic circle is one of the circles of latitude. It is located at the northern region if the equator and passes through more than one continent. These continent includes: North Asia,
Northern America, and
Europe.
In these continent, it passes through eight known counties. These countries includes: Canada, Russia, Norway, Sweden, Finland, Denmark, Alaska and Iceland. It covers 4% of the Earth surface and it's climatic conditions are extreme.
Deposition is the process in which sediments, soil and rocks are added to a landform or landmass. When previous weathers surface material , is deposited to a building layer of sediment .
Answer:
(a) The energy of the photon is 1.632 x 
J.
(b) The wavelength of the photon is 1.2 x 
m.
(c) The frequency of the photon is 2.47 x 
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
- = -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x 
)
- = 1.632 x Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x 
Js), c is the speed of light (3 x 
m/s) and λ is the wavelength.
10.20(1.6 x ) = (6.6 x * 3 x )/ λ
λ = 
= 1.213 x
Wavelength of the photon is 1.2 x m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x = 6.6 x x f
f = 
= 2.47 x Hz
Frequency of the emitted photon is 2.47 x Hz.
