Considering Conservation of Momentum, the momentum

before and after must remain the same:
so:
before:
Answer:
α = -π/3 rad/s²
θ = 1.5π rad ≈ 4.71 rad
θ = 0.75 rev
Explanation:
30 rev/min (2π rad/rev) / (60 s/min) = π rad/s
α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²
θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad
θ = 1.5π rad / 2π rad/rev = 0.75 rev
Answer:
0.453 m/s
Explanation:
Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is
where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is 
The rate of rotation will then be 1.81/1.256=1.441 rev/s
Similarly, circumference of inner part will be
where subscript i represent inner. Substituting 0.1 for inner diameter then

The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s
The football and air resistance between contact
Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = 
Let the height of the bin be 'h'
Therefore the total area, 
The cost is:
C = 2sh
Volume of the box, V =
(1)
Total cost,
(2)
From eqn (1):

Using the above value in eqn (1):


Differentiating the above eqn w.r.t 'a':

For the required solution equating the above eqn to zero:


a = 4
Also

The path in order to minimize the cost must be a rectangle.