Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.