Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = - 1.4 m
Firstly models are just predictors, they are not exact models. Exponential growth <span>models are good when populations are small relative to the amount of resources available. ... This is where the </span>exponential<span> model breaks down. The </span>logistic<span> function tries to compensate for this with the carrying capacity.</span>
Answer:
The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J
Explanation:
When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.
The relationship between the maximum kinetic energy (
) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:
= h(f − f₀)
= hf - hf₀
E is the energy of the absorbed photons: E = hf
ϕ is the work function of the surface: ϕ = hf₀
= E - ϕ
Frequency f = 8.12×10¹⁸ Hz
Maximum kinetic energy = 4.16×10⁻¹⁷ J
Speed of light c = 3 x 10⁸ m/s
Planck's constant h = 6.63 × 10⁻³⁴ Js
E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸
E = 53.8356 x 10⁻¹⁶ J
from = E - ϕ ;
ϕ = E -
ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷
ϕ = 53.4196 x 10⁻¹⁶ J
The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J
travel through a vacuum at the speed of light. Other waves need a medium; sound waves need molecules that vibrate.
