The weight is not coming from the center of the mass because the force that act on it is not is equal is side.(2) section B donot have weight because the ruler bend down and section be raise up so no weight.
Answer:
a) X = 17.64 m
b) X = 17.64 + 4∆t^2 + 16.8∆t
c) Velocity = lim(∆t→0)〖∆X/∆t〗 = 16.8 m/s
Explanation:
a) The position at t = 2.10s is:
X = 4t^2
X = 4(2.10)^2
X = 17.64 m
b) The position at t = 2.10 + ∆t s will be:
X = 4(2.10 + ∆t)^2
X = 17.64 + 4∆t^2 + 16.8∆t m
c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,
∆X= 4∆t^2 + 16.8∆t
Divide by ∆t on both sides:
∆X/∆t = 4∆t + 16.8
Taking the limit as ∆t approaches to zero we get:
Velocity =lim(∆t→0)〖∆X/∆t〗 = 4(0) + 16.8
Velocity = 16.8 m/s
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.
The answer is A.
p=m/v
p= 240/60
p= 4 g/cm^3
The answer is C. an electron in an orbit has a fixed energy.