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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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What is the mass of a falling rock if it produces a force of 50 N?
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××F=m \times a×

50N is your force and the acceleration is -9.8m/s^2 due to gravity. 

So, you just plug that in. 

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3 years ago
When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it fr
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i) 3.514 s, ii) 5.692 m/s

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Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

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Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

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7 0
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Answer:

The bulb B glows brighter.

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Given that,

A glows brightly and B glows dimly.

According to ohm's law,

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Then bulb B glows brighter because the resistance is high in bulb A so the current will be low.

The resistance is low in bulb B so the current will be high.

Hence, The bulb B glows brighter.

6 0
3 years ago
Read 2 more answers
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