A=vf-vi / time
* vi here is zero as the runner started from rest*
so
a= 10-0/5 which equals 2 m/s^2
Explanation:
1 nano-meter radiations are very difficult to observe from the ground based telescope because most of this range of radiation is absorbed through ozone layer. A very small amount of this range escape out of the ozone layer. This remaining few radiations are very difficult to track from the ground base telescope.
1 millimeter range of radiation comprises of infrared. It has range from 710 nano-meter to 1 millimeter. Infrared radiation can be easily absorbed from water and carbon di oxide molecules present in the atmosphere. So, it is absorbed by water and carbon di oxide molecules in the atmosphere. Thus, it is difficult to observe from the ground based telescope.
100 meter radiations are are radio-waves. The charged particle present in the uppermost layer of atmosphere absorbs these radio waves. So, these waves are absorbed by charged particle in the upper atmosphere. Thus, it is difficult to observe from the ground based telescope.
It shows that acceleration of particle M is directly proportional to its displacement and its direction is opposite to that of displament. Thus particle M performs simple harmonic motion but M is projection of particle performing U.C.M. hence S.H.M. is projection of U.C.M. along a diameter, of circle.
Using the following formula for linear-motion, the missing variable can be solved:
s = Vi * t + 1/2 (a * t^2)
Where: s = displacement = 29.7 m
Vi = initial velocity = 8.5 m/s
a = acceleration = 9.8
t = time = ?
Substituting:
29.7 = 8.5t + 1/2 (9.8*t^2)
29.7 = 8.5t + 4.9t^2
Dividing both sides by 4.9:
6.06 = 1.73t + t^2
t^2 + 1.73t - 6.06 = 0
(t - 1.74)(t + 3.48) = 0
t = 1.74s
From the above values, the correct answer is 1.74 seconds.
