find the weight of a solid iron cylinder of height 25 cm and the radius of the base 14 cm if one cubic centimetre of iron weighs 7.8 grams
Answers
Answer:
120.12 kg
Step-by-step explanation:
Volume of a cylinder = πr²h
In this case :
height = 25 cm
radius = 14 cm
Doing the substitution we have :
Volume = 22/7 × 14² × 25 = 15400 cm³
The density of the solid is :
7.8 grams/cm³
Mass = Volume × Density
Since we have the density and the volume we can get the mass from the formula above.
Mass = 15400 × 7.8 = 120120 grams
We can convert this to kilograms as follows:
In kilograms this is given as :
120120/1000 = 120.12 Kgs
Step-by-step explanation:
(i)
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = l×b×h
=(12×8×4.5) cm3= 432 cm3
Total Surface area = 2(lb + lh+ bh)
=2(12×8 + 12×4.5 +8×4.5) cm2=2(96 +54 + 36) cm2=2× 186 cm2=372 cm 2
Lateral surface area = 2(l+b)×h
=[2(12+8)×4.5] cm2=[2(20)×4.5] cm2=40×4.5 cm2=180 cm2
(ii)
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid = l×b×h
=(26×14×6.5)m3=2366 m3
Total surface area = 2(lb + lh+ bh)
=2(26×14+26×6.5+6.5×14) m2=2(364+169+91) m2=2×624 m2=1248 m2
Lateral surface area = 2(l+b)×h
=[2(26+14)×6.5] m2=[2×40×6.5] m2=520 m2