Answer:
decibels (dB)
Explanation:
The sound intensity level is a quantity derived from the sound intensity.
The intensity of a wave is defined as the power of the source of the wave divided by the area through which the power of the wave is spread, mathematically:
where
P is the power of the source
is the surface area over which the wave spreads (assuming that the wave propagates in all directions, it corresponds to the surface area of a sphere of radius 
, where r is the distance between the source of the wave and the observer)
For sound waves, the intensity is often expressed using another unit, called decibel (dB), defined as follows:
where
is the sound intensity level in decibels
I is the intensity of the sound wave
is the threshold intensity of a sound that a person can normally hear.
... then your weight is <em>25.2 lbf</em> on the moon.
Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
