Imma take a guess and say it’s a project portfolio but I wouldn’t put that bc I don’t see an image
Answer:
a. is correct.
Explanation:
During servicing you should examine each bearing carefully for chips, pits, and scratches, so Technician A is correct.
Discoloration of a bearingbis not acceptable wear, so Technician B is false.
Combine the answers: Only Technician A is correct therefore answer a. is the right answer.
Answer:
Change in length = 0.1257 mm
Change in diameter= -0.03771mm
Explanation:
Given
Diameter, d = 15 mm
Length of rod, L = 200mm
F = Force= 300N
d = 0.015m
Ep=2.70 GPa, np=0.4.
First, we have to calculate the normal stress using
σ = F/A where F = Force acting on the Cross-sectional area
A = Area
Area is calculated as πd²/4 where d = 0.015m
A = 22/7 * 0.015²/4
A = 0.000176785714285m²
A = 1.768E-4m²
So, stress. σ = 300N/1.768E-4m²
σ = 1696832.579185520Pa
σ = 1.697MPa
Calculating E(long)
E(long) = σ /Ep
E(long) = 1.697E-3/2.70
E(long) = 0.0006285
At this point, we fan now calculate the change in length of the element;
∆L = E(long) * L
∆L = 0.0006285 * 200mm
∆L = 0.1257mm
Calculating E(lat)
E(lat) = -np * E(long)
E(lat) = -4 * 0.0006285
E(lat) = -0.002514
At this point, we can now calculate the change in diameter of the element;
∆D = E(lat) * D
∆L = -0.002514 * 15mm
∆L = -0.03771mm
Answer:
Explanation:
Using equation of pure torsion
where
T is the applied Torque
is polar moment of inertia of the shaft
t is the shear stress at a distance r from the center
r is distance from center
For a shaft with
Outer Diameter
Inner Diameter
Applying values in the above equation we get
x 
Thus from the equation of torsion we get
Applying values we get
T =829.97Nm
Answer:
Normal force = 0.326N
Explanation:
Given that:
mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg
height of the mass = 1.1 m
radius = 0.2 m
acceleration due to gravity = 9.8 m/s²
We are to determine the normal force pressing on the track at A.
To to that;
Let consider the conservation of energy relation; which says:
mgh = mgr + 1/2 mv²
gh = gr + 1/2 v²
gh - gr = 1/2v²
g(h-r) = 1/2v²
v² = 2g(h-r)
However; the normal force will result to a centripetal force; as such, using the relation
N =mv²/r
replacing the value for v² = 2g(h-r) in the above relation; we have:
Normal force = 2mg(h-r)/r
Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2
Normal force = 0.065268/0.2
Normal force = 0.32634 N
Normal force = 0.326N
