Respuesta :
Answer: The density of glycerine is [tex]78.589\frc{pounds}{foot^3}[/tex]
Explanation: We are given density of glycerine in [tex]gram/cm^3[/tex] which is [tex]1.26g/cm^3[/tex]
To convert t into [tex]pounds/foot^3[/tex], we are given conversion rates:
454 grams = 1 pound
So, 1 gram will be equal to [tex]\frac{1}{454}pounds[/tex]
Similarly,
[tex]28317cm^3=1foot^3[/tex]
So, [tex]1cm^3[/tex] will be equal to [tex]\frac{1}{28317}foot^3[/tex]
Converting the density of glycerin from [tex]g/cm^3[/tex] to [tex]pounds/foot^3[/tex] using above conversion rates, we get:
[tex]\frac{1.26g}{cm^3}=(\frac{1.26}{454}\times \frac{28317}{1})\frac{pounds}{foot^3}[/tex]
[tex]1.26g/cm^3=78.589pounds/foot^3[/tex]
Density of glycerin is [tex]78.589\frc{pounds}{foot^3}[/tex]
The density of glycerine has been 78.589 [tex]\rm \bold{pounds/foot^3}[/tex].
The density can be defined as the mass of a substance per unit volume. The density of glycerine has been 1.26 [tex]\rm \bold{g/cm^3}[/tex].
The unit can be converted as:
454 grams = 1 pound
1 grams = [tex]\rm \dfrac{1}{454}[/tex] pounds.
28,317 [tex]\rm cm^3[/tex] = 1 [tex]\rm foot^3[/tex]
1 [tex]\rm cm^3[/tex] = [tex]\rm \dfrac{1}{28,317}\;cm^3[/tex]
The units have been converted by the conversion of mass as well grams in the desired unit. Since 1 gram has been x pounds then y grams has been x times of y.
The density of glycerine has been:
[tex]\rm \dfrac{1.26\;g}{1\;cm^3}[/tex] = [tex]\rm \dfrac{1.26\;g}{1\;cm^3}\;\times\;\dfrac{\frac{1}{454}\;pounds }{\frac{1}{28,317}\;foot^3 }[/tex]
= [tex]\rm \dfrac{1.26\;\times\;454}{28,317}\;pounds/foot^3[/tex]
= 78.589 [tex]\rm \bold{pounds/foot^3}[/tex]
The density of glycerine has been 78.589 [tex]\rm \bold{pounds/foot^3}[/tex].
For more information about conversion of units, refer to the link:
https://brainly.com/question/25426050
