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klemol [59]
2 years ago
10

A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

mmersed in water, the tension is 31.6 N . When the rock is totally immersed in an unknown liquid, the tension is 11.4 N. What is the Density of the unknown liquid. -When I looked at this problem, I though we needed to know the volume of the rock. Can someone show me how to do it without the volume of this rock?

Physics
1 answer:
Luden [163]2 years ago
5 0

Answer:

\rho _{liquid}=1995.07kg/m^{3}

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have T_{3}+B = W_{rock}

T_{3}+\rho _{Liquid}V_{rock}g=W_{rock}.....(\alpha)

When the rock is suspended in air for equilibrium we have

T_{1}=W_{rock}....(\beta)

When the rock is suspended in water for equilibrium we have

T_{2} + \rho _{water}V_{rock}g=W_{rock}.....(\gamma)

Using the given values of tension and solving α,β,γ simultaneously for \rho _{Liquid} we get

W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\

Solving for density of liquid we get

\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000

\rho _{liquid}=1995.07kg/m^{3}

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3 0
3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
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Answer:

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To find the magnitude of electric field at the mid point,

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Electric field at the mid-point due to charge Q' is given by:

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\vec{E'} = 418.84 N/C

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8 0
3 years ago
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