Question

calculate specific gravity, density and specific gravity of one litre of liquid which weighs 7n

Answer 1

Answer:

After evaluating all the options we have:

A. From all of the options of specific weight, density, and specific gravity of 1 liter of liquid, the correct is option 1: 7000 N/m³, 713.5 kg/m³, 0.7135.  

B. The correct option of the multiplying factor for converting one stoke into m²/s is 4: 10⁻⁴.  

A. Let's calculate the specific weight, density, and specific gravity of 1 L of the liquid that weighs 7 N.

The specific weight is given by:

\gamma = dgγ=dg   (1)

Where:

γ: is the specific weight

d: is the density

g: is the gravity = 9.81 m/s²

We need to find the density which is:

d = \frac{m}{V}d=Vm   (2)

Where:

m: is the mass

V: is the volume = 1 L = 0.001 m³

The mass can be found knowing that the liquid weighs (W) 7 N, so:

W = mgW=mg  

m = \frac{W}{g}m=gW   (3)  

By entering equations (3) and (2) into (1) we have:

\gamma = dg = \frac{mg}{V} = \frac{W}{V} = \frac{7 N}{0.001 m^{3}} = 7000 N/m^{3}γ=dg=Vmg=VW=0.001m37N=7000N/m3

Hence, the specific weight is 7000 N/m³.

The density can be found as follows:

d = \frac{m}{V} = \frac{W}{gV} = \frac{7 N}{9.81 m/s^{2}*0.001 m^{3}} = 713.5 kg/m^{3}d=Vm=gVW=9.81m/s2∗0.001m37N=713.5kg/m3

Then, the density is 713.5 kg/m³.

The specific gravity (SG) of a liquid can be calculated with the following equation:

SG = \frac{d}{d_{H_{2}O}} = \frac{713.5 kg/m^{3}}{1000 kg/m^{3}} = 0.7135SG=dH2Od=1000kg/m3713.5kg/m3=0.7135

Hence, the specific gravity is 0.7135.

Therefore, the correct option is 1: 7000 N/m³, 713.5 kg/m³, 0.7135.

B. A Stokes is a measurement unit of kinematic viscosity.

One m²/s is equal to 10⁴ stokes, so to convert 1 stokes to m²/s we need to multiply for 10⁻⁴.

Hence, the correct option is 4: 10⁻⁴.

                                                                                                                                             

You can find more about specific weight here: https://brainly.com/question/13178675?referrer=searchResults    

           

I hope it helps you!

Answer 2

Given,

The volume of the liquid = 1 L,

Weight = 7 N.

To find,

Specific gravity,

Density.

Solution,

The specific gravity of a liquid is defined as the ratio of the density of a liquid to the density of the reference substance, that is water in the case of liquids. It is given by,

$SG=\frac{\rho _{liquid}}{\rho _{H_2O}}$     ...(1)

where,

SG = specific gravity of the liquid, and

$\rho_{liquid} \hspace{3} and \hspace{3} \rho_{H_2O}$ are the densities of the liquid and water respectively.

The density is defined as mass per unit volume of a liquid or substance, and is given by,

$\rho =\frac{mass(m)}{volume(V)}$     ...(2)

Here, the given volume is 1 L, and 1 L = 0.001 m³. So,

V = 0.001 m³.

The weight is given as 7 N. Since,

$weight =m\times g$

⇒ $m=\frac{W}{g}$

⇒ $m=\frac{7}{9.8}$ kg

⇒ m = 0.714 kg

Now, from eq. (2),

$\rho=\frac{m}{V} =\frac{0.714}{0.001}$

⇒ ρ = 714 kg/m³.

Using the above-calculated value of density and eq. (1), for specific gravity,

As $\rho _{H_2O}=1000 \hspace{3} kg/m^3$

⇒ $SG=\frac{714}{1000}$

⇒ SG = 0.714.

Therefore, the specific gravity of the given liquid is 0.714, and its density is 714 kg/m³.