Answer:
a) 3.37 x [tex]10^{3} kg/m^3[/tex]
b) 6.42kg/[tex]m^{3}[/tex]
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 [tex]10^{-4}[/tex] [tex]kg/m^{3}[/tex]. So density of metal = mass of metal / volume of metal = 5 / 14 x [tex]10^{-4}[/tex] [tex]kg/m^{3}[/tex] = 3.37 x [tex]10^{3} kg/m^3[/tex]
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/[tex]m^{3}[/tex]
(a) The required density of metal piece is [tex]3.37 \times 10^{3} \;\rm kg/m^{3}[/tex].
(b) The density of the unknown liquid is [tex]6.42 \;\rm kg/m^{3}[/tex].
Given data:
The weight of metal in air is, W = 50.0 N.
The weight of metal in water is, W' = 36.0 N.
The weigh of unknown liquid is, W'' = 41.0 N.
(a)
Calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density . Since, Weight of metal in air = 50N = mg implies the mass of metal is 5kg.
Then, the difference of weight of the metal in air and water = upthrust acting on it .
[tex]W- W' = V \times \rho \times g\\\\50-36 = V \times 1000 \times 9.8\\\\V \approx 14 \times 10^{-4}\;\rm m^{3}[/tex]
Then the density of metal piece is,
[tex]\rho' = \dfrac{m}{V}\\\\\rho' = \dfrac{5}{14 \times 10^{-4}}\\\\\rho'=3.37 \times 10^{3} \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the required density of metal piece is [tex]3.37 \times 10^{3} \;\rm kg/m^{3}[/tex].
(b)
Since, water exerts a buoyant force to the metal, whose value is given as,
Fb = 50−36
Fb = 14N
This is equal to the weight of water displaced.
The mass of water displaced is,
= 14/10 = 1.4 kg.
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 [tex]6.42 \;\rm kg/m^{3}[/tex].
Thus, we can conclude that the density of the unknown liquid is [tex]6.42 \;\rm kg/m^{3}[/tex].
Learn more about the Buoyancy here:
https://brainly.com/question/19256828
