Question

A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of the water? b) What is the buoyancy force on the rock? c) What is the mass of the water that the rock displaces? d) What is the weight of the rock under water?

Answer 1

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Step-by-step explanation:

a) Mass of the rock out of the water = Volume x Density

Volume = 470 cm³

Density = 3.55 g/cm³

Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

Volume = 470 cm³

Density of liquid = 1 g/cm³

`\texttt{Buoyancy force}= (470* 1* 9.81)/(1000) = 4.61 N`

c) Mass of the water displaced = Volume of body x Density of liquid

Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

Weight of rock under water = 16.37 - 4.61 =11.76 N