Let suppose, you are going to develop a web-application for school management system. Then what architectural pattern will you u
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Answer:
The shear stress will be 80 MPa
Explanation:
Here we have;
τ = (T·r)/J
For rectangular tube, we have;
Average shear stress given as follows;
Where;
= 100 mm × 200 mm = 20000 mm² = 0.02 m²
t = Thickness of the shaft in question = 2 mm = 0.002 m
T = Applied torque
Therefore, 50 MPa = T/(2×0.002×0.02)
T = 50 MPa × 0.00008 m³ = 4000 N·m
Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m
Therefore, = 0.05 m × 0.25 m = 0.0125 m².
Therefore, from the following average shear stress formula, we have;
Plugging in then values, gives;
The shear stress will be 80,000,000 Pa or 80 MPa.
Answer:
a) (Ω-m)^{-1}
b) Resistance = 121.4 Ω
Explanation:
given data:
diameter is 7.0 mm
length 57 mm
current I = 0.25 A
voltage v = 24 v
distance between the probes is 45 mm
electrical conductivity is given as
(Ω-m)^{-1}[/tex]
b)
Resistance = 121.4 Ω