Answer:
A) Emin = eV
B) Vo = (E_light - Φ) ÷ e
Explanation:
A)
Energy of electron is the product of electron charge and the applied potential difference.
The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;
Emin = eV
B)
The maximum stopping potential energy is eVo,
The energy of the electron due to the light is E_light.
If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy
Φ = E_light - eVo
Therefore,
eVo = E_light - Φ
Vo = (E_light - Φ) ÷ e
Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.
From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as
We are given the second smallest nonzero thickness at which destructive interference occurs.
This corresponds to, m = 2, therefore
The index of refraction of soap is given, then
Combining the results of all steps we get
Rearranging, we find
Answer:
400 W/m^2 and 31℃
Explanation:
The output heat flux q"= 20 W/m^2 (geven)
The output heat flux from.the wall to the air by convection
q"conv = h(ts - t∞)
q"conv = 20(50-30) = 400 W/m^2
Therefor, this case is unsteady and the wall temperature changes with time till the energy balance exist.
ENERGY BALANCE
The input energy must be equal to the output energy for steady state condition. If not the state will be unstaidy or transient.
2. Its noticed that the output heat flux is not that the I put heat flux, therefore the wall tempers will be decreased till the output heat flux is reduced to the value of the given input heat flux
T steady = T∞ +q"/h
= 30 + 20/20 = 31℃
It pushes the currents to opposite sides
Answer:
115 kPa
Explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
Assuming no elevation change, h₁ = h₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
Plugging in values:
(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²
P = 115,000 Pa
P = 115 kPa
