A ball is dropped from a tower that is 206 m high, How long does it take to reach the ground?
1 answer:
It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.
From the question given above, the following data were obtained:
Height (H) = 206 m
<h3>Time (t) =? </h3>NOTE: Acceleration due to gravity (g) = 10 m/s²
The time taken for the ball to get to the ground can be obtained as follow:
H = ½gt²
206 = ½ × 10 × t²
206 = 5 × t²
Divide both side by 5
Take the square root of both side
Therefore, it will take 6.42 s for the ball to get to the ground.
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1) Medium "b" has more optical density
2) Light must hit the interface between the two mediums perpendicularly
Explanation:
1)
Refraction occurs when light propagates from a medium into a second medium.
The optical density of a medium is given by its index of refraction, which is defined as:
where
c is the speed of light in a vacuum
v is the speed of light in a medium
Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.
In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.
2)
We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:
where
are the index of refraction of the two mediums
are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)
Here we want the direction of propagation of the light ray not to change: this means that it must be
(1)
However, here we have two mediums "a" and "b" with different index of refraction, so
Therefore the only angle that can satisfy eq.(1) is
So, the light must hit the surface perpendicular to the interface between the two mediums.
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