Answer:
The new angular speed of the merry-go-round is 8.31 rev/min.
Explanation:
Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):
(1)
The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:
(2)
with I the moment of inertia and ωi the initial angular speed of the merry-go-round
The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):
(3)
The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:
(4)
with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:
(5)
(note that the angular speed is the same as the merry-go-round)
using (5) on (4), and (4) on (3):
(6)
By (5) and (2) on (1):
Solving for ωf (12.0 rev/min = 1.26 rad/s):
![\omega_f= \frac{I\omega_i}{]I+mR^2}=\frac{(260)(1.26)}{260+(24.0)(2.20)^2}](https://tex.z-dn.net/?f=%5Comega_f%3D%20%5Cfrac%7BI%5Comega_i%7D%7B%5DI%2BmR%5E2%7D%3D%5Cfrac%7B%28260%29%281.26%29%7D%7B260%2B%2824.0%29%282.20%29%5E2%7D%20)
Since g is constant, the force the escaping gas exerts on the rocket will be 10.4 N
<h3>
What is Escape Velocity ?</h3>
This is the minimum velocity required for an object to just escape the gravitational influence of an astronomical body.
Given that the velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in 0.60s. The gravitational field intensity is 9.8N/kg.
To calculate the force the escaping gas exerts of the rocket, let first highlight all the given parameters
- Mass (m) of the rocket 0.25 Kg
- Initial velocity u = 15 m/s
- Final Velocity v = 40 m/s
- Gravitational field intensity g = 9.8N/kg
The force the gas exerts of the rocket = The force on the rocket
The rate change in momentum of the rocket = force applied
F = ma
F = m(v - u)/t
F = 0.25 x (40 - 15)/0.6
F = 0.25 x 41.667
F = 10.42 N
Since g is constant, the force the escaping gas exerts on the rocket is therefore 10.4 N approximately.
Learn more about Escape Velocity here: brainly.com/question/13726115
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Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
Weight is the force exerted on an object by gravity. So, weight of any object on the moon is 1/6 that on Earth.
