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
The velocity of the girl is -4.8 m/s.
Using the principle of conservation of linear momentum, The total momentum of bodies before and after collision is constant. Since the two objects are stationary, the initial momentum of each body is zero.
Thus;
0 = (80 × 3) + (50 × v)
0 = 240 + 50 v
-240 = 50 v
v = -240/50
v = -4.8 m/s
Note that the negative sign shows that the velocity of the girl is in opposite direction that that of the girl.
Learn more about momentum: brainly.com/question/904448
Answer: 
Explanation:
Acceleration 
is expressed in the following formula:
Where:
is the final velocity of the projectile
is the initial velocity of the projectile
is the time
Solving:
This is the acceleration of the projectile
Explanation:
Momentum Is defined as the product of of mass and its velocity
Momentum (M) =mass *velocity
SI unit of momentum is kgm/s
The rate of change in momentum
=change in momentum / time
=(mv-mu)/t
Answer:(a)360N,(b)171N,(c)2.702m
Explanation:
(a)Maximum Friction Force =
=360 N
(b)Moment about Ground Point
(c)
Here maximum friction force can be 360 N
Therefore 
Where x is the maximum distance moved by man along the ladder
740x=2000
x=2.702m
