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
Answer:
fit of the continents, paleoclimate indicators, truncated geologic features, and fossils.
Explanation:
Answer:

Explanation:
Since the two charged bodies are symmetric, we can calculate the electric field taking both of them as point charges.
This can be easily seen if we use Gauss's law, 
We take a larger sphere of radius, say r, as the Gaussian surface. Then the electric field due to the charged sphere at a distance r from it's center is given by,

which is the same as that of a point charge.
In our problem the charges being of opposite signs, the electric field will add up. Therefore,

where,
= distance between the center of one sphere to the midpoint (between the 2 spheres)
Answer:
The toy car
Explanation:
the real car is parked so yeah but maybe in some way technically the real car has more "momentum"
100 cm is 1 meter. So your answer would be 0.362 meters.