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4 years ago

11

A metal ball-bearing with a circumference of 45.1 mm weighs 11.6 g. what is the density of the metal in g/cm3 (v of a sphere = (

4/3)Ïr3; circumference of a circle = 2Ïr)?

1 answer:

scoundrel [369]4 years ago

4 0

1) First of all, let's calculate the radius r of the ball-bearing. We know the value of the circumference:
$c=2\pi r = 45.1~mm=4.51~cm$
From this, we can find the radius:
$r= \frac{c}{2 \pi} = \frac{4.51~cm}{2 \pi}=0.72~cm$

2) Now, let's find the volume:
$V= \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.72~cm)^3 = 1.56~cm^3$

3) Finally, we can calculate the density. We know the mass, $m=11.6~g$ , therefore we have
$d= \frac{m}{V}= \frac{11.6~g}{1.56~cm^3}=7.44~g/cm^3$

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