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3 years ago

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A certain quantity of a liquid has a volume of 10cm at 20°C. Calculate its volume at 50°C, if it’s cubic expansivity is 10^(-3)

K^(-1)

1 answer:

Pavel [41]3 years ago

6 0

Answer:

10.3 cm³

Explanation:

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

Original volume (V₁) = 10 cm³

Initial temperature (θ₁) = 20 °C

Final temperature (θ₂) = 50 °C

Cubic expansivity (γ) = 10¯³ K¯¹

Final volume (V₂) =?

γ = V₂ – V₁ / V₁(θ₂ – θ₁)

10¯³ = V₂ – 10 / 10( 50 – 20)

10¯³ = V₂ – 10 / 10(30)

10¯³ = V₂ – 10 / 300

Cross multiply

10¯³ × 300 = V₂ – 10

0.3 = V₂ – 10

Collect like terms

0.3 + 10 = V₂

10.3 = V₂

V₂ = 10.3 cm³

Thus, the volume at 50 °C is 10.3 cm³

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Answer

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inclined at an angle = 28°

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h = L (1 - cos θ)

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$v_f = \sqrt{2gh_i}$

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$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

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