The length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.
<h3>Extension of the spring 2</h3>
The extension of the spring 2 is determined from the net force acting on spring 2.
F(net) = Kx
<em>Net force on spring 2 = sum of forces acting downwards - sum of forces acting upwards.</em>
Net force on spring 2 = (force on 2 + force on 3) - (force on 1)
F(net) = (m₂ + m₃)g - (m₁)g
F(net) = (6.67 + 6.67)9.8 - (6.67)9.8
F(net) = 130.732 - 65.366
F(net) = 65.366 N
x = F/k
x = (65.366)/(8130)
x = 0.00804 m
The length of spring 2 = x + 0.23 m
= 0.00804 m + 0.23 m = 0.238 m
Thus, the length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.
Learn more about extension of elastic material here: brainly.com/question/26838571
<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car
change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2)
where v(f) and v(i) are the final and initial speeds
change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2)
185,000=1/2 m(255m^2/s^2)
solving for m
m=1451kg</span>
Answer:
<em>His angular velocity will increase.</em>
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = 
'ω'
where
' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = 
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to .
From
'ω' = ω
since is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
