Answer: 1.33
Explanation:
We would apply Snell's law which is expressed as
niSinθi = nrSinθr
where
θi = angle of incidence
θr = angle of refraction
ni = index of refraction of the incident medium(air)
nr = index of refraction of the refractive medium(liquid in this case)
From the information given,
ni = 1(index for air)
θi = 37
θr = 27
By substituting these values into the formula, we have
1 * sin37 = nr * sin27
nr = sin37/sin27
nr = 1.33
The index of the liquid is 1.33
I think <span>its C
hope it helps
</span>
Answer:
??????¿?????????????????????¿
Answer:
The size of the hole will increase
Explanation:
Heat is a form of energy that depends to a great extent on the temperature of the system,when a solid is heated there will be increases in it's length,area and volume otherwise known as linear,beta and superficial expansivities respectively.When Annette increased the temperature of the ring she supplied heat to the ring that led to the expansion of the ring,what really happened was that she increased the area expansivities of the ring which is the increase in area per unit area per degree rise in temperature,at such the interior diameter of the ring increase to such an extent that the ball passes through it.
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad
