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)
= 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.- t = interval of the time = 54.9 s.
What was asked :
= angular frequency = ... rad/s
Step by step :



<h3>Conclusion :</h3>
So, the angular frequency of the blades approximately 36.43π rad/s.
Answer:
Explanation:
Given

Frictional Force is balanced by force due to car acceleration
Frictional force 




The answer is C)the rating that the golfers give Callaway clubs
Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
Answer:
distance between both oasis ( 1 and 2) is 27.83 Km
Explanation:
let d is the distance between oasis1 and oasis 2
from figure
OC = 25cos 30
OE = 25sin30
OE = CD
Therefore BC = 30-25sin30
distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem
in

PUTTING ALL VALUE IN ABOVE EQUATION


d = 27.83 Km
distance between both oasis ( 1 and 2) is 27.83 Km