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:
36.408cm3
Explanation:
Since we acknowledge that density is d= m/v, once we switch it up to maintain v as the number to be found it will change to v=m/d. Therefore, 275.32/7.562 is 36.408 and the unit is cm cube!
Hope that helped!!
Answer:
A. 181.24 N
Explanation:
The magnitude of hte electrostatic force between two charged objects is given by the equation

where
k is the Coulomb's constant
q1, q2 are the magnitudes of the two charges
r is the separation between the charges
In this problem, we have:
is the magnitude of the 1st charge
is the magnitude of the 2nd charge
r = 2.5 cm = 0.025 m is the separation between the charges
Therefore, the magnitude of the electric force is:

So, the closest answer is
A) 181.24 N
Answer:
No, neutrons have about the same mass as a proton, but both have more mass than electrons.
Hope this helps a bit,
Flips
Answer:
Vi = 5 m/s
Explanation:
let (a) acceleration = 0.75 m/s²
(t) time = 20 seconds
Vf = final velocity = 72 km/hr (convert to m/s to units consistency = 20 m/s)
find Initial velocity (Vi)
Vf - Vi
a = -----------
t
Vi = Vf - (a * t) = 20 - (0.75 * 20)
Vi = 5 m/s