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3 years ago

An unknown fluid has a specific gravity of 0.750. What is the volume of 22.5 kg of this fluid?

Physics

1 answer:

Andru [333]3 years ago

5 0

<h2>Option C is the correct answer.</h2>

Explanation:

Specific gravity of fluid = 0.750

Density of fluid = Specific gravity of fluid x Density of water

Density of fluid = 0.750 x 1000

Density of fluid = 750 kg/m³

Mass of fluid = 22.5 kg

We have

Mass = Volume x Density

22.5 = Volume x 750

Volume = 0.03 m³ = 30 L

Option C is the correct answer.

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ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

$U'_{el} = 414kN\cdot m/m^3$

Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

$U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})$

$U_{el} = 2.663N\cdot m$

Therefore the net strain energy of the sample is $2.663N\codt m$

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insens350 [35]

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s−15

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Xenon and,
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