Question

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may want to review (Pages 370 - 372) . For help with math skills, you may want to review: Conversion Factors Part A What is the tension in the string

Answer 1

Answer:

The tension on the string is  $T = 43.302 \ N$

Explanation:

From the question we are told that

    The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

       The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

       

Generally the volume of the rock is mathematically evaluated as

          $V = \frac{m_r}{\rho}$

substituting values

        $V = \frac{5000}{4.3}$

       $V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

      $V_w = \frac{V}{2}$  

substituting values

     $V_w = \frac{1162.7 }{2}$

     $V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

     $m_w = V_w$     According to Archimedes principle

=>   $m_w = 581.4 \ g$

     $m_w = 0.5814 \ kg$

The weight of the water displace is  

      $W _w = m_w * g$

      $W _w = 0.5814 * 9.8$

      $W _w = 5.698 \ N$

The actual weight of the rock is  

      $W_r = m_r * g$

     $W_r = 5.0 * 9.8$

     $W_r = 49.0 \ N$

The tension on the string is

       $T = W_r - W_w$

substituting values

       $T = 49.0 - 5.698$

       $T = 43.302 \ N$