I think it is c density and temperature
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:
0.4
Explanation:
F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.
Since the mass is moving with a constant velocity, there's no acceleration hence
where N is the weight of object and \mu is coefficient of kinetic friction.
the subject

Substituting F for 8 N and N for 20 N

Therefore, coefficient of kinetic friction is 0.4
Answer:
Balanced force
Explanation:
Balanced Forces, When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. ... Although friction is acting on the person, there is no change in velocity and friction is not a net force in this case. Friction is only a net force if it changes the velocity of a mass.