Question

A 55-liter tank is full and contains 40kg of fuel. Find using Sl units: • Density p. • Specific Weight y • Specific Gravity Answer tolerance = 1%. Be sure to include units. The sign of the answers will not be graded, use a positive value for your answer. Your answers: p= (Enter a positive value) y = (Enter a positive value) SG = (Enter a positive value)

Answer 1

Answer:

Density of the fuel is 727.3 kilograms per cubic meter.

Specific weight of the fuel is 7127.3 Newtons per cubic meter.

Specific gravity of the fuel is 0,727.

Explanation:

In order to use SI units, we have to convert liters to cubic meters. Knowing that a liter is a cubic decimeter and a cubic decimeter is $1*10^{-3}$ cubic meters, we know that the tank has 0,055 cubic meters of fuel (because it is full).

Now that we have things in SI units, we calculate density:

$p_{fuel}= \frac{mass}{volume} = \frac{40 kg}{0.055 m^{3} } =727.3 \frac{kg }{m^{3} }$

Knowing the mass per unit of volume, we can calculate weight per unit of volume thanks to Newton's second law (mass times acceleration, g in this case, equals force (weight)), i.e. specific weight:

$y=p*g=727,3 \frac{kg}{m^{3}}*9.8\frac{m }{s^{2}}=7127,3 \frac{N}{m^{3}}$

With density we can also calculate how dense the fuel is related to a reference (water), i.e. specific gravity. SG is a dimensionless number that tell us how much denser (SG>1) or lighter per unit of volume (SG<1) a substance is than water. We use water as a reference because it is one of the most used substances in our life, and it is a standard density (1000 kg per cubic meter at 4°C and 1 atm).

$SG=\frac{p_{fuel} }{p_{water} } =\frac{727.3 \frac{kg }{m^{3} }}{1000 \frac{kg }{m^{3} }} =0,727$