Answer:
33.429 N-m
Explanation:
Given :
Inclination angle of two shaft, α = 20°
Speed of shaft A, 
= 1000 rpm
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Now we know that for maximum velocity,
= 1064.1 rpm
Now we know
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Therefore moment of inertia of flywheel, I = m.
=30 X 
= 0.3 kg-
Now torque on the output shaft
T₂ = I x ω
= 0.3 X 1064.2 rpm
= 
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m
Answer:
A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.
Explanation:
“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.
The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
<em>"Your</em><em> </em><em>question is not complete, it seems to be missing the following information;"</em>
The sequence of operation is A - E - D - C - B - A - F
The given parameters;
- <em>number of pieces that will flow from the first machine A to machine F, = 2,000 pieces</em>
- <em>initial unit load specified in the first machine, L₁ = 50</em>
- <em>final unit load, L₂ = 100 </em>
- <em>the capacity of the vehicle = 1 unit load</em>
<em />
The given sequence of operation of the vehicle;
A - E - D - C - B - A - F
<em>the vehicle makes </em><em>6 trips</em><em> for </em><em>100</em><em> unit </em><em>loads</em>
The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.
100 unit loads ----------------- 6 trips
2000 unit loads --------------- ?
Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.
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