Answer:
Explanation:
If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.
n = c / v
If light travels faster in warmer air, it will have a lower refractive index
nh < nc
Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
sin(θ2) = sin(θ1) * n1/n2
If blue light from the sky passing through the hot air will cross to the cold air, then
n1 = nh
n2 = nc
Then:
n1 < n2
So:
n1/n2 < 1
The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.
55 meters. If she started at 10 meters and ran 45 more, 10+45=55.
Answer:
Explanation:
<u>Diagonal Launch
</u>
It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.
The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is
Where vo is the magnitude of the initial velocity, 
is the angle, t is the time and g is the acceleration of gravity
The maximum height the object can reach can be computed as
There are two times where the value of y is 
when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making 
Removing and dividing by t (t different of zero)
Then we find the total flight as
We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is
Answer:
Inverted (displaced downwards)
Explanation:
The pulse becomes INVERTED upon reflecting off the boundary with the wall. That is, an upward-displaced pulse will reflect off the end and return with a downward displacement. This inversion behavior will always be observed when the end of the medium is fixed, like this wall in this instance. This INVERSION BEHAVIOR can also be observed when the medium is connected to another more heavy or more dense medium. And in this case, when the pulse reaches the end of the medium, a portion of the pulse will reflect off the end and return with an inverted displacement. The heavier medium acts like a fixed end to cause the pulse to be inverted.
Summary: a pulse reaching the end of a medium becomes inverted whenever it either:
i. reflects off a fixed end,
ii. is moving in a less dense medium and reflects off a more dense medium.
