# Bricks

Openings in perforated bricks occupy 10% and brick has dimensions 30 cm, 15 cm and 7.5 cm. Calculate
a) the weight of a perforated bricks, if you know that the density of the full brick material is p = 1800 kg/m3 (1.8 kg/dm3)
b) the number of perforated bricks, which allowed the driver to load on the truck with a load capacity of 5 tons and avoid the car overload.

Result

m =  5.468 kg
n =  914

#### Solution:

$m=3 \cdot \ 1.5 \cdot \ 0.75 \cdot \ 0.9 \cdot \ 1.8=\dfrac{ 2187 }{ 400 }=5.468 \ \text{kg}$
$n=5000/(3 \cdot \ 1.5 \cdot \ 0.75 \cdot \ 0.9 \cdot \ 1.8)=914$

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