Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

What is the weight of the piece of aluminum shown in the picture at 0.0974 lbs/cubic inch? (Hint: subtract the volume of the cylindrical hole) Round to the nearest tenth as needed

What Is The Weight Of The Piece Of Aluminum Shown In The Picture At 00974 Lbscubic Inch Hint Subtract The Volume Of The Cylindrical Hole Round To The Nearest Te class=

Sagot :

Weight of the piece of aluminium is 5.3 lbs

Explanation:

We consider the aluminium as a rectangular prism with a cylindrical hole

The formula for volume of a rectangular prism:

[tex]Volume\text{ = width }\times length\text{ }\times\text{ height}[/tex]

When we look at the aluminium, we see that has inner opening aside the cylindrical hole. The volume of the aluminium without the cylindrical hole:

[tex]\begin{gathered} \text{Volume of the opened prism= volume of outer rectangular prism - Volume of the inner rectangular prism} \\ \end{gathered}[/tex][tex]\begin{gathered} \text{outer prism: 6 in by 6 in by 71/2 in} \\ \text{length = 6 in, width = 6 in, height = 7 1/2 in} \\ \text{Volume of the outer prism = }6\times6\times\frac{15}{2} \\ \text{Volume of the outer prism = 270 }in^3 \\ \\ In\text{ner prism: } \\ width\text{ = 5 in, length = 6 in, } \\ \text{height = 7 in} \\ \text{Volume of inner prism = 5}\times6\times7 \\ \text{Volume of inner prism =}210\text{ }in^3 \end{gathered}[/tex][tex]\begin{gathered} \text{Volume of the opened prism without the hole = 270 - 210} \\ \text{Volume of the opened prism without the hole = }60in^3 \end{gathered}[/tex]

The opened aluminuim piece with the cylindrical hole:

Volume of aluminium piece = volume of opened prism - volume of cylindrical hole

[tex]\begin{gathered} \text{volume of cylinder = }\pi r^{2}h \\ \text{diameter of hole = 3 in} \\ \text{radius = diameter/2 = }\frac{3}{2\text{ }} \\ \text{radius = 1.5 in} \\ \text{height = 1/2 in = 0.5 in} \\ \text{let }\pi\text{ = 3.14} \\ \text{Volume of cylinder = }3.14\times1.5^2\times0.5 \\ \text{Volume of cylinder = 3.5325 in}^3 \end{gathered}[/tex][tex]\begin{gathered} \text{Volume of aluminium piece = 60 - 3.5325} \\ \text{Volume of aluminium piece = }54.4675\text{ in}^3 \end{gathered}[/tex]

Density of the aluminium piece = 0.0974 lbs/cubic inch

Volume of the aluminium piece = 54.4675 in³

weight = ?

[tex]\begin{gathered} \text{Density = }\frac{mass}{\text{volume}} \\ let\text{ mass = weight} \\ 0.0974lbs/in^3\text{ = }\frac{\text{weight}}{54.4675in^3} \\ \text{weight = }0.0974\frac{lbs}{in^3\text{ }}\text{ }\times\text{ }54.4675in^3 \\ \text{weight = 5.3051 lbs} \\ \\ To\text{ the nearest tenth, weight of the aluminium piece = 5.3 lbs} \end{gathered}[/tex]