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:
a) 107.1875 Hz
b) 214.375 Hz
c) 321.5625 Hz
Explanation:
L = length of the open organ pipe = 1.6 m
v = speed of sound = 343 m/s
f = fundamental frequency
fundamental frequency is given as
inserting the values
Hz
b)
first overtone is given as
f' = 2f
f' = 2 (107.1875)
f' = 214.375 Hz
c)
first overtone is given as
f'' = 3f
f'' = 3 (107.1875)
f'' = 321.5625 Hz
Answer:
Explanation:
Let the charge particle have charge equal to +q .
force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )
force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.
F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )
= (Bqv) - j
= - Bqv j
The force will be along - ve y - direction .
If we take charge as negative or - q
force due to electric field will be along - y axis .
magnetic force = F = -q ( v i x B k )
= + Bqv j
magnetic force will be along + y axis
So it is difficult to find out the nature of charge on the particle from this experiment.
