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labwork [276]
2 years ago
7

Oil of specific gravity 0.75 flows through a smooth contraction in a pipe at a volumetric flow rate of 3.2 cu ft / sec.Find the

force required to hold the contraction in place. Indicate the direction of this force.
Physics
1 answer:
pogonyaev2 years ago
6 0

Answer:

The force required to hold the contraction in place is 665.91 N ↑

Explanation:

Given;

specific gravity of oil, γ = 0.75

Volumetric flow rate, V 3.2 Ft³/s = 0.0906 m³/s

\gamma =\frac{\rho_o}{\rho_w}

where;

\rho_o is the density of oil

\rho_w is the density of water = 1000 kg/m³

∴density of oil (\rho_o) = γ × density of water(\rho_w)

                               = 0.75 × 1000 kg/m³

                                = 750kg/m³

Buoyant Force = ρVg

                         = 750 × 0.0906 × 9.8

                         = 665.91 N ↑

This force acts upward or opposite gravitational force.

Therefore, the force required to hold the contraction in place is 665.91 N ↑

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Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

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0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

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alexandr402 [8]

Answer:

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