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
(D) The gravitational force between the astronaut and the asteroid.
Reason :
All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.
Answer: search it on browser
The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.
Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).
At the beginning ... time=0 ... the position is 1 meter.
At the end ... time=5 ... the position is zero.
The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.
Honed I don’t know where the question is
