An 16.00-kg object is suspended by a rope at some height, h if the object has a potential energy of 900.01j what is the objects
2 answers:
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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.
Learn more about refraction:
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Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
