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1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown
liquid. Find the densities of the following:
a. the metal
b. the unknown liquid

Sagot :

Zezima

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]

snehashish65

(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

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