What is the magnitude of the torque that the axle must apply to prevent the disk from rotating?
1 answer:
The required torque at the axle, is given by the difference between the
moments of the applied forces.
The torque required is <u>19.62 N·m counterclockwise</u>
Reasons:
The given parameters are;
Mass of the disk, m = 5.0 kg
Location of the axle = Half the radius of the disk
Diameter of the disk, D = 40 cm = 0.4 m
Applied mass, 0.1 m from the axle = 15 kg
Applied mass, 0.3 m from the axle = 10 kg
Required:
Magnitude of torque at the axle that prevent the disk from rotating
Solution:
Torque needed = Clockwise moment - Counterclockwise moment
Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m
Counterclockwise moment = 15 kg × 0.1 m × 9.81 m/s² = 14.715 N·m
τ + Counterclockwise moment = Clockwise moment
τ + 14.715 N·m = 34.335 N·m
Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m
Torque required, τ = <u>19.62 N·m counterclockwise</u>
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<em>The probable question drawing obtained from a similar question online is attached</em>
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Answer:
20 seconds
Explanation:
The equation for Power is:
Which means...
We can plug in the given values into the equation:
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v + xv
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v + x₂v
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ