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Find the weight of the steel rivet shown in the figure. (Steel weighs 0.0173 |b/cu cm.) (Round to the nearest tenth as needed.)

Find the weight of the steel rivet shown in the figure. (Steel weighs 0.0173 |b/cu-example-1

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Given that the density of steel, we can find the weight using the formula:


\rho=(m)/(V)

We want the weight (m), then


m=\rho V

We know that


\rho=0.0173\text{ lb/cm}^3

Then we need to calculate the volume of the rivet.

As we can see it's compounded by a cylinder and a truncated cone, the formula to calculate the volume for these figures are


V=\pi r^2h

For the cylinder

h → height

r → radius

and for the truncated cone


V=(\pi h)/(3)(R^2+Rr+r^2)

R → Big radius

r → Small radius

h → height

Now we have the volume formulas, the total volume will be the sum of these two volumes, then we can say that


m=\rho(V_1+V_2)

Where V1 is the cylinder volume and V2 is the truncated cone volume.

Let's find V1:


\begin{gathered} V_1=\pi r^2h \\ \\ V_1=\pi(1.7)^2\cdot10.7 \\ \\ V_1=97.1475_{} \end{gathered}

Result in cubic centimeters.

And V2 is


\begin{gathered} V_2=(\pi h)/(3)(R^2+Rr+r^2) \\ \\ V_2=(\pi(2.1))/(3)((3.4)^2+3.4\cdot1.7+(1.7)^2) \\ \\ V_2=44.4880 \end{gathered}

Result also in cubic centimeters.

Therefore, the total volume is


\begin{gathered} V=V_1+V_2 \\ V=$$141.6356$$ \end{gathered}

In cubic centimeters as well.

Now we have the volume we can just apply


m=\rho V

Therefore


\begin{gathered} m=0.0173\cdot141.6356 \\ \\ m=2.4503\text{ lb} \end{gathered}

Then, the weight of the steel rivet is 2.45lb.

If we round it to the nearest tenth it will be 2.5lb

Therefore, the final answer, rounded to the nearest tenth is


m=2.5\text{ lb}

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