You said that she's losing 1.9 m/s of her speed every second.
So it'll take
(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)
to lose all of her initial speed, and stop.
Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:

Our values are given by,

As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:


Skate 2:


Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:




Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Answer: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec