Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density 
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
Apply conservation of angular momentum:
L = Iw = const.
L = angular momentum, I = moment of inertia, w = angular velocity, L must stay constant.
L must stay the same before and after the professor brings the dumbbells closer to himself.
His initial angular velocity is 2π radians divided by 2.0 seconds, or π rad/s. His initial moment of inertia is 3.0kg•m^2
His final moment of inertia is 2.2kg•m^2.
Calculate the initial angular velocity:
L = 3.0π
Final angular velocity:
L = 2.2w
Set the initial and final angular momentum equal to each other and solve for the final angular velocity w:
3.0π = 2.2w
w = 1.4π rad/s
The rotational energy is given by:
KE = 0.5Iw^2
Initial rotational energy:
KE = 0.5(3.0)(π)^2 = 14.8J
Final rotational energy:
KE = 0.5(2.2)(1.4)^2 = 21.3J
There is an increase in rotational energy. Where did this energy come from? It came from changing the moment of inertia. The professor had to exert a radially inward force to pull in the dumbbells, doing work that increases his rotational energy.
Answer:
If research results can be replicated, it means they are more likely to be correct.
