Question

a liquid weighs 15 kn and occupies 3.75m³ find its specific weight mass density and specific gravity​

Answer 1

W=mg

5x2=[L/2-2] 10x10

L/2=1+2

L=6m

please mark as brainliest answer

Answer 2

Answer:

The liquid's specific weight, γ, calculated is $4\times 10^{3} N/m^{3}$.

The liquid's mass density, ρ, calculated is $407.8 kg/m^{3}$.

The liquid's specific gravity, S, calculated is $0.407$.

Explanation:

Data given,

The given weight of the liquid, W = $15kN$ = $15\times 10^{3}N$

The volume occupied by the liquid, V = $3.75m^{3}$

a) Specific weight ( γ ):

As we know,

• Specific weight = the density of the substance × acceleration due to gravity.
• γ = $\rho \times g$

Also,  $\rho = \frac{mass}{volume}$

And mass = $\frac{weight}{gravity}$    ( W = $mg$ )

Now, after substituting all the equations, we get:

• γ = $\frac{W}{V}$ = $\frac{15\times 10^{3} }{3.75}$ = $4\times 10^{3} N/m^{3}$

Therefore, the specific weight of the liquid, γ, calculated is $4\times 10^{3} N/m^{3}$.

b) Mass density ( ρ ):

As mentioned above;

• $\rho = \frac{mass}{volume}$

And mass = $\frac{weight}{gravity}$    ( W = $mg$ )

Therefore,

• $\rho = \frac{W}{gV}$      
• $\rho = \frac{15\times 10^{3} }{9.81\times 3.75}$        ( g = $9.81 m/s^{2}$ )
• ρ = $407.8kg/m^{3}$

Hence, the mass density of the liquid, ρ, calculated is $407.8 kg/m^{3}$.

c) Specific gravity ( S ):

As we know,

• S = $\frac{Density of liquid}{Density of water}$

Here, the water's density = $1000kg/m^{3}$

And as calculated above,

• The liquid's density = $407.8kg/m^{3}$

Therefore,

• S = $\frac{407.8}{1000}$ = $0.407$

Hence, the specific gravity of the liquid, S, calculated is $0.407$.