Answer:
Light's angle of refraction = 37.1° (Approx.)
Explanation:
Given:
Index of refraction = 1.02
Base of refraction = 1
Angle of incidence = 38°
Find:
Light's angle of refraction
Computation:
Using Snell's law;
Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction
Sin38 / Light's angle of refraction = 1.02 / 1
Sin[Light's angle of refraction] = Sin 38 / 1.02
Sin[Light's angle of refraction] = [0.6156] / 1.02
Sin[Light's angle of refraction] = 0.6035
Light's angle of refraction = 37.1° (Approx.)
Answer:
The temperature is 2541.799 K
Explanation:
The formula for black body radiation is given by the relation;
Q = eσAT⁴
Where:
Q = Rate of heat transfer 56.6
σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)
A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²
e = emissivity = 0.288
T = Temperature
Therefore, we have;
T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴
T = 2541.799 K
The temperature = 2541.799 K.
Answer:
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)
A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.
Answer:
The charge resides on the outer surface = 
C
Explanation:
Surface area of cell 
Separation between two plate 
Dielectric constant 
Potential difference 
The capacitance of parallel plate capacitor in free space is given by,
Where 
permittivity of free space = 
The Capacitance of capacitor is increase by 
times when it placed in dielectric medium.
And we know that, 
So charge on the outer surface is given by,
