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nasty-shy [4]
3 years ago
15

What is the difference between liquid water and water vapor?

Physics
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

Explanation:

Liquid water changes to water vapor when it evaporates or boils. The gas inside the bubbles of boiling water is water vapor. Water vapor can change back into liquid water when it cools down. Water vapor is always invisible.

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Intermolecular forces hold together which of the following?
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Intermolecular forces exist between the molecules of a substance.
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An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a
Blababa [14]

Answer:

\frac{\Delta L}{L} =5.37\times 10^{-4}

Explanation:

Given:

  • cross sectional area of the wire, A=5.75\times 10^{-6}\ m^2
  • density of aluminium wire, \rho=2.7\times 10^3\ kg.m^{-3}
  • young's modulus of the material, E=7\times 10^{10}\ N.m^{-2}
  • wave speed, v=118\ m.s^{-1}

<u>We have mathematical expression for strain as:</u>

\frac{\Delta L}{L} =\frac{\sigma}{E} ...............................(1)

and since, \sigma =\frac{T}{A}

where, T = tension force in the wire

equation (1) becomes:

\frac{\Delta L}{L} =\frac{T}{A.E} ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

v=\sqrt{\frac{T}{\mu} } ...................................(3)

where: \mu= linear mass density of the wire

\therefore \mu=\rho \times A

Now, equation (3) becomes

v=\sqrt{\frac{T}{\rho \times A} }

T=v^2.\rho \times A ............................(4)

Using eq. (2) & (4) for tension T

v^2.\rho \times A=A.E\times \frac{\Delta L}{L}

\frac{\Delta L}{L} =\frac{v^2.\rho}{E}

putting the respective values

\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}

\frac{\Delta L}{L} =5.37\times 10^{-4}

6 0
3 years ago
Work done= ________ transferred
mamaluj [8]

Answer:

Work done= Energy transferred

Explanation:

Work is the transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. If you put energy into an object, then you do work on that object (mass).

4 0
3 years ago
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