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:
[tex] \gamma = dg [/tex] (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:
[tex] d = \frac{m}{V} [/tex] (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:
[tex] W = mg [/tex]
[tex] m = \frac{W}{g} [/tex] (3)
By entering equations (3) and (2) into (1) we have:
[tex] \gamma = dg = \frac{mg}{V} = \frac{W}{V} = \frac{7 N}{0.001 m^{3}} = 7000 N/m^{3} [/tex]
Hence, the specific weight is 7000 N/m³.
- The density can be found as follows:
[tex] d = \frac{m}{V} = \frac{W}{gV} = \frac{7 N}{9.81 m/s^{2}*0.001 m^{3}} = 713.5 kg/m^{3} [/tex]
Then, the density is 713.5 kg/m³.
- The specific gravity (SG) of a liquid can be calculated with the following equation:
[tex] SG = \frac{d}{d_{H_{2}O}} = \frac{713.5 kg/m^{3}}{1000 kg/m^{3}} = 0.7135 [/tex]
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!
