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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

User Adamax
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Answer:

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

User Christoph Wolk
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