Question

1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown

Answer 1

Answer:

a) 3.37 x $10^{3} kg/m^3$

b) 6.42kg/$m^{3}$

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x $10^{-4}$  $kg/m^{3}$. So density of metal = mass of metal / volume of metal = 5 / 14 x $10^{-4}$  $kg/m^{3}$ = 3.37 x $10^{3} kg/m^3$

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/$m^{3}$