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3 years ago

The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03

Engineering

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

Answer is c 0.500

Explanation:

$SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}$

We know that $\rho_{water}=62.42lb/ft^{3}$

Applying values we get

$SpecificGravity=\frac{31.2}{62.4}=0.5$

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu

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b) 3366.33 ft³/sec

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Explanation:

Given:

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Solving for V2, we have:

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e) for critical depth, we use :

$y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3$

= 3.80 ft

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