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.
Use this site is better https://www.cymath.com/
To solve this problem we will apply the concepts related to the Force of gravity from Newtonian theory for which it is necessary to
Where,
G = Gravitational universal constant
M = Mass of Earth
m = mass of Object
x = Distance between center of mass of the objects.
From this equation we can observe that the Force is inversely proportional to the squared distance between the two objects. The greater the distance, the lower the force of gravity and vice versa.
If you want to increase the force of gravity, you need to reduce the distance of the two. Therefore the correct option is B. Talk to the distance between them.
Scientists study fossils of plants, animals, and other organisms in order to better understand what life was like on Earth many years ago and how it has changed over time. Fossils are important evidence for the theory of evolution.
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
<h3>Magnitude of net force on the crate</h3>
F(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
<h3>Net work done on the crate</h3>
W = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
<h3>Acceleration of the crate</h3>
a = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
<h3>Speed of the crate</h3>
v² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
Learn more about work done here: brainly.com/question/8119756
#SPJ1
