Answer :
Answer:
a
The tension in the string is [tex]T = 0.85 N[/tex]
b
The new balance reading is [tex]M_b = 885.86 g[/tex]
Explanation:
From the question we are told that
The mass of the beaker of water is [tex]m = 875 .0g[/tex]
The diameter of the copper ball is [tex]d = 2.75 cm = \frac{2.75}{100} = 0.0275 m[/tex]
There are two forces acting on the copper ball
The first is the Buoyant force of the water pushing it up which is mathematically represented as
[tex]F = \rho V g[/tex]
Where [tex]\rho[/tex] is the density of water which has value of [tex]\rho = 1000 kg/m^3[/tex]
g is the acceleration due to gravity [tex]g= 9.8 \ m/s^2[/tex]
[tex]V[/tex] is the volume of water displaced by the copper ball which is mathematically evaluated as
[tex]V = \frac{4}{3} \pi r^3[/tex]
The radius r is [tex]r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m[/tex]
Substituting value
[tex]V = \frac{4}{3} * 3.142 * (0.01375)^3[/tex]
[tex]V = 1.08 9 *10^{-5 } m^3[/tex]
Substituting for F
[tex]F = 1000 * 1.089 *10^{-5} * 9.8[/tex]
[tex]F = 0.1067 N[/tex]
The second force is the weight of the copper ball which is mathematically represented as
[tex]W_c = mg[/tex]
Now m is the mass which can be mathematically evaluated as
[tex]m = \rho_c * V[/tex]
Where is the density of copper with value of [tex]\rho_c = 8960 kg /m^3[/tex]
So [tex]m = 8960 * 1.089 *10^{-5}[/tex]
[tex]m = 0.0976[/tex]
So the weight of copper is
[tex]W_c = 0.09756 * 9.8[/tex]
[tex]W_c = 0.956 N[/tex]
Now the tension the string would be mathematically evaluated as
[tex]T = W_c - F[/tex]
So [tex]T = 0.956 - 0.1067[/tex]
[tex]T = 0.85 N[/tex]
From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance
Now the mass equivalent of this weight is mathematically evaluated as
[tex]m_z = \frac{1.0645 }{9.8 }[/tex]
[tex]m_z = 0.01086 kg[/tex]
Converting to grams
[tex]m_z = 10.86 g[/tex]
So the new balance reading is
[tex]M_b = 875.0 +10.86[/tex]
[tex]M_b = 885.86 g[/tex]