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Greeley [361]
3 years ago
8

If a certain mass of mercury has a volume of 0.002 m^3 at a temperature of 20°c, what will be the volume at 50°c

Physics
2 answers:
V125BC [204]3 years ago
5 0
Change in volume = mass x coefficient of linear expansion x change in temperature
.002 x .0001802 x 30 = .000010812

initial volume + change in volume = Final volume
.002 + .000010812 = .002010812 m cubed
eimsori [14]3 years ago
5 0
<h3><u>Answer;</u></h3>

<em><u>= 0.0020108 m³</u></em>

<h3><u>Explanation;</u></h3>
  • Using the concept on thermal expansion of matter, in this case thermal expansion of liquid mercury.

  V₁=0.002 m³, T₂ =20 °C

   V₂= ?   T₂ = 50° C

  • Thus, change in temperature will be; T₂ - T₁

That is, 50°C - 20°C = 30 °C

  • The coefficient (γ) of volume expansion of Mercury at 20 °C is 0.00018 per centigrade.
  • Therefore, the volume of mercury at 50° C will be given by;

       V₂ =V₁ [1 + γ(50-30)]

              =  0.002 [1 + 0.00018(30)]

              = 0.0020108 m³

Therefore; the volume of mercury at 50°C is  0.0020108 m³

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The force of attraction between a ball is F=.........×10^-¹¹
DIA [1.3K]

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4.45×10¯¹¹ N

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From the question given above, the following data were obtained:

Mass of ball (M₁) = 4 Kg

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