Question

Height Width Height/h = 100mm Length / Width / d = 25mm of a Prism

Can someone help me find the following answers.

1. What's is the volume of the square prism in mm3?

2. What is the volume of the cylinder in mm3?

3. What is the volume of the waste material in mm3 when the cylinder is turned from the square prism?

4. What is the percentage of material that has been wasted in mm3?

5. What is the weight of the cylinder if the wood used weighs 15g per cm3?

6.If we cut the cylinder into 10 disks and laid them out flat next to each other, how long would the line of disks be in cm?

7. If you think of the disks as circles, what is the total surface area of all of them added together in mm2?

8. If it takes 45 seconds to cut one disk by hand, how long would it take to cut the cylinder into 10 disks? Give your answer in minutes and seconds.

9.If an automated system could cut one disk every 15 seconds, how many disks could it cut in the time we cut our original 10 by hand

10.If we wanted to make the same sized cylinder from steel and steel weighs 150g per cm3, what is the percentage increase in weight from the wooden cylinder to the steel version?​

Answer 1

1. Volume of the square prism:
`\(62,500 \ \text{mm}^3\)`, 2. Volume of the cylinder:
`\(12,269.59 \ \text{mm}^3\)`, 3. Volume of waste material:
`\(50,230.41 \ \text{mm}^3\)`, 4. Percentage of material wasted:
`\(80.37\%\)`, 5. Weight of the cylinder (wood):
`\(184.04 \ \text{g}\)` 6. Length of the line of disks:
`\(39.27 \ \text{cm}\)`, 7. Total surface area of disks:
`\(4908.73 \ \text{mm}^2\)`, 8. Time to cut 10 disks by hand: 7 minutes and 30 seconds, 9. Number of disks cut by an automated system: 30 disks, 10. Percentage increase in weight from wood to steel: 902.29%.

The given variables are:

- h: Height of the square prism

- w: Width of the square prism

- d: Depth (or length) of the square prism

- r: Radius of the cylinder (half of the width of the square prism)

`\[d = 25 \ \text{mm}, \quad h = 100 \ \text{mm}, \quad \text{density of wood} = 15 \ \text{g/cm}^3, \quad \text{density of steel} = 150 \ \text{g/cm}^3\]`

1. Volume of the Square Prism:

`\[V_{\text{prism}} = d * d * h`

`= 25 \ \text{mm} * 25 \ \text{mm} * 100 \ \text{mm}`

`= 62,500 \ \text{mm}^3`

2. Volume of the Cylinder:

`\[V_{\text{cylinder}} = \pi * \left((d)/(2)\right)^2 * h\]`

`= \pi * \left(\frac{25 \ \text{mm}}{2}\right)^2 * 100 \ \text{mm}`

`= 12,269.59 \ \text{mm}^3`

3. Volume of Waste Material:

`\[V_{\text{waste}} = V_{\text{prism}} - V_{\text{cylinder}}`

`= 62,500 \ \text{mm}^3 - 12,269.59 \ \text{mm}^3`

`= 50,230.41 \ \text{mm}^3`

4. Percentage of Material Wasted:

`\[\text{Percentage wasted} = \left(\frac{V_{\text{waste}}}{V_{\text{prism}}}\right) * 100`

`= \left((50,230.41)/(62,500)\right) * 100`

`\approx 80.37\%`

5. Weight of the Cylinder:

`\[W_{\text{wood}} = V_{\text{cylinder}} * \text{density of wood}\]`

`= 12,269.59 \ \text{mm}^3 * 15 \ \text{g/cm}^3`

`= 184,043.85 \ \text{g}`

`= 184.04 \ \text{g}`

6. Length of the Line of Disks:

Calculate the circumference of the circle:

`\[C = 2 * \pi * \left((d)/(2)\right)\]`

`\[C = 2 * \pi * \left(\frac{25 \ \text{mm}}{2}\right)`

`\approx 39.27 \ \text{mm}`

Total length
`= \(10 * C`

`= 10 * 39.27 \ \text{mm}`

`= 392.7 \ \text{mm}`

`= 39.27 \ \text{cm}\)`

7. Total Surface Area of Disks:

`\[A_{\text{total}} = 2 * A_{\text{circle}} + A_{\text{lateral}}\]`

`\[A_{\text{circle}} = \pi * \left((d)/(2)\right)^2`

`= \pi * \left(\frac{25 \ \text{mm}}{2}\right)^2 \approx 490.87 \ \text{mm}^2`

`\[A_{\text{lateral}} = 2 * \pi * \left((d)/(2)\right) * h`

`= 2 * \pi * \left(\frac{25 \ \text{mm}}{2}\right) * 100 \ \text{mm} \approx 3926.99 \ \text{mm}^2`

`\[A_{\text{total}} = 2 * 490.87 \ \text{mm}^2 + 3926.99 \ \text{mm}^2`

`= 4908.73 \ \text{mm}^2`

8. Time to Cut 10 Disks by Hand:

If it takes 45 seconds to cut one disk, then the total time is

10 x 45 seconds

= 450 seconds

= 7 minutes and 30 seconds.

9. Automated System's Efficiency:

If an automated system could cut one disk every 15 seconds, it could cut:

=
`\( \frac{\text{Total Time by Hand}}{15} \)` disks.

`= \(\frac{450 \ \text{seconds}}{15 \ \text{seconds/disk}}`

= 30 disks

10. Percentage Increase in Weight (Wood to Steel):

Calculate the weight of the steel cylinder using the density of steel (150g per cm³):

`\[W_{\text{steel}} = V_{\text{cylinder}} * \text{density of steel}\]`

`= 12,269.59 \ \text{mm}^3 * 150 \ \text{g/cm}^3`

`= 1,840,438.5 \ \text{g}`

`= 1840.44 \ \text{g}`

Percentage increase
`= \(\left(\frac{W_{\text{steel}} - W_{\text{wood}}}{W_{\text{wood}}}\right) * 100\)`

`= \left((1840.44 - 184.04)/(184.04)\right) * 100`

`\approx 902.29\%\]`