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.

Correct answer:

m =  5.4675 kg
n =  914

Step-by-step explanation:

a=30 cm dm=30/10  dm=3 dm b=15 cm dm=15/10  dm=1.5 dm c=7.5 cm dm=7.5/10  dm=0.75 dm  V=a b c=3 1.5 0.75=278=3.375 dm3 ρ=1.8 kg/dm3  m1=ρ V=1.8 3.375=24340=6.075 kg  q=10%=110100=0.9  m=q m1=0.9 6.075=5.4675 kg
M=5 t kg=5 1000  kg=5000 kg  n1=M/m=5000/5.4675914.4947  n=n1=914.4947=914

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