Answer:
21.5°
Explanation:
Given,
Refractive index of water, n₁ = 1.33
Refractive index of polystyrene, n₂ = 1.49
Angle of reflection = ?
Angle of refraction = 19.1°
Using Snell's law
n₁ sin θ₁ = n₂ sin θ₂
1.33 x sin θ₁ = 1.49 x sin 19.1°
sin θ₁ = 0.366
θ₁ = 21.5°
According to law of reflection angle of incidence is equal to angle of reflection.
Angle of reflection = 21.5°
Answer:
Distance is 50m
Displacement is 0m
Explanation:
Distance is based on the amount of length you covered, regardless of where you end.
Displacement only considered where you started and where you ended, which is at the same spot in this case. Therefore, no displacement.
Answer:
486nm
Explanation:
in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that
![\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bwavelength%7D%3DR.%5B%5Cfrac%7B1%7D%7Bn_%7Bf%7D%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7Bn_%7Bi%7D%5E%7B2%7D%20%7D%5D%20%5C%5Cn_%7Bf%7D%3D2%5C%5Cn_%7Bi%7D%3D4%5C%5CR%3DRydberg%20constant%20%3D1.097%2A10%5E%7B7%7Dm%5E%7B-1%7D%5C%5Csubtitiute%20%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D1.097%2A10%5E%7B7%7D%5B%5Cfrac%7B1%7D%7B2%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%7D%7B4%5E%7B2%7D%7D%5D%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%201.097%2A10%5E%7B7%7D%2A0.1875%5C%5C%5Cfrac%7B1%7D%7Bwavelength%7D%3D%202.06%2A10%5E%7B6%7D%5C%5Cwavelength%3D4.86%2A10%7B-7%7Dm%5C%5Cwavelength%3D%20486nm%5C%5C)
Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state
Answer:
Hz
Explanation:
We know that
1 cm = 0.01 m
= Length of the human ear canal = 2.5 cm = 0.025 m
= Speed of sound = 340 ms⁻¹
= First resonant frequency
The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as
for first resonant frequency, we have n = 1
Inserting the values
Hz
Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
