Questions: 5. V starih časih so žgali opeke iz apnenca. a. Koliko kg bi tehtala opeka s stranicami 15 cm, 30 cm in 20 cm? b. Nariši skico opeke. c. Opeko postavimo na mizo z največjo ploskvijo. Kolikšen je tlak po opeko?

Solution Steps

To find the weight of the brick, we first need to calculate its volume. The dimensions of the brick are given as \(15 \, \text{cm} \times 30 \, \text{cm} \times 20 \, \text{cm}\).

The volume \(V\) is calculated as: \[ V = \text{length} \times \text{width} \times \text{height} = 15 \, \text{cm} \times 30 \, \text{cm} \times 20 \, \text{cm} = 9000 \, \text{cm}^3 \]

Assuming the density of the brick material (e.g., limestone) is approximately \(2.3 \, \text{g/cm}^3\), we can calculate the mass \(m\) of the brick.

Convert the volume to cubic meters for standard density units: \[ V = 9000 \, \text{cm}^3 = 0.009 \, \text{m}^3 \]

Calculate the mass: \[ m = \text{density} \times V = 2.3 \, \text{g/cm}^3 \times 9000 \, \text{cm}^3 = 20700 \, \text{g} = 20.7 \, \text{kg} \]

The pressure \(P\) exerted by the brick when placed on the table with its largest face is calculated using the formula: \[ P = \frac{F}{A} \] where \(F\) is the force (weight of the brick) and \(A\) is the area of the largest face.

The largest face has dimensions \(30 \, \text{cm} \times 20 \, \text{cm}\), so: \[ A = 30 \, \text{cm} \times 20 \, \text{cm} = 600 \, \text{cm}^2 = 0.06 \, \text{m}^2 \]

The force \(F\) is the weight of the brick: \[ F = m \times g = 20.7 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 203.067 \, \text{N} \]

Calculate the pressure: \[ P = \frac{203.067 \, \text{N}}{0.06 \, \text{m}^2} = 3384.45 \, \text{Pa} \]

Final Answer