Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
The part of the system that is considered the resistance force is;
B
Explanation:
The simple machine is a system of pulley that has two pulleys
The effort, which is the input force at A gives the value of the tension at C and D which are used to lift the load B
Therefore, we have;
A = C = D
B = C + D = C + C = 2·C
∴ C = B/2
We have;
C = B/2 = A
Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B
The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done
It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force
Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.
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average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.
Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT
ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system
For ideal gas
Δh = CpΔT
by putting the values
Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding
