Question

s 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)

Answer 1

Answer:

a  

The tension in the string is  $T = 0.85 N$

 b

 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 $g= 9.8 \ m/s^2$

          $V$ 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  

                  $M_b = 875.0 +10.86$

                  $M_b = 885.86 g$