In order to know how long it has been falling for you take the final velocity "49m/s" and divide it by the acceleration "9.8m/s" and get 5, since you have been using seconds in the calculations the answer is 5 seconds. (Fun fact, it is actually 9.82m/s per second since it accelerates and they rounded it down.)
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x
m/s)
f= 3 x
/ 2.4
f=1.25 x
hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x
=> 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m
Because, they have a strong argument regarding it and, the judges, so it makes sense of what the society thinks, and what they are representing.
Answer:
density
Explanation:
mecurys is significantly smaller and less massive than earth
KE= .5*M*V^2
.5*.06*50^2
=75 joules