Answer:
The coefficient of kinetic friction = 0.026
Explanation:
An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N.
Here we have to note that the weight is pulled at a constant speed . This means that the net force acting on the weight is zero.
The external force acting on the body is in the forward direction and the friction acts in the backward direction.
Friction increases as the mass of the body increases.
Friction = 
We now equate this to the external force of 15 N.
15 = 
= 
= 0.026
The coefficient of kinetic friction = 0.026
The netro level 8 and 4 is the hydrogen atom 2. Hope u enjoy your grade
The maximum speed of the face is determined as 6.75 x 10⁻⁵ m/s.
<h3>
Maximum speed of the face</h3>
v = fλ
where;
- f is frequency of the wave
- λ is the wavelength
<h3>Angular speed of the wave</h3>
ω = 2πf
v(max) = ωA
v(max) = 2πfA
where;
- A is amplitude of the wave
v(max) = 2π(215)(50 x 10⁻⁹ m)
v(max) = 6.75 x 10⁻⁵ m/s
Thus, the maximum speed of the face is determined as 6.75 x 10⁻⁵ m/s.
Learn more about maximum speed here: brainly.com/question/4931057
#SPJ1
It might be radiation and reflection but I’m not sure
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
