kaylaamberd

01.12.2020 •

Engineering

With a hub or fixed drum, the drum has a hub for inner and outer bearings and is retained by a

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Ответ:

ehuntsman8221

21.09.2022

Engineering

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.

$s.g =\frac{The\ specific\ weight\ of\ the\ substance}{The\ specific\ weight\ of\ the\ water}$

The specific weight of a substance is defined as its weight per unit volume

$s.w.= \frac{Weigth }{Volumen}$

Let

$\the\ specific\ weight\ of\ the\ water\ ={62.4 \frac{lbf}{ft^{3} }}$

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

$\\volume\ of\ the\ water\ = \frac{weight\ of\ the\ water\ }{62.4 \frac{lbf}{ft^{3} }} \\\\\\volume\ of\ the\ water\ = \frac{41\ lb }{62.4 \frac{lbf}{ft^{3} }}\\\\\\volume\ of\ the\ water\ =0.65 feet^{3} \\$

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

$s.w.= \frac{Weight }{Volumen}\\s.w.= \frac{50\lb}{0.65ft^{3} }\\s.w=76.097 \frac{lb}{ft^{3} }$

Step 5

find the specific gravity if the glycerin

$s.g(gly) =\frac{The\ specific\ weight\ of\ the\ substance}{The\ specific\ weight\ of\ the\ water}\\s.g(gly) =\frac{76.097 \frac{lb}{ft^{3} }}{62.4 \frac{lbf}{ft^{3}} }\\S.g(gly)=1.185$

have a great day

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