Answer:
V₂ = 1.5 m/s
Explanation:
given,
speed of the first piece = 6 m/s
speed of the third piece = 3 m/s
speed of the second fragment = ?
mass ratios = 1 : 4 : 2
fragment break fly off = 120°
α = β = γ = 120°
sin α = sin β = sin γ = 0.866
using lammi's theorem

A,B and C is momentum of the fragments

4 x V₂ = 2 x 3
V₂ = 1.5 m/s
To solve this exercise it is necessary to apply the kinematic equations of angular motion.
By definition we know that the displacement when there is constant angular velocity is

From our given data we know that,



Moreover we know that

Therefore for time t=8.1s we have,



That number in revolution is:


Here, we see that there are 15 complete revolutions
And 0.108 revolutions i not complete, so the tunable rotation is

Therefore the angle of the speck at a time 8.1s is 
Answer:
F = 39.2 N (hand force) and N = 68.6 N (shoulder force)
Explanation:
In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole
Let's set the reference system at the fit point of the shoulder
∑ τ = 0
We will assume that the counterclockwise turns are positive
w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0
all distances are measured from the support of the man (x₀ = 0.60 m)
F = (w₁ 0.60 + W 0.1) / 0.4
F = (m₁ 0.6 + m 0.1) g / 0.4
let's calculate
F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4
F = 39.2 N
this is the force that the hand must exert to keep the system in balance
We apply the translational equilibrium condition
-w₁ -W + N - F = 0
N = w₁ + W + F
N = (m₁ + m) g + F
let's calculate
N = (2.6 + 0.4) 9.8 + 39.2
N = 68.6 N