Question

Archimedes needs your help determining whether the new crown made for the king by a local jeweler is authentic. What is the buoyant force that he should expect from a 0.50 kg solid gold crown when it is immersed in water? The density of gold is 19,300 kg/m3 and the density of water is 1000 kg/m3.

Answer 1

Buoyant force is equal to 0.253 N

Step-by-step explanation:

We have given mass of the gold m = 0.50 kg

Density of gold
`d_g=19300kg/m^3`

Volume displaced
`V_(dis)=(mass)/(density)=(0.5)/(19300)=2.59* 10^(-5)m^3`

Density of water
`d_w=1000m^3`

So buoyant force of water is equal to
`F_b=d_g* v_(dis)* g`

`F_b=1000* 2.59* 10^(-5)* 9.8=0.253N`

Buoyant force is equal to 0.253 N