Answer:
The answer is "[tex]\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}[/tex]"
Explanation:
Given:
[tex]\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\[/tex]
calculation:
[tex]Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978[/tex]
The specific gravity , specific weight, and density of the are 0.978, 9600 N/m^2, 978.59 Kg/m^3 respectively.
The specific weight of any substance is defined as the weight of the substance per unit volume.
[tex]\rm Specific \ weight= \dfrac{w}{v}[/tex]
Weight of crude oil is w= 9.6 N
putting the value in the equation we will get
[tex]\rm Specific weight = \dfrac{9.6}{10^{-3}} =9600\ \dfrac{N}{m^2}[/tex]
The Density of any substance is defined as the mass of any substance per unit volume.
[tex]\rm Density = \dfrac{m}{v}[/tex]
Now mass will be w=mg so
[tex]m=\dfrac{w}{g}=\dfrac{9.6}{9.81}=0.9785\ kg[/tex]
[tex]\rm Density = \dfrac{0.9785}{10^{-3}}=978.59\ \ \dfrac{kg}{m^3}[/tex]
The specific gravity is defined as the ratio of the density of any fluid to the density of a standard fluid or water.
[tex]\rm Specific \ gravity= \dfrac{\rho_f}{\rho_w}[/tex]
[tex]\rm Specific \ gravity= \dfrac{\ 978}{1000}=0.978[/tex]
Thus the specific gravity , specific weight, and density of the are 0.978, 9600 N/m^2, 978.59 Kg/m^3 respectively.
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