The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.
<h3> What is speed?</h3>
Speed is defined as the rate of change of the distance or the height attained.
The given data in the problem is;
The initial diameter is,
initial radius,

The initial crossection area;

The final crossection area;

The initial flow rate is;
R = density ×velocity ×area

The speed of the water in the wider part will be;
From the continuity equation;

Hence, the speed of the water in the wider part will be 1.194 m/sec.
To learn more about the speed, refer to the link;
brainly.com/question/7359669
#SPJ1
Answer:
I'm not really sure but I think it is
four
Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
The ball is travelling faster when the two objects hits the level ground below.
<h3>Time of motion of the objects</h3>
The time of motion of the objects depends on height and initial velocity of projection of the objects.
The stone has no initial vertical velocity while the ball has initial vertical velocity.
Thus, the ball is travelling faster when the two objects hits the level ground below.
Learn more about time of motion here: brainly.com/question/2364404
#SPJ1