Physics, 06.05.2020 03:46 kingdesto3481

A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)

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Answer from: expeditionofsin

The tension in the string is T = 0.85 N

The new balance reading is M_b = 885.86 g

Explanation:

From the question we are told that

The mass of the beaker of water is m = 875 .0g

The diameter of the copper ball is $d = 2.75 cm = \frac{2.75}{100} = 0.0275 m$

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

$F = \rho V g$

Where $\rho$ is the density of water which has value of $\rho = 1000 kg/m^3$

g is the acceleration due to gravity $g= 9.8 \ m/s^2$

is the volume of water displaced by the copper ball which is mathematically evaluated as

$V = \frac{4}{3} \pi r^3$

The radius r is $r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m$

Substituting value

$V = \frac{4}{3} * 3.142 * (0.01375)^3$

$V = 1.08 9 *10^{-5 } m^3$

Substituting for F

$F = 1000 * 1.089 *10^{-5} * 9.8$

F = 0.1067 N

The second force is the weight of the copper ball which is mathematically represented as

W_c = mg

Now m is the mass which can be mathematically evaluated as

$m = \rho_c * V$

Where is the density of copper with value of $\rho_c = 8960 kg /m^3$

So $m = 8960 * 1.089 *10^{-5}$

m = 0.0976

So the weight of copper is

W_c = 0.09756 * 9.8

W_c = 0.956 N

Now the tension the string would be mathematically evaluated as

T = W_c - F

So T = 0.956 - 0.1067

T = 0.85 N

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

$m_z = \frac{1.0645 }{9.8 }$

m_z = 0.01086 kg

Converting to grams

m_z = 10.86 g

So the new balance reading is

M_b = 875.0 +10.86

M_b = 885.86 g

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A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper...