Answer:
Time =t2=58.4 h
Explanation:
Since temperature is the same hence using condition
x^2/Dt=constant
where t is the time as temperature so D also remains constant
hence
x^2/t=constant
2.3^2/11=5.3^2/t2
time=t^2=58.4 h
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Answer:
ideal fluid follow Newtonian law
that is, shear stress is directly proportional to rate change of shear strain.
watch handwritten explanation
Answer:
The theoretical maximum specific gravity at 6.5% binder content is 2.44.
Explanation:
Given the specific gravity at 5.0 % binder content 2.495
Therefore
95 % mix + 5 % binder gives S.G. = 2.495
Where the binder is S.G. = 1, Therefore
Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495
(95 +5)/(Vx +5) = 2.495
2.495 × (Vx + 5) = 100
Vx =35.08 to 95
Or density of mix = Mx/Vx = 95/35.08 = 2.7081
Therefore when we have 6.5 % binder content, we get
Per 100 mass unit
93.5 Mass unit of Mx has a volume of
Mass/Density = 93.5/2.7081 = 34.526 volume units
Therefore we have
At 6.5 % binder content.
(100 mass unit)/(34.526 + 6.5) = 2.44
The theoretical maximum specific gravity at 6.5% binder content = 2.44.