a content creator because if i was a rapper i probably wouldn't make good songs lol
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:
The correct option is;
c. 22.6
Explanation:
The given parameters are;
The hypotenuse of the vector = 32
The angle of the vector = 45°
Therefore, the vector component in the y-axis is given as follows;

Substituting the values from the question gives;

The vector component in the y-axis,
, is approximately 22.6.
Given Information:
Angular displacement = θ = 51 cm = 0.51 m
Radius = 1.8 cm = 0.018 m
Initial angular velocity = ω₁ = 0 m/s
Angular acceleration = α = 10 rad/s
²
Required Information:
Final angular velocity = ω₂ = ?
Answer:
Final angular velocity = ω₂ = 21.6 rad/s
Explanation:
We know from the equations of kinematics,
ω₂² = ω₁² + 2αθ
Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.
ω₂² = (0)² + 2αθ
ω₂² = 2αθ
ω₂ = √(2αθ)
We know that the relation between angular displacement and arc length is given by
s = rθ
θ = s/r
θ = 0.51/0.018
θ = 23.33 radians
finally, final angular velocity is
ω₂ = √(2αθ)
ω₂ = √(2*10*23.33)
ω₂ = 21.6 rad/s
Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.