Answer:
b). Occurs at the outer surface of the shaft
Explanation:
We know from shear stress and torque relationship, we know that
where, T = torque
J = polar moment of inertia of shaft
τ = torsional shear stress
r = raduis of the shaft
Therefore from the above relation we see that
Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.
When r= 0, then τ = 0
and when r = R , τ is maximum
Thus, torsional shear stress is maximum at the outer surface of the shaft.
Answer:
False ( B )
Explanation:
considering that the wind turbine is a horizontal axis turbine
Power generated/extracted by the turbine can be calculated as
P = n * 1/2 *<em> p</em> *Av^3
where: n = turbine efficiency
<em>p = air density </em>
<em> </em>A = πd^2 / 4
v = speed
From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine
Answer:
Midnight on the birthday. Hope this helps!
9514 1404 393
Answer:
see attached
Explanation:
Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.
(5 cm³/s)(1.06 g/cm³) = 5.3 g/s
If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:
(5.3 g/s)/2 = 2.65 g/s
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The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...
(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s
Downstream, we have half the volumetric flow and a smaller area.
(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s
