Question

A thief plans to steal a gold sphere with the radius of 28.9cm from a museum. If the gold has a density of 19.3 g/cm^3 what is the mass of the sphere in pounds? (Volume of a sphere is V= (4/3) pi r^3) Is the thief likely to be able to walk off with gold sphere unassisted?

Answer 1

Answer: The mass of gold in pounds is 4299.8 pounds.

Explanation:

To calculate the volume of sphere, we use the formula:

$V=\frac{4}{3}\pi r^3$

where,

r = radius of sphere

We are given:

Radius of osmium = 28.9 cm

Volume of osmium = $\frac{4}{3}\times 3.14\times (15)^3=101056cm^3$

Density of gold = $19.3g/cm^3$

To calculate mass of a substance, we use the equation:

$Density=\frac{Mass}{Volume}$

Putting values in above equation, we get:

$19.3g/cm^3=\frac{\text{Mass of gold}}{101056cm^3}\\\\\text{Mass of gold}=1950380.8g$

To convert the given mass into pounds, we use the conversion factors:

1 pound = 453.592 g

So, $1950380.8g=1950380.8g\times \frac{1pound}{453.592g}=4299.8pounds$

Hence, the mass of gold in pounds is 4299.8 pounds.

Answer 2

First you have to plug the radius (28.9 cm) in the equation for volume, so you would have V = (4/3)*pi*(28.9)^3(which means cubed) to get 101107.2126 cm^3 for volume, next use that in the formula m = d*V = 19.3g/cm^3 * 101107.2126 to get 1,951,369.203g to convert to pounds you multiply that by 0.00220462262 or 0.002 for short to get 4,302.03 pounds
Unless he can carry 4k pounds, I don't think he getting away.