Equation of uniformly accelerated motion
1 answer:
You might be interested in
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
To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as
Where,
d = Path difference
= wavelength
n = Any integer which represent the number of repetition of the spectrum
In this question the distance between the two source will be minimum for the case of minimum path difference, then n= 1
Therefore the minimum distance that should you separate two sources emitting the same waves is 2.5mm