Classical physics considered light to behave as a wave in all environments; it had a set amplitude, frequency etc. The problem was that this meant that there was a continuous variation in its properties, hence if the amplitude of the light was incresed by a bit, a phenomenon like the phhotoelectric one would become only marginally more apparent. However, in this case, there is a cutoff point which means that the only-wave theory had to be wrong.
Answer:
The average velocity is 2 m/s.
Explanation:
The velocity of the particle is the time derivative of its position 
![$v =\frac{dx(t)}{dt} = \frac{d}{dt}[t^2+2] $](https://tex.z-dn.net/?f=%24v%20%3D%5Cfrac%7Bdx%28t%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%5Bt%5E2%2B2%5D%20%24)

Now the average from
and
is

Thus, the average velocity is 2 m/s.
It will be 80 miles and it can be done only in 16 min
Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :

Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :

For the position variation we use the vertical position equation :

Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is

For the second time interval :
t1 = 4 s → t2 = 9 s


Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :

Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then


The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is

When an earthquake strikes usually seismic waves carry the energy outward (a.) With the greatest power at the focus and as the seismic waves travel outward they become more and more weak until they disappear.<span />