Primary mental skills needed for riding a motorcycle
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Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
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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
