https://file.aiquickdraw.com/mathbot/file/1726228043732pq8jlpdp.jpg

Let's go through both parts of the problem step by step.

### Part (i): Calculate the mass of a single can in grams.

The can is modeled as a hollow, open-ended cylinder with:
- Height \( h = 90 \, \text{mm} = 0.09 \, \text{m} \)
- External diameter \( D = 52 \, \text{mm} = 0.052 \, \text{m} \), so the external radius \( r_{\text{ext}} = \frac{D}{2} = 0.026 \, \text{m} \)
- Material thickness \( t = 0.1 \, \text{mm} = 0.0001 \, \text{m} \), giving an internal radius \( r_{\text{int}} = r_{\text{ext}} - t = 0.026 - 0.0001 = 0.0259 \, \text{m} \)
- Density of aluminium \( \rho = 2700 \, \text{kg/m}^3 \)

The mass of the can will be the volume of aluminium used multiplied by the density.

#### Step 1: Surface area of the cylinder
The can is open-ended, so the surface area consists of:
1. The outer cylindrical wall area:
   \[
   A_{\text{outer}} = 2\pi r_{\text{ext}} h
   \]
2. The inner cylindrical wall area:
   \[
   A_{\text{inner}} = 2\pi r_{\text{int}} h
   \]
3. The circular base thickness (since the can is open):
   \[
   A_{\text{base}} = \pi \left( r_{\text{ext}}^2 - r_{\text{int}}^2 \right)
   \]

#### Step 2: Volume of aluminium used
The volume is essentially the difference between the external and internal volume of the can, calculated using the surface areas.

The volume \( V \) of aluminium used is:
\[
V = (A_{\text{outer}} - A_{\text{inner}}) \cdot h + A_{\text{base}}
\]

#### Step 3: Calculate the mass
The mass is given by:
\[
\text{Mass} = V \times \rho
\]

After calculating this, we'll express the mass in grams, where \( 1 \, \text{kg} = 1000 \, \text{g} \).

### Part (ii): Estimate the value added from one tonne of aluminium into cans.

Given:
- Mass of one can = 7.5 g
- Selling price per can = 15 pence (or £0.15)
- Price of aluminium sheets = £2300 per tonne
- \( 1 \, \text{tonne} = 1000 \, \text{kg} \)

#### Step 1: Calculate the number of cans produced from 1 tonne of aluminium
First, convert the tonne into grams:
\[
1 \, \text{tonne} = 1000 \times 1000 = 1,000,000 \, \text{grams}
\]
The number of cans that can be produced from 1 tonne is:
\[
\text{Number of cans} = \frac{1,000,000 \, \text{g}}{7.5 \, \text{g/can}} = 133,333.33 \, \text{cans}
\]

#### Step 2: Calculate the revenue from selling all cans
The total revenue is:
\[
\text{Revenue} = \text{Number of cans} \times \text{price per can} = 133,333.33 \times 0.15 = £20,000
\]

#### Step 3: Calculate the value added
The value added is the difference between the selling price and the cost of raw aluminium:
\[
\text{Value added} = £20,000 - £2300 = £17,700
\]

### Would you like to go through the detailed calculations for Part (i) now?

Take the density of aluminium as 2700 kg/m^3, calculate the mass of a single can in grams to 2 decimal places. A different sized can has a mass of 7.5g and each one is sold for 15 pence. Assuming the price of aluminium sheets is £2300 per tonne, and that there is no waste in the manufacturing process, estimate the value added from the transformation of one tonne of aluminium sheet into cans.

Learn how to calculate the mass of a hollow aluminium can using geometric and density principles. Estimate the value added by manufacturing cans from one tonne of aluminium, incorporating cost and revenue analysis. Detailed step-by-step breakdown for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation