What is the weight of the piece of aluminum shown below at 0.0974 lbs/in.^3 (hint:subtract the volume of the cylindrical hole.) Round to the nearest tenth as needed.

Question

asked Mar 22, 2023 521 views

1 Answer

Linus Caldwell · Answer 1 · 2023-03-28T22:37:51+0000

We will have the following:

First, we will determine the volume of the side plate:

$V_1=0.5\cdot4\cdot5\Rightarrow V_1=10$

Now, we find the volume of the front plate:

$V_2=0.5\cdot4\cdot4-\pi(2)^2(0.5)\Rightarrow V_2=8-2\pi$

Now, we calculate the total volume:

$V_t=2V_1+V_2\Rightarrow V_t=2(10)+(8-2\pi)$
$\Rightarrow V_t=28-2\pi$

Now, we calculate its weight:

$w=0.0974\cdot(b)/(in^3)\cdot(28-2\pi)\cdot in^3\Rightarrow w=2.115217751\ldots$
$\Rightarrow w\approx2.1$

So, it's weight is approximately 2.1 Lb.