To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.
Torque in a body is defined as,

And in angular movement like

Where,
F= Force
d= Distance
I = Inertia
Acceleration Angular
PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

On the other hand we have the speed data expressed in RPM, as well


Acceleration can be calculated by



In the case of Inertia we know that it is equivalent to


Matching the two types of torque we have to,




PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,



Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
Answer:
Explanation:
Given
Length of rope 
Weight of rope 
weight density
Work done to lift rope 33 m


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Answer: Work can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton • meter (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.
Explanation:
Development of the Embryo. The next stage in development is the embryo, which develops within the amniotic sac, under the lining of the uterus on one side. This stage is characterized by the formation of most internal organs and external body structures.