Answer:
This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc
Explanation:
The integration of the differential mass of the hollow right circular cone yields 

and for a uniform disc
I = 1/2πρtr⁴ = 1/2Mr².
 
        
             
        
        
        
Volume = 873 - 50 = 723 cm^3
        
             
        
        
        
So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.
<h3>Introduction</h3>
Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

With the following condition :
 = angular frequency (rad/s) = angular frequency (rad/s)
 = change of angle value (rad) = change of angle value (rad)
- t = interval of the time (s)
<h3>Problem Solving</h3>
We know that :
 = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s. = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
- t = interval of the time = 54.9 s.
What was asked :
 = angular frequency = ... rad/s = angular frequency = ... rad/s
Step by step :



<h3>Conclusion :</h3>
So, the angular frequency of the blades approximately 36.43π rad/s.
 
        
             
        
        
        
Answer:
The force pulling the roller along the ground is 128.55 N
Explanation:
A force of 200 N acting at an angle of 50° with the ground level
This force is pulled a garden roller
We need to find the force pulling the roller along the ground 
The force that pulling the roller along the ground is the horizontal 
component of the force acting 
→ The force acting is 200 N at direction 50° with ground (horizontal)
→ The horizontal component = F cosФ
→ F = 200 N , Ф = 50
→ The horizontal component = 200 cos(50) = 128.55 N
128.55 N is the horizontal component of the force that pulling the 
roller along the ground
<em>The force pulling the roller along the ground is 128.55 N</em>