Answer:
a} <u>Infrared radiant heaters.</u>
<u>b} Fan heaters.</u>
Explanation:
I hope this is okay, should I give more.
The refractive index for glycerine is

, while for air it is

.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:

where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
NOTE: The given question is incomplete.
<u>The complete question is given below.</u>
A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.
Solution:
Speed of yellow light in water (v) = 2.00 x 10⁸ m/s
Refractive Index of water with respect to air (μ) = 4/3
Refractive Index = Speed of yellow light in air / Speed of yellow light in water
Or, The speed of yellow light in air = Refractive Index × Speed of yellow light in water
or, = (4/3) × 2.00 x 10⁸ m/s
or, = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s
Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.
Answer:
A
Explanation:
its oxygen I hope i am correct have a great day.
F = m₁ a₁ = m₂ a₂
if m₁ = m and m₂ = 2m :
F = ma₁ = 2m a₂ ⇒ a₁ = 2 a₂
since v = at + v₀ with t = 3, v₀ = 0 ⇒ v = 3a:
v₁ = 2 v₂
since p = vm with v₁ = 2v and v₂ = v :
p₁ = v₁m₁ = 2v ⁻ m
p₂ = v₂m₂ = v ⁻ 2m
p₁ = p₂