Question

A thief uses a can of sand to replace a solid gold cylinder that sits on a weight sensitive, alarmed pedestal. the can of sand and the gold cylinder have exactly the same dimensions (length = 22 cm and radius = 3.8 cm).

a.calculate the mass of each cylinder (ignore the mass of the can itself). (density of gold = 19.3 g/cm3 and density of sand = 3.00g/cm3)

b.did the thief set off the alarm?

Answer 1

just looking at the density of the sand and gold you can see the gold is heavier but here are calculations for proof:

calculate volume of cylinder

V=PI x r^2 x h = 3.14 x 3.8^2 x 22 = 997.52 cubic cm

now calculate mass of the gold and the sand

gold = 19.3 x 997.52 = 19252 grams (19.25 kilograms)

sand = 3.00 x 997.52 = 2992.56 grams (2.99 kilograms)

so yes he set the alarm off

Answer 2

Answer:

a)Mass of gold cylinder is 19,261.80 grams and mass of can of sand is 2,994.063 grams.

b) Thief will not able to set off the alarm as mass of gold can is larger than the mass of can filled with sand.

Step-by-step explanation:

a) Radius of the gold can = r = 3.8 cm

Height of the gold can = h = 22 cm

Volume of the cylinder =$\pi r^2h$

Volume of the gold can = V

$V=\pi r^2 h=3.14\times (3.8 cm)^2\times 22cm=998.021 cm^3$

Can of sand and the gold cylinder have exactly the same dimensions so it will also have same volume.

Volume of can of sand = V' = V =$998.021 cm^3$

Mass of gold cylinder = M

Mass of can of sand = M'

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

$M=D\times V=19.3 g/cm^3\times 998.021 cm^3=19,261.80 g$

Density of the sand= $D=3.00 g/cm^3$

$M'=D'\times V=3.00 g/cm^3\times 998.021 cm^3=2,994.063 g$

b) As we can see that ; M>M'

Since, the gold cylinder sits on a weight sensitive, alarmed pedestal which will easily detect the change in mass of object kept on it.

Thief will not able to set off the alarm as mass of gold can is larger than the mass of can filled with sand.