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

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

Step-by-step 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 = (m_r)/(\rho)`

substituting values

`V = (5000)/(4.3)`

`V = 1162.7 \ dm^3`

The volume of the rock immersed in water is

`V_w = (V)/(2)`

substituting values

`V_w = (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`