Respuesta :
Answer:
1.185
Explanation:
Hello
The specific gravity, consists of the ratio that exists between the density of one substance and the density of another reference substance(water) ,the water is used in this case.
[tex]s.g =\frac{The\ specific\ weight\ of\ the\ substance}{The\ specific\ weight\ of\ the\ water}[/tex]
The specific weight of a substance is defined as its weight per unit volume
[tex]s.w.= \frac{Weigth }{Volumen}[/tex]
Let
[tex]\the\ specific\ weight\ of\ the\ water\ ={62.4 \frac{lbf}{ft^{3} }} [/tex]
Step 1
find the weight of the water
(1)weight of the container= 8 lb
(2)weight of the container + weight of the water=49 lb
from(2)
weight of the water=49 lb -weight of the container
weight of the water=49 lb -8 lb
weight of the water=41 lb
Step 2
find the volume of the container using the weight of the water
[tex]\\volume\ of\ the\ water\ = \frac{weight\ of\ the\ water\ }{62.4 \frac{lbf}{ft^{3} }} \\\\\\volume\ of\ the\ water\ = \frac{41\ lb }{62.4 \frac{lbf}{ft^{3} }}\\\\\\volume\ of\ the\ water\ =0.65 feet^{3} \\[/tex]
the volume of the water = volumen of the container
Step 3
Find the weight of the glycerin
(1)weight of the container= 8 lb
(3)weight of the container + weight of the glycerin=58 lb
from(3)
weight of the glycerin=58 lb -weight of the container
weight of the glycerin=58 lb -8 lb
weight of the glycerin=50 lb
Step 4
Find the weight of the glycerin
[tex]s.w.= \frac{Weight }{Volumen}\\s.w.= \frac{50\lb}{0.65ft^{3} }\\s.w=76.097 \frac{lb}{ft^{3} }[/tex]
Step 5
find the specific gravity if the glycerin
[tex]s.g(gly) =\frac{The\ specific\ weight\ of\ the\ substance}{The\ specific\ weight\ of\ the\ water}\\s.g(gly) =\frac{76.097 \frac{lb}{ft^{3} }}{62.4 \frac{lbf}{ft^{3}} }\\S.g(gly)=1.185[/tex]
have a great day
