Respuesta :
Answer:
0.21 mm
Step-by-step explanation:
Method 1
To find the thickness (t) of the steel, we first need to find the volume of the steel used to make the steel can.
The volume of the steel can calculated as the difference between the volume of the outer cylinder (the whole can) and the volume of the inner cylinder (the empty space inside the can).
The formula for the volume of a cylinder is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume of a Cylinder}}\\\\V=\pi r^2 h\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius of the circular base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the height.}\end{array}}[/tex]
The radius of the outer cylinder is 6.1 cm and its height is 6.3 cm.
The radius of the inner cylinder is (6.1 - t) cm and its height is (6.3 - 2t) cm, where t is the thickness of the steel.
Therefore, the volume of the steel in cm³ can be expressed as
[tex]V =\pi \cdot 6.1^2 \cdot 6.3 - (\pi \cdot (6.1 - t)^2 \cdot (6.3 - 2t))\\\\V =234.423\pi - (\pi \cdot (t^2-12.2t+37.21)(6.3 - 2t))\\\\V =234.423\pi - \pi (-2t^3+30.7t^2-151.28t+234.423)\\\\V=\pi(234.423-(-2t^3+30.7t^2-151.28t+234.423))\\\\V=\pi(234.423+2t^3-30.7t^2+151.28t-234.423)\\\\V=\pi(2t^3-30.7t^2+151.28t)[/tex]
The formula for volume in terms of mass and density is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume}}\\\\V=\dfrac{m}{\rho}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$m$ is the mass.}\\\phantom{ww}\bullet\;\textsf{$\rho$ is the density.}\end{array}}[/tex]
Given that the mass of the empty steel can is 76 g and the density of steel is approximately 7.8 grams per cubic centimeter, then the volume of the steel used to make the can is:
[tex]V = \rm \dfrac{76\;g}{7.8\; g/cm^3}=\dfrac{380}{39}\; cm^3[/tex]
Set the two volume expressions equal to each other and solve for t:
[tex]\pi(2t^3-30.7t^2+151.28t) = \dfrac{380}{39}\\\\\\2t^3-30.7t^2+151.28t = \dfrac{380}{39\pi}\\\\\\2t^3-30.7t^2+151.28t - \dfrac{380}{39\pi}=0\\\\\\t=0.0205874902...\; \rm cm[/tex]
To convert centimeters to millimeters, multiply by 10:
[tex]t=0.0205874902... \times 10\\\\t=0.205874902...\\\\t=0.21\; \rm mm[/tex]
Therefore, the thickness of the steel is:
[tex]\Large\boxed{\boxed{0.21 \; \rm mm}}[/tex]
[tex]\dotfill[/tex]
Method 2
(This is a slightly less accurate method, but still returns the same approximate result).
To estimate the thickness (t) of the steel, we first need to calculate the surface area of the steel can.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Surface Area of a Cylinder}}\\\\SA=2\pi rh+2\pi r^2\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$SA$ is the surface area.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius of the circular base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the height.}\end{array}}[/tex]
Given values:
- r = 6.1 cm
- h = 6.3 cm
Therefore:
[tex]SA=2\cdot \pi \cdot 6.1 \cdot 6.3 + 2 \cdot \pi \cdot 6.1^2\\\\SA=76.86\pi+ 74.42\pi \\\\SA=76.86\pi+ 74.42\pi \\\\SA=151.28\pi\; \rm cm^2[/tex]
Let t be the thickness of the steel in centimeters.
Therefore, the volume of the steel in cubic centimeters is:
[tex]V=151.28\pi t\; \rm cm^3[/tex]
The formula for volume in terms of mass and density is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume}}\\\\V=\dfrac{m}{\rho}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$m$ is the mass.}\\\phantom{ww}\bullet\;\textsf{$\rho$ is the density.}\end{array}}[/tex]
Given that the mass of the steel can is 76 g and the density of steel is approximately 7.8 grams per cubic centimeter, then the volume of the steel used to make the can is:
[tex]V = \rm \dfrac{76\;g}{7.8\; g/cm^3}=\dfrac{380}{39}\; cm^3[/tex]
Set the two volume expressions equal to each other and solve for t:
[tex]151.28\pi t=\dfrac{380}{39}\\\\\\ t=\dfrac{380}{39\cdot 151.28\pi}\\\\\\t=0.020501592...\rm \; cm[/tex]
To convert centimeters to millimeters, multiply by 10:
[tex]t=0.020501592... \times 10\\\\t=0.20501592...\\\\t=0.21\; \rm mm[/tex]
Therefore, the thickness of the steel is:
[tex]\Large\boxed{\boxed{0.21 \; \rm mm}}[/tex]
