The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
<h3> Constant angular acceleration</h3>
Apply the following kinematic equation;
ωf² = ωi² - 2αθ
where;
- ωf is the final angular velocity when the centrifuge stops = 0
- ωi is the initial angular velocity
- θ is angular displacement
- α is angular acceleration
ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s
θ = 52 rev x 2π rad/rev = 326.7 rad
0 = ωi² - 2αθ
α = ωi²/2θ
α = ( 356.05²) / (2 x 326.7)
α = 194.02 rad/s²
Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
Learn more about angular acceleration here: brainly.com/question/25129606
#SPJ1
Answer:
hence the y - component of the velocity is 50m/s
<span>Without friction, there will be undamped simple harmonic motion. The force of the spring is proportional to the distance from the equilibrium point. The period of oscillation will be independent of the amplitude.
I hope my answer has come to your help. God bless and have a nice day ahead!</span>
Answer:
He is gaining kinetic energy and losing potential energy
Explanation:
Given:
The magnitude of each charge is q1 = q2 = 1 C
The distance between them is r = 1 m
To find the force when distance is doubled.
Explanation:
The new distance is
The force can be calculated by the formula
Here, k is the constant whose value is
On substituting the values, the force will be
