Answer with Explanation:
We are given that
Weight of an ore sample=17.5 N
Tension in the cord=11.2 N
We have to find the total volume and the density of the sample.
We know that
Tension, T=[tex]W-F_b[/tex]
[tex]F_b[/tex]=buoyancy force
T=Tension force
W=Weight
By using the formula
[tex]11.2=17.5-F_b[/tex]
[tex]F_b=17.5-11.2=6.3[/tex] N
[tex]F_b=V_{object}\times \rho_{water}\cdot g[/tex]
Where
[tex]V_{object}[/tex]=Volume of object
[tex]\rho_{water}=1000 kgm^{-3}[/tex]=Density of water
[tex]g=9.8 ms^{-2}[/tex]=Acceleration due to gravity
Substitute the values then we get
[tex]6.3=9.8\times 1000\times V_{object}[/tex]
[tex]V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3[/tex]
Volume of sample=[tex]6.43\times 10^{-4} m^3[/tex]
Density of sample,[tex]\rho_{object}=\frac{Mass}{volume_{object}}[/tex]
Where mass of ore sample=1.79 kg
Substitute the values then, we get
[tex]\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3[/tex]
Density of the sample=[tex]2.78\times 10^{3} kgm^{-3}[/tex]
(a) The total volume of the sample in the given water is [tex]6.43 \times 10^{-4} \ m^3[/tex]
(b) The density of the given sample is 2,777.6 kg/m³.
The given parameters;
The mass of the object is calculated as;
[tex]m = \frac{W}{g} = \frac{17.5}{9.8} = 1.786 \ kg[/tex]
The upthrust on the object when immersed in water is calculated as follows;
[tex]F_n = 17.5 N - 11.2 \ N\\\\F_n = 6.3 \ N[/tex]
The volume of the object is calculated as follows;
[tex]F_n= \rho Vg\\\\V= \frac{F_n}{\rho g} \\\\V = \frac{6.3}{9.8 \times 1000} \\\\V = 6.43 \times 10^{-4} \ m^3[/tex]
The density of the sample is calculated as follows;
[tex]\rho = \frac{nass}{volume} \\\\\rho = \frac{1.786}{6.43\times 10^{-4}} \\\\\rho = 2,777.6 \ kg/m^3[/tex]
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