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
Length of the pipe = 0.39 m
Number of harmonics = 3
Now there are 3 loops so here we can say


now here at the center of the pipe it will form Node
we need to find the distance of nearest antinode
So distance between node and its nearest antinode will be


So the distance will be 6.5 cm
Answer:
The first
Explanation:
Cause it will continue in motion till another force is applied
Answer:
no where is the main part of the question dude
I’m not sure but I think it’s
△ m=5 and △= -3 and so
Answer: 5/△-3 m/s
So sorry if it’s wrong