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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 glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the wa

Furkat [3]

Answer:

$h=0.46m$

Explanation:

From the question we are told that:

Velocity of water $V=3m/s$

Height=?

Generally, the equation for Water Velocity is mathematically given by

$V=\sqrt{2gh}$

Therefore Height h is given as

$h=\frac{v}{2g}$

$h=\frac{3^2}{2*9.81}$

$h=0.46m$

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3 years ago

As Becky was driving "Old Betsy," the family station wagon, the engine finally quit, being worn out after 171,000 miles. It can

andriy [413]

Answer:

1) 4.361 x 10 raised to power 8 revolutions

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Explanation:

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4 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$