15. Calculate the critical angle of glass and water combination. Show your calculation. 16. What is the critical angle for the interface between Mystery A and glass

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j