find the volume of iron required to make an open box use external dimensions are 36 cm and 25 cm into 16.5 centimetre barkbox bring 1.5 CM thick throughout if 1 cm cube of side and weighs 8.5 grams find the weight of the empty box in kilograms
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the external dimensions of the rectangular box =36cm*25cm*16.5cm
external volume of the box = 14850 cubic cm
since the thickness of the box is 1.5 cm and box is open.
internal length = (36-1.5*2)cm= 36-3=33 cm
internal breadth = 25-1.5*2=25-3=22 cm
internal height = 16.5-1.5 = 15 cm [since box is open]
therefore the internal dimensions of the rectangular box= 33cm*22cm*15cm
internal volume of the box = 10890 cubic cm
therefore the volume of the iron = external volume of the box - internal volume of the box
= 14850-10890
=3960 cubic cm
since 1 cubic cm of iron weighs 8.5 grams.
therefore weight of 3960 cubic cm = 3960*8.5=33660 gm
i.e. the weight of the empty box is 33.66 kg.
hope this helps you.
cheers!!
external volume of the box = 14850 cubic cm
since the thickness of the box is 1.5 cm and box is open.
internal length = (36-1.5*2)cm= 36-3=33 cm
internal breadth = 25-1.5*2=25-3=22 cm
internal height = 16.5-1.5 = 15 cm [since box is open]
therefore the internal dimensions of the rectangular box= 33cm*22cm*15cm
internal volume of the box = 10890 cubic cm
therefore the volume of the iron = external volume of the box - internal volume of the box
= 14850-10890
=3960 cubic cm
since 1 cubic cm of iron weighs 8.5 grams.
therefore weight of 3960 cubic cm = 3960*8.5=33660 gm
i.e. the weight of the empty box is 33.66 kg.
hope this helps you.
cheers!!
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