Answer:
1.78 rad/s
1.70344 rad/s
Explanation:
v = Velocity = 0.8 m/s
m = Mass of person = 60 kg
= Distance between center of mass of person and pole = 0.45 m
= New distance between center of mass of person and pole = 0.46 m
I = Moment of inertia
Angular speed is given by
The angular speed is 1.78 rad/s
In this system the angular momentum is conserved
The new angular speed is 1.70344 rad/s
Answer:
To establish this relationship we must examine the potentials that these forces create. The electrical potential is described by
Ve = k q / r
The potential for strong nuclear force is
Vn (r) = - gs / 4pir exp (-mrc / h)
Where gs is the stacking constant and r the distance between the nucleons,
We can compare these potentials where the force is derived from the relationship
E = -dU / dr
F = q E
Explanation:
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:
Resultant force, R = 10 N
Explanation:
It is given that,
Force acting along +x direction, 
Force acting along +y direction, 
Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,
Here 
R = 10 N
So, the resultant force on the object is 10 N. Hence, this is the required solution.
